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1.
Similarity PROPERTIES OF AREAS OF TRIANGLES
PROPERTY - I The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights. In ABC, P seg AD is the height and A seg BC is the base. In PQR, seg PS is the height and seg QR is the base.
A (ABC) BC AD A (PQR) QR PS
B
D
S
C
Q
R
To learn the next property, we have to first understand the meaning of Triangles with equal heights. In theorems and problems we will come across three situations where two or more triangles have equal heights.
1.
In the adjoining figure,
A
P
seg AD and seg PS are the heights of ABC and PQR respectively. If AD = PS then ABC and PQR are said to have equal heights. 2.
B
D
In the adjoining figure,
C
S
A
Q P
R l
line l || line m ABC and PQR lie between the same two parallel lines l and m, they are said to have
B
D
C
S
Q
R
m
equal heights. 3.
In the adjoining figure,
A
ABD, ADC and ABC have common vertex A and their bases BD, DC and BC lie on the same line BC.
B E Also seg AE line BC. seg AE is the height of ABD, ADC and ABC. These three triangles have same height. S C H O O L S E C TI O N
D
C
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PROPERTY - II The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases. In the adjoining figure, ABC
1.
A
P
l
and PQR lie between the same two parallel lines l and m. 2.
Their heights are equal. A (ABC) BC A (PQR) QR
C
B
In the adjoining figure, ABD, ADC and ABC have common vertex A, and their bases BD, CD and BC lie on the same line BC. Hence they have equal heights. Considering two triangles at a time we get the following results.
A (ABD) BD (i) A (ADC) CD
A (ABD) BD (ii) A (ABC) BC
m
R
Q
A
B
D
C
A (ADC) DC (iii) A (ABC) BC
PROPERTY - III The ratio of areas of two triangles having equal bases, is equal to the ratio of their corresponding heights. D C In the adjoining figure, ABC and ABD have the same base AB. A (ABC) CP A (ABD) DQ A P B Q
PROPERTY - IV Areas of two triangles having equal bases and equal heights, are equal. In the adjoining figure, A ABD and ACD have common vertex A and their bases BD and CD lie on the same line BC. Their heights are equal. Also, D is the midpoint of seg BC. D C B BD = CD. Their bases are equal. A (ABD) = A (ACD) 2
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EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5) 1.
In the adjoining figure, E seg BE seg AB and seg BA seg AD. If BE = 6 and AD = 9 (1 mark) A (ABE) find A (ABD)
A (ABE) BE A (ABD) = AD
Sol.
A (ABE) 6 A (ABD) = 9
A (ABE) 2 A (ABD) = 3
B
A
D
[Triangles with common base]
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5) 2.
S
In the adjoining figure, seg SP side YK and seg YT seg SK. If SP = 6, YK = 13, YT = 5 and TK = 12 then find : A (SYK) : A (YTK). (2 marks)
A ( SYK) YK SP = A (YTK) TK YT
Sol.
A ( SYK) 13 6 = A (YTK) 12 5
A ( SYK) 13 A (YTK) = 10
T
P
K
Y
[ The ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights ]
[Given]
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5) 3.
In the adjoining figure, RP : PK = 3 : 2 then find the values of : (i) A (TRP) : A (TPK) (ii) A (TRK) : A (TPK) (iii) A (TRP) : A (TRK) (3 marks)
Sol.
RP PK Let RP RK RK RK
T
R
P
3 = [Given] 2 the common multiple be x = 3x and PK = 2x = RP + PK [ R - P - K] = 3x + 2x = 5x
K
TRP, TPK and TRK have a common vertex T and their bases RP, PK and RK lie on the same line RK Their heights are equal S C H O O L S E C TI O N
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(i)
A A A A
(TRP) RP (TPK) = PK (TRP) 3x (TPK) = 2x
[Triangles having equal heights]
A (TRP) 3 A (TPK) = 2
Similarly, A ( TRK) RK (ii) A (TPK) = PK A ( TRK) 5x A (TPK) = 2x
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A ( TRK) 5 = A (TPK) 2
(iii)
A (TRP) RP A ( TRK) = RK
A (TRP) 3x = A ( TRK) 5x
A (TRP) 3 A ( TRK) = 5
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5) 4. Sol.
The ratio of the areas of two triangles with the common base is 6 : 5. Height of the larger triangle is 9 cm. Then find the corresponding height of the smaller triangle. (2 marks) Let the areas of the larger and the smaller triangle be A1 and A2 respectively. Let their heights be h1 and h2 respectively. A1 6 [Given] A2 = 5 and h1 = 9 cm The two triangles have a common base [Given] A1 h1 A = h [Triangles with common base] 2 2 9 6 = h 5 2
59 6 15 = 2 = 7.5
h2 = h2 h2
The corresponding height of the smaller triangle is 7.5 cm. EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5) 5.
In the adjoining figure, seg PR seg BC, seg AS seg BC and seg QT seg BC. Find the following ratios : A (ABC) A (ABS) (i) A (PBC) (ii) A (ASC)
A (PRC) A (BPR) (iii) A (BQT) (iv) A (CQT) Sol.
(i)
A (ABC) AS A (PBC) = PR
A (ABS) (ii) A ( ASC) 4
=
BS SC
(3 marks) B
A
P
Q
R
S
T C
[Triangles with common base] [Triangles having equal heights] S C H O O L S E C TI O N
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A (PRC) RC PR (iii) A (BQT) = BT QT
[ The ratio of the areas of two triangles equal to the ratio of the products of their bases and corresponding heights ]
Similarly, A (BPR) BR PR (iv) A (CQT) = CT QT EXERCISE - 1.1 (TEXT BOOK PAGE NO. 5) 6.
Sol.
In the adjoining figure, seg DH seg EF and seg GK seg EF. If DH = 12 cm, GK = 20 cm and A (DEF) = 300 cm2 then find (i) EF (ii) A (GEF) (iii) A (DFGE). (3 marks) 1 E (i) Area of triangle = × base × height 2 1 A (DEF) = × EF × DH 2 1 300 = × EF × 12 2 300 = EF × 6 300 EF = 6
D
K
H
F
G
EF = 50 cm
A (DEF) DH A (GEF) = GK 300 12 = A (GEF) 20
(ii)
[Triangles with common base]
300 20 12
A (GEF) =
A (GEF) = 500 cm2
(iii)
A (DFGE) = A (DEF) + A (GEF) = 300 + 500
A (DFGE) = 800 cm2
[Area addition property]
EXERCISE - 1.1 (TEXT BOOK PAGE NO. 6) 7.
In the adjoining figure, seg ST || side QR Find the following ratios : (i)
P
A (PST) A (PST) A (QST) (ii) (iii) A (QST) A (RST) A (RST) (3 marks)
S
PST and QST have a common vertex T Q and their bases PS and QS lie on the same line PQ.
Sol.
A ( PST) PS = A ( QST) QS
S C H O O L S E C TI O N
T
R
[Triangles having equal heights] 5
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(ii) PST and RST have a common vertex S and their bases PT and RT lie on the same line PR. A ( PST) PT A ( RST) = [Triangles having equal heights] RT (iii)seg ST || side QR [Given] QST and RST lie between the same two parallel lines ST and QR Their heights are equal Also, they have a common base ST A (QST) = A (RST) [Areas of two triangles having equal bases and equal heights are equal] A ( QST) A ( RST) = 1
BASIC PROPORTIONALITY THEOREM (B.P.T.) Statement : If a line parallel to a side of a triangle intersects other sides in two distinct points, then the line divides those sides in proportion. (4 marks) A Given : In ABC, (i) line DE || side BC (ii) Line DE intersects sides AB and AC E D at points D and E respectively.
AD AE = DB EC Construction : Draw seg BE and seg CD. To Prove :
B
C
Proof : ADE and BDE have a common vertex E and their bases AD and BD lie on the same line AB.
Their heights are equal
A (ADE) AD A (BDE) = DB
6
.......(i) [Triangles having equal heights]
ADE and CDE have a common vertex D and their bases AE and EC lie on the same line AC. Their heights are equal.
A (ADE) AE = .......(ii) [Triangles having equal heights] A ( CDE) CE line DE || side BC [Given] BDE and CDE are between the same two parallel lines DE and BC. Their heights are equal. Also, they have same base DE. [ Areas of two triangles having equal bases and equal heights are equal ]
A(BDE) = A(CDE)
A (ADE) A (ADE) = A (BDE) A (CDE) ......(iv) [From (i), (ii) and (iii)]
AD AE = DB EC
......(iii)
[From (i), (ii) and (iv)] S C H O O L S E C TI O N
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EXERCISE - 1.2 (TEXT BOOK PAGE NO. 11) 1. Sol.
Find the values of x in the following figures, if line l is parallel to one of the sides of the given triangles : (2 marks) (i) In ABC, B [Given] line l || side BC
l
AP AY = PB YC 3 6
=
[By B.P.T.]
A
65 3 x = 10
5
Y
x
C
x =
SQ SP = RQ TP
x 1.3 = 4.5 3.9
S
[Given]
x
l
[By B.P.T.]
1.3 Q
P
4.5 3.9 T
4.5 1.3 x = 3.9
x =
x = 1.5
R
4.5 3
(iii)In LMN, line l || side LN
P 3
5 x
(ii) In STR, line l || side TR
6
MQ MP = NQ LP
L
[Given]
2 P
[By B.P.T.]
N 8
8 x = 2 3
x
M
Q
3
l
83 2
x =
x = 12 EXERCISE - 1.2 (TEXT BOOK PAGE NO. 12)
6.
B
In the adjoining figure, ML || BC and NL || DC. Then prove that :
Proof :
AM AN = . AB AD
In ABC, seg ML || side BC
S C H O O L S E C TI O N
M
(3 marks)
A
L
C
N
[Given]
D
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BM AM BM AM AM AB AM AM AB In ADC, seg NL || DN AN DN AN AN AD AN AN AD AM MB
CL AL CL AL = AL AC = AL AL = AC =
side DC CL = AL CL AL = AL AC = AL AL = AC AN = AD
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[By B. P. T] [By Componendo] [ A - M - B and A - L - C] ......(i)
[By Invertendo] [Given] [By B. P. T.] [By Componendo] [ A - N - D and A - L - C]
......(ii)
[By Invertendo] [From (i) and (ii)]
CONVERSE OF BASIC PROPORTIONALITY THEOREM Statement : If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. (4 marks) A Given :In ABC, line l intersects sides AB and AC at points F D and E respectively such that D E l AD AE = DB EC To prove : Line l || side BC C B Proof : (Indirect method) Let us suppose that line DE is not parallel to side BC We can draw a line DF parallel to side BC, such that A-F-C. In ABC, line DF || side BC AD AF = .........(i) [By B.P.T.] DB FC AD AE = .........(ii) [Given] But, DB EC AF AE = [From (i) and (ii)] FC EC AF FC AE EC = [By componendo] FC EC AC AC = [ A - F - C, A - E - C] FC EC FC = EC F and E are not two different points. Line DF and line DE coincide line DE || side BC line l || side BC 8
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EXERCISE - 1.2 (TEXT BOOK PAGE NO. 11) 2.
E and F are points on the side PQ and PR respectively of PQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 1.3 cm, PF = 3.6 cm and FR = 2.4 cm. (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm. (iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.
Sol.
(i)
PE 3.9 = EQ 1.3 PE 3 = EQ 1
P 3.9
......(i)
E
PF 3.6 = FR 2.4
R
Q
P 4
8
E
F 9
4.5 Q
......(i)
PF 8 = ......(ii) FR 9 In PQR, PE PF = EQ FR line EF || side QR
R
[From (i) and (ii)] [By converse of B.P.T.]
PR = 2.56 = FR = FR =
PF + FR 0.36 + FR 2.56 – 0.36 2.20 cm
[ P - F - R]
P 0.
18
[ P - E - Q] E 1.28 Q
36
PE + EQ 0.18 + EQ 1.28 – 0.18 1.10 cm
0.
(iii) PQ = 1.28 = EQ = EQ =
2.4
PE 4 = EQ 4.5 4 10 PE = 4.5 10 EQ PE 40 = EQ 45 PE 8 = EQ 9
F
1.3
PF 3 = .....(ii) FR 2 In PQR, PE PF [From (i) and (ii)] EQ FR line EF is not parallel to side QR. (ii)
3.6
F 2.56 R
PE 0.18 = EQ 1.10 0.18 100 PE = 1.10 100 EQ
S C H O O L S E C TI O N
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PE 18 = EQ 110 PE 9 = EQ 55
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......(i)
PF 0.36 = FR 2.20 0.36 × 100 PF = 2.20 × 100 FR PF 36 = FR 220 PF 9 = FR 55 In PQR,
.....(ii)
PE PF = EQ FR line EF || side QR
[From (i) and (ii)] [By converse of B.P.T.]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) 15.
B
D and E are the points on sides AB and AC such that AB = 5.6, AD = 1.4, AC = 7.2 and AE = 1.8. Show that DE || BE. (3 marks)
D
A
Proof :
AB 5.6 DB DB AC 7.2 EC EC
= = = = = = = =
AD + DB 1.4 + DB 5.6 – 1.4 4.2 units AE + EC 1.8 + EC 7.2 – 1.8 5.4 units
E
C
[ A - D - B]
[ A - E - C]
AD 1.4 = DB 4.2 1.4 × 10 AD = 4.2 × 10 DB
AD 14 = DB 42
AD 1 = DB 3
.......(i)
AE 1.8 = EC 5.4 10
1.8 × 10 AE = 5.4 × 10 EC S C H O O L S E C TI O N
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AE 18 = EC 54 AE 1 = EC 3
.......(ii)
In ABC,
AD AE = DB EC seg DE || side BC
[From (i) and (ii)] [By converse of B.P.T.]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) 9.
In the adjoining figure, points A, B and C are on seg OP, seg OQ and seg OR respectively P such that AB || PQ and AC || PR. Then show that BC || QR. (3 marks) Proof : In POQ, seg AB || side PQ [Given]
OB OA = BQ AP
......(i)
In POR, seg AC || side PR
OA OC = AP CR
Q B A
O C R
[By B.P.T.] [Given]
.....(ii)
[By B.P.T.]
In OQR, OB OC = BQ CR seg BC || side QR
[From (i) and (ii)] [By converse of B.P.T.]
EXERCISE - 1.2 (TEXT BOOK PAGE NO. 12) 7.
In the adjoining figure, seg ED || seg QO and seg DF || seg OR. Prove that seg EF || side QR. Proof : In PQO, seg ED || side QO PE PD = EQ DO In PRO, seg DF || seg OR
PF PD = FR DO In PQR,
PE PF = EQ FR seg EF || side QR S C H O O L S E C TI O N
P E
D
F
(3 marks)
......(i)
O
[Given]
Q
R
[By B.P.T.] [Given]
.....(ii)
[By B.P.T.]
[From (i) and (ii)] [By converse of B.P.T.] 11
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PROBLEM SET - 1 (TEXT BOOK PAGE NO. 191) Two triangles ABC and DBC lie on the same side of the base BC. A From a point P on BC, PQ || AB and PR || BD are drawn. They Q intersect AC at Q and DC at R. Prove that QR || AD. (3 marks) B Proof : In ABC, [Given] seg PQ || side AB CQ CP = .......(i) [By B.P.T.] AQ BP
D
19.
In BCD, seg PR || side BD
CP CR = BP DR In ACD, CQ CR = AQ DR seg QR || side AD
R
C
P
[Given] .......(ii)
[By B.P.T.]
[From (i) and (ii)] [By converse of B.P.T.]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 191) 21.
BD AB = , prove that AD is the DC AC bisector of A. (Hint : Produce BA to E such that AE = AC. Join EC). (4 marks) In a ABC, if D is a point on BC such that
E
A
Construction : Take a point E on ray BA, such that AE = AC. Draw seg EC Proof : In AEC, ......(i) seg AE seg AC AEC ACE .....(ii)
BD AB = DC AC
BD AB = DC AE In BEC,
BD AB = DC AE seg AD ll side EC On transversal AC, DAC ACE On transversal BE, BAD AEC
C
D
[Construction] [Isosceles triangle theorem] [Given]
.....(iii)
[From (i)]
[From (iii)] [By converse of B.P.T.] .....(iv)
[Converse of alternate angles test]
......(v)
[Converse of corresponding angles test] [From (ii), (iv) and (v)]
DAC BAD Ray AD is the bisector of A. 12
B
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PROPERTY OF INTERCEPTS MADE BY THREE PARALLEL LINES The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines. y x line l || line m || line n and lines l, m and n cut the transversal x E P in points P, Q and R respectively. l Q F and lines l, m and n cut the transversal y m in point E,F and G respectively. G R n PQ EF = QR FG
EXAMPLE :
Sol.
p
q
In the adjoining figure, line l || line m ||line n. A R l Lines p and q are transversals. 8 From given information m B S find ST. (2 marks) 10 line l || line m || line n [Given] n C T On transversals p and q, [By Property of Intercepts made AB RS = by threeparallel lines ] BC ST 8 12 = [Given] 10 ST 12 ×10 ST = 8
ST
= 15 units
PROPERTY OF AN ANGLE BISECTOR OF A TRIANGLE Statement : In a triangle, the angle bisector divides the side opposite to (4 marks) the angle in the ratio of the remaining sides. E Given : In ABC, ray AD is the bisector of BAC A such that B - D - C. xx BD AB = DC AC B C D Construction : Draw a line passing through C, parallel to line AD and intersecting line BA at point E, B - A - E. Proof : In BEC, line AD || side CE [Construction]
To Prove :
BD AB = .........(i) DC AE line CE || line AD On transversal BE, BAD AEC ........(ii)
S C H O O L S E C TI O N
[By B.P.T.] [Construction] [Converse of corresponding angles test] 13
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Also, On transversal AC, DAC ACE ........(iii) But, BAD DAC ........(iv) In AEC, AEC ACE seg AC seg AE AC = AE ........(v)
BD = DC
AB AC
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[Converse of alternate angles test] [ ray AD bisects BAC] [From (ii), (iii) and (iv)] [Converse of Isosceles triangle theorem]
[From (i) and (v)]
EXERCISE - 1.2 (TEXT BOOK PAGE NO. 12) 5.
P
Ray PT is the angle bisector of QPR. 5.6 cm Find the value of x and the perimeter of PQR. (2 marks) In PQR, ray PT bisects QPR
Sol.
PQ QT = PR TR
5.6 4 = x 5
Q
• • x
4 cm
[Given]
T
R
5 cm
[Property of angle bisector of a triangle]
5 5.6 4 x= 7 PR = 7 cm x=
QR = QT + TR [ Q - T - R] QR = 4 + 5 QR = 9 cm Perimeter of PQR = PQ + QR + PR = 3.6 + 9 + 7 Perimeter of PQR = 19.6 cm EXERCISE - 1.2 (TEXT BOOK PAGE NO. 12) 3.
Sol.
Point Q is on the side MP such that MQ = 2 and MP = 5.5. Ray NQ is the angle bisector MNP of MNP. (2 marks) Find MN : NP. MP = MQ + QP [M - Q - P] 5.5 = 2 + QP QP = 5.5 – 2 QP = 3.5 units In MNP, ray NQ bisects MNP
14
N ××
M
Q
P
[Given] S C H O O L S E C TI O N
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MQ MN = QP NP
[Property of angle bisector of a triangle]
MN 2 = NP 3.5 2 10 MN = 3.5 10 NP
MN 20 = NP 35
MN 4 = NP 7
MN : NP = 4 : 7 EXERCISE - 1.2 (TEXT BOOK PAGE NO. 12) 4.
Ray YM is the angle bisector X of XYZ, where XY = YZ. Find the relation between XM and MZ.
(2 marks)
M
Sol. In XYZ, ray YM bisects XYZ
XY XM = YZ MZ
XM MZ XM = MZ 1=
[Given]
Y
•
•
Z
[Property of angle bisector of a triangle] [ XY = YZ]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) 16.
In PQR, if QS is the angle bisector of Q then, show that
A (PQS) PQ = A (QRS) QR Proof :
P
(3 marks) Q
S R
In PQR ray QS bisects PQR [Given] [ By property of angle bi sec tor PQ PS = ......(i) QR of a triangle ] SR PQS and QRS have a common vertex Q and their bases PS and SR lie on the same line PR. Their heights are equal [ The ratio of the areas of A (PQS) PS A (QRS) = two triangles having equal SR
A (PQS) PQ = A (QRS) QR
S C H O O L S E C TI O N
heights is equal to the ratio of their corresponding bases ] [From (i)] 15
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SIMILARITY OF TRIANGLES Definition : For a given one - to - one correspondence between the vertices of two triangles, if (i) their corresponding angles are congruent. (ii) their corresponding sides are in proportion, then the correspondence is known as Similarity and the triangles are said to be Similar Triangles. A
×
4
P
2 2
B
•
C
6
Q
•
In above figure, for correspondence ABC PQR (i) A P, B Q, C R and AB BC AC 2 (ii) = = = PQ QR PR 1 Hence ABC and PQR are similar triangles.
×
1 R
3
ABC is similar to PQR under ABC PQR, this statement is written symbolically as ABC ~ PQR.
NOTE If two triangles are similar, then i) their corresponding angles are congruent ii) their corresponding sides are in proportion. If ABC ~ PQR, then i) A P, B Q, C R AB BC AC ii) PQ QR PR
TEST OF SIMILARITY
1.
When two triangles are similar, then three pairs of corresponding angles are congruent and three pairs of corresponding sides are in proportion. But to prove that two triangles are similar, we select only three conditions taken in proper order. These conditions are called Tests of similarity. There are 3 tests of similarity : A - A - A test (A - A test) : For a given one - to - one correspondence between the vertices of two triangles, if the corresponding angles are congruent, then the two triangles C are similar. R In fig. ABC PQR, if, A P, B Q and C R, then ABC ~ PQR A
×
B
P
×
Q
We know that sum of measures of three angles of a triangle is 180º. Because of this if two pairs of corresponding angles of two given triangles are 16
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congruent then remaining pair is also congruent, and thus the triangles become similar triangles. This is known as A - A test. A - A Test : For a given one-one correspondence between the vertices of two triangles, if two angles of one triangle are congruent with the corresponding two angles of other triangle, then the two triangles are similar. 2.
S - A - S Test : For a given one-one correspondence between the vertices of two triangles, if two sides of one triangle are proportional to the corresponding sides of the other triangle and angles included by them are congruent, then the two triangles are similar. A In the adjoining figure, under the correspondence ABC PQR, 6 AB BC 2 = = PQ QR 40º 1 B and B Q, 5 then ABC ~ PQR by S - A - S test of similarity.
3.
P 3 C Q
40º 2.5
R
S - S - S Test : For a given one-one correspondence between the vertices two triangles, if three sides of one triangle are proportional to the three corresponding sides of other triangle, then the two triangles are similar. A In the adjoining figure, under the correspondence P ABC PQR, 10 9 4.5 5 AB BC AC 2 = = = PQ QR PR 1 B
then ABC ~ PQR by S - S - S test of similarity. 8
C Q
4
R
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 188) 1. Sol.
In each of the following figures you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer. (2 marks) AB 4.6 A (i) = PQ 2.3 AB 2 P = ......(i) PQ 8 1 4.6 4 2.3 BC 10 = QR 5 R Q 5 BC 2 C B = ......(ii) 10 QR 1
AC 8 = PR 4 AC 2 = ......(iii) PR 1 In ABC and PQR, AB BC AC = = PQ QR PR ABC PQR S C H O O L S E C TI O N
[From (i), (ii) and (iii)] [By SSS test of similarity] 17
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(ii)
AO 5 = ......(i) CO 3 BO 5 = ......(ii) DO 3 In AOB and COD,
B 5
5 3
O
3 D
C
AO BO = CO DO AOB COD AOB COD
[Vertically opposite angles] [By SAS test of similarity]
(iii)m ABC = m ADE = 60º
[Given]
ABC ADE
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[From (i) and (ii)]
A
......(i)
D 60º
In ABC and ADE, ABC ADE
[From (i)]
BAC DAE
[Common angle] B
ABC ADE AB 24 = PQ 12 AB 2 = ......(i) PQ 1 BC 7 = ......(ii) QR 5 AC 25 = .....(iii) PR 13 In ABC and PQR,
E
60º
C
[By AA test of similarity] A
(iv)
P 25
24
13
12 B
7
CQ
5
R
BC AB AC [From (i), (ii) and (iii)] QR PQ PR ABC is not similar to PQR.
(v)
18
AB 2.5 = BC 2.5 AB = 1 ......(i) BC AD 2.5 = CD 2.5 AD = 1 ......(ii) CD BD = BD BD = 1 ......(iv) BD In ABD and CBD, AB BD AD = = BC BD CD ABD CBD
2.5
A
D 2.5
2.5 B
C
2.5
[Common side]
[From (i), (ii) and (iii)] [By SSS test of similarity] S C H O O L S E C TI O N
MT
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(vi) In AOB and DOC, AOB DOC [Common angles] mBAO = 52º m CDO = 42º BAO is not congruent to CDO AOB is not similar to DOC.
A 52º
O
B
C 42º
D
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 16) 1.
Study the following figures and find out in each case whether the triangles are similar. Give reason. (2 marks) (i) MN = MT + TN [ M - T - N] M MN = 2 + 4 3 MN = 6 units 2 MK = MP + PK [ M - P - K] P T MK = 3 + 6 MK = 9 units 6 4 MN 6 = MT 2 K N
MN 3 = MT 1
......(i)
MK 9 = MP 3
MK 3 = .....(ii) MP 1 In MNK and MTP,
MN = MT NMK MNK (ii) PX PX PX
MK MP TMP ~ MTP
[From (i) and (ii)] [Common angle] [By SAS test of similarity] [ P - R - X]
= PR + RX = a + 2a = 3a
PX PR
=
3a a
PX PR
=
3 1
P
R
......(i)
X
XS 3b = RT 2b
2b
3b
T
S
XS 3 = .....(ii) RT 2 In PXS and PRT,
PX XS [From (i) and (ii)] PR RT PXS is not similar to PRT S C H O O L S E C TI O N
19
MT
GEOMETRY
EDUCARE LTD.
D
(iii)
70º
m M = m Q = 55º ......(i) m N = m R = a .....(ii) In DMN and AQR, M Q N R DNM ~ AQR
A Q
55º a
[From (i)] 55º M [From (ii)] [By AA test of similarity]
R a
N
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 189)
AB 3.8 = DE 7.6
Sol.
AB 1 = DE 2
B 60º
12 cm
F
80º
3 3 cm 6 3 cm C
6 cm
D
........(i)
BC 6 = EF 12 BC 1 = EF 2 3 3 AC = DF 6 3 AC 1 = DF 2 In ABC and DEF, AB BC AC = = DE EF DF ABC ~ DEF C F In ABC, m A = 80º m B = 60º m C = 40º
E
7.6 cm
Using the information given in the adjoining figure, find F. (3 marks) A 3.8 cm
9.
.......(ii)
.......(iii)
[From (i), (ii) and (iii)] [By SSS test of similarity] [c.a.s.t]
.....(iv)
[Given] ......(v)
[Remaining angle]
m F = 40º
[From (iv) and (v)]
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 17) 5. Sol.
A vertical pole of a length 6m casts a shadow of 4 m long on the ground. At the same time a tower casts a shadow 28 m long. Find the height of the tower. (2 marks) P In the adjoining figure, seg AB and seg PQ A represents the pole and tower respectively and ? 6m seg BC and seg QR represents the shadow cast by them respectively.
B
20
4m
C Q
28 m
R
S C H O O L S E C TI O N
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Now, ABC ~ PQR, AB BC = PQ QR 6 4 = PQ 28 6 28 PQ = 4 PQ = 42
[c.s.s.t.] [Given]
The height of the tower is 42 m. PROBLEM SET - 1 (TEXT BOOK PAGE NO. 189) 10. Sol.
A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the shadow 40 m on the ground. Determine the height of the tower. (2 marks) P In the adjoining figure, seg AB and seg PQ A represents the vertical stick ? and the tower respectively 12 m and seg BC and seg QR represents the shadow cast by them respectively. B 8m C Q 40 m R ABC ~ PQR
AB BC = PQ QR
[c.s.s.t.]
12 8 = PQ 40
[Given]
40 × 12 8 PQ = 60 PQ =
Height of the tower is 60 m. PROBLEM SET - 1 (TEXT BOOK PAGE NO. 188) 2.
A triangle ABC with sides AB = 6 cm, BC = 12 cm and AC = 8 cm is enlarged to PQR such that its longest side is 18 cm. Find the ratio and (3 marks) hence find the lengths of the remaining sides of PQR.
Sol.
P A
B
12 cm
ABC and the AB BC = = PQ QR 6 12 = = PQ 18 S C H O O L S E C TI O N
?
?
8 cm
6 cm
C
Q
18 cm
R
enlarged PQR are similar. AC [c.s.s.t.] PR 8 [Given] PR 21
MT
GEOMETRY
6 2 8 = = PQ 3 PR 6 2 = [From (i)] PQ 3
......(i)
8 PR
63 2 PQ = 9 cm
=
2 3
[From (i)]
83 2 PR = 12 cm
PQ =
The ratio is
EDUCARE LTD.
PR =
2 and PQ = 9 cm, PR = 12 cm. 3
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 17) 6.
Sol.
Triangle ABC has sides of length 5, 6 and 7 units while PQR has perimeter of 360 units. If ABC is similar to PQR then find the sides of PQR. (3 marks) In ABC, AB = 5 units BC = 6 units [Given] AC = 7 units Perimeter of PQR = 360 units [Given] PQ + QR + PR = 360 ......(i) ABC ~ PQR [Given]
AB BC AC = = PQ QR PR
5 6 7 = = PQ QR PR
567 5 6 7 = = = PQ QR PR [By theorem on equal ratios] PQ QR PR
5 6 7 18 = = = PQ QR PR 360
5 6 7 1 = = = PQ QR PR 20 5 1 = PQ 20
[c.s.s.t.]
[From (i)] .....(ii) [From (ii)]
PQ = 100 units 6 1 = QR 20
[From (ii)]
QR = 120 units
7 1 = PR 20
[From (ii)]
PR = 140 units 22
S C H O O L S E C TI O N
MT
GEOMETRY
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EXERCISE - 1.3 (TEXT BOOK PAGE NO. 17) 3.
A
In the adjoining figure, E is a point on side CB produced of an isosceles ABC with AB = AC. If AD BC and EF AC, prove that ABD ~ ECF. (2 marks) In ABC, seg AB seg AC ABC ACB ......(i) In ABD and ECF, ABD FCE
Proof :
ADB EFC ABD ~ ECF
F
E
[Given] D B [Isosceles triangle theorem]
C
[From (i) and B - D - C, A - F - C , C - B - E] [ each is 90º] [By AA test of similarity]
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 16) 2.
Sol.
ABC is a right angled at B. D is any point on AB. DE AC. If AD = 6 cm, AB = 12 cm, AC = 18 cm, find AE. (2 marks) In ABC and AED, BAC DAE ABC AED ABC ~ AED
D
C
[Common angle] [ Each is 90º] [By AA test of similarity]
AB AC = AE AD
[c.s.s.t.]
12 18 = AE 6
[Given]
AE
E
B
AE
A
12 × 6 18 = 4 units =
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 189) 8.
Sol.
In the adjoining figure, D C AB || DC. 5 3 – x Using the information given 9 O x 1 –3 – (3 marks) find the value of x. x 3 seg AB || seg DC [Given] B A On transversal BD, CDB ABD [Converse of alternate angles test] CDO ABO .......(i) [D - O - B] In DOC and BOA, CDO ABO [From (i)] DOC BOA [Vertically opposite angles] DOC ~ BOA [By AA test of similarity]
S C H O O L S E C TI O N
23
MT
GEOMETRY
DO BO 3 x–3 3(3x – 19) 9x – 57 9x – 57 2 x – 8x – 9x + 15 +57 x2 – 17x + 72 x2 – 9x – 8x + 72 x (x – 9) – 8 (x – 9) (x – 9) (x – 8) x – 9 = 0 or x – 8 = 0 x = 9 or x = 8
= = = = = = = = = =
EDUCARE LTD.
OC [c.s.s.t.] OA x –5 3x – 19 (x – 5) (x – 3) x2 – 3x –5x + 15 x2 – 8x + 15 0 0 0 0 0
EXERCISE - 1.3 (TEXT BOOK PAGE NO. 17) 4.
D is a point on side BC of ABC such that ADC = BAC. A Show that AC2 = BC × DC. (2 marks)
In ABC and DAC, BAC ADC ACB ACD ABC ~ DAC
Proof :
AC BC = DC AC AC2 = BC × DC
[Given] B D [Common angle] [By AA test of similarity]
C
[c.s.s.t.]
EXERCISE - 1.2 (TEXT BOOK PAGE NO. 13) 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect AO CO = . (3 marks) each other at the point O. Show that BO DO B A Proof : ABCD is a trapezium side AB || side DC [Given] O On transversal AC, BAC DCA [Converse C of alternate angles test] D BAO DCO ......(i) [ A - O - C] In AOB and COD, BAO DCO [From (i)] AOB COD [Vertically opposite angles] AOB ~ COD [By AA test of similarity] AO BO = [c.s.s.t.] CO DO AO CO = [By Alternendo] BO DO
24
S C H O O L S E C TI O N
MT
GEOMETRY
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EXERCISE - 1.2 (TEXT BOOK PAGE NO. 13) 10.
Using the basic proportionality theorem, prove that the line joining the mid points of any two sides of a triangle is parallel to the third side and A half of it. (5 marks) Given : In ABC, D and E are the midpoints of sides AB E D and AC respectively. To prove : (a) seg DE || sideBC C B 1 (b) DE = BC 2 Proof :(a) Let us assume that seg DE is not parallel to side BC. We can draw a line passing through D parallel to side BC intersecting side AC at F such that A - F - E - C. In ABC, line DF || side BC
AD AF = DB FC But, AD = DB AD =1 DB
[By B.P.T.] [ D is the midpoint of side AB]
......(ii)
AF =1 [From (i) and (ii)] FC AF = FC F is the midpoint of side AC But, E is also the midpoint of side AC [Given] But, a segment can have one and only midpoint F and E are not different points. seg DE || side BC ......(iii)
(b) line DE || line BC On transversal AB, ADE ABC In ADE and ABC, ADE ABC DAE BAC ADE ~ DABC
........(i)
AD DE AE = = AB BC AC But, AD =
AD 1 = AB 2
DE 1 = BC 2
DE = S C H O O L S E C TI O N
[From (iii)] ......(iv)
[Converse of corresponding angles test] [From (iv)] [Common angle] [By AA test of similarity]
.......(v)
1 AB 2
[c.s.s.t.] [ D is the midpoint of side AB]
......(vi) [From (v) and (vi)]
1 BC 2 25
MT
GEOMETRY
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 189) 7.
Sol.
EDUCARE LTD.
A
In the adjoining figure, DE || BC and AD : DB = 5 : 4, D E Find (i) DE : BC (ii) DO : DC (iii) A (DOE) : A (DCE) (5 marks) O seg DE || seg BC [Given] B On transversal AB, ADE ABC ......(i) [By converse of corresponding angles test] In ADE and ABC, ADE ABC [From (i)] DAE BAC [Common angle] ADE ~ ABC [By AA test of similarity] AD DE = .....(ii) [c.s.s.t] AB BC AD 5 = [Given] DB 4 DB 4 = [By Invertendo] AD 5 DB + AD 4+5 = [By componendo] AD 5 AB 9 = [A - D - B] AD 5 AD 5 = ......(iii) AB 9 DE 5 = [From (ii) and (iii)] BC 9
DE : BC = 5 : 9
(ii) seg DE || seg BC On transversal DC, EDC BCD EDO BCO In DOE and COB, EDO BCO DOE BOC DOE ~ COB DO DE = CO BC DO 5 = CO 9 CO 9 = DO 5 CO + DO 9+5 = DO 5 DC 14 = DO 5 DO 5 = DC 14
DO : DC = 5 : 14
C
......(iv) [Given] .......(v)
[Converse of alternate angles test] [D - O - C] [From (v)] [Vertically opposite angles] [By AA test of similarity] [c.s.s.t] [From (iv)] [By Invertendo] [By Componendo] [ D - O - C] [By Invertendo]
......(vi)
(iii)DOE and DCE have a common vertex E and their bases DO and DC lie on the same line DC. 26
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
Their heights are equal A (DOE) DO A (DCE) = DC A (DOE) 5 A (DCE) = 14
[Triangles having equal heights] [From (vi)]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) 18.
Let X be any point on side BC of ABC, XM and XN are drawn parallel to BA and CA. MN meets produced CB in T. Prove that TX2 = TB . TC. (5 marks) T
Proof :
seg AB || seg MX seg NB || seg MX On transversal TX, TBN TXM In TBN and TXM, TBN TXM BTN XTM TBN TXM TB TN = TX TM seg NX ll seg AC seg NX ll seg MC On transversal TC, TXN TCM In TXN and TCM, TXN TCM NTX MTC TXN TCM TX TN = TC TM TB TX = TX TC TX2 = TB × TC
A M N B
[Given] [ A - N - B] ......(i)
X
C
[Converse of corresponding angles test] [From (i)] [Common angle] [By AA test of similarity]
.....(ii)
[c.s.s.t.] [Given] [ A - M - C]
......(iii)
[Converse of corresponding angles test] [From (iii)] [Common angle] [By AA test of similarity]
.....(iv)
[c.s.s.t.] [From (ii) and (iv)]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 192) 25.
In the adjoining figure, A DEFG is a square and BAC = 90º Prove that : F G (ii) AGF ~ EFC (i) AGF ~ DBG (iii) DBG ~ EFC (iv) DE2 = BD . EC Proof : B C E D (i) DEFG is a square [Given] DE = EF = GF = DG .......(i) [Sides of a square] seg GF || seg DE [Opposite side of a square] seg GF || seg BC [B - D - E - C) On transversal AB, AGF ABC .......(ii) [Converse of corresponding angles test] On transversal AC, AFG ACB ......(iii) S C H O O L S E C TI O N
27
MT
GEOMETRY
In AGF and DBG, AGF GBD GAF BDG AGF DBG (ii) In AGF and EFC, AFG FCE GAF FEC AGF ~ EFC
EDUCARE LTD.
......(iv)
[From (ii) and A - G - B, B - D - C] [ each is 90º] [By AA test of similarity]
......(v)
[From (iii) and A - F - C, C - E - A] [ Each is 90º] [By AA test of similarity]
(iii) DBG ~ EFC
BD DG = EF EC EF × DG = BD × EC DE × DE = BD × EC DE2 = BD × EC
(iv)
[From (iv) and (v)] [c.s.s.t.] [From (i)]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 191) 20.
A
In the adjoining figure, ADB and CDB have the same base DB. If AC and BD intersect at O
A (ADB) AO then prove that A (CDB) = CO . (4 marks) Proof :
D
N
O
M
B
ADB and CDB have a common base BD, C A(ADB) AN = ....(i) [Triangles with common base] A(CDB) CM In ANO and CMO, ANO CMO [ Each is 90º] AON COM [Vertically opposite angles] ANO ~ CMO [By AA test of similarity] AN AO = ......(ii) [c.s.s.t.] CM CO A (ADB) AO = [From (i) and (ii)] A (CDB) CO PROBLEM SET - 1 (TEXT BOOK PAGE NO. 191)
S Two poles of height ‘a’ meters and ‘b’ meters are ‘p’ meters apart. Prove that the height ‘h’ drawn from N b of the point of intersection N of the lines joining the top of each pole to h the foot of the opposite pole is A ab y x T (5 marks) a + b meters. p Proof : AB = AT + TB [ A - T - B] AB = (x + y) = p ......(i) In ATN and ABR, TAN BAR [Common angle] ATN ABR [ each is 90º] ATN ABR [By AA test of similarity] AT TN = [c.s.s.t] AB BR
24.
28
R a B
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
x h = p a In BTN and BAS, TBN ABS BTN BAS BTN BAS BT TN = AB AS y h = p b Adding (ii) and (iii), x y h h + + = p p a b x+y 1 1 = h + p a b p p
a + b = h ab
......(ii)
[Given and from (i)] [Common angle] [ each is 90º] [By AA test of similarity] [c.s.s.t]
......(iii)
[Given and from (i)]
[From (i)]
a + b 1 = h ab ab a+b = h ab h = a + b metres PROBLEM SET - 1 (TEXT BOOK PAGE NO. 191)
The bisector of interior A K • of ABC meets BC in D. The bisector of exterior A meets BC produced in E. A × BD CD = • •× Prove that . F BE CE [Hint : For the bisector of A which B AB BE E D = C ] (5 marks) is exterior of BAC, AC CE Construction : Draw seg CF || seg AE such that B - F - A - K Proof : In ABC, ray AD bisects BAC [Given] BD AB = ......(i) [By property of angle bisector of a triangle] CD AC In ABE, seg CF || side AE [Construction] BF BC = [By B.P.T.] AF CE BF + AF BC + CE = [By Componendo] AF CE AB BE = ......(ii) [B - F - A and B - C - E] AF CE seg CF || seg AE [Construction] on transversal AF, KAE AFC ......(iii) [Converse of corresponding angles test] on transversal AC, EAC ACF ......(iv) [Converse of alternate angles test] 22.
S C H O O L S E C TI O N
29
MT
GEOMETRY
But, KAE EAC AFC ACF In AFC, AFC ACF AF = AC AB BE = AC CE BD BE = CD CE
EDUCARE LTD.
.......(v) ......(vi)
[ ray AE is the bisector of KAC] [From (iii), (iv) and (v)]
......(vii)
[From (vi)] [Converse of Isosceles triangle theorem]
......(viii) [From (ii) and (vii)] [From (i) and (viii)]
BD CD = BE CE
[By Alternendo]
AREAS OF SIMILAR TRIANGLES Statement : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (4 marks) Given : ABC ~ PQR. A A (ABC) BC2 AB2 P AC2 To Prove : A (PQR) = = = 2 2 2 QR PQ PR Construction : (i) Draw seg AD side BC, B-D-C (ii) Draw seg PS side QR, B D CQ S R Q-S-R [ The ratio of the areas of two triangles A (ABC) BC AD Proof : A (PQR) = QR PS ......(i) is equal to ratio of the products of a base and its corresponding height ] ABC ~ PQR [Given] AB BC AC ......(ii) [c.s.s.t.] PQ QR PR Also, B Q .....(iii) [c.a.s.t.] In ADB and PSQ, ADB PSQ [Each is a right angle] B Q [From (ii)] ADB ~ PSQ [By A-A test of similarity] AD BD AB .....(iv) [c.s.s.t.] PS QS PQ AD BC ......(v) [From (ii) and (iv)] PS QR A (ABC) BC AD A (PQR) = [From (i)] QR PS A (ABC) BC BC A (PQR) = [From (v)] QR QR
30
A (ABC) BC2 = A (PQR) QR 2
A (ABC) BC² AB² AC² A (PQR) = QR² = PQ² = PR² [From (ii) and (vi)]
.......(vi)
S C H O O L S E C TI O N
MT
GEOMETRY
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EXERCISE - 1.3 (TEXT BOOK PAGE NO. 21) 1.
Sol.
ABC ~ DEF (i) If A (ABC) = 9 cm2, A (DEF) = 64 cm2, DE = 5.6 cm then find AB. (ii) If A (ABC) : A (DEF) = 16 : 25, BC = 2.2 cm then find EF. (iii)If AB = 2.4 cm, DE = 1.6 cm, find the ratio of the area of ABC and DEF. (2 marks) (i) ABC ~ DEF [Given] A (ABC) AB2 = [Areas of similar triangles] A (DEF) DE2
9 64
=
AB2 (5.6)2
[Given]
3 8
=
AB 5.6
[Taking square roots]
AB
=
AB
3 5.6 8 = 3 × 0.7
AB
= 2.1 cm
(ii) ABC ~ DEF
A (ABC) BC2 = A (DEF) EF 2
[Given] [Areas of similar triangles]
16 25
=
(2.2)2 EF 2
[Given]
4 5
=
2.2 EF
[Taking square roots]
EF
=
2.2 5 4
EF
=
5.5 2
EF
= 2.75 cm
(iii)ABC ~ DEF
[Given]
A (ABC) AB2 = A (DEF) DE2
[Areas of similar triangles]
A (ABC) (2.4)2 = A (DEF) (1.6)2
[Given]
A (ABC) 5.76 A (DEF) = 2.56
A (ABC) 576 A (DEF) = 256
A (ABC) 9 A (DEF) = 4
A (ABC) : A (DEF) = 9 : 4. S C H O O L S E C TI O N
31
MT
GEOMETRY
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PROBLEM SET - 1 (TEXT BOOK PAGE NO. 188) 3.
Sol.
A model of a ship is made in ratio 1 : 200 (i) The length of the model is 4 m. Calculate the length of the ship. (ii) The area of the deck of the ship is 16,0000 m2. Find the area of the deck of the model. (2 marks) (i) A model of a ship is made in the ratio 1 : 200. Length of the model is 4 m [Given] Length of the ship = 4 × 200 Length of the ship = 800 m (ii) Area of the deck of the ship = 160000 m2 Area of the deck of the mod el (length of the mod el)2 Area of the deck of the ship = (length of the ship)2 2
Area of the deck of the mod el 1 = 160000 200 Area of the deck of the mod el 1 = 160000 40000 1 160000 Area of the deck of the model = 40000 Area of the deck of the model = 4 m2 PROBLEM SET - 1 (TEXT BOOK PAGE NO. 188)
4.
Sol.
BC (ii)
DC 3
DC DC DC 32
cm
4c m
3
A In the adjoining figure, D is a point on BC such that ABD = CAD. If AB = 5 cm, cm AD = 4 cm and AC = 3 cm. Find 5 (i) BC (ii) DC (iii) A (ACD) : A (BCA). C (3 marks) B In ACD and BCA, D CAD ABC [Given and B - D - C] ACD ACB [Common angle] ACD BCA [By AA test of similarity] AC DC AD = = [c.s.s.t] BC AC AB 3 DC 4 = = ......(i) [Given] BC 3 5 3 4 = [From (i)] (i) BC 5 3×5 BC = 4 15 BC = 4
= 3.75 cm
4 5 4×3 = 5 12 = 5 = 2.4 cm
=
[From (i)]
S C H O O L S E C TI O N
MT
GEOMETRY
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(iii) ACD BCA A(ACD) AD2 A(BCA) = AB2
A(ACD) 42 A(BCA) = 52 A(ACD) 16 = A(BCA) 25
[Areas of similar triangles] [Given]
A(ACD) : A(BCA) = 16 : 25 EXERCISE - 1.4 (TEXT BOOK PAGE NO. 21) 5.
Sol.
In ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC. If PQ divides ABC into two equal parts means equal in BP . (5 marks) area, find AB seg PQ divides ABC into two parts of equal areas [Given] A (APQ) =
B
1 A (ABC) 2
A (APQ) 1 ......(i) A (ABC) = 2 seg PQ || side BC On transversal AC, AQP ACB ......(ii) In APQ and ABC, AQP ACB PAQ BAC APQ ~ ABC
P
A (APQ) AP 2 = A (ABC) AB2 1 2
1
2
[Taking square roots]
AB BP – = AB AB
1–
BP AB
S C H O O L S E C TI O N
1 2
[Areas of similar triangles]
AP AB
= =
BP = AB
C
[From (ii)] [Common angle] [By AA test of similarity]
=
=
Q
[Converse of corresponding angles test]
[From (i)]
AB – BP AB
1–
A
AP 2 AB2
[Given]
=
AP = AB
1 2
1 2
[ A - P - B]
1 2
1 2
BP AB 2 –1 2 33
MT
GEOMETRY
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PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) 17.
Sol.
In the adjoining figure, XY || AC and XY divides the triangular region ABC into two equal areas. Determine AX : AB. (4 marks) seg XY divides ABC in two C parts of equal areas 1 A(XYB) = A(ACB) 2 A (XYB) 1 A (ACB) = .......(i) 2 seg XY || side AC On transversal BC, XYB ACB .....(ii) In XYB and ACB, XYB ACB XBY ABC XYB ~ ACB A (XYB) XB2 A (ACB) = AB2 1 XB2 = 2 AB2 1 XB = 2 AB 1 XB = 2 AB 1 AB – AX = 2 AB 1 AB AX – = 2 AB AB 1 AX 1– = 2 AB
1–
1 2
=
AX AB
=
A X
Y
B
[Given] [Converse of corresponding angles test] [From (ii)] [Common angle] [By AA test of similarity] [Areas of similar triangles] [From (i)] [Taking square roots]
[ A - X - B]
AX AB 2–1 2
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) F In the adjoining figure, ABCD is a square. The BCE on C side BC and ACF on the diagonal AC D are similar to each other. Then show 1 A (ACF) that A (BCE) = (3 marks) 2 Proof : BCE ~ ACF [Given] A B 2 A (BCE) BC A (ACF) = ......(i) [Areas of similar triangles] AC2 ABCD is a square [Given]
23.
34
E
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
AB = BC In ABC, m ABC AC 2 AC 2 AC 2 A (BCE) A (ACF) A (BCE) A (ACF)
= CD = AD
......(ii)
90º AB2 + BC2 BC2 + BC2 2BC2 .....(iii) 2 BC = 2BC2 1 = 2 1 A (BCE) = A (ACF) 2
= = = =
[Sides of a square] [Angle of a square] [By Pythagoras theorem] [From (i)] [From (i) and (iii)]
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 21) 4.
Sol.
B In the adjoining figure, DE ||BC, (i) If DE = 4 cm, BC = 8 cm, D A (ADE) = 25 cm2, find A (ABC) (ii) If DE : BC = 3 : 5, then A (ADE) : A (DBCE). (5 marks) C seg DE || seg BC [Given] A E On transversal AC, AED ACB ......(i) [Converse of corresponding angles test] In ADE and ABC, AED ACB [From (i)] DAE BAC [Common angle] ADE ~ ABC .....(ii) [By AA test of similarity] 2 A (ADE) DE A (ABC) = [Areas of similar triangles] BC2 25 42 A (ABC) = 2 [Given] 8 25 16 A (ABC) = 64 64 25 A (ABC) = 16
A (ABC) = 100 cm2
(ii) ADE ABC A (ADE) DE2 A (ABC) = BC2 2 A (ADE) DE = A (ABC) BC 2 A (ADE) 3 A (ABC) = 5 A (ADE) 9 A (ABC) = 25 A (ABC) 25 A (ADE) = 9 A (ABC) – A (ADE) 25 – 9 = A (ADE) 9 S C H O O L S E C TI O N
[From (ii) [Areas of similar triangles]
[Given]
[By Invertendo] [By Dividendo] 35
MT
GEOMETRY
A (DBCE) 16 = A (ADE) 9 A (ADE) 9 = A (DBCE) 16 A (ADE) : A (DBCE) = 9 : 16
EDUCARE LTD.
[Area addition property] [By Invertendo]
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 21) 6.
In ABC, seg DE || side BC. If 2A (ADE) = A (DBCE), find AB : AD and
A 3 .DE (4 marks) seg DE || side BC [Given] On transversal AB, D E ABC ADE .....(i) [Converse of corresponding angles test] In ABC and ADE, ABC ADE [From (i)] BAC DAE [Common angle] B ABC ~ ADE [By AA test of similarity] AB BC AC = = .....(ii) [c.s.s.t.] AD DE AE A (ABC) AB2 = .....(iii) [Areas of similar triangles] A (ADE) AD2 A (ABC) = A (ADE) + A (DBCE) [Area addition property] = A (ADE) + 2A (ADE) [ A (DBCE) = 2A (ADE)] A (ABC) = 3A (ADE) A (ABC) 3 .....(iv) A (ADE) = 1 3 AB2 = [From (iii) and (iv)] 1 AD2 AB 3 = ......(v) [Taking square roots] AD 1 AB : AD = 3 : 1
show BC = Sol.
BC AB = DE AD BC 3 = DE 1 BC = 3 × DE
C
[From (ii)] [From (v)]
EXERCISE - 1.4 (TEXT BOOK PAGE NO. 21) 2.
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas. (3 marks) Given : (i) In ABC, P seg AD side BC A (ii) In PQR, seg PS side QR (iii) ABC ~ PQR (iv) AD = 6 cm, PS = 9 cm C Q B D S R
36
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
A (ABC) To find : A (PQR) Sol. ABC ~ PQR A (ABC) AD2 A (PQR) = PS2 A (ABC) 62 A (PQR) = 2 9 2 A (ABC) 6 A (PQR) = 9 2 A (ABC) 2 A (PQR) = 3 A (ABC) 4 A (PQR) = 9
[Given] [Altitudes of similar triangles] [Given]
A (ABC) : A (PQR) = 4 : 9 EXERCISE - 1.4 (TEXT BOOK PAGE NO. 21) 3.
The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians ? (3 marks) P
A
B
D
E
C
Q
S
T
R
Given :(i) ABC ~ PQR (ii) In ABC, seg AD side BC and seg AE is the median (iii)In PQR, seg PS side QR and seg PT is the median (iv) A (ABC) = 81cm2, A (PQR) = 49 cm2 AD AE To find : (i) (ii) PS PT Sol. ABC ~ PQR [Given] 2 A (ABC) AD A (PQR) = [Altitudes of similar triangles] PS2 81 AD2 = [Given] 49 PS2 AD 9 = [Taking square roots] PS 7 AD : PS = 9 : 7 ABC ~ PQR [Given] A (ABC) AE2 A (PQR) = [Medians of similar triangles] PT 2
81 AE2 = 49 PT 2 AE 9 = PT 7
[Given] [Taking square roots]
AE : PT = 9 : 7
S C H O O L S E C TI O N
37
MT
GEOMETRY
EDUCARE LTD.
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 189) 6.
Sol.
P In the adjoining figure, ABCD is a parallelogram whose diagonals intersect at O. P is a point on the diagonal AC such that PA : AO = 1 : 2. A Q D BP meets DA produced at Q. Then find O (i) PQ : QB (ii) A (PQA) : A (PBC) (iii) A (PQA) : A (QBCA) C PA PA 1 PA 1 1 1 B = = × = × = ] (5 marks) [Hint : AC 2AO 2 AO 2 2 4 (i) ABCD is a parallelogram [Given] seg AD || seg BC ......(i) [By definition] In PBC, [From (i) and Q - A - D] seg QA || side BC PQ PA = ....(ii) [By B.P.T.] QB AC [ Diagonals of paralle log ram PA PA = bisec t each other ] AC 2AO
PA AC PA AC PA AC PQ QB
1 PA × 2 AO 1 1 = × 2 2 1 = 4 1 = 4 =
PA 1 AO 2 .......(iii) [From (ii) and (iii)]
PQ : QB = 1 : 4 (ii) seg QA || seg BC On transversal PB, PQA PBC In PQA and PBC, PQA PBC QPA BPC PQA ~ PBC
38
A (PQA) PA 2 = A (PBC) PC 2 2 A (PQA) PA = A (PBC) PC PA 1 = AC 4 AC 4 = PA 1 AC + PA 4 +1 = PA 1 PC 5 = PA 1
[From (i) and Q - A - D] .......(iv)
[Converse of corresponding angles test] [From (iv)] [Common angles] [By AA test of similarity] [Areas of similar triangles]
.......(v) [From (iii)] [By Invertendo] [By Componendo] [ P - A - C] S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
PA 1 = PC 5 2 A (PQA) 1 A (PBC) = 5
.......(vi)
[By Invertendo] [From (v) and (vi)]
A (PQA) 1 A (PBC) = 25
A (PQA) : A (PBC) = 1 : 25
A (PQA) 1 A (PBC) = 25 A (PBC) 25 A (PQA) = 1
(iii)
A (PBC) – A (PQA) 25 – 1 = A (PQA) 1 A (QBCA) 24 = A (PQA) 1
A (PQA) 1 A (QBCA) = 24
[By Invertendo] [By Dividendo] [Area addition property] [By invertendo]
A (PQA) : A (QBCA) = 1 : 24
CENTROID OF A TRIANGLE :
A
In ABC, E seg AD, seg BE and seg CF are the medians F G G is the centroid of ABC 2 B AG = AD D 3 1 GD = AD [ Centroid of a triangle trisects each 3 AG = 2GD median]
GD =
C
1 AG 2 PROBLEM SET - 1 (TEXT BOOK PAGE NO. 188)
A G is the centroid of ABC. GE and GF are drawn parallel to AB and AC respectively. Find A (GEF) : A (ABC). G (Hint : Draw the median AD. Then GD : AD = 1 : 3. B E D F C A (GED) : A (ABD) = 1 : 9) (5 marks) Construction : Draw seg AG and let it intersect side BC at point D. Sol. seg GE ll side AB [Given] On transversal BC, GEC ABC [Converse of corresponding angles test] GED ABD ......(i) [E - D - C and B - D - C]
5.
S C H O O L S E C TI O N
39
MT
GEOMETRY
In GED and ABD GED ABD GDE ADB GED ABD A (GED) GD2 A (ABD) = AD2
[From (i)] [Common angle] [By AA test of similarity] [Areas of similar triangles]
2 A (GED) GD A (ABD) = AD 1 AD But, GD = 3 GD 1 = AD 3 2 A (GED) 1 A (ΑΒD) = 3
......(ii)
[ Centriod of a triangle trisec ts each median] .....(iii) [From (ii) and (iii)]
A (GED) 1 A (ΑΒD) = 9 Similarly we can get,
.....(iv)
A (GFD) 1 A (ΑCD) = 9
.....(v)
A (GED) A (GFD) 1 = = A (ΑΒD) A (ΑCD) 9 A (GED) A (GFD) 1 A (ΑΒD) + A (ΑCD) = 9 A (GEF) 1 A (ΑBC) = 9
A (GEF) : A (ABC)
EDUCARE LTD.
[From (iv) and (v)] [Theorem on equal ratio] [Area addition property]
= 1:9
ALTERNATIVE METHOD :
seg GE ll side AB On transversal BC, GEC ABC seg GF ll side AC On transversal BC, GFB ACB In GEF and ABC, GEF ABC GFE ACB GEF ABC A (GEF) GE2 = A (ABC) AB2 2 A (GEF) GE A (ABC) = AB
In GED and ABD, GED ABD GDE ADB GED ABD 40
[Given] ......(i)
[Converse of corresponding angles test] [Given]
......(ii)
[Converse of corresponding angles test] [From (i) and E - F - C] [From (ii) and E - F - C] [By AA test of similarity] [Areas of similar triangles]
......(iii)
[From (i) and E - D - C] [Common angle] [By AA test of similarity] S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
GE GD = AB AD 1 AD But, GD = 3
......(iv)
[ Centriod of a triangle trisec ts each median ]
GD 1 = AD 3
......(v)
GE 1 = AB 3
.....(vi)
2 A (GEF) 1 = A (ΑΒC) 3
[c.s.s.t.]
[From (iv) and (v)] [From (iii) and (vi)]
A (GEF) 1 = A (ΑΒC) 9
A (GEF) : A (ABC) = 1 : 9
SIMILARITY IN RIGHT ANGLED TRIANGLES In a right angled triangle, if the perpendicular is drawn from the vertex of the right angle to the hypotenuse, then the triangles on either side of the perpendicular are similar to the original triangle and to each other. In ABC,
A
m ABC = 90º
D
seg BD hypotenuse AC ABC ~ ADB ~ BDC B
C
PROPERTY OF GEOMETRIC MEAN In a right angled triangle, the length of perpendicular segment drawn on to the hypotenuse from the opposite vertex, is the geometric mean of the length of the segments into which the hypotenuse is divided. P In PQR, m PQR = 90º S seg QS hypotenuse PR QS² = PS × RS Q
R
EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) ABD is a triangle in which A = 90º and seg AC hypotenuse BD.Show that : (i) AB2 = BC . BD (ii) AD2 = BD . CD (3 marks) (iii)AC2 = BC . CD Proof :(i) In ABD, [Given] m BAD = 90º 2.
S C H O O L S E C TI O N
A
B
C
D
41
MT
GEOMETRY
seg AC hypotenuse BD BAD ~ BCA ~ ACD .....(i) BAD ~ BCA AB BD = BC AB AB2 = BC × BD
EDUCARE LTD.
[Given] [Similarity in right angled triangles] [From (i)] [c.s.s.t.]
(ii) BAD ~ ACD AD BD = CD AD AD2 = BD × CD
[From (i)]
(iii)BCA ~ ACD AC BC = CD AC AC2 = BC × CD
[From (i)]
[c.s.s.t.]
[c.s.s.t.]
EXERCISE 1.7 (TEXT BOOK PAGE NO. 41) M
4.
In the adjoining figure, LMN = 90º and LKN = 90º, seg MK seg LN. Prove that R is the midpoint of seg MK. (3 marks) L N R Proof : In LMN, [Given] m LMN = 90º seg MR hypotenuse LN [Given] K MR2 = LR × RN .....(i) [By property of geometric mean] In LKN, m LKN = 90º [Given] seg KR hypotenuse LN [Given] KR2 = LR × RN .....(ii) [By property of geometric mean] MR2 = KR2 [From (i) and (ii)] MR = KR [Taking square roots] R is the midpoint of seg MK
THEOREM OF PYTHAGORAS Statement : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides. (4 marks) Given : In ABC, m ABC = 90º A To Prove : AC² = AB² + BC² D Construction : Draw seg BD side AC such that A - D - C. Proof : In ABC, B m ABC = 90º [Given] C seg BD hypotenuse AC [Construction] ABC ~ ADB ~ BDC .....(i) [Similarity in right angled triangles] ABC ~ ADB [From (i)] AB AC = [Corresponding sides of similar triangles] AD AB AB² = AC × AD ........(ii) ABC ~ BDC [From (i)] BC AC = [Corresponding sides of similar triangles] DC BC BC² = AC × DC ........(iii) 42
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
Adding AB² + AB² + AB² + AB² +
(ii) and BC² = BC² = BC² = BC² = AC² =
(iii) we get, AC × AD + AC × DC AC (AD + DC) AC × AC [ A - D - C] AC² AB² + BC²
EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) 3. Sol.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall. (2 marks) A In the adjoining figure, seg AB represents the wall seg AC represents the ladder seg BC represents the distance of the foot of the ladder from the base of the wall C B AC = 10 m AB = 8 m In ABC, [Given] m ABC = 90º AC 2 = AB2 + BC2 (10)2 = (8)2 + BC2 100 = 64 + BC2 BC2 = 100 – 64 BC2 = 36 [Taking square roots] BC = 6 m
The distance of the foot of the ladder from the base of the wall is 6 m. PROBLEM SET - 1 (TEXT BOOK PAGE NO. 189) 11.
Sol.
A In the adjoining figure, 4 an altitude is drawn to the hypotenuse. The lengths D of different segments are marked x 5 y in each figure. Determine the values of x, y, z in each case. (3 marks) B C z (i) In ABC, mABC = 90º [Given] [Given] seg BD hypotenuse AC 2 BD = AD × DC [By property of geometric mean] y2 = 4 × 5 y = [Taking square roots] 4×5
y = 2 5 In ADB, m ADB = 90º AB 2 = AD2 + BD2 AB 2 = 42 + y2 x2 = 42 + (2 5 )2 x2 = 16 + 20 x2 = 36 x = 6
S C H O O L S E C TI O N
[Given] [By Pythagoras theorem]
[Taking square roots] 43
MT
GEOMETRY
In BDC, m BDC = 90º BC2 = BD2 + CD2 z 2 = y2 + 52
EDUCARE LTD.
[By Pythagoras theorem]
z2 = (2 5 )2 + 52 z2 = 20 + 25 z2 = 45
z =
z = 3 5
9×5
(ii) In PSQ, m PSQ = 90º PQ 2 = PS2 + QS2 6 2 = 42 + y2 36 = 16 + y2 y2 = 36 – 16 y2 = 20 y = 45
[Taking square roots]
P
[Given] [By Pythagoras theorem]
4 S
6
X
y Q
z
R
[Taking square roots]
y = 2 5 In PQR, m PQR = 90º seg QS hypotenuse PR QS 2 = PS × SR y2 = 4 × x ( 2 5 )2 = 4 × x 20 = 4 × x 20 x = 4 x = 5 In QSR, m QSR = 90º QR 2 = QS2 + SR2 z 2 = y2 + x 2
[Given] [Given] [By property of geometric mean]
[Given] [By Pythagoras theorem]
z2 = ( 2 5 )2 + (5)2 z2 = 20 + 25 z2 = 45
z =
z = 3 5
9×5
[Taking square roots]
EXERCISE 1.6 (TEXT BOOK PAGE NO. 36) 2.
Find the side of square whose diagonal is 16 2 cm . Given : ABCD is a square. A AC = 16 2 cm To find : Side of a square x Sol. ABCD is a square [Given] Let the sides of the square be x cm In ABC, B m ABC = 90º [Angle of a square] 44
x
(2 marks)
16 2 cm
D
x
x
C
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
AC 2
= AB2 + BC2
[By Pythagoras theorem]
2
16 2 = x2 + x2 256 × 2 = 2x2 256 2 x2 = 2 x2 = 256 x = 16
[Taking square roots]
The side of a square is 16 cm. EXERCISE 1.6 (TEXT BOOK PAGE NO. 36) 4.
Find the perimeter of an isosceles right triangle with each of its (2 marks) congruent sides is 7 cm. A Given : In ABC, m ABC = 90º 7 cm AB = BC = 7 cm To find : Perimeter of ABC Sol. In ABC, B C 7 cm m ABC = 90º [Given] AC2 = AB2 + BC2 [By Pythagoras theorem] 2 2 2 AC = (7) + (7) AC2 = 49 + 49 AC2 = 98 AC = 49 2 [Taking square roots] AC = 7 2 cm Perimeter of ABC = AB + BC + AC
= 7 7 7 2 = 14 7 2
Perimeter of ABC = 7 2
2 cm
EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) 4.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. (4 marks) A
D
Given :
(i) ABCD is a rhombus (ii) Diagonals AC and BD O intersect each other at point O. To prove : AB2 + BC2 + CD2 + AD2 = AC2 + BD2 B Proof : ABCD is a rhombus [Given] AB = BC = CD = AD .....(i) [Sides of a rhombus] m AOB = 90º ....(ii) AO =
1 AC 2
1 BO = BD 2 In AOB, m AOB = 90º AB2 = AO2 + BO2 S C H O O L S E C TI O N
....(iii)
C
[Diagonals of a rhombus are perpendicular bisectors of each other]
.....(iv) [From (ii)] [By Pythagoras theorem] 45
MT
GEOMETRY 2
EDUCARE LTD.
2
1 1 AB = AC BD [From (iii) and (iv)] 2 2 1 1 AB2 = AC2 + BD2 4 4 Multiplying throughout by 4, 4AB2 = AC2 + BD2 AB2 + AB2 + AB2 + AB2 = AC2 + BD2 AB2 + BC2 + CD2 + AD2 = AC2 + BD2 [From (i)] 2
EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) 5.
The perpendicular AD on the base BC of ABC intersects A BC at D so that BD = 3 CD. Prove that 2AB2 = 2AC2 + BC2. (3 marks) Proof : BC = BD + CD [B - D - C] BC = 3CD + CD [ BD = 3CD] BC = 4CD .....(i) D B In ADB, m ADB = 90º [Given] AB 2 = AD2 + BD2 [By Pythagoras theorem] AB 2 = AD2 + (3CD)2 [ BD = 3CD] AB 2 = AD2 + 9CD2 AB 2 = AD2 + CD2 + 8CD2 .....(ii) In ADC, m ADC = 90º [Given] AC 2 = AD2 + CD2 .....(iii) [By Pythagoras theorem] AB2 = AC2 + 8CD2 [From (ii) and (iii)]
C
2
BC AB = AC + 8 4 BC2 AB2 = AC2 + 8 × 16 2 BC AB2 = AC2 + 2 2AB2 = 2AC2 + BC2 2
2
[From (i)]
[Multiplying throughout by 2]
EXERCISE 1.5 (TEXT BOOK PAGE NO. 31) 8.
ABC is a triangle in which AB = AC and D is any point on BC. Prove that : AB2 – AD2 = BD. CD. (4 marks)
Proof : 46
A
In AEB, B m AEB = 90º [Given] D E AB2 = AE2 + BE2 ......(i) [By Pythagoras theorem] In AED, m AED = 90º [Given] 2 2 2 AD = AE + DE .....(ii) [By Pythagoras theorem] Subtracting equation (ii) from (i), AB2 – AD2 = AE2 + BE2 – (AE2 + DE2) AB2 – AD2 = AE2 + BE2 – AE2 – DE2 AB2 – AD2 = BE2 – DE2 AB2 – AD2 = (BE + DE) (BE – DE) AB2 – AD2 = (BE + DE) × BD .......(iii) [ B - D - E]
C
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
In AEB and AEC, m AEB = m AEC = 90º hypotenuse AB hypotenuse AC seg AE seg AE AEB AEC seg BE seg CE .....(iv) AB2 – AD2 = (CE + DE) × BD AB2 – AD2 = CD × BD
[Given] [Given] [Common side] [By hypotenuse side theorem] [c.s.c.t.] [From (iii) and (iv)] [ D - E - C]
EXERCISE 1.7 (TEXT BOOK PAGE NO. 41) A Seg AD is the median of ABC, and AM BC. Prove that : 2 BC (i) AC2 = AD2 + BC × DM + 2 2 BC (ii) AB2 = AD2 – BC × DM + (4 marks) C 2 B M D Proof :(i) In AMD, m AMD = 90º AD2 = AM2 + DM2 ......(i) [By Pythagoras theorem] In AMC, m AMC = 90º [Given] 2 2 2 AC = AM + MC [By Pythagoras theorem] AC2 = AM2 + (DM + DC)2 [ M - D - C] 2 2 2 AC = AM + DM + 2DM × DC + DC2 AC2 = AM2 + DM2 + 2DC × DM + DC2 2 [ From (i) and D is the midpo int BC 2 2 AC = AD + BC × DM + of side BC ] 2
5.
(ii) In AMB, m AMB = 90º [Given] 2 2 2 AB = AM + BM [By Pythagoras theorem] AB2 = AM2 + (BD – DM)2 AB2 = AM2 + BD2 – 2BD × DM + DM2 AB2 = AM2 + DM2 – 2BD × DM + BD2 2 [ From (i) and D is the midpo int BC AB2 = AD2 – BC × DM + of side BC ] 2 EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) 6.
ABC is a triangle where C = 90º. Let BC = a, CA = b, AB = c and let ‘p’ be the length of the perpendicular from C on AB. Prove that (i) cp = ab, (ii)
Proof :
1 1 1 = 2 + 2 2 p a b
Area of a triangle = A (ABC) =
S C H O O L S E C TI O N
A (ABC) =
(5 marks) A
1 × base × height 2 1 × AB × CD 2 1 ×c×p 2
D
b
c
p
......(i)
C
a
B
47
MT
GEOMETRY
Also,
A (ABC) =
EDUCARE LTD.
1 × BC × AC 2
1 ×a×b 2 From (i) and (ii) we get, A (ABC) =
......(ii)
1 1 ×c×p = ×a×b 2 2 cp = ab = ab
(ii) cp 1 1 cp = ab 1 1 2 2 = 2 2 c p a b
1 c2 = 2 p a 2 b2 In ACB, m ACB = 90º AB 2 = AC2 + BC2 c 2 = b2 + a2 1 b2 a 2 p2 = a 2 b2
1 b2 a2 p2 = a 2 b 2 a 2 b2
1 1 1 + 2 2 = 2 p a b
[By Invertendo] [Squaring both sides] .....(iii) [Given] [By Pythagoras theorem] .....(iv) [From (iii) and (iv)]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) C ABC is a right angled triangle with A = 90º. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm M 6 cm and 8 cm. Find the radius of the circle. (4 marks) Construction : Let the sides AB, BC and AC N O touch the circle at points L, M and N respectively. Draw seg OA, seg OB, A B L seg OC, seg OL, seg OM and ON. 8 cm Sol. In ABC, m BAC = 90º [Given] BC2 = AC2 + AB2 [By Pythagoras theorem] BC2 = 62 + 82 BC 2 = 36 + 64 BC 2 = 100 BC = 10 cm .......(i) [Taking square roots] Let the radius of the circle be r OL = OM = ON = r .......(ii) [Radii of the same circle] seg OL side AB seg OM side BC [Radius is perpendicular to the tangent] seg ON side AC
12.
48
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
1 × base × height 2 1 A (AOB) = × AB × OL 2 1 A (AOB) = ×8×r .....(iii) 2 A (AOB) = 4r Similarly, A (BOC) = 5r .....(iv) .....(v) A (AOC) = 3r Area of a triangle =
Adding (iii), (iv) and (v), A (AOB) + A (BOC) + A (AOC) A (ABC) 1 × AB × AC 2 1 ×8×6 2 24 r
[From (ii)]
= 4r + 5r + 3r [From (iii), (iv) and (v)] = 12r [Area addition property) = 12r = 12r = 12r = 2
The radius of the circle is 2 cm.
CONVERSE OF THEOREM OF PYTHAGORAS Statement : In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the angle opposite to the first side (4 marks) is a right angle and the triangle is right angled triangle. P A Given : In ABC, AC² = AB² + BC² To Prove : m ABC = 90º Construction : Draw PQR such that PQ = AB, QR = BC and R C Q B m PQR = 90º Proof : QR = BC .......(i) [Construction] PQ = AB ......(ii) In PQR, ......(iii) [Construction] m PQR = 900 PR² = PQ² + QR² [By Pythagoras theorem] PR² = AB² + BC² .......(iv) [From (i) and (ii)] But, AC² = AB² + BC² .......(v) [Given] PR² = AC² [From (iv) and (v)] PR = AC .......(vi) [Taking square roots] In ABC and PQR, [From (ii)] side AB side PQ [From (i)] side BC side QR [From (vi)] side AC side PR ABC PQR [SSS test of congruence] ABC PQR [c.a.c.t] m ABC = m PQR = 90º [From (iii)] m ABC = 90º S C H O O L S E C TI O N
49
MT
GEOMETRY
EDUCARE LTD.
EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) 1. (i) Sol.
Sides of triangles are given below. Determine which of them are right (2 marks) angled triangles. 8, 15, 17 (17)2 = 289 ......(i) 2 2 (8) + (15) = 64 + 225 = 289 ......(ii) (17)2 = (8)2 + (15)2 [From (i) and (ii)] The given sides form a right angled triangle. [By Converse of Pythagoras theorem]
(ii) Sol.
9, 40, 41 (41)2 = 1681 ......(i) 2 2 (9) + (40) = 81 + 1600 = 1681 ......(ii) 2 2 2 (41) = (9) + (40) [From (i) and (ii)] The given sides form a right angled triangle. [By Converse of Pythagoras theorem]
(iii) Sol.
40, 20, 30 (40)2 = 1600 ......(i) (20)2 + (30)2 = 400 + 900 = 1300 ......(ii) 2 2 2 (40) (20) + (30) The given sides do not form a right angled triangle.
[From (i) and (ii)] [By Converse of Pythagoras theorem]
(iv) Sol.
11, 60, 61 (61)2 = 3721 ......(i) (11)2 + (60)2 = 121 + 3600 = 3721 ......(ii) (61)2 = (11)2 + (60)2 [From (i) and (ii)] The given sides form a right angled triangle. [By Converse of Pythagoras theorem]
(v) Sol.
11, 12, 15 (15)2 = 225 ......(i) (11)2 + (12)2 = 121 + 144 = 265 ......(ii) (15)2 (11)2 + (12)2 The given sides do not form a right angled triangle.
(vi) Sol.
50
[From (i) and (ii)] [By Converse of Pythagoras theorem]
12, 35, 37 (37)2 = 1369 ......(i) (12)2 + (35)2 = 144 + 1225 = 1369 ......(ii) (37)2 = (12)2 + (35)2 [From (i) and (ii)] The given sides form a right angled triangle. [By Converse of Pythagoras theorem] S C H O O L S E C TI O N
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THEOREM OF 300 - 600 - 900 TRIANGLE If the angles of a triangle are 300, 600 and 900, then the side opposite to 300 is half of the hypotenuse and the side opposite to 600 is A 3 times the hypotenuse. 30 0 2 In ABC, m A = 30º, m C = 60º and m B = 90º
BC =
1 3 AC and AB = AC. 2 2
B
60 0
C
THEOREM OF 450 - 450 - 900 TRIANGLE If the angles of a triangle are 45º - 45º - 90º then the length of the 1 perpendicular sides are times the hypotenuse. 2 In ABC, A m A = 45º 45 0 m C = 45º m B = 90º 1 AB = BC = AC. 2 45 0 C B EXERCISE 1.6 (TEXT BOOK PAGE NO. 36) 1.
Find the length of the altitude of an equilateral triangle, each side measuring (3 marks) ‘a’ units. A Given : ABC is an equilateral triangle. AB = BC = AC = a seg AD side BC a a To find : AD Sol. ABC is an equilateral triangle AB = BC = AC = a ......(i) [Given] C B In ADB, D a [Given] m ADB = 90º [Angle of an equilateral triangle] m ABD = 60º m BAD = 30º [Remaining angle] ADB is a 30º - 60° - 90º triangle By 30º - 60º - 90º triangle theorem, 3 × AB 2
[Side opposite to 60º]
AD =
3 ×a 2
[From (i)]
AD =
3 a units. 2
AD =
S C H O O L S E C TI O N
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EXERCISE 1.6 (TEXT BOOK PAGE NO. 36) 5.
Sol.
P In the adjoining figure, PQR = 60º and A ray QT bisects PQR. T seg BA ray QP and B seg BC ray QR. If BC = 8 find the perimeter of ABCQ. Q (4 marks) R C Ray QT is the angle bisector of PQR [Given] B lies on ray QT seg BA ray QP [Given] seg BC ray QR BA = BC [Angle bisector theorem] But, BC = 8 units .......(i) [Given] BA = 8 units ......(ii) m PQR = 60º [Given] 1 m PQT = m RQT = × 60º [ Ray QT bisects QR] 2 m PQT = m RQT = 30º ......(iii) In BAQ, m BAQ = 90º [Given] m AQB = 30º [From (iii), P - A - Q and Q - B - T] m ABQ = 60º [Remaining angle] BAQ is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem 1 AB = BQ [Side opposite to 30º] 2 1 8 = BQ [From (ii)] 2 BQ = 16 units
AQ =
3 × BQ 2
3 × 16 2 AQ = 8 3 units
[Side opposite to 60º]
AQ =
......(iv)
Similarly, QC = 8 3 cm ......(v) Perimeter of ABCQ = AB + BC + QC + AQ = 888 3 8 3 [From (i), (ii), (iv) and (v)] = 16 16 3
Perimeter of ABCQ = 16 1
3
units
EXERCISE 1.6 (TEXT BOOK PAGE NO. 37) 6.
In the adjoining figure,
45º
if LK = 6 3 find MK, ML, KN, MN and the perimeter of MNKL. (4 marks) Sol. 52
In MLK, m MLK = 90º
[Given]
N
M
L
30º 6 3
K
S C H O O L S E C TI O N
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m MKL = 30º [Given] m LMK = 60º [Remaining angle] MLK is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem, LK
=
6 3 =
MK
=
3 MK 2
[Side opposite to 60º]
3 × MK 2 6 3 2
[Given]
3 = 12 units ......(i) 1 ML = × MK [Side opposite to 30º] 2 1 ML = × 12 2 ML = 6 units ......(ii) In MKN, [Given] m MKN = 90º m MNK = 45º [Given] m NMK = 45º [Remaining angle] MKN is a 45º - 45º - 90º triangle By 45º - 45º - 90º triangle theorem, 1 MK = KN = × MN ......(iii) 2 1 × MN [From (iii)] MK = 2 1 12 = × MN 2 MN = 12 2 units .....(iv) KN = 12 units ......(v) [From (i) and (iii)] Perimeter of MNKL = MN + KN + KL +ML = 12 2 12 6 3 6 [From (ii), (iv) and (v) and given] = 18 12 2 6 3
MK
Perimeter of MNKL
=
6 32 2
3
units
EXERCISE 1.6 (TEXT BOOK PAGE NO. 37) 7.
Sol.
In the adjoining figure, PQRV is a trapezium in which seg PQ || seg VR. SR = 4 and PQ = 6. Find VR. (4 marks) V In QSR, m QSR = 90º m SQR = 45º m QRS = 45º QSR is a 45º - 45º - 90º triangle. QS = SR
S C H O O L S E C TI O N
P
6
60º
T
Q 45º
S
4
R
[Given] [Given] [Remaining angle] [Congruent sides of 45º - 45º - 90º triangle] 53
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GEOMETRY
But, SR = 4 units QS = 4 units In PQTS, seg PQ || seg TS m T = m S = 90º PQTS is a rectangle PQ = TS = 6 units QS = PT = 4 units
......(i) .....(ii)
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[Given]
[Given and V - T - S - R] [Given] .....(iii) .....(iv)
[Opposite sides of a rectangle and from (ii) and given]
In PTV, m PTV = 90º [Given] [Given] m VPT = 60º m PVT = 30º [Remaining angle] PTV is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem, 1 × PV [Side opposite to 30º] PT = 2 1 4 = × PV [From (iv)] 2 PV = 8 units ......(v) VT =
3 PV 2
[Side opposite to 60º]
3 ×8 2 VT = 4 3 units VR = VT + TS + SR
VT =
[From (v)] .....(vi)
VR = 4 3 6 4
[V - T - S - R] [From (i), (iii) and (vi)]
VR = 4 3 10
VR = 2 2 3 5 units EXERCISE 1.5 (TEXT BOOK PAGE NO. 30) 7.
Prove that three times the square of any side of an equilateral triangle A (3 marks) is equal to four times the square of an altitude.
To prove : 3AB2 = 4AD2 Proof : ABC is an equilateral triangle In ADB, m ADB = 90º m ABD = 60º
[Given] [Given] [Angle of an B equilateral triangle] [Remaining angle]
D
C
m BAD = 30º ADB is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem, AD =
3 AB 2
[Side opposite to 60º]
2AD =
3 AB 4AD2 = 3AB2 3AB2 = 4AD2 54
[Squaring both sides]
S C H O O L S E C TI O N
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CONVERSE OF 30º - 60º - 90º TRIANGLE THEOREM 3 time 2 the hypotenuse, then the measure of the angle opposite to the side is 60º. In ABC, A m ABC = 90º [Given] 1 AB = AC 10 5 2 m C = 30º
Statement : In a right angled triangle, if the length of one side is
3 AC 2 m A = 60º
C
B
BC =
5 3
EXERCISE 1.6 (TEXT BOOK PAGE NO. 36) 3.
If the sides of a triangle measure 18 cm, 18 3 cm and 36 cm, show that it is a 30º - 60º - 90º triangle. (3 marks) A Given : In ABC, 18 cm AC = 36 cm, AB = 18 cm, BC = 18 3 cm To prove : ABC is a 30º - 60º - 90º triangle. 36 cm Proof : AC2 = (36)2 2 AC = 1296 .......(i)
AB2 + BC2 = (18)2 + 18 = 324 + 972 AB2 + BC2 = 1296 AC 2 = AB2 + BC2 ABC is a right angled triangle at point B 1 1 × AC = × 36 2 2 1 × AC = 18 2 1 × AC = AB 2 m C = 30º
3
2
B
......(ii) .....(iii)
18 3 cm
C
[From (i) and (ii)] [Converse of Pythagoras theorem] [ AC = 36]
[ AB = 18] .....(iv)
[By converse of 30º - 60º - 90º triangle theorem]
In ABC, [From (iii)] m B = 90º m C = 30º [From (iv)] m A = 60º [Remaining angle] ABC is a 30º - 60º - 90º triangle.
APPOLLONIUS THEOREM Appollonius theorem is a theorem relating the length of a median of a triangle to the lengths of its sides. Let us study how to make use of Appollonius theorem in the given figure. A In ABC, seg AD is a median, By Appollonius theorem, AB² + AC² = 2AD² + 2BD² or C B D AB² + AC² = 2ADB + 2DC² S C H O O L S E C TI O N
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EXERCISE 1.7 (TEXT BOOK PAGE NO. 40) 1. Sol.
In ABC, AP is a median. If AP = 7, AB2 + AC2 = 260 then find BC. (2 marks) In ABC, seg AP is the median AB2 + AC2 = 2AP2 + 2BP2 260 = 2 (7)2 + 2BP2 260 = 2 (49) + 2BP2 260 = 98 + 2BP2 260 – 98 = 2BP2 2BP2 = 162 162 BP 2 = 2 BP 2 = 81 BP = 9 units 1 BP = BC 2 1 9 = BC 2
A
[Given] B P [By Appollonius theorem] [Given]
C
[Taking square roots] [ P is the midpoint of seg BC]
BC = 18 units EXERCISE 1.7 (TEXT BOOK PAGE NO. 41)
2.
Sol.
In the adjoining figure, AB2 + AC2 = 122, BC = 10. Find the length of the median (2 marks) on side BC. In ABC, seg AQ is the median 1 BQ = QC = × BC 2 1 BQ = QC = × 10 2 BQ = QC = 5 units ......(i) AB2 + AC2 = 2AQ2 + 2BQ2 122 = 2AQ2 + 2 (5)2 122 = 2AQ2 + 2 (25) 122 = 2AQ2 + 50 2AQ2 = 122 – 50 2AQ2 = 72 AQ2 = 36
AQ = 6 units
A
[Given]
B
Q
C
[Given] [By Appollonius theorem] [From (i) and given]
[Taking square roots]
PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190) 13. Sol.
56
In PQR, seg PM is the median. If PM = 9 and PQ2 + PR2 = 290. Find QR. (2 marks) In PQR, P seg PM is the median [Given] PQ2 + PR2 = 2PM2 + 2QM2 [By Appollonius theorem] 9 290 = 2 (9)2 + 2QM2 [Given] 290 = 2 (81) + 2QM2 R Q 290 = 162 + 2QM2 M
S C H O O L S E C TI O N
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290 – 162 = 2QM2 128 = 2QM2 128 QM2 = 2 QM2 = 64 QM = 8 units 1 QM = QR 2 1 8 = QR 2 8 × 2 = QR
[Taking square roots] [ M is midpoint of side QR]
QR = 16 units PROBLEM SET - 1 (TEXT BOOK PAGE NO. 190)
From the information given in the figure show that PM = PN = 3 a . (3 marks) P Proof :
a
a
14.
MQ = QR = a [Given] Q is the midpoint of seg MR .......(i) M S a Q a R In PMR, a seg PQ is the median [From (i) and by definition] 2 2 2 2 PM + PR = 2PQ + 2MQ [By Appollonius theorem] PM2 + a2 = 2 (a)2 + 2 (a)2 PM2 + a2 = 2a2 + 2a2 PM2 + a2 = 4a2 PM2 = 4a2 – a2 PM2 = 3a2 PM = .......(ii) [Taking square roots] 3a Similarly we can prove, PN = ......(iii) 3a PM = PN =
3a
N
[From (ii) and (iii)]
EXERCISE 1.7 (TEXT BOOK PAGE NO. 41) 3.
Adjacent sides of a parallelogram are 11 cm and 17 cm. If the length of one of its diagonals is 26 cm. Find the length of the other. (3 marks) A 11 cm
Sol.
ABCD is a parallelogram 1 OB = OD = × BD .....(i) 2
1 × 26 2 OB = OD = 13 cm OB = OD =
S C H O O L S E C TI O N
[Given]
B
17 cm
D
O C
[ Diagonals of paralle log ram bisec t each other] [Given]
57
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GEOMETRY
In ADB, seg AO is the median AB2 + AD2 = 2AO2 + 2OB2 (11)2 + (17)2= 2AO2 + 2 (13)2 121 + 289 = 2AO2 + 2 (169) 410 = 2AO2 + 338 410 – 338 = 2AO2 72 = 2AO2 AO2 = 36 AO = 6 cm 1 AO = × AC 2
EDUCARE LTD.
[From (i) and by definition] [By Appollonius theorem]
[Taking square roots]
[ Diagonals of paralle log ram bisec t each other]
1 × AC 2 AC = 12 cm 6 =
Length of other diagonal is 12 cm. EXERCISE 1.7 (TEXT BOOK PAGE NO. 41) 6.
In the adjoining figure, PQR = 90º. T is the mid point of the side QR. (3 marks) Prove that PR2 = 4PT2 – 3PQ2.
Proof :
In PQR, seg PT is the median
P
Q
[Given]
T
PQ2 + PR2 = 2PT2 + 2QT2 ......(i)
[By Appollonius theorem]
In PQT, m PQT = 90º PT2 = PQ2 + QT2 QT2 = PT2 – PQ2
[Given] [By Pythagoras theorem]
PQ2 + PR2 = PQ2 + PR2 = PR 2 = PR 2 =
R
.....(ii)
2PT2 + 2 (PT2 – PQ2) 2PT2 + 2PT2 – 2PQ2 4PT2 – 2PQ2 – PQ2 4PT2 – 3PQ2
[From (i) and (ii)]
ALTERNATIVE METHOD :
58
In PQR, m PQR = 90º PR 2 = PQ2 + QR2 PR 2 = PQ2 + (2QT)2 PR 2 = PQ2 + 4QT2 .....(i) In PQT, m PQT = 90º PT2 = PQ2 + QT2 QT2 = PT2 – PQ2 .....(ii) PR 2 = PQ2 + 4 (PT2 – PQ2) PR 2 = PQ2 + 4PT2 – 4PQ2 PR 2 = 4PT2 – 3PQ2
[Given] [By Pythagoras theorem] [ T is the midpoint of side QR]
[Given] [By Pythagoras theorem] [From (i) and (ii)]
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HOTS PROBLEM (Problems for developing Higher Order Thinking Skill) In ABCD, side BC || side AD. side AC and side BD intersect in 1 point Q. If AQ = AC then 3 1 show that DQ = BQ. (5 marks) 2 Proof : side AD || side BC On transversal BD ADB CBD ADQ CBQ .........(i) In AQD and CQB, AQD CQB ADQ CBQ AQD ~ CQB AQ DQ = .........(ii) QC BQ 1 But, AQ = AC 3 3AQ = AC 3AQ = AQ + QC 3AQ – AQ = QC 2AQ = QC AQ 1 = .........(iii) QC 2 DQ 1 = BQ 2 1 DQ = BQ 2
A
1.
2.
Q B
C
[Given] [Converse of alternate angles test] [ B - Q - D] [Vertically opposite angles] [From (i)] [By A-A test of similarity] [c.s.s.t.] [Given] [ A - Q - C]
[From (ii) and (iii)]
A
A line cuts two sides AB and side AC of ABC in points P and Q respectively then show that
A (APQ) AP × AQ = A (ABC) AB × AC
D
Q P
(2 marks) B
C
Construction : Draw seg BQ Proof : APQ and ABQ have a common vertex Q and their bases AP and AB lie on the same line AB.
A (APQ) AP ......(i) [Triangles having equal heights] A (ABQ) = AB ABQ and ABC have a common vertex B and their bases AQ and AC lie on the same line AC Their heights are equal A (ABQ) AQ ......(ii) A (ABC) = AC Multiplying (i) and (ii),
S C H O O L S E C TI O N
[Triangles having equal heights]
59
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A (APQ) A (ABQ) AP AQ × = × A (ABQ) A (ABC) AB AC
A (APQ) AP × AQ = A (ABC) AB × AC
3.
In the adjoining figure, AD is the bisector of the exterior A of ABC. seg AD intersects the side BC produced in D.
K A• •
E
BD AB = . (5 marks) B CD AC C Construction : Draw seg CE || seg AD such that B - E - A. Proof : In ABD, seg EC || side AD [Construction] Prove that
BC BE = CD EA
[By B.P.T.]
BC + CD BE + EA = CD EA
[By Componendo]
BD AB = CD AE seg EC || seg AD on transversal EK, KAD AEC on transversal AC, DAC ACE But, KAD DAC AEC ACE In AEC, AEC ACE AE = AC BD AB = CD AC
......(i)
D
[B - C - D and B - E - A] [Construction]
......(ii)
[Converse of corresponding angles test]
......(iii)
[Converse of alterante angles test]
.......(iv) ......(v)
[ ray AD is the bisector of KAC] [From (ii), (iii) and (iv)]
......(vi)
[From (v)] [Converse of isosceles triangle theorem] [From (i) and (vi)]
C In ABC, ACB = 90º, seg CD seg AB E seg DE seg CB show that : CD2 × AC = AD × AB × DE. (2 marks) Proof : In ACB, m ACB = 90º [Given] D A seg CD hypotenuse AB CD2 = AD × DB .....(i) [Property of geometric mean] In ACB and DEB, ACB DEB [ Each is 90º] ABC DBE [Common angle] ACB DEB [By AA test of similarity] AC AB = [c.s.s.t.] DE DB AC × DB = AB × DE
4.
60
B
S C H O O L S E C TI O N
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AB × DE .....(ii) DB Multiplying (i) and (ii), AC =
AB × DE DB CD2 × AC = AD × AB × DE
CD2 × AC = AD × DB × 5.
In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9AD2 = 7AB2. (Hint : Draw AE BC) (2 marks)
A
Construction :Draw seg AE side BC, such that B - D - E - C. B D E C Proof : ABC is an equilateral triangle. [Given] AB = BC = AC ........(i) [Sides of an equilateral triangle] In AED, m AED = 90º [Construction] AD² = AE² + DE² .......(ii) [By Pythagoras Theorem] In AEB, m AEB = 90º [Construction] m ABE = 60º [Angle of an equilateral triangle] m BAE = 30º [Remaining angle] AEB is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem, 3 AB 2 1 BE = AB 2 DE = BE – BD
AE =
DE =
DE =
......(iii) ......(iv)
1 1 AB – BC 2 3
1 1 AB – AB 2 3 3AB 2AB DE = 6 1 DE = AB .......(v) 6 2 2 3 1 AB + AB AD² = 6 2 3 1 AD² = AB² + AB² 4 36 27 AB² + AB² AD² = 36 28AB² AD² = 36 7 AD² = AB² 9 9AD² = 7AB²
S C H O O L S E C TI O N
[Side opposite to 60º] [Side opposite to 30º] [ B - D - E] [From (iv) and Given] [From (i)]
[From (ii), (iii) and (v)]
61
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GEOMETRY
6.
Sol.
EDUCARE LTD.
In the adjoining figure, S R each of segments PA, QB, RC and Q SD is perpendicular to line l. P If AB = 6, BC = 9, CD = 12 and PS = 36, then determine PQ, QR and RS. l seg PA line l A B C D seg QB line l [Given] seg RC line l seg SD line l seg PA || seg QB || seg RC || seg SD ......(i) [Perpendiculars drawn to the same line are parallel to each other] AD = AB + BC + CD [A - B - C - D] AD = 6 + 9 + 12 AD = 27 units .....(ii) seg PA || seg QB || seg SD [From (i)] On transversals AD and PS, [By property of int ercepts made AB PQ = by three parallel lines] AD PS 6 PQ = [From (ii) and given] 27 36 36 × 6 PQ = 27 PQ = 8 units ......(iii)
seg PA|| seg QB || seg RC On transversals AD and PS, PQ AB = QR BC 8 6 = QR 9 9×8 QR = 6
QR
PS 36 36 RS
RS = 16 units
= 12 units = = = =
PQ + QR + RS 8 + 12 + RS 20 + RS 36 – 20
[From (i)]
[By property of int ercepts made by three parallel lines] [Given and from (iii)]
......(iv) [ P - Q - R - S] [From (iii), (iv) and given]
7. In ABC, ABC = 135º. Prove that AC2 = AB2 + BC2 + 4A (ABC). Proof : m ABC + m ABD = 180º [Angles forming linear pair] 135º + m ABD = 180º A m ABD = 180º – 135º m ABD = 45º .....(i) In ADB, 135º m ADB = 90º [Given] C B D m ABD = 45º [From (i)] m BAD = 45º ......(ii) [Remaining angle] In ABD, ABD BAD [From (i) and (ii)] seg AD seg DB .....(iii) [Converse isosceles triangle theorem]
62
S C H O O L S E C TI O N
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8.
GEOMETRY
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In ADB, m ADB = 90º [Construction] 2 2 2 AB = AD + DB .....(iv) [By Pythagoras theorem] In ADC, m ADC = 90º [Construction] AC2 = AD2 + DC2 [By Pythagoras theorem] AC2 = AD2 + (DB + BC)2 [ D - B - C] AC2 = AD2 + DB2 + 2 × DB × BC + BC2 AC2 = AB2 + BC2 + 2 × DB × BC [From (iv)] AC2 = AB2 + BC2 + 2 × AD × BC .....(v) [From (iii)] 1 × base × height Area of triangle = 2 1 A (ABC) = × BC × AD 2 1 4A (ABC) = 4 × × BC × AD [Multiplying throughout by 4] 2 4A (ABC) = 2 × AD × BC ......(vi) AC2 = AB2 + BC2 + 4A (ABC) [From (v) and (vi)]
PQR is a right triangle, right angled at Q such that QR = b and a = A (PQR). If QN PR then 2a . b show that QN = b4 + 4a2
Proof :
Area of triangle =
A (PQR) =
a =
Also, A (PQR) =
N
1 × base × height Q 2
R
1 × QR × PQ 2 1 × b × PQ 2
2a = PQ b
P
[Given] .....(i)
1 × PR × QN 2
a
=
1 × PR × QN 2
QN
=
2a PR
In PQR, m PQR = 90º PR2 = PQ2 + QR2
PR =
PQ2 QR 2
PR =
2a 2 b b
[Given]
.....(ii) [Given] [By Pythagoras theorem] [Taking square roots]
2
S C H O O L S E C TI O N
[From (i) and given] 63
MT
GEOMETRY
PR =
4a 2 b2 b2
PR =
4a 2 b4 b2
PR =
QN =
QN =
b4 4a 2 b 2a
EDUCARE LTD.
......(iii)
b4 4a 2 b 2a . b
[From (ii) and (iii)]
b4 + 4a2
15.
ABCD is an quadrilateral M is the midpoint of diagonal AC and N is the midpoint of diagonal BD. Prove that : AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + D A 4MN 2. Proof : N
M
C
B
In ABC, seg BM is the median AB2 + BC2 = 2BM2 + 2CM2 In ADC, seg DM is the median CD2 + DA2 = 2DM2 + 2CM2
[ M is midpoint of side AC] [By Appollonius theorem]
......(i)
[ M is midpoint of side AC] ......(ii) [By Appollonius theorem]
Adding (i) and (ii), AB2 + BC2 + CD2 + DA2 = 2BM2 + 2CM2 + 2DM2 + 2CM2 AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + 4CM2 2
1 AB + BC + CD + DA = 2BM + 2DM + 4 AC 2 [ M is the midpoint of seg AC] 2
2
2
2
2
2
1 AC2 4 AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + AC2 ......(iii) In BMD, seg MN is the median [ N is the midpoint of side BD] 2 2 2 2 BM + DM = 2MN + 2BN [By Appollonius theorem] 2 2 2 2 2BM + 2DM = 4MN + 4BN [Multiplying throughout by 2] AB2 + BC2 + CD2 + DA2 = 2BM2 + 2DM2 + 4 ×
2
64
2BM + 2DM = 4MN
2BM2 + 2DM2 = 4MN2
2BM2 + 2DM2 = 4MN2 AB2 + BC2 + CD2 + DA2 = AC2 +
2
2
2
1 + 4× BD [ N is the midpoint of side BD] 2 1 + 4 × BD2 4 + BD2 ......(iv) 2 2 BD + 4MN [From (iii) and (iv)] S C H O O L S E C TI O N
MT 17.
GEOMETRY
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Through the midpoint M of the side CD of parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL. D
A L
Proof :
In ELA and BLC, AEL CBL ALE CLB ELA ~ BLC
B
M C
[From (i) and E - L - B] [Vertically oppsoite angles] [By AA test of similarity]
EL EA = .....(i) BL BC ABCD is a parallelogram seg AD || seg BC seg AE || seg BC On transversal BE, AEB CBE .....(ii) In DME and CMB, side DM side CM DME CMB DEM CBM DME CMB DE = BC .....(iii) But, AD = BC .....(iv) DE = BC = AD .....(v)
[c.s.s.t.] [Given] [By definition] [ A - D - E] [Converse of alternate angles test] [Given] [Vertically opposite angles] [From (ii) and A - D - E, B - M - E] [By SAA test of congruence] [c.s.c.t.] [Opposite sides of a parallelogram] [From (iii) and (iv)]
EL ED DA = BL BC
[From (i) and E - D - A]
EL BC BC = BL BC
[From (v)]
EL 2BC = BL BC
E
EL 2 = BL 1 EL = 2BL
MCQ’s 1.
In ABC, AB = 3 cm, BC = 2 cm and AC = 2.5 cm. DEF ~ ABC, EF = 4 cm. What is the perimeter of DEF ? (a) 30 cm (b) 22.5 cm (c) 15 cm (d) 7.5 cm
2.
The sides of two similar triangles are 4 : 9. What is the ratio of their area ? (a) 2 : 3 (b) 4 : 9 (c) 81 : 16 (d) 16 : 81
3.
The areas of two similar triangles are 18 cm 2 and 32 cm 2 respectively. What is the ratio of their corresponding sides ? (a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9
S C H O O L S E C TI O N
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4.
ABC ~ PQR, AB = 6 cm, BC = 8 cm, CA = 10 cm and QR = 6 cm. What is the length of side PR ? (a) 8 cm (b) 10 cm (c) 4.5 cm (d) 7.5 cm
5.
In XYZ, ray YM is the bisector of XYZ where XY = YZ and X - M - Z, then which of the relation is true ? (a) XM = MZ (b) XM MZ (c) XM > MZ (d) None
6.
ABC is an equialteral triangle. seg AD side BC which of the following relations is true ? (a) 2AB2 = 3AD2 (b) 3AB2 = 2AD2 (c) 3AB2 = 4AD2 (d) 4AB2 = 3AD2
7.
In ABC, AB = 10 cm, BC = 8 cm and AC = 6 cm. What is the type of ABC ? (a) Scalene (b) Isosceles (c) Right angled (d) Equilateral
8.
ABC ~ PQR, A = 47º, Q = 83º. What is the measure of C ? (a) 50º (b) 55º (c) 60º (d) 65º
9.
The sides of a triangle are 24 cm, 32 cm and 40 cm. What is the length of the median on the longest side ? (a) 12 cm (b) 16 cm (c) 18 cm (d) 20 cm
10.
In PQR, Q = 90º, P = 60º, PR = 36 cm. Find PQ ? (a) 12 cm (b) 15 cm (c) 18 cm
(d)
18 3 cm
11.
In ABC, AB = 6 cm, BC = 8 cm and AC = 10 cm. ABC is enlarged to PQR such that the largest side is 12.5 cm. What is the length of the smallest side of PQR ? (a) 7.5 cm (b) 9 cm (c) 8 cm (d) 10 cm
12.
In ABC, B - D - C and BD = 6 cm, DC = 4 cm. What is the ratio of A (ABC) to A (ACD) ? (a) 2 : 3 (b) 5 : 2 (c) 3 : 2 (d) 5 : 3
13.
In MNP, N = 90º, seg NQ side MP. If MQ = 12 cm and PQ = 27 cm. What is the length of side NQ ? (a) 6 cm (b) 13.5 cm (c) 19.5 cm (d) 18 cm
14.
In XYZ, PQ || YZ, X - P - Y and X - Q - Z. If What is XZ ? (a) 15.6 cm (c) 7.8 cm
66
(b) (d)
XP 4 = and XQ = 4.8 cm. PY 13
20.4 cm 10.2 cm S C H O O L S E C TI O N
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15.
In ABC, P is a point on side BC such that BP = 4 cm and PC = 7 cm. A (APC) : A (ABC) = .................... . (a) 11 : 7 (b) 7 : 11 (c) 4 : 7 (d) 7 : 4
16.
The length of the altitude of an equilateral triangle whose side is 6 cm is .............. . (a) (b) 5 3 3 (c) 2 3 (d) 3 3
17.
Among the following group which of them form the sides of a right angled triangle. (a) 20, 30, 40 (b) 11, 12, 15 (c) 9, 40, 41 (d) 10, 15, 17
18.
In PQR, seg RS is the bisector, of PRQ, PS = 8, SQ = 6, PR = 20 then QR = ................. . (a) 10 (b) 15 (c) 30 (d) 40
19.
A man goes 15 m due east and then 20 m due north. His distance from the starting point will be ................ . (a) 35 m (b) 5 m (c) 25 m (d) 15 m
20.
In ABC, C = 90, AC = 6, BC = 8, then median CD = ............ . (a) 3.5 (b) 5 (c) 4 (d) 4.5
21.
In ABC, perpendicular AD from A meets BC at D. If BD = 8 cm, DC = 2 cm, AD = 4 cm then .............. . (a) ABC is isosceles (b) AC = 2AB (c) ABC is equilateral (d) ABC is right angled
22.
If PQR is an equilateral triangle such that PS QR, then PS2 is ............ . 3 SR2 (b) 2SR2 (a) 2 (c) 3SR2 (d) 4SR2
23.
In ABC, line PQ || side BC, AP = 3, BP = 6, AQ = 5 then the value of CQ is ............ . (a) 20 (b) 10 (c) 5 (d) 16
24.
In the adjoining figure, AB2 + AC2 = 122, BC = 10 then the length of AQ is ............... .
(a) 3 (c) 12 25.
(b) (d)
A
6 B 36
Q
C
In XYZ, Y = 90º, Z = aº, X = a + 30º then the value of ‘a’ is ............. . (a) 30 (b) 45 (c) 60 (d) 90
S C H O O L S E C TI O N
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: ANSWERS : 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25.
(c) (a) (a) (c) (d) (a) (d) (b) (c) (c) (d) (b) (a)
15 cm 3:4 XM = MZ Right angled 20 cm 7.5 cm 18 cm 7 : 11 9, 40, 41 25 m ABC is right angled 10 30
2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22. 24.
(d) (d) (c) (a) (c) (b) (b) (d) (b) (b) (c) (b)
16 : 81 7.5 cm 3AB2 = 4AD2 50º 18 cm 5:2 20.4 cm
3 3 15 m 5 3SR2 6
1 Mark Sums 1. Sol.
In ABC, D is a point on side BC such that BD = 6 cm and DC = 4 cm. Find A (ABD) : A (ADC). A ABD and ADC have a common vertex A and their bases BD and DC lie on the same line BC their heights are equal. [Triangles having A (ABD) BD A (ADC) = equal heights] DC C B A (ABD) 6 4 6 D A (ADC) = 4 A (ABD) 3 A (ADC) = 2
2. Sol.
3. Sol.
For ABC ~ PQR, state all the corresponding congruent angles. ABC ~ PQR [Given] A P B Q [c.a.s.t.] C R
A (ABC) BC 1 ABC ~ APQ, if A (APQ) = , find . PQ 4 ABC ~ APQ [Given] A (ABC) BC2 A (APQ) = [Areas of similar triangles] PQ2 2 BC 1 = [Given] PQ2 4
68
A (ABC) : A (ADC) = 3 : 2
BC 1 = PQ 2
[Taking square roots] S C H O O L S E C TI O N
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Sol.
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In the adjoining figure, M ray NQ is the bisector of MNP. If MN = 25, Q NP = 40, MQ = 12.5, find the length of seg PQ. × In MNP, P N × ray NQ bisects MNP [Given] MQ MN = [Property of angle bisector of a triangle] PQ NP 12.5 25 = PQ 40 12.5 5 = PQ 8 8 12.5 PQ = 5 PQ = 8 × 2.5
5. Sol.
Sol.
PQ = 20 units
The ratio of the areas of two triangles with the common base is 6 : 5. Height of the larger triangle is 9 cm. Then find the corresponding height of the smaller triangle. Let the areas of larger and smaller triangle be A1 and A2 respectively. Let their respective heights be h1 and h2 A1 6 = and h1 = 9 cm [Given] A2 5 The two triangles have common base [Given] A1 h1 A = h [Triangles having equal bases] 2 2 9 6 = h 5 2 59 h2 = 6 15 h2 = 2 h2 = 7.5
6.
GEOMETRY
Corresponding height of the smaller triangle is 7.5 cm
In the adjoining figure, A ABCD is a trapezium. 10 seg AD || seg PQ || seg BC P AP = 10, PB = 12, DQ = 15. 12 Find the value of QC. seg AD || seg PQ || seg BC [Given] B On transversals AB and DC, [Pr operty of int ercepts made DQ AP = QC by three parallel lines] PB 15 10 = QC 12 15 5 = QC 6
S C H O O L S E C TI O N
D 15 Q ? C
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7.
8. Sol. 9. Sol.
11. Sol.
70
[c.s.s.t.]
In PQR, m P = 90º. S is the midpoint of side QR. If QR = 10 cm, what is the length of PS ? Q
Sol.
ABC ~ PQR AB BC AC = = PQ QR PR
AB . PQ
If two triangles are similar, then what is relation between their sides and corresponding altitudes ? If two triangles are similar, then the ratio of corresponding altitudes is equal to the ratio of their corresponding sides.
10.
6 15 5 QC = 18 units QC =
ABC ~ PQR. State which ratio of sides are equal to
Sol.
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In QPR, m QPR = 90º seg PS is the median on hypotenuse QR 1 PS = QR 2
S
[Given] P
R
[In a right angled triangle, the length of the median drawn to the hypotenuse is half of the hypotenuse]
1 10 2 PS = 5 cm PS =
In ABC, m A = 60º, m C = 30º and AC = 6 cm. What is the length of side AB ? In ABC, m A = 60º [Given] m C = 30º m B = 90º [Remaining angle] ABC is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem, 1 AC [Side opposite to 30º] AB = 2 1 6 AB = 2 AB = 3 cm In ABC, ray AD is the bisector of BAC, B - D - C. If AB = 7.5 cm and AC = 4.5 cm find BD : DC. In ABC, ray AD bisects BAC [Given] AB BD = [Property of angle bisector of a triangle] AC DC S C H O O L S E C TI O N
MT
12.
B
C
D
BD : DC = 5 : 3
A (PQR) =
24 =
1 base height 2
1 × PR × QS 2
1 PR 8 2 24 = PR × 4 24 PR = 4 PR = 6 m
Q
8 cm P
S
R
In XYZ, m X = 90º, m Y = 60º. If XZ = 5 3 cm, what is the length of YZ ? In XYZ, m X = 90º [Given] m Y = 60º m Z = 30º [Remaining angle] XYZ is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem, 3 YZ 2
XZ =
Sol.
4.5
7.5
Area of a triangle =
14.
A
7.5 BD = 4.5 DC BD 75 = DC 45 BD 5 = DC 3
A (PQR) = 24 cm2, the height QS is 8 cm. What is the length of side PR ?
Sol.
13. Sol.
GEOMETRY
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[Side opposite to 60º]
3 YZ 2 5 32
5 3 =
YZ =
YZ = 10 cm
3
If ABC ~ DEF, A (ABC) = 36 cm2, A (DEF) = 64 cm2, what is the ratio of the length of sides BC and EF ? ABC ~ DEF [Given] 2 A (ABC) BC [Areas of similar triangles] A (DEF) = EF 2
36 BC2 = 64 EF 2
BC 6 = EF 8
BC 3 = EF 4
S C H O O L S E C TI O N
[Taking square roots]
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15. Sol.
PQRS is a trapezium. seg PQ || seg SR. A (QSR) = 10 cm2. FindA (PSR). PQRS is a trapezium [Given] Q P seg PQ || seg PR QSR and PSR lie between the same two parallel lines PQ and SR their heights are equal and they S R have a common base RS A (QSR) = A (PSR) [Triangles having equal base and equal height] 2 But, A (QSR) = 10 cm [Given]
16.
Sol.
A (PSR) = 10 cm2
If DEF ~ MNK, DE = 5, MN = 6, find the value of
Sol.
17.
EDUCARE LTD.
DEF ~ MNK A (DEF) DE2 = A (MNK) MN2 A (DEF) 52 A (MNK) = 62 A (DEF) 25 A (MNK) = 36
A (DEF) A (MNK) .
[Given] [Area of similar triangles] [Given]
In ABC, AB = 3 cm, BC = 5 cm and AC = 4 cm. State the vertex of the triangle which contains the right angle. AB2 + AC2 = 32 + 42 AB2 + AC2 = 9 + 16 AB2 + AC2 = 25 ......(i) BC2 = 52 BC2 = 25 .....(ii) AB2 + AC2 = BC2 [From (i) and (ii)] ABC is a right angled triangle [By converse of Pythagoras theorem] The vertex of ABC containing right angle is A.
18. Sol.
In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR. In PQR, m P = 30º m R = 60º [Given] m Q = 90º PQR is a 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem, 1 PR [Side opposite to 30º] QR = 2
19. Sol.
In DEF, m D = 90º, m E = 45º, m F = 45º. If EF = 8 2 cm, find DE. In DEF, D = 90º, E = 45º, F = 45º [Given] DEF is 45º - 45º - 90º triangle 1 EF DE = 2 1 8 2 DE = 2
72
DE = 8 cm S C H O O L S E C TI O N
MT 20. Sol.
A (ABC) The height and the base of ABC and PQR are equal. Find A (PQR) . ABC and PQR have equal base and equal height [Given] A (ABC) = A (PQR) [Triangles with equal base and equal height]
21. Sol.
[Taking square roots]
m X = 70º
If ABC ~ DAC and AC = 3, then find BC × CD. ABC ~ DAC AC BC = [c.s.c.t.] CD AC AC2 = BC × CD 32 = BC × CD
25.
AB = 4 units
If ABC ~ DEF and DEF ~ XYZ and m A = 70º, then find m X. ABC ~ DEF [Given] DEF ~ XYZ ABC ~ XYZ A = X [c.a.c.t.] But, m A = 70º [Given]
24. Sol.
m Y = 60º
If ABC ~ DBA and BC × DB = 16, find AB. ABC ~ DBA [Given] BC AB = [c.s.s.t.] AB DB 2 AB = BC × DB AB2 = 16 [Given]
23. Sol.
A (ABC) A (PQR) = 1
PQR ~ XYZ. If m Q = 60º then find m Y. PQR ~ XYZ [Given] Q Y [c.a.c.t.] But, m Q = 60º [Given]
22. Sol.
GEOMETRY
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BC × CD = 9
PR 2 = and PQ = 12, then find XY. XZ 3 PQR ~ XYZ [Given] PQ PR = [c.s.c.t.] XY XZ 12 2 = XY 3 12 3 XY = 2 XY = 6 × 3
XY = 18 units
If PQR XYZ,
Sol.
S C H O O L S E C TI O N
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3.
Geometric Constructions
Construction of various geometrical figures is a very important part of the study of geometry for understanding the concepts learnt in theoretical geometry. BASIC CONSTRUCTIONS (i)
To draw a perpendicular bisector of a given line segment.
A
B
T
(ii)
To draw an angle bisector of a given angle. A•
• D
B
(iii)
• •
• C
To draw a perpendicular to a line at a given point on it.
P
Q
R
S C H O O L S E C TI O N
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(iv)
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To draw a perpendicular to a given line from a point outside it. P
B
A
m
(v)
To draw an angle congruent to a given angle. L
M
(vi)
P
Q
N
R
To draw a line parallel to a given line through a point outside it.
R
m
130
S C H O O L S E C TI O N
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) 1.
Draw perpendicular bisector of seg AB of length 8.3 cm.
A
(2 marks)
B
8.3 cm
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) 2.
(2 marks)
Draw an angle of 125º and bisect it. P
M
• • 125º Q
R
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) 3.
Construct LMN, such that LM = 6.2 cm, MN = 4.9 cm, LN = 5.6 cm. (2 marks) N
(Rough Figure) N 5.6 cm
5.6 cm
L S C H O O L S E C TI O N
4.9 cm
6.2 cm
L
4.9 cm
6.2 cm
M
M
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) 4.
Construct PQR such that PQ = 5.7 cm, P = Q = 50º.
(Rough Figure)
R
(2 marks) R
50º
50º
P
5.7 cm
Q
50º
50º P
5.7 cm
Q
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) Construct DEF such that, DE = 6.5 cm, E = 50º, F = 30º; and draw EM DF, measure the length EM. (3 marks) Analysis : In DEF, m D + m E + m F = 180º F m D + 50 + 30 = 180º m D + 80 = 180º m D = 180º – 80º 30º m D = 100º 5.
(Rough Figure) F 30º
D
50º 100º 6.5 cm
E
M
100º
50º 6.5 cm
D
E
M
132
S C H O O L S E C TI O N
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TYPE : 1 [A] Constructing tangents to a circle from a point on the circle. Example : Draw a tangent to a circle of radius 2 cm at a point on it.
M
1. 2. 3. 4.
2 cm
P
Steps of construction : Draw a circle with radius 2 cm. Let ‘M’ be the centre of the circle. Take any point ‘P’ on the circle Draw ray MP. Draw the line ‘l’ perpendicular to the ray MP at point ‘P’. Line ‘l’ is the required tangent to the circle. EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93)
1.
Draw a tangent at any point ‘M’ on the circle of radius 2.9 cm and centre ‘O’. (2 marks) (Rough Figure) O 2.9 cm M
O
S C H O O L S E C TI O N
2.9 cm
M
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EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 2.
Draw a tangent at any point R on the circle of radius 3.4 cm and centre ‘P’. (2 marks)
(Rough Figure)
P 3.4 cm
P
3.4 cm
R
R
EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 3.
Draw a circle of radius 2.6 cm. Draw tangent to the circle from any point on the circle using centre of the circle. (2 marks)
(Rough Figure)
O 2.6 cm
O
134
2.6 cm
P
P
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 12.
Draw a tangent to a circle of a radius 3.1 cm and centre O at any point ‘R’ on the circle. (2 marks) (Rough Figure)
O 3.1 cm
O
R
R
3.1 cm
TYPE : 1 [B] Constructing tangents to a circle from a point on the circle without using centre. Example : Given a circle, with a point P on it. Draw a tangent to the circle without using its centre. •
R
Q X N
1. 2. 3. 4. 5. 6. 7.
•
P
M
Steps of construction : Draw the required circle. Take any point ‘P’ on it. Draw chord PQ. Take any point ‘R’ on the alternate arc of arc PXQ other points than P and Q. Join QR and RP. Draw a ray PN making an angle congruent to QRP, taking QP as one side and point P as vertex. The line containing ray PN is the required tangent.
S C H O O L S E C TI O N
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EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 6.
Draw a circle of radius 2.7 cm and draw chord PQ of length 4.5 cm. Draw tangents at P and Q without using centre. (3 marks)
(Rough Figure)
R •
R • P
•
4.5 cm T•
P •
Q
•S
• Q
4.5 cm T
•
S
EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 7.
Draw a circle having radius 3 cm draw a chord XY = 5 cm. Draw tangents at point X and Y without using centre. (3 marks)
(Rough Figure)
Z •
Z • X •
5 cm B•
X •
136
•A
• Y
5 cm
B
•Y
A
S C H O O L S E C TI O N
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 13.
Draw a circle of radius 3.6 cm, take a point M on it. Draw a tangent to the circle at M without using centre of the circle. (2 marks)
(Rough Figure)
L •
L •
•M N
•A • M
N
A
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 14.
Draw a circle of suitable radius and draw a chord XY of length 4.6 cm. Draw tangents at points X and Y without using centre. (3 marks) Z •
(Rough Figure)
Z •
X • 4.6 cm • Y P X •
4.6 cm
• Y
P S C H O O L S E C TI O N
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TYPE : 2 [A] Constructing tangents to a circle from a point outside the circle. Example : Draw a tangent to the circle of radius 1.7cm from a point at a distance of 5.2 cm from the centre.
1. 2. 3. 4. 5. 6.
1.7 cm
A Steps of construction : Draw a circle with radius 1.7 cm. O M Let O be the centre of the circle. 5.2 cm Take a point P such that OP = 5.2cm. Draw perpendicular bisector of seg OP and mark the midpoint of seg OP as ‘M’. With ‘M’ as a centre and radius MP draw a semicircle . Let ‘A’ be the point of intersection of semicircle and the circle. Draw a line joining P and A. Line PA is the required tangent.
P
EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 8.
Draw a tangent to the circle from the point B, having radius 3.6 cm and centre ‘C’. Point B is at a distance 7.2 cm from the centre. (3 marks) A
3.6 cm
(Rough Figure)
C
7.2 cm
B
3.6 cm
A
C
138
M 7.2 cm
B
S C H O O L S E C TI O N
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EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 9.
Draw a tangent to the circle from the point L with radius 2.8 cm. Point ‘L’ is at a distance 5 cm from the centre ‘M’. (3 marks)
(Rough Figure)
A
5.2 cm
P
1.7
cm
O
1.7 cm
A
O
M 5.2 cm
P
EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 10.
Draw a tangent to the circle with centre O and radius 3.3 cm from a point A such that d (O, A) = 7.5 cm. Measure the length of tangent segment. (3 marks)
(Rough Figure)
3.3 cm
B
C
A
3.3 cm
B
7.5 cm
C
M 7.5 cm
A
The length of tangent segnment AB is 6.7 cm. S C H O O L S E C TI O N
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 15.
Construct tangents to the circle from point B with radius 3.5 cm and centre A. Point B is at a distance 7.3 cm from the centre. (3 marks)
(Rough Figure) 3.5 cm
C
A
C
7.3 cm
B
3.5 cm
cm 3.5
D
A
M 7.3 cm
B
cm 3.5
D
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 16.
Draw tangents to the circle with centre P and radius 2.9 cm. From a point Q which is at a distance 8.8 cm from the centre. (3 marks)
(Rough Figure) 2.9 cm
A
P cm 2.9
8.8 cm
A
Q
2.9 cm
B
P cm 2.9
M 8.8 cm
Q
B
140
S C H O O L S E C TI O N
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TYPE : 3 Constructing circumcircle of triangles 1. 2. 3.
(i) (ii) (iii)
A circle passing through the vertices of the triangle is called the circumcircle of a triangle. Circumcentre can be obtained by drawing perpendicular bisectors of any two sides of a triangle. The point of intersection of the perpendicular bisectors is called circumcentre and it is equidistant from the vertices of the triangle. The position of circumcentre depends upon the type of a triangle. If the triangle is an obtuse angled triangle, the circumcentre lies outside the triangle. If the triangle is an acute angled triangle, the circumcentre lies inside the triangle. If the triangle is a right angled triangle, the circumcentre lies on the midpoint of the hypotenuse.
Example : Draw ABC, with AB = 4.1 cm, BC = 6.5 cm and AC = 5 cm. Construct circumcircle of ABC. Measure the radius of the circle. m
A
cm
4.
1
cm
5
B
Steps of construction :
O 6.5 cm
C
1.
Construct ABC, with AB = 4.1 cm, BC = 6.5 cm and AC = 5 cm.
2.
Draw perpendicular bisectors of any two sides of ABC and let them intersect at point O.
3.
Draw a circle with centre O and radius OA.
4.
This circle is the circumcircle of ABC.
S C H O O L S E C TI O N
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 1.
Draw the circumcircle of PMT such that, PM = 5.4 cm, P = 60º, M = 70º. (3 marks)
(Rough Figure)
T
T 70º
60º P
M
5.4 cm
O
60º P
70º 5.4 cm
M
EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 3.
Construct the circumcircle of KLM in which KM = 7 cm, K = 60º, M = 55º. (3 marks)
(Rough Figure)
L
L
K
60º 7 cm
55º
M
O
60º K
142
7 cm
55º M
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 4.
Construct a right angled triangle PQR where PQ = 6 cm, QPR = 40º, PRQ = 90º. Draw circumcircle of PQR. (3 marks)
(Rough Figure) Q
Q R
6 cm
40º
P
O
R
6 cm
40º
P
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) 6.
Construct LEM such that, LE = 6cm, LM = 7.5 cm, LEM = 90º and draw its circumcircle. (3 marks)
(Rough Figure) M M
7.5 cm
E
7.5 cm O
E
S C H O O L S E C TI O N
6 cm
6 cm
L
L
143
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) In PQR, Q = circumcircle of Analysis : PQ2 + QR2 But, PQ2 + QR2 PR 2 PR 26.
90º, seg QM is the median. PQ2 + QR2 = 169. Draw a PQR. (4 marks) = 169 ......(i) [Given]
= PR 2 = 169 = 13 1 PR PM = MR = 2 1 × 13 = 2 PM = MR = 6.5 cm In PQR, m PQR = 90º 1 QM = PR 2 1 × 13 = 2 QM = 6.5 cm
......(ii)
[By Pythagoras theorem]
[By definition of median]
[Median drawn to the hypotenuse is half of hypotenuse]
(Rough Figure) Q
P Q
P
M 13 cm
M 13 cm
R
R
Q
144
S C H O O L S E C TI O N
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 25.
Construct a circumcircle of ABC such that AB = 5 cm, AC = 12 cm, BAC = 90º. (3 marks)
(Rough Figure)
5 cm
B
A
C
12 cm
5 cm
B
A
S C H O O L S E C TI O N
O
12 cm
C
145
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 2.
Construct the circumcircle of SIM in which SI = 6.5 cm, I = 125º, IM = 4.4 cm. (3 marks)
(Rough Figure)
M 4 4. cm
I
125º 6.5 cm
S
O M
4 4. cm
125º I
146
6.5 cm
S
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 196) 7.
Construct DCE, such that, DC = 7.9 cm, C = 135º, D = 20º and draw circumcircle. (3 marks)
(Rough Figure)
E 135º
C
20º 7.9 cm
D
O
E
135º C
20º 7.9 cm
D
Note : This figure is drawn proportionally and not with given measurements. S C H O O L S E C TI O N
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TYPE : 4 Constructing incircle of triangles 1. 2. 3.
A circle which touches all the sides of a triangle is called the incircle of the triangle. The centre of the incircle is called incentre. Incentre is obtained by drawing angle bisectors of the triangle. The angle bisectors are concurrent and their point of intersection is equidistant from the sides of the triangle.
Example : Construct SRP such that RP = 6 cm, R = 750 and P = 550. S
1. 2. 3. 3. 4.
Steps of construction : Draw SRP with RP = 6 cm, R = 75º and P = 55º Draw angle bisectors of R and P. • 75º • Let ‘I’ be the point of intersection R of these angle bisectors . Draw seg IM side RP. Draw a circle with centre I and radius IM. The circle so obtained is the incircle of SRP.
I
55º × M
×
6 cm
P
EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 5.
Construct the incircle of RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm. (3 marks) R
(Rough Figure)
R 6.5 cm
6 cm
6 cm
S
148
• •
O
M 7 cm
S
6.5 cm
× ×
7 cm
T
T
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 7.
Construct the incircle of DEF in which DE = DF = 5.8 cm, EDF = 65º. (3 marks)
(Rough Figure) F
F
5.8 cm
D
5.8 cm
65º
E
5.8 cm
O
D
× ×
65º M 5.8 cm
• •
E
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 8.
Construct incircle of SGN such that SG = 6.7 cm, S = 70º,G = 50º and draw incircle of SGN. (3 marks)
(Rough Figure) N N
S
O
S
••70º
S C H O O L S E C TI O N
M 6.7 cm
50º ××
70º
50º 6.7 cm
G
G
149
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 8.
Construct any right angled triangle and draw incircle of that triangle. (3 marks) ABC is the required right angled triangle. Such that AB = 5 cm, BC = 7 cm and m ABC = 90º
(Rough Figure) A A 5 cm B
7 cm
C
5 cm O
× B
×
M
• •
7 cm
C
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 9.
Construct the incircle of SRN, such that RN = 5.9 cm, RS = 4.9 cm, R = 95º. (3 marks)
(Rough Figure) S
4.9 cm S
95º R
5.9 cm
N
4.9 cm O
R
150
•• 95º
M
× × 5.9 cm
N
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 6.
Construct the incircle of STU in which, ST = 7 cm, T = 120º, TU = 5 cm.
(3 marks) (Rough Figure) U
5 cm 120º
U T
S
7 cm
5 cm O 120º •• M T
× ×
S
7 cm
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 10.
Construct DAT such that DA = 6.4 cm, D = 120, A = 25 and draw incircle of DAT. (3 marks)
(Rough Figure) T
120º
T
25º
D
6.4 cm
A
O 120º •• D M
S C H O O L S E C TI O N
25º ×× 6.4 cm
A
151
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EXERCISE - 3.1 (TEXT BOOK PAGE NO. 84) 9.
Construct the circumcircle and incircle of an equilateral XYZ with side 6.3 cm. (3 marks) X
(Rough Figure)
X ••
6.3 cm
6.3 cm
6.3 cm
6.3 cm
Z
6.3 cm
O
× ×
Y
Y
6.3 cm
Z
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 11.
Draw the circumcircle and incircle of an equilateral triangle ABC with side 6.6 cm. (3 marks) A
(Rough Figure)
6.6 cm
152
B
• • 6.6 cm
cm 6.6
cm 6.6
O
B
A
6.6 cm
××
6.6 cm
C
C
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TYPE : 5 To construct an arc having the given segment as its chord and subtending a given angle at any point on the arc. Example : Draw an arc such that seg AB of length 5.4 cm subtends an AQB of 50º on it. Q 50 0
O 1000 40 0
40 0 A
B
5.4 cm
Step of construction : 1. 2. 3. 4. 5.
Draw seg AB of length 5.4 cm. Draw rays AO and BO making an angle of 400 with seg AB on the same side. Draw an arc with O as the centre and radius OA. Take any point Q on the arc. Draw AQB. Arc AQB is the required arc. EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93)
11.
Draw an arc with seg AB = 6.3 cm, inscribing ACB = 65º.
(3 marks)
(Rough Figure) C 65º
C 65º
A
6.3 cm
B
O 130º A
25º
S C H O O L S E C TI O N
25º 6.3 cm
B
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EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 13.
Draw an arc such that chord ST = 5.6 cm, inscribing SVT = 80º. (3 marks)
(Rough Figure)
V 80º
S
V
T
5.6 cm
80º
O 10º
S
160º
10º
5.6 cm
T
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 17.
Construct an arc PQM such that seg PM of length 6.2 cm subtends an angle of 40º on it. (3 marks)
(Rough Figure)
Q 40º
Q P 40º
6.2 cm
M
O 80º
50º P
154
50º 6.2 cm
M
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EXERCISE - 3.2 (TEXT BOOK PAGE NO. 93) 12.
Draw an arc with seg MN = 8.9 cm, inscribing MPN = 125º. (2 marks)
(Rough Figure)
P 125º
M
N
8.9 cm
P 125º
M
35º
N
35º
8.9 cm
110º O
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 18.
Construct an arc DCV such that seg DV of length 9.5 cm subtends an angle of 135º on it. (2 marks)
(Rough Figure)
C 135º
D
9.5 cm
V
C 135º V
D 45º
9.5 cm
45º
O
S C H O O L S E C TI O N
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TYPE : 6 Constructing triangles with a given base, angle opposite to the base and median. EXERCISE - 3.4 (TEXT BOOK PAGE NO. 101) 1.
Construct LMN such that LM = 6.6 cm, LNM = 65º and ND is median ND = 5 cm. (4 marks)
(Rough Figure) N
65º
m 5c
N
N L
65º
D 6.6 cm
M
m 5c
O 130º
25º
25º
D 6.6 cm
L
M
EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 2.
Construct GHI such that GI = 5.4 cm, GHI = 75º. HR is median. HR = 3.2 cm. (4 marks)
(Rough Figure) H
75º
3. 2
H
H 75º
G
156
3. 2 15º
cm
O 150º R 5.4 cm
G
cm
R 5.4 cm
I
15º
I
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) Construct SAB such that SB = 7.6 cm, SAB = 50º seg AD is median and AD = 5 cm. (4 marks)
19.
(Rough Figure) A 50º
5 cm
S
D 7.6 cm
B
O 100º A
A
50º
5 cm
40º
40º
S
D 7.6 cm
B
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) Construct DEF such that DF = 8.1 cm, DEF = 140º and median EM = 2.5 cm. (4 marks)
21.
(Rough Figure) E 140º 2.5 cm E
D 50º
D
E
2.5 cm 140º
M 8.1 cm
F
F
M 8.1 cm
50º
80º O
S C H O O L S E C TI O N
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EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 3.
Construct ABC such that BC = 7.8 cm, BAC = 100º and median AM = 3.5 cm. (4 marks)
(Rough Figure) A 100º
3.5 cm B A
A 100º
B
3. 5 10º
C
M 7.8 cm
cm 7.8 cm M 160º
10º
C
O
EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 4.
Construct XYZ such that XY = 9.5 cm, XZY = 115º, ZP is median. ZP = 3.3 cm. (4 marks)
(Rough Figure)
Z
115º 3.3 cm
Z
Z
P 9.5 cm
X
115º
Y
3.3 cm
X
25º
P 9.5 cm
25º
Y
130º O
158
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TYPE : 7 Constructing triangles with a given base, angle opposite to the base and an altitude. EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) Construct DEF such that DF = 6.2 cm, DEF = 60º, EM DF and EM = 4.4 cm. (4 marks)
5.
(Rough Figure)
E
4.4 cm
60º
D
M 6.2 cm
E
E
F B
4.4 cm
60º
4.4 cm
O 120º
D
30º
30º
M
A
F
6.2 cm
EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 6.
Construct RST such that RT = 5.7 cm, RST = 55º, SD RT, SD = 3.4 cm. S (4 marks) (Rough Figure) 3.6 cm
55º
R
D 5.7 cm S
S
T
B
55º
DR
S C H O O L S E C TI O N
3.4 cm
3.4 cm
O 110º
35º
35º 5.7 cm
T
A
159
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EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 9.
Construct KLM such that KM = 7.2 cm, KLM = 72º, LA KM, KA = 4.8 cm. (4 marks) L
(Rough Figure)
72º
L 72º
K
4.8 cm
A 7.2 cm
M
O 144º K
18º 4.8 cm
7.2 cm
18º
A
M
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 22.
Construct LAC such that LC = 6.7 cm, LAC = 72º and altitude AB has length 4 cm. (4 marks)
(Rough Figure)
A
4 cm
72º
L
A
A
B 6.7 cm
C
N
4 cm
4 cm
72º
O 144º L
160
18º
B
18º 6.7 cm
C
M
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PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 20.
Construct KPM such that KM = 7 cm, KPM = 90º and length of altitude PS is 2.9 cm. (4 marks) P 2.9 cm
(Rough Figure)
K
M
S 7 cm
P
B
2.9 cm
2.9 cm
P
K
S
A
M
7 cm
EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 8.
Construct CVX such that CX = 9.1 cm, CVX = 130º, VD CX and V VD = 1.7 cm. (4 marks) (Rough Figure) 1.7 cm
130º
C V
V
X
D 9.1 cm
B
130º 1.7 cm
1.7 cm C 40º
D
9.1 cm
X 40º
A
100º O S C H O O L S E C TI O N
161
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EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) 10.
Construct SPQ such that SQ = 8.3 cm, SPQ = 127º, PM SQ, PM = 1.6 cm. (4 marks) P
(Rough Figure)
1.6 cm
127º
S
P
M 8.3 cm
Q
B
P
127º
1.6 cm
1.6 cm
S
M
37º
Q A
37º
8.3 cm
106º O
EXERCISE - 3.3 (TEXT BOOK PAGE NO. 101) Construct PQR such that PQ = 9.2, PRQ = 112º, RK is an attitude, RK = 2.4 cm. (4 marks)
7.
(Rough Figure)
R
2.4 cm
112º
P R
K 9.2 cm
R
Q B
112º 2.4 cm P
2.4 cm Q
22º
K
9.2 cm
22º
A
136º O
162
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TYPE : 8
Constructing similar triangles EXERCISE - 3.4 (TEXT BOOK PAGE NO. 105)
1.
BC 2 : ; EF 3 (4 marks)
ABC ~ DEF, In ABC, AB = 5.2 cm, BC = 4.6 cm, B = 45º and
construct DEF. Analysis :ABC ~ DEF
[Given]
AB BC AC 2 = = = DE EF DF 3 B = E = 45º
AB 2 = DE 3
5.2 2 = DE 3
......(i)
[c.s.s.t.] [c.a.s.t.]
[From (i)]
15.6 = DE 2 DE = 7.8 cm
BC EF
=
2 3
4.6 EF
=
2 3
13.8 2 EF
[From (i)]
= EF = 6.9 cm
Information for constructing DEFis complete.
(Given triangle)
5. 2
cm
A
45º B
4.6 cm
C
(Required triangle)
7. 8
cm
D
45º E S C H O O L S E C TI O N
6.9 cm
F
163
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EXERCISE - 3.4 (TEXT BOOK PAGE NO. 105) 2.
LMN ~ XYZ, In LMN, LM = 6 cm, MN = 6.8 cm, LN = 7.6 cm and
LM 4 = ; construct XYZ. XY 3 Analysis : LMN ~ XYZ
LM MN LN 4 = = = XY YZ XZ 3
LM XY
=
4 3
[From (i)]
6 XY
=
4 3
= XY
18 4 XY
(4 marks) [Given] ...... (i)
MN 4 = YZ 3 6.8 YZ
[c.s.s.t.]
[From (i)]
4 3
=
20.4 = YZ 4 YZ = 5.1 cm
= 4.5 cm
LN 4 = [From (i)] XZ 3
7.6 4 = XZ 3
22.8 = XZ 4 XZ = 5.7 cm
Information for constructing XYZ is complete.
(Given triangle) L
cm
6 cm
6 7.
M
N
6.8 cm
(Required triangle)
164
cm
Y
7 5.
4.5 cm
X
5.1 cm
Z S C H O O L S E C TI O N
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EXERCISE - 3.4 (TEXT BOOK PAGE NO. 105) 3.
RHP ~ NED, In NED, NE = 7 cm, D = 30º, N = 20º and
construct RHP. Analysis :RHP ~ NED
HP 4 = ; ED 5 (4 marks)
[Given]
RH HP RP 4 = = = NE ED ND 5
......(i)
[c.s.s.t.]
R = N = 20º P = D = 30º
[c.a.s.t.]
H = E = 130º
RH 4 = NE 5
RH 4 = 7 5
RH =
28 5
[From (i)
= 5.6 cm
Information for constructing RHP is complete.
(Given triangle) D 30º
130º 20º
N
7 cm
E
(Required triangle)
P 30º
R
S C H O O L S E C TI O N
130º
20º 5.6 cm
H
165
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EXERCISE - 3.4 (TEXT BOOK PAGE NO. 105) LTR ~ HYD, In HYD, HY = 7.2 cm, YD = 6 cm, Y = 40º and
4.
construct LTR. Sol.
(4 marks)
RHP ~ NED [Given] LT TR LR 5 = = = .....(i) [c.s.s.t.] HY YD HD 6 T = Y = 40º [c.a.s.t.] LT 5 TR 5 = [From (i)] = HY 6 YD 6 LT 5 TR 5 = = 7.2 6 6 6 36 30 LT = TR = 6 6 LT = 6 TR = 5 Information for constructing HYD is complete. L
(Given triangle)
6
T
LR 5 = ’ HD 6
[From (i)]
H
(Required triangle) 2 7.
cm
cm
40º
40º
5 cm
Y
R
6 cm
D
EXERCISE - 3.4 (TEXT BOOK PAGE NO. 105) 5.
AMT ~ AHE, In AMT, MA = 6.3 cm, MAT = 120º, AT = 4.9 cm and MA 7 = ’ construct AHE. (4 marks) HA 5 T
E cm 4.9
120º
H
×
A
×
6.3 cm
M
A1 A2 A3 A4
•
A5
A6
•
A7
166
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EXERCISE - 3.4 (TEXT BOOK PAGE NO. 105) 6.
SHR ~ SVU, In SHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and
SH 3 = ; construct SVU. SV 5
(4 marks)
U
(Rough Figure) U
R
S
cm
4.5 cm
H
V
5.8 cm
cm 5.2 ×
S
5. 2
5. 8
cm
R
4.5 cm
×
H
V
S1 S2 •
S3 S4 •
S5
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 23.
ABC ~ LMN. In ABC, AB = 5.1 cm, B = 55º, C = 65º and
Sol.
then construct LMN. ABC ~ LMN
AB LM
=
BC AC 3 = = MN LN 5
AB LM
=
3 5
5.1 LM
=
3 5
AC 3 = . LN 5 (4 marks)
[Given] .....(i)
[c.s.s.t] [From (i)]
25.5 = LM 3 LM = 8.5 Information for constructing LMN is complete.
S C H O O L S E C TI O N
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(Given triangle)
(Required triangle) C
N
65º
65º
55º
55º
60º A
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B
5.1 cm
L
M
8.5 cm
PROBLEM SET - 3 (TEXT BOOK PAGE NO. 197) 24. Ans.
XY 6 = DE 5 (4 marks)
XYZ ~ DEF, in DEF, DE = 5.5 cm, E = 40º, EF = 4 cm and then construct XYZ. XYZ ~ DEF [Given] XY YZ XZ 6 = = = ......(i) [c.s.s.t.] DE EF DF 5 XY 6 YZ 6 = [From (i)] = DE 5 EF 5 XY 6 YZ 6 = = 5.5 5 4 5 33 24 xy = YZ = 5 5 XY = 6.6 YZ = 4.8 Information for constructing XYZ is complete.
[From (i)]
(Given triangle) D
5 5.
cm
40º E
4 cm
F
(Required triangle)
X
6 6.
cm
40º Y
168
6.6 cm
Z S C H O O L S E C TI O N
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HOTS PROBLEM (Problems for developing Higher Order Thinking Skill) 14.
To draw seg AB of length
65 without using Pythagoras theorem. (4 marks)
Analysis : In ABC, ABC = 90º seg BD hypotenuse AC ABC ~ ADB [Theorem on similarity of right angled triangle]
AB AD AB2 AB2 AB2
AC AB = AD × AC = 5 × 13 = 65 =
AB =
A
65 B
A
B
[c.s.s.t.]
5 cm
5 cm
(Analytical Figure)
C
D 13 cm
C
D 13 cm
S C H O O L S E C TI O N
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OR Analysis : In CAD, m CAD = 90º seg AB hypotenuse CD AB2 = CB × BD AB2 = 5 × 13 AB2 = 65 AB =
[Property of Geometric mean]
[Taking square roots]
65
A
C
5 cm B
(Analytical Figure)
D
13 cm
A
C
5 cm
D
B 18 cm
Note : This figure is drawn proportionally and not with given measurements. 170
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16.
Draw segment AB of any length. Take point D on AB such that AD2 = 3BD2. (4 marks) Analysis : In CDB, (Analytical Figure) tan CBD
=
tan 60
=
AD BD
=
AD BD
3
C
CD BD
[By definition] [ CD = AD]
2
AD BD2
3
=
AD2
= 3BD2
60º
• A
D
B
[Squaring both sides]
C
60º A
22.
D
B
l
Draw a triangle ABC with side BC = 6 cm, B = 45º and A = 100º, then 4 construct a triangle whose sides are times the corresponding sides 7 of ABC. (4 marks)
Analysis : In ABC, m A = 100º m B = 45º m C = 35º
[Given] [Remaining angle]
(Rough Figure) A 100º P
B
S C H O O L S E C TI O N
45º
Q 6 cm
35º
C
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A
P
100º
45º B
•
6 cm
35º •
Q
C
B1 B2 B3
× B4 B5 B6
× B
7 PBQ is the required triangle 4 whose sides are times the corresponding sides of ABC 7
23.
Construct a triangle ABC, in which BC = 3.5 cm, B = 60º and altitude AD = 2.5 cm and draw its incircle and measure its radius. (4 marks)
2.5 cm
(Rough Figure) A
B
60º
D 3.5 cm
C
2.5 cm
2.5 cm
A
B
• •
I
60º
D M 3.5 cm
× ×
C
Note : This figure is drawn proportionally and not with given measurements. 172
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Construct an isosceles triangle whose base is 8 cm and altitude 4 cm. Draw its circumcircle and measure its radius. (4 marks) Analysis : ABC is an isosceles triangle with AB = AC seg AD side BC 1 BC [Perpendicular drawn to the base, bisects the base] BD = DC = 2 1 (Rough Figure) A ×8 BD = DC = 2 BD = DC = 4 AD = BD = DC = 4 cm A 4 cm
24.
B
D 8 cm
C
4 cm
B
C
D 8 cm
In PQR, QR = 7.5 cm, QPR = 110º and PQ + PR = 8.3 cm then construct PQR and measure PQR. Construct its circumcircle. (5 marks) Analysis : line l is perpendicular bisector of side TR PT = PR .......(i) [Perpendicular bisector theorem] QT = 8.3 cm PQ + PT = 8.3 [Q - P - T] PQ + PR = 8.3 [From (i)] In PTR, [From (i)] side PT side PR PTR PRT [Isosceles triangle theorem] Let, PTR = PRT = x Now, QPR is an exterior angle of PTR, QPR = PTR + PRT [Remote interior angles theorem] 110 = x + x (Rough Figure) 110 = 2x P x = 55 110º PTR = PRT = 55º Information to draw RQT is complete. R 25.
Q
S C H O O L S E C TI O N
7.5 cm
173
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(Analytical Figure)
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T 55º
8. 3
cm
P
Q
110º 55º
7.5 cm
R
R
l
55º
7.5 cm
O
110º Q
P
8.3 cm
S
Construct LMN, such that LN = 8 cm and LMN = 80º and LM – MN = 3cm. Construct its circumcircle. (5 marks) Analysis : Line l is a perpendicular bisector of side TN TM = MN .......(i) [Perpendicular bisector theorem] LM = LT + TM [L - T - M] (Rough Figure) LM = 3 + MN [From (i)] M LM – MN = 3 cm In MTN, 80º side MT side MN [From (i)] MTN MNT [Isosceles triangle theorem] Let, MTN = MNT = x L N 8 cm x + x + M = 180 x + x + 80 = 180 2x = 180 – 80 2x = 100 (Analytical Figure) M x = 50 80º T m MTN = MNT = 50º c 3 130º LTN + MTN = 180 N L 8 cm LTN + 50 = 180 LTN = 180– 50 LTN = 130º Information for drawing LTN is complete. 26.
174
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N
m 8c O
130º L
80º
3 cm
M
T
ConstructXYZ such that, YZ = 6.2 cm, Z = 65º and XY – XZ = 2.4 cm and draw incircle of it. (4 marks) Analysis : Line l is a perpendicular bisector of side YW XY = XW .......(i) [Perpendicular bisector theorem] XW = XZ + ZW [X - Z - W] XY = XZ + 2.4 [From (i)] XY – XZ = 2.4 27.
l
(Rough Figure)
X
X •• 65º Y ×
Y
6.2 cm
65º ×
Z
6.2 cm
Z cm 2.4
l
(Analytical Figure) X
W
Y
65º
Z cm 2.4
6.2 cm
W
S C H O O L S E C TI O N
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In RST, RS = 5 cm, RT = 6.8 cm and median RM = 5.3 cm construct a circumcircle of RST. (4 marks) Analysis : In RST extend median RX to point P such that R - X - P and RX = XP also SX = XT (Rough Figure) PSRT is a parallelogram R Information to constructing parallelogram PSRT is complete and RST can be obtained. 6. 8 Hence draw its circumcircle. c 5c m
28.
m 5.3 cm
S S
R
M 10.6 cm
R
T
M
6.8 cm
m 5c
P cm 6.8
6.8 cm
S
T
M
T
(Analytical Figure) P
In ABC, BC = 6 cm and median AM = 5.1 cm. G is the centroid of ABC and BGC = 130º. Construct ABC. (4 marks) Analysis : In ABC, G is the centroid on median AM 1 (Rough Figure) GM = AM [Centroid bisects each median] 3 A 1 GM = × 5.1 = 1.7 cm 3 Also, BGC = 130º and BC = 6 cm Information for constructing BGC is complete. G Position of A can be obtained an line GM. 130º Hence draw ABC. 29.
cm 5.1
B A
A
M 6 cm
C
1 5. cm G
G
7 1. 40º
176
cm
B
M 6 cm
C 40º
S C H O O L S E C TI O N
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Draw a triangle ABC, right angled at B such that, AB = 3 cm and BC = 4 cm. Now construct a triangle similar to ABC, each of whose 7 times the corresponding side of ABC. (4 marks) sides is 5 P
(Rough Figure) P A A 3 cm
3 cm B
B
C
4 cm
4 cm
C
R
R
B1 B2 B3 B4 B5 B6 B7
MCQ’s 1.
What is the point of concurrence of the medians of a triangle called ? (a) Circumcentre (b) Incentre (c) Orthocentre (d) Centroid
2.
What is the point of concurrence of the altitudes of a triangle called ? (a) circumcentre (b) incentre (c) orthocentre (d) centroid
3.
What is the point of concurrence of the angle bisectors of a triangle called ? (a) circumcentre (b) incentre (c) orthocentre (d) centroid
4.
An arc of a circle containing an angle of 70º is to be drawn on the upper side of seg AB. What are the measures of the angles to be drawn at points A and B ? (a) 20º on the upper side of seg AB (b) 70º on the upper side of seg AB (c) 20º on the lower side of seg AB (d) 70º on the lower side of seg AB
5.
An arc of a circle containing an angle of 140º is to be drawn on the upper side of seg AB. What are the measures of the angles to be drawn at points A and B. (a) 70º on the upper side of seg AB (b) 50º on the upper side of seg AB (c) 50º on the lower side of seg AB (d) 70º on the lower side of seg AB
6.
To find the circumcentre of ABC, we bisect .............. of ABC. (a) side AB (b) all sides (c) any two sides (d) any two angles
S C H O O L S E C TI O N
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7.
To find incentre of a given triangle, we bisect .............. (a) any two angles (b) all sides (c) all angles (d) one side and one angle
8.
From a point outside a circle, .................. tangents can be drawn (a) one (b) two (c) at the most two (d) none of these
9.
The circumcentre of an acute angled triangle is ................. of the triangle. (a) on one side (b) in the interior (c) in the exterior (d) none of these
10.
If the circumcentre lies in the exterior of the triangle, then it is .......... triangle. (a) a right angled (b) an acute angled (c) an isosceles (d) an obtuse angled
11.
Tangent drawn from a point M on the circle is perpendicular to the ............. . (a) chord MP (b) diameter MN (c) chord AB (d) radius OP
12.
To draw arc of measure 120º on seg AB, we first construct isosceles triangle with base angle of .............. . (a) 30º (b) 60º (c) 90º (d) 120º
13.
Three sides of ABC are given. To construct similar PQR, at least .................. of PQR must be given. (a) one angle (b) any two angles (c) any one side (d) all sides
14.
The circumcentre and incentre of ............... triangle are at the same point. (a) a scalene (b) an isosceles (c) an equilateral (d) an acute angled
15.
To construct ABC of base AB = 5 cm and height CP = 6 cm, we draw parallel line at a distance of ................ cm. (a) 1 (b) 5 (c) 6 (d) 11
16.
The sides of ABC are 6 cm, 8 drawn. What is the radius of the (a) 5 cm (c) 4 cm
17.
ABC ~ XYZ .......... .............. . (a) AB, XY (b) (c) AC, AZ (d)
cm, 10 cm. A circumcentre of ABC is circumcircle ? (b) 10 cm (d) 24 cm BC, YZ B, Y
18.
To draw a tangent at point be on arc ABC .............. must be given. (a) centre (b) none (c) diameter (d) length of chord AC
19.
ABC ~ XYZ and
AB 2 XY 1
m ABC m XYZ = ............. . 178
S C H O O L S E C TI O N
MT
1 2
(b)
2
(c) 1
(d)
1 3
(a)
20.
GEOMETRY
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O is the centre of a circle with radius 5 cm, the length of the tangent segment drawn from the point 13 cm from centre O is .......... cm. (a) 5 (b) 13 (c) 12 (d) 18
: ANSWERS : 1.
(d) Centroid
2.
(c) orthocentre
3.
(b) incentre
4.
(a) 20º on the upper side of seg AB
5.
(b) 50º on the upper side of seg AB 6.
(c) any two sides
7.
(a) any two angles
(b) two
9.
(b) in the interior
10.
(d) an obtuse angled
11.
(b) diameter MN
12.
(a) 30º
13.
(c) any one side
14.
(c) an equilateral
15.
(c) 6
16.
(a) 5 cm
17.
(d) B, Y
18.
(a) none
19.
(c) 1
20.
(c) 12
8.
S C H O O L S E C TI O N
179
4. `
(i) (ii) (iii)
(i) (ii) (iii)
Trigonometry
Introduction : The word TRIGONOMETRY is derived from Greek words Tri meaning three, gona meaning sides and metron meaning measure. Thus, Trigonometry deals with measurements of sides and angles of a triangle. In our syllabus, we restrict our learning to right angled triangles. In a right angled triangle, The side opposite to right angle is called the hypotenuse. For any acute angle, the side opposite to it is called the opposite side. For any acute angle, the side adjacent to it other than the hypotenuse is called the adjacent side. In ABC, m ABC = 90º seg AC is the hypotenuse. For ACB, seg AB is the opposite side. For ACB, seg BC is the adjacent side.
A
B
C
TRIGONOMETRIC RATIOS OF AN ACUTE IN A RIGHT ANGLED TRIANGLE For any acute angle in a right angled triangle, the three above mentioned sides, can be arranged two at a time, in six different ratios. These ratios are called Trigonometric ratios. A
In ABC, m ABC = 90º m ACB =
B
180
C
Sine ratio of
= sin
=
Opposite side Hypotenuse
=
AB AC
Cosine ratio of
= cos
=
Adjacent side Hypotenuse
=
BC AC
Tangent ratio of
= tan
=
Opposite side Adjacent side =
AB BC
Cosecant ratio of
Hypotenuse = cosec = Opposite side
=
AC AB
Secant ratio of
= sec
Hypotenuse = Adjacent side
=
AC BC
Cotangent ratio of = cot
Adjacent side = Opposite side
=
BC AB S C H O O L S E C TI O N
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RELATIONS BETWEEN TRIGONOMETRIC RATIOS 1.
cosec =
1 sin
1 2. sec = cos
4.
tan
sin cos
cos 5. cot = sin
=
1 3. cot = tan
TRIGONOMETRIC IDENTITIES (i) sin2 + cos2 = 1
`
(ii) 1 + tan2 = sec2
(iii) 1 + cot2 = cosec2
Angle in Standard position or standard angle : Directed angle : Consider the ray OA and rotate it in anti-clockwise rm direction about O, the final position of the ray OA is la a in ray OB. In the adjoining figure, the rotation from rm e the ray OA to ray OB defines an AOB. It gives T the direction from ray OA to ray OB. Vertex O It is called as directed angle.
B
Initial arm
A
•
Initial arm : The initial position of the ray is called the initial arm. In the above figure OA is the initial arm.
•
Terminal arm : The final position of the ray after rotation is called terminal arm. In the above figure OB is the terminal arm.
•
Vertex : The point of rotation is called the vertex. In the above figure O is the vertex. In directed angle we take the following into consideration : B A Initial arm m r a m Terminal arm l ar nal mi itia r n The amount and the sense e I T of rotation of the initial ray. O O The rotation may be in Te r Ini mi tial anticlockwise or clockwise direction. nal ar
(i) (ii) (iii) (iv)
m
A
arm
B
NOTE 1. 2.
If the rotation of the initial ray is anti-clockwise then the directed angle is positive. If the rotation of the initial ray is clockwise then the directed angle is negative.
•
Standard angle : In rectangular co-ordinate system a directed angle with its vertex at the origin O and the initial ray along positive X-axis is called Standard Angle or Angle in Standard Position.
•
Measure of standard angle : The amount of rotation of the ray from the initial position to terminal position is the measure of standard angle.
S C H O O L S E C TI O N
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Angle in Quadrant : A directed angle in standard position is said to be in particular quadrant if its terminal arm lies in that quadrant. Quadrantal Angle : A directed angle in standard position whose terminal arm lies along co-ordinate axes is called a quadrantal angle. EXERCISE - 4.1 (TEXT BOOK PAGE NO. 112) 1. (i) Sol.
Draw the figure and write the answers : For the angle in standard position if the initial arm rotates 220º in clockwise direction then terminal arm is in which quadrant ? (1 mark) Y Since the initial arm rotates in clockwise direction and the angle is more than – 180º but less than – 270º, the terminal arm lies in II quadrant.
O
X
X
220º
Y
(ii) Sol.
For the angle in standard position if the initial arm rotates 25º in anticlockwise direction then terminal arm is in which quadrant ? (1 mark) Y Since the initial arm rotates in anticlockwise direction and the angle is more than 0º but less than 90º, 25º terminal arm lies in I quadrant. X X
O
Y
(iii) Sol.
For the angle in standard position if the initial arm rotates 305º in anticlockwise direction then terminal arm is in which quadrant ? (1 mark) Y Since the initial arm rotates in anticlockwise direction and the angle is more than 270º but less than 360º, 305º terminal arm lies in IV quadrant.
X
O
X
Y
EXERCISE - 4.1 (TEXT BOOK PAGE NO. 112) 2. Sol.
The terminal arm is in II quadrant, what are the possible angles ?(1 mark) Y The terminal arm is in II quadrant, the angle is in between 90º and 180º if the initial arm rotates anticlockwise direction or the angle is between O –270º and – 180º if the initial arm X X rotates clockwise.
Y
182
S C H O O L S E C TI O N
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EXERCISE - 4.1 (TEXT BOOK PAGE NO. 112) 3. Sol.
The terminal arm is on negative Y-axis, what are the possible angles ? What can you say about this angle ? (1 mark) Y The terminal arm is on negative y-axis, the possible angles are 270º and – 90º. These angles are called quadrantal angles. O
X
X
Y
We define the trigonometric ratios in terms of co-ordinates of a point P (x, y) as follows : (i)
sin =
y r
(ii)
cos =
x r
(iii)
tan =
y , where x 0 x
(iv)
cosec =
(v)
sec =
r , where x 0 x
(v)
cot =
r , where y 0 y
x , where y 0 y
Signs of trigonometric ratios in different quadrants : If x is positive, cosine is positive. If x is negative, cosine is negative. If y is positive, sine is positive. If y is negative, sine is negative. EXERCISE - 4.2 (TEXT BOOK PAGE NO. 116) 1. (i) Sol.
Find the trigonometric ratios in standard position whose terminal arm passes through the points : (4, 3) (2 marks) The terminal arm passes through P (4, 3) x = 4 and y = 3 r =
x2 + y2
=
(4)2 + (3)2
=
16 + 9
=
25 r = 5 units Let the angle be sin
=
y 3 = r 5
cosec
r 5 = y = 3
cos
=
x 4 = r 5
sec
=
tan
=
y 3 = x 4
cot
x 4 = y = 3
S C H O O L S E C TI O N
r 5 = x 4
183
MT
GEOMETRY
(iii) Sol.
x2 + y2
=
(– 24)2 + (– 7)2
=
576 + 49
= 625 r = 25 units Let the angle be y –7 = = sin r 25
Sol.
(2 marks)
(– 24, – 7) The terminal arm passes through P (– 24, – 7) x = – 24 and y = – 7 r =
(iv)
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cosec
r – 25 = y = 7
cos
=
x – 24 = r 25
sec
=
tan
=
–7 y 7 = – 24 = x 24
cot
x – 24 24 = y = –7 = 7
(–1 ,
3
r – 25 = x 24
)
(2 marks)
The terminal arm passes through P x = – 1 and y = r =
–1, 3
3
x2 + y2
=
1
=
1 3
2
3
2
=
4 r = 2 units Let the angle be
sin
=
cos
=
tan (ii) Sol.
y = r
3 2
x –1 = r 2 3 y = = =– 3 –1 x
cosec
r = y =
sec
=
cot
3
r –2 = x 1 x –1 = y = 3
(5, – 12) The terminal arm passes through P (5, – 12) x = 5 and y = – 12 r =
2
(2 marks)
x2 + y2
=
(5)2 + (–12)2
=
25 + 144
= 169 r = 13 units 184
S C H O O L S E C TI O N
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Let the angle y = sin r x cos = r y tan = x (v) Sol.
be –12 = 13 5 = 13 –12 = 5
cosec sec cot
r –13 = y = 12 r 13 = = x 5 x –5 = y = 12 (2 marks)
(1, – 1) The terminal arm passes through P (1, – 1) x = 1 and y = – 1 r =
x2 + y2
=
(1)2 + (–1)2
=
11
=
2
r = 2 units Let the angle be sin cos tan (vi) Sol.
–1 y = 2 r 1 x = = 2 r y –1 = = =–1 x 1
=
cosec
r 2 = y = = – 2 –1
sec
=
cot
r 2 = = 2 x 1 x 1 = y = –1 = – 1
(– 2, – 3) The terminal arm passes through P (– 2, – 3) x = – 2 and y = – 3 r =
x2 + y2
=
(– 2)2 (– 3)2
=
49
=
(2 marks)
13
r = 13 units Let the angle be y sin = = r x cos = = r y tan = = x
–3 13 –2 13 –3 3 –2 = 2
cosec sec cot
r 13 – = y = = –3 13 r – = = = –2 x x –2 2 = y = –3 = 3
13 3 13 2
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 1. Sol.
If the terminal arm passes through the point (1, –1) making an angle find the value of sec . (2 marks) The terminal arm passes through point (1, – 1) x = 1 and y = – 1
S C H O O L S E C TI O N
185
MT
GEOMETRY
r
=
x2 y2
r
=
(1)2 (–1)2
r
=
11
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r = 2 units Let the angle be
r x
sec
=
sec
=
2 1
sec
=
2 EXERCISE - 4.2 (TEXT BOOK PAGE NO. 116)
3.
Find where the angle lies if the terminal arm passes through :
(i) Sol.
(5, – 7) (1 mark) (5, – 7) x is positive and y is negative The terminal arm is in IV quadrant.
(iii) Sol.
(– 3, – 3) (1 mark) (iv) (– 3, – 3) Sol. x is negative and y is negative The terminal arm is in III quadrant.
(ii) Sol.
(– 8, 1) (1 mark) (– 8, 1) x is negative and y is positive The terminal arm is in II quadrant. (0, 2) (1 mark) (0, 2) x is zero and y = 2 The terminal arm is on positive Y-axis.
EXERCISE - 4.2 (TEXT BOOK PAGE NO. 116) 2. Sol.
If the angle = – 60º, find the value of sin , cos , sec and tan .(2 marks) = – 60º sin (– ) = – sin sin (– 60) = – sin 60 sin (– 60) = –
3 2
sec (– ) = sec sec (– 60)= sec 60 sec (– 60)= 2 cos (– ) = cos cos (– 60) = cos 60 cos (– 60) =
1 2
tan (– ) = – tan tan (– 60)= – tan 60 tan (– 60)= – 3 186
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PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 3. Sol.
If the angle = – 60º, find cos and cosec . = – 60º cos (– ) = cos cos (– 60) = cos 60
(1 mark)
1 2 cosec (– ) = – cosec cosec (– 60) = – cosec 60
cos (– 60) =
cosec (– 60) =
–
2 3
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) 1.
5 . where is an acute angle, find the value of other 13 trigonometric ratios using identities. (3 marks)
If sin =
sin =
Sol.
cosec = =
cosec =
5 13 1 sin 1 5 13
[Given]
13 5
sin2 + cos2 = 1
5 13
2
+ cos2 = 1
25 + cos2 = 1 169 25 169
cos2 = 1 –
cos2 =
169 – 25 169
cos2 =
144 169
cos = sec = =
S C H O O L S E C TI O N
sec =
12 13 1 cos
[Taking square roots]
1 12 13
13 12 187
MT
GEOMETRY
tan = =
tan = cot = =
cot =
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sin cos 5 13 12 13 5 12 1 tan 1 5 12 12 5
EXERCISE - 4.2 (TEXT BOOK PAGE NO. 116) 4. Sol.
188
7 and is in fourth quadrant, find the other five trigonometric 25 ratios. (3 marks) 7 cos = 25 1 25 sec = sec cos 7 sin2 + cos2 = 1 sin2 = 1 – cos2 2 7 = 1– 25 49 = 1– 625 625 – 49 = 625 576 = 625 24 sin = [Taking square roots] 25 1 25 cosec = co sec sin 24 sin cos = tan 24 25 = tan 7 25 24 tan = 7 1 7 cot = cot tan 24 is in the fourth quadrant sin , cosec , tan and cot are negative – 24 – 25 – 24 –7 sin = , cosec = , tan = and cot = 25 24 7 24 If cos =
S C H O O L S E C TI O N
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EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) 2.
7 , then find the values of sin and sec , is in IV quadrant. 24 (3 marks) 7 cot = – 24 1 24 tan = – tan cot 7 1 + tan2 = sec2
If cot = –
Sol.
2
24 1 + – 7 576 1+ 49 49 576 49 625 49
= sec2 = sec2 = sec2 = sec2
25 7 7 25
sec =
cos = sin cos = tan sin = tan × cos – 24 7 × sin = 7 25
sin =
1 cos sec
– 24 25
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) 3. Sol.
3 sin – 4 cos = 0, then find the values of tan , sec and cosec , (3 marks) where is an acute angle. 3 sin – 4 cos = 0 3 sin = 4 cos sin 4 cos = 3
4 3 1 + tan2 = sec 2 2 4 1 + = sec 2 3 16 1+ = sec 2 9 9 16 = sec 2 9 25 = sec 2 9 5 sec = 3
S C H O O L S E C TI O N
tan =
[Taking square roots] 189
MT
GEOMETRY
1 tan 1 = 4 3 3 cot = 4 1 + cot2 = cosec2 2 3 1 + = cosec2 4 9 1+ = cosec2 16 16 9 = cosec2 16 25 = cosec2 16 5 cosec = 4
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cot =
4. Sol.
[Taking square roots]
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) sin + cos If tan = 1, then find the value of sec + cosec , where is an acute angle. (3 marks) tan = 1 sin cos = 1 sin = cos .......(i) 1 + tan2 = sec2 1 + (1)2 = sec2 1 + 1 = sec2 2 = sec2 sec = [Taking square roots] 2 1 cos = sec 1 cos = 2
sin = cosec = =
cosec =
sin cos sec cos ec =
1 2 1 sin 1 1 2
[From (i)]
2 1 1 2 2 2 2 2
= 190
2 2 2 S C H O O L S E C TI O N
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= = =
sin cos sec cos ec =
2 2 2 2 2 22 2 4 1 2
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) 5.
If sec =
Sol.
1 – cosec , then find the value of 1 + cosec , where is in IV quadrant. 3 (3 marks) 2 sec = 3 1 cos = sec 1 = 2 3
2
3 2 sin2 + cos2 = 1 2 3 2 = 1 sin + 2 3 sin2 + = 1 4
cos =
3 4 4 – 3 sin2 = 4 1 sin2 = 4 1 sin = 2 1 cosec = sin 1 = 1 2 cosec = 2 is in IV quadrant cosec = – 2 1 – (– 2) 1 – cos ec = 1 (– 2) 1 cos ec 1 2 1 – cos ec = 1–2 1 cos ec sin2 = 1 –
1 – cos ec 1 cos ec
S C H O O L S E C TI O N
[Taking square roots]
= –3 191
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PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 14.
If
3 tan = 3 sin , find the value of sin2 – cos2 , where 0. (3 marks)
Sol.
3 tan sin 3 cos
= 3 sin = 3 sin
3 cos
= 3
cos
=
cos2
=
cos2
=
sin + cos = sin2 = 2
2
= =
sin2
=
3 3 3 9 1 3 1 1 – cos2 1 1– 3 3 –1 3 2 3
sin2 – cos2 =
2 1 – 3 3
=
2 –1 3
sin2 – cos2 =
.......(i)
[From (i)]
......(ii) [From (i) and (ii)]
1 3
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 122) 8. (i) Sol.
Eliminate , if x = a sec , y = b tan x = a sec
(2 marks)
x ......(i) a y = b tan y tan = ......(ii) b 2 2 1 + tan = sec
sec
y 1+ b
192
2
y2 1+ 2 b
x2 y2 – 2 a2 b
=
x = a
2
[From (i) and (ii)]
x2 = 2 a =1 S C H O O L S E C TI O N
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x = 2 cos – 3 sin , y = cos + 2 sin x = 2 cos – 3 sin ......(i) y = cos + 2 sin ......(ii) Multiplying (ii) by 2, 2y = 2 cos + 4 sin .....(iii) Subtracting (iii) from (i), x – 2y = 2 cos – 3 sin – (2 cos + 4 sin ) x – 2y = 2 cos – 3 sin – 2 cos – 4 sin x – 2y = – 7 sin – (x – 2y) ......(iv) sin = 7 x 2y in equation (ii) Substituting sin – 7 x 2y y = cos + 2 7 2 x 2y y = cos 7 2 x 2y y+ = cos 7 7y 2 x 2y = cos 7 7y 2 x 4y = cos 7 2 x 3y cos = 7 We know, sin2 + cos2 = 1 2
(iii) Sol.
(3 marks)
2
– (x – 2y) 2x 3y = 1 7 7
(x – 2y)2 (2x 3y)2 = 1 49 49 Multiplying throughout by 49, (x – 2y)2 + (2x + 3y) = 49
x = 3 cosec + 4 cot , y = 4 cosec – 3 cot x = 3 cosec + 4 cot ......(i) y = 4 cosec – 3 cot .....(ii) Multiplying (i) by 4, 4x = 12 cosec + 16 cot .....(iii) Multiplying (ii) by 3, 3y = 12 cosec – 9 cot .....(iv) Subtracting (iv) from (iii), 4x – 3y = 12 cosec + 16 cot – (12 cosec – 9 cot ) 4x – 3y = 12 cosec + 16 cot – 12 cosec + 9 cot 4x – 3y = 25 cot 4x – 3y cot = 25 4x – 3y Substituting cot = in equation (i) 25
S C H O O L S E C TI O N
(3 marks)
193
MT
GEOMETRY
4x – 3y = 3cosec + 4 25 16x – 12y = 3cosec + 25
x
x 16x – 12y x– 25 25x – 16x 12y 25 9x 12y 25 3 (3x 4y) 3 25
cosec
We know, cosec2 – cot2
3x 4y 25
2
4x – 3y – 25
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= 3cosec = 3cosec = 3cosec = cosec =
3x 4y 25
= 1 2
= 1
(3x 4y)2 (4x – 3y)2 – = 1 625 625 Multiplying throughout by 625, (3x + 4y)2 – (4x – 3y)2 = 625 EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121)
6. Sol.
Find the possible values of sin x if 8 sin x – cos x = 4. 8 sin x – cos x = 4 8 sin x – 4 = cos x ..... (i) sin2 x + cos2 x = 1 sin2 x + (8 sin x – 4)2 = 1 [from (i) sin2 x + 64 sin2 x – 64 sin x + 16 = 1 sin2 x + 64 sin2 x – 64 sin x + 16 – 1 = 0 65 sin2 x – 64 sin x + 15 = 0 2 64 sin x + sin2 x – 64 sin x + 16 – 1 = 0 65 sin2 x – 64 sin x + 15 = 0 2 65 sin x – 39 sin x – 25 sin x + 15 = 0 13 sin x (5 sin x – 3) – 5 (5 sin x – 3) = 0 (5 sin x – 3) (13 sin x – 5) = 0 5 sin x – 3 = 0 or 13 sin x – 5 = 0 5 sin x = 3 or 13 sin x = 5 3 5 sin x = or sin x = 5 13
(3 marks)
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 8. Sol.
194
Find the possible value of cos x if cot x + cosec x = 5. cot x + cosec x = 5 cos x 1 sin x sin x = 5 cos x 1 = 5 sin x cos x 1 = sin x ....... (i) 5 2 2 sin x + cos x = 1
(3 marks)
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cos x 1 5
2
+ cos2 x = 1
[from (i)]
cos x 12
+ cos2 x = 1 25 (cos x + 1)2 + 25cos2 x = 25 [Multiplying throughout by 25] cos2 x + 2 cos x + 1 + 25cos2 x – 25 = 0 26cos2 x + 2 cos x – 24 = 0 2 (13 cos2 x + cos x – 12) = 0 13 cos2 x + cos x – 12 = 0 2 13 cos x + 13 cos x – 12 cos x – 12 = 0 13 cos x (cos x + 1) – 12 (cos x + 1) = 0 (cos x + 1) (13 cos x – 12) = 0 cos x + 1 = 0 or 13 cos x – 12 = 0 cos x = – 1 or 13 cos x = 12 12 cos x = – 1 or cos x = 13
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) a2 b2 11. If x = a sin , y = b tan then prove that 2 – 2 = 1. (2 marks) x y Proof : x = a sin 1 a sin = x 1 a cosec = ......(i) cos ec sin x y = b tan 1 b = tan y b 1 cot cot = .....(ii) ta n y We know, 1 + cot2 = cosec 2 cosec2 – cot2 = 1 2
2
b a – x y 2 a b2 – 2 2 x y
= 1
[From (i) and (ii)]
= 1
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) If a cos + b sin = m and a sin – b cos = n, prove that a2 + b2 = m2 + n2. (3 marks) Proof : a cos + b sin = m (a cos + b sin )2 = m 2 .......(i) [Squaring both sides] (a sin – b cos ) = n (a sin – b cos )2 = n 2 ......(ii) [Squaring both sides] Adding (i) and (ii), (a cos + b sin )2 + (a sin – b cos )2 = m2 + n2 a2 cos2 + 2ab cos . sin + b2 sin2 + a2 sin2 – 2ab sin . cos + b2 cos2 = m2 + n2 a2 cos2 + b2 cos2 + a2 sin2 + b2 sin2 = m2 + n2 cos2 (a2 + b2) + sin2 (a2 + b2) = m2 + n2 (a2 + b2) (cos2 + sin2 ) = m2 + n2 a2 + b2 = m2 + n2 [ sin2 + cos2 = 1] 12.
S C H O O L S E C TI O N
195
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PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) If tan + sin = m and tan – sin = n, show that m2 – n2 = 4 mn . (4 marks) Proof : tan + sin = m tan – sin = n m2 – n2 = (tan + sin )2 – (tan – sin )2 = tan2 + 2 tan .sin + sin2 – [tan2 – 2 tan sin + sin2 ] = tan2 + 2 tan .sin + sin2 – tan2 + 2 tan .sin – sin2 = 4 tan .sin .......(i) 9.
4 mn
= 4
tan
sin tan – sin
= 4 tan2 – sin2 = 4
sin2 – sin2 cos 2
1 2 – 1 = 4 sin 2 cos
2 2 = 4 sin sec – 1
2
2
= 4 ta n . sin = 4 × sin × tan From (i) and (ii), m2 – n2 = 4 mn
1 tan2 sec2 2 2 tan sec – 1 .....(ii)
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 10.
Proof : 196
p2 – 1 = sin . (3 marks) p2 + 1 sec + tan = p 1 sin cos cos = p 1 sin = p cos (1 sin )2 = p2 cos 2 sin2 cos2 1 1 sin 2 2 = p 2 2 1 – sin2 cos 1 – sin 1 sin 2 2 1 sin 1 – sin = p 1 sin 2 1 – sin = p 1 sin + 1 – sin p2 1 [By Componendo-Dividendo] 1 sin – 1 + sin = p2 – 1 2 2 p 1 2 sin = p2 – 1 1 p2 1 = sin p2 – 1 p2 – 1 [By Invertendo] p2 + 1 = sin
If sec + tan = p, show that
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PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 13. If sin + sin2 = 1, prove that cos2 + cos4 = 1. Proof : sin + sin² = 1 [Given] sin = 1 – sin² sin2 + cos 2 = 1 2 sin = cos 2 2 1 – sin = cos sin2 = cos4 [Squaring both sides]
(2 marks)
sin2 + cos 2 = 1 2 2 1 – cos = sin
1 – cos2 = cos4
cos² + cos4 = 1
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) 7.
Prove the following : 1 1 (x) If tan A + tan A = 2, show that tan2 A + tan2 A = 2. 1 Proof : tan A + tan A = 2
(2 marks)
2
1 tan A tan A 1 1 tan2 A + 2 tan A . tan A tan2 A 1 tan2 A + 2 + tan2 A 1 tan2 A + tan2 A 1 tan2 A + tan2 A
= 4 = 4 = 4 = 4–2 = 2
(iii) sec2 + cosec2 = sec2 . cosec2 Proof : L.H.S. = sec2 + cosec2 1 1 = cos2 sin2 sin2 + cos 2 = cos 2 . sin2 1 = cos2 . sin2 = sec2 . cosec2 = R.H.S. sec2 + cosec2 = sec2 . cosec2 (viii) Proof :
[Squaring both sides]
(3 marks) 1 1 sec cos , cos ec sin
[ sin2 + cos2 = 1]
sec2 + cosec2 = tan + cot L.H.S. = = = =
S C H O O L S E C TI O N
(3 marks)
sec 2 cos ec 2
1 tan2 1 cot 2 tan2 2 cot 2 (tan cot )2
1 tan cot tan cot 1 197
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= tan + cot = R.H.S.
sec2 + cosec2 = tan + cot
tan sec + 1 + = 2 cosec sec + 1 tan
(v) Proof :
L.H.S. =
(3 marks)
tan sec 1 sec 1 tan
=
tan2 (sec + 1)2 (sec 1) tan
=
tan2 sec 2 2sec 1 (sec 1) tan
=
sec 2 sec 2 2sec (sec 1) tan
=
2sec 2 2sec (sec 1) tan
=
2sec (sec + 1) (sec 1) tan
=
2sec tan
1 cos = 2 cos sin
[ 1 + tan2 = sec2 ]
sin 1 tan cos , sec cos
2 sin = 2 cosec = R.H.S. =
(i) Proof :
tan sec + 1 + = 2 cosec sec + 1 tan
1 – cos A 1 + cos A = cosec A – cot A L.H.S. = =
= =
1 – cos A 1 cos A
1 – cos A 1 – cos A 1 cos A 1 – cos A 1 – cos A
2
1 – cos 2 A
1 – cos A
2
sin2 A 1 – cos A = sin A 1 cos A = sin A – sin A 198
(3 marks)
sin2 A cos2 A 1 2 2 sin A 1 – cos A
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= cosec A – cot A = R.H.S.
1 cos , cot cosec = sin sin
1 – cos A = cosec A – cot A 1 + cos A
cosec x – 1 1 cosec x + 1 sec x + tan x
(ii) Proof :
L.H.S. = =
=
= = =
= = = = = = =
(4 marks)
cos ec x – 1 cos ec x 1
cos ec x – 1 cos ec x cos ec x 1 cos ec x cos ec
x – 1
– 1 – 1
2
cos ec 2 x – 1
cos ec
x – 1
2
cot2 x
1 cot2 x cos ec 2 x 2 2 cot x cos ec x – 1
cos ec x – 1 cot x cos ec x 1 – cot x cot x 1 sin x – tan x cos x sin x 1 – tan x cos x sec x – tan x (sec x – tan x) (sec x tan x) (sec x tan x) sec 2 x – tan2 x sec x tan x 1 tan2 x sec2 x 1 2 2 sec x tan x sec x – tan x 1 R.H.S.
1 cosec x – 1 = sec x + tan x cosec x + 1
x = 1 + 3 sec2 x.tan2 x (3 marks) sec6 x – tan6 x (sec2 x)3 – (tan2 x)3 (sec2 x – tan2 x)[(sec2 x)2 + sec2 x . tan2 x + (tan2 x)2] 1 × [(sec2 x)2 + (tan2 x)2 + sec2 x . tan2 x] (sec2 x – tan2 x)2 + 2 sec2 x . tan2 x + sec2 x . tan2 x [(a – b)2 = a2 – 2ab + b2, a2 + b2 = (a – b)2 + 2ab] 2 = 1 + 3 sec x.tan2x. = R.H.S. 6 sec x – tan6 x = 1 + 3 sec2 x . tan2 x
(iv) sec6 x – tan6 Proof : L.H.S. = = = = =
S C H O O L S E C TI O N
199
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1 1 1 1 – = – cosec A – cot A sin A sin A cosec A + cot A
(ix)
1 1 1 1 i.e. cosec A – cot A + cosec A + cot A = sin A + sin A Proof :
L.H.S. =
(3 marks)
1 1 + cosec A – cot A cosec A + cot A
cosec A + cot A + cosec A – cot A (cosec A – cot A) (cosec A + cot A) 2cosec A = cosec 2 A – cot2 A 1 cot2 A cos ec 2 A 2 cos ec A = 2 2 1 cos ec A – cot A 1 = cosec A + cosec A 1 1 = sin A sin A = R.H.S. 1 1 1 1 + = + cosec A – cot A cosec A + cot A sin A sin A 1 1 1 1 – = – cosec A – cot A sin A sin A cosec A + cot A =
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 6.
cos 2 sin3 + = 1 + sin . cos 1 – tan sin – cos cos2 sin3 L.H.S. = 1 – tan sin – cos
(3 marks)
Show that :
Proof :
cos 2 sin3 sin – cos = 1 – sin cos cos 2 sin3 sin – cos = cos – sin cos cos 3 sin3 = cos – sin sin – cos
=
cos3 sin3 – cos – sin cos – sin
=
cos 3 – sin3 cos – sin
=
200
= = = 2 cos 1 – tan
(cos – sin ) (cos2 + cos . sin + sin2 ) (cos – sin ) 2 2 cos + sin + sin . cos 1 + sin . cos [ sin2 + cos2 = 1] R.H.S. sin3 + sin – cos = 1 + sin . cos S C H O O L S E C TI O N
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PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 4.
Show that :
Proof :
tan cot + = 1 + sec . cosec . 1 – cot 1 – tan
(4 marks)
tan cot L.H.S. = 1 – cot 1 – tan sin
cos cos sin = 1 – cos 1 – sin sin cos
sin
cos cos sin cos – sin = sin – cos sin cos =
sin2 cos2 cos (sin – cos ) sin (cos – sin )
sin2 cos2 = cos (sin – cos ) – sin (sin – cos ) =
sin2 cos 2 1 – sin – cos cos sin
1 sin3 – cos 3 = sin – cos cos sin
sin – cos sin2 sin cos cos 2 1 = sin – cos cos sin
sin2 sin . cos cos2 = cos . sin =
1 sin . cos cos . sin
[ sin2 + cos2 = 1]
1 sin . cos = cos . sin cos . sin 1 1 = cos × sin 1 = sec . cosec + 1
1 1 sec cos , cos ec sin
= R.H.S. tan cot + 1 – cot 1 – tan = 1 + sec . cosec .
EXERCISE - 4.3 (TEXT BOOK PAGE NO. 121) 7.
Prove the following :
(vi)
1 + sin A 1 + sin A + cos A = cos A 1 + cos A – sin A
(3 marks)
Proof : 1 – sin2 A = cos2 A (1 – sin A) (1 + sin A) = cos A . cos A S C H O O L S E C TI O N
201
MT
GEOMETRY
1 sin A cos A = cos A 1 – sin A 1 sin A cos A 1 sin A cos A = = cos A 1 – sin A cos A 1 – sin A 1 + sin A 1 + sin A + cos A = 1 + cos A – sin A cos A
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[By theorem on equal ratios]
tan A tan A + sec A + 1 = (3 marks) sec A – 1 tan A + sec A – 1 Proof : 1 + tan2 A = sec2 A tan2 A = sec2 A – 1 tan A . tan A = (sec A – 1) (sec A + 1) tan A sec A 1 = sec A – 1 tan A tan A sec A 1 tan A sec A 1 sec A – 1 = tan A = sec A – 1 tan A [By theorem on equal ratios] tan A tan A + sec A + 1 = sec A – 1 tan A + sec A – 1 (vii)
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198)
sin – cos + 1 1 Show that : sin + cos – 1 = sec – tan (4 marks) Proof : sin2 + cos2 = 1 cos2 = 1 – sin2 cos . cos = (1 – sin ) (1 + sin ) cos 1 sin = 1 – sin cos By theorem on equal ratios, 1 + sin – cos cos 1 sin cos – (1 – sin ) = 1 – sin = cos 1 sin – cos cos cos – (1 – sin ) = 1 – sin Dividing the numerator and denominator of R.H.S. by cos cos cos 1 + sin – cos cos – 1 + sin = (1 – sin ) cos 1 1 + sin – cos 1 sin cos – 1 + sin = – cos cos sin – cos + 1 1 sin + cos – 1 = sec – tan 5.
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 7.
If 3 tan2 – 4 3 tan + 3 = 0, find the value of .
Sol. 202
3 tan2 – 4 3 tan + 3
= 0
3 tan – 4 tan
3
3 tan – 4 tan
3
= 0 = 0
3 tan – 3 tan – tan + 3
= 0
3
2
2
2
(3 marks)
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3 tan tan –
3 – 1 tan –
tan – 3
3
3 tan – 1
= 0
tan –
= 0
=
3
But, tan 60 =
3
3 tan
tan
= 0
OR
= tan 60 = 60
3 tan – 1 = 0 3 tan = 1 1 tan = 3 1 But, tan 30 = 3 tan = tan 30
`
= 30
Heights and Distances: Many times , we require to find the height of a tower, building, tree or distance of a ship from the lighthouse or width of the river etc. We cannot measure them actually, we can find the heights and distances with the help of trigonometric ratios.
(i) Line of vision : The line connecting the eye of the observer and the object is called as the Line of vision. B
(ii) Angle of Elevation : If A, B are two points such that B is at higher level than A and AM is horizontal line through A, then MAB is the angle of elevation of B with respect to A.
n Li
A (iii) Angle of Depression : If A, B are two A points such that B is at lower level than A and AM is the horizontal line through A, then MAB is the angle of depression of B with respect to A.
on isi V f eo
Angle of Elevation Horizontal Line M Horizontal Line
M
Angle of Depression
Lin eo fV isi on
B EXERCISE - 4.4 (TEXT BOOK PAGE NO. 127) 1. Sol.
For a person standing at a distance of 80 m from a church, the angle of elevation of its top is of measure 45º. Find the height of the church. (3 marks) seg AB represents the church C represents the position of observer. A BC = 80 m ACB is the angle of elevation m ACB = 45º In right angled ABC, AB tan 45º = [By definition] 45º C BC B 80 m AB 1 = 80 AB = 80 m
The height of the church is 80 m. S C H O O L S E C TI O N
203
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EXERCISE - 4.4 (TEXT BOOK PAGE NO. 127) 6.
Sol.
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to the ground. The inclination of the string with the ground is 60º. Find the length of the string, assuming that there is no slack in the string. ( 3 = 1.73) (3 marks) seg AB represents the distance of a kite from ground. A AB = 60 m seg AC represents the length of the string m ACB = 60º 60 m In right angled ABC,
sin 60º =
AB AC
3 2
=
60 AC
AC =
120 3
AC =
[By definition]
60º
B
C
120 3 3 AC = 40 3 m AC = 40 × 1.73 AC = 69.2 m
The length of the string, assuming that there is no slack in the string is 69.2 m. PROBLEM SET - 4 (TEXT BOOK PAGE NO. 199) 19.
Sol.
A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied to a peg at the ground. The height of the pole is 12 m and the angle made by the rope with the ground level is 30º. Calculate the distance covered by the artist in climbing to the top of the pole. (3 marks) A seg AB represents the height of the pole. seg BC represents the distance of the pole from where the rope is tied to the ground. 12 m ACB is angle made by rope with ground seg AC represents the length of rope 30º B C In right angle ABC,
sin 30º =
AB AC
[By definition]
1 12 = 2 AC AC = 24 m
The distance covered by the artist in climbing the top of the pole is 24 m. 204
S C H O O L S E C TI O N
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EXERCISE - 4.4 (TEXT BOOK PAGE NO. 127) 3.
Sol.
Two buildings are in front of each other on either side of a road of width 10 metres. From the top of the first building, which is 30 metres high, the angle of elevation of the top of the second is 45º. What is the height of the second building ? (4 marks) C seg AB and seg CD represents the two buildings AB = 30 m seg BD represents the width of the road A 45º E BD = 10 m A represents the position of observer. 30 m CAE is the angle of elevation m CAE = 45º B D 10 m ABDE is a rectangle AB = DE = 30 m [Opposite sides of rectangle] BD = AE = 10 m In right angled CEA, CE [By definition] tan 45º = AE CE 1 = 10 CE = 10 m CD = CE + ED [ C - E - D] CD = 10 + 30 CD = 40 m
The height of second building is 40 m. EXERCISE - 4.4 (TEXT BOOK PAGE NO. 127) 5.
Sol.
A tree is broken by the wind. The top struck the ground at an angle of 30º and at a distance of 30 m from the root. Find the whole height of the tree. ( 3 = 1.73) seg AB represents the height of the tree The tree breaks at point D seg AD is the broken part of tree which then takes the position of DC AD = DC mDCB = 30º BC = 30 m In right angled DBC, tan 30º =
DB BC
1 3
DB = 30 30 DB = 3
(5 marks)
D
[By definition] B
30º 30 m
C
30 3 3
DB =
DB = 10 3 m
S C H O O L S E C TI O N
A
205
MT
GEOMETRY
BC DC
cos 30º = 3 2
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[By definition]
30 DC 30 2 3
=
DC =
DC =
30 3 2 3
DC =
20 3 m
AD = AB =
DC = 20 3 m AD + DB
AB =
20 3 + 10 3
AB = AB = AB =
30 3 m 30 × 1.73 51.9 m
[ A - D - B]
The height of tree is 51.9 m. PROBLEM SET - 4 (TEXT BOOK PAGE NO. 198) 15.
Sol.
A tree 12m high, is broken by the wind in such a way that its top touches the ground and makes an angle 60º with the ground. At what height from the bottom, the tree is broken by the wind ? ( 3 = 1.73) seg AB represents the tree A AB = 12 m The tree breaks at point D seg AD is the broken part of tree which then takes the position of DC D AD = DC 12 m m DCB = 60º Let DB = x m AD + DB = AB [ A - D - B] AD + x = 12 B AD = (12 – x) m DC = (12 – x) m In right angled DBC, DB [By definition] sin 60º = DC x 3 = 12 – x 2
(5 marks)
206
C
3 12 – x = 2x
12 3 –
60º
3 x = 2x
12 3
x 2
3
= 2x
3x
= 12 3
x =
12 3 2 3 S C H O O L S E C TI O N
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GEOMETRY
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DB
=
DB
=
DB
=
DB
=
DB
=
12 3 m 2 3
3 3 2 – 3
12 3 2 –
2
24 3 – 12 (3)
DB DB
(2)2 –
3
2
24 3 – 36 4–3
24 (1.73) – 36 1 = 41.52 – 36 = 5.52 m
The height at which the tree is broken from the bottom by the wind is 5.52 m. PROBLEM SET - 4 (TEXT BOOK PAGE NO. 199) 16.
Sol.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60º. When he moves 40 m away from the bank, he finds the angle of elevation to be 30º. Find the height of the tree and the width of the river. ( 3 = 1.73) (5 marks) Let seg AB represents the tree A seg BC represents width of river Let BC = x m C and D represents the initial and final positions of the observer DC = 40 m ACB and ADB are the angles of elevation 30º 60º B m ACB = 60º and m ADB = 30º D C 40 m In right angled ACB, AB [By definition] tan 60º = BC AB = 3 x
AB
=
.....(i) 3xm In right angled ADB,
tan 30º =
1 3
AB
AB DB
[By definition]
AB = 40 x 40 x
m .....(ii) 3 From (i) and (ii) we get,
S C H O O L S E C TI O N
=
207
MT
GEOMETRY
3x
=
3x = 3x – x = 2x = x = BC =
AB AB AB
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40 x 3 40 + x 40 40 20 20 m
= 20 3 m = 20 × 1.73 = 34.6 m
[From (i)]
Height of tree is 34.6 m and width of river is 20 m. EXERCISE - 4.4 (TEXT BOOK PAGE NO. 127) 4.
Sol.
Two poles of height 18 metres and 7 metres are erected on the ground. A wire of length 22 metres tied to the top of the poles. Find the angle made by the wire with the horizontal. (4 marks) seg AB and CD represents two poles. AB = 18 m, CD = 7 m seg AC represent the length of the wire. AC = 22 m EBDC is a rectangle EB = CD = 7 m [Opposite sides of rectangle] AB = AE + EB [ A - E - B] A 18 = AE + 7 22 m 18 – 7 = AE AE = 11 m 18 m E C In right angled AEC, 7m AE [By definition] sin C = AC D B 11 sin C = 22 1 sin C = ......(i) 2 But, 1 ......(ii) sin 30º = 2 sin C = sin 30º C = 30º
The angle made by the wire with horizontal is 30º. EXERCISE - 4.4 (TEXT BOOK PAGE NO. 127) 2.
Sol.
208
From the top of a lighthouse, an observer looks at a ship and find the angle of depression to be 60º. If the height of the lighthouse is 90 metres then find how far is that ship from the lighthouse ? ( 3 = 1.73) (3 marks) E seg AB represents the lighthouse A 60º C represents the position of ship. A represents the position of observer. 90 m EAC is the angle of depression. AB = 90 m 60º B C m EAC = 60º S C H O O L S E C TI O N
MT
GEOMETRY
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EAC ACB [Converse of alternate angle test] m ACB = 60º In right angled ABC, AB [By definition] tan 60º = BC 90 = 3 BC 90 BC = 3 90 3 BC = 3 BC = 30 3 m BC = 30 × 1.73 BC = 51.9 m The ship is 51.9 m far from the lighthouse. PROBLEM SET - 4 (TEXT BOOK PAGE NO. 199) 17.
The angle of elevation of a cloud from a point 60 m above a lake is 30º and the angle of depression of the reflection of cloud in the lake is 60º. Find the height of the cloud. (5 marks) E
Sol.
Let E be the position of the cloud 30º A 60º and let BC represent the surface of the lake. 60 m Let A be the point of observer and let F be the reflection of the cloud B EC = CF Let EC = CF = x m ABCD is a rectangle [By definition] AB = CD = 60 m [Opposite sides of rectangle] EC = ED + DC [E - D - C] x = ED + 60 ED = (x – 60)m Also, DF = DC + CF [D - C - F] DF = (60 + x) DF = (x + 60) m In right angled ADE, ED [By definition] tan 30º = AD 1 x – 60 = 3 AD
AD
=
x
60 m C
x
F
3 x – 60 m
In right angled ADF, DF tan 60º = AD x + 60 = 3 3 (x – 60) S C H O O L S E C TI O N
D
[By definition]
209
MT
GEOMETRY
3 (x – 60) 3x – 180 3x – x 2x x
= = = = =
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x + 60 x + 60 60 + 180 240 120
The height of the cloud above the lake is 120 m. PROBLEM SET - 4 (TEXT BOOK PAGE NO. 199) 18.
A man on cliff observes a boat at an angle of depression 30º, which is sailing towards the point of the shore immediately beneath him. Three minutes later the angle of depression of the boat is found to be 60º. Assuming that the boat sails at a uniform speed, determine how much more time it will take to reach the shore. (5 marks)
210
E
A
Sol.
D and C are the initial and 30º final positions of the ship. 60º A represents the position of observer. EAD and EAC are the angles of depression. 60º 30º B m EAD = 30º, m EAC = 60º C m EAD = m ADB = 30º [By converse of alternate angles test] m EAC = m ACB = 60º The ship took 3 mins to travel from D to C Let the speed of the boat be x units/minute Distance = Speed × Time CD = x × 3 CD = 3x units ......(i) In right angled ABC, AB [By definition] tan 60º = BC AB 3 = BC ......(ii) 3 BC = AB In right angled ABD, AB tan 30º = BD 1 AB = 3 BD BD = AB ......(iii) 3 From (ii) and (iii) we get, BD 3 BC = 3 3BC = BD 3BC = BC + CD 3BC – BC = CD 2BC = CD CD BC = 2 3x BC = 2 BC = 1.5 x units
D
[By definition]
[ B - C - D]
S C H O O L S E C TI O N
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GEOMETRY
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Dis tan ce Speed BC = x 1.5x = x = 1.5 minutes = 1.5 × 60 seconds = 90 seconds The time taken by the ship to reach the shore is 90 seconds.
Time =
PROBLEM SET - 4 (TEXT BOOK PAGE NO. 199) 20.
Sol.
A bird was flying in a line parallel to the ground from north to south at a height of 2000 metres. Tom, standing in the middle of the field, first he observed the bird in the north at an angle of 30º. After 3 minutes, he again observed it in the south at an angle of 45º. Find the speed of the bird in kilometers per hour. ( 3 = 1.73) (5 marks) C A A and C represents the first and the second position of the bird at north and south respectively. 200m seg BD is the distance covered by 45º 30º B bird in 3 mins. E D seg AB and seg CD represents the height at which the bird is flying. AB = CD = 2000 m AEB and CED are the angles of elevation m AEB = 30º and m CED = 45º In right angled ABE, AB [By definition] tan 30º = BE 1 2000 = 3 BE BE = 2000 3 m In right angled CDE, CD [By definition] tan 45º = ED 2000 1 = ED
ED BD BD
= 2000 m = BE + ED = 2000 3 2000
BD
= 2000
BD
=
BD
= 2
2000
3 1 km
[ B - E - D]
3 1 m
1000
[ 1 km = 1000 m]
3 1 km
Dis tance Speed = Time 2 3 1 Speed = 3
[ 1 hour = 60 minutes]
60 S C H O O L S E C TI O N
211
MT
GEOMETRY
Speed =
Speed Speed Speed Speed
= = = =
2
EDUCARE LTD.
3 1 60
3 40 3 1 40 (1.73 + 1) 40 (2.73) 109.2 km/hr.
Speed of the bird is 109.2 km/hr.
HOTS PROBLEM (Problems for developing Higher Order Thinking Skill) 18.
P is the circumcentre of an acute angled triangle ABC with cirumradius R. Midpoint of BC is D. Show that the perimeter of ABC is 2R (sin A + sin B + sin C). (5 marks) A Proof :
R
Let, m BAC = x m BPC = 2x
In BDP and CDP, seg BD seg CD seg DP seg DP seg PB seg DC BDP CDP BPD CPD
......(i) .....(ii)
B
P D
R C
[ An angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle] [ D is midpoint of side BC] [Common side] [Radii of the same circle] [By SSS test of congruence] [c.s.c.t.]
1 m BPC 2 1 m BPD = m CPD = × 2x [From (ii)] 2 m BPD = m CPD = x .....(iii) In right angled BDP, BD sin BPD = BP BD sin x = ......(iv) [From (i)] R 2BD sin x = 2R [Multiplying and dividing the R.H.S. by 2] BC sin x = 2R [ D is midpoint of seg BC] BC sin A = 2R 2R . sin A = BC .......(v) Similarly, 2R . sin B = AC ......(vi) 2R . sin C = AB .....(vii)
m BPD = m CPD =
212
S C H O O L S E C TI O N
MT
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GEOMETRY
Adding (v), (vi) and (vii), 2R . sin A + 2R . sin B + 2R . sin C = BC + AC + AB 2R . (sin A + sin B + sin C) = Perimeter of ABC Perimeter of ABC = 2R (sin A + sin B + sin C) 20. Sol.
Find the area of regular polygon having 15 sides which is inscribed in a circle of radius 4 cm. (sin 24 = 0.407) (4 marks) A regular polygon having 15 sides is inscribed in a circle 360 Measure of arc () = = 24º 15 Radius (r) = 4 cm 1 2 r sin Area of OAB = 2 O 1 × 4 × 4 × sin 24 = 2 24º = 8 × 0.407 4 4 Area of OAB = 3.256 cm2 A B Area of a regular polygon = 15 × 3.256 2 = 48.84 cm
Area of a regular polygon is 48.84 cm2. 21. Sol.
Prove that (1 + tan )2 + (1 + cot )2 = (sec + cosec )2. (4 marks) L.H.S. = (1 + tan )2 + (1 + cot )2 = 1 + 2 tan + tan2 + 1 + 2 cot + cot2 = 1 + tan2 + 1 + cot2 + 2 tan + 2 cot = sec2 + cosec2 + 2 (tan + cot ) [ 1 + tan2 = sec2 , 1 + cot2 = cosec2 ] sin cos = sec2 + cosec2 + 2 cos sin
sin2 + cos 2 = sec2 + cosec2 + 2 cos × sin 1 = sec2 + cosec2 + 2 × cos × sin [ sin2 + cos2 = 1] = sec2 + cosec2 + 2 × sec × cosec = (sec + cosec )2 = R.H.S. (1 + tan )2 + (1 + cot )2 = (sec + cosec )2 31. Sol.
If sec – tan = P. Obtain the values of tan, sec and sin in terms of P. (4 marks) sec – tan = p [Given] sec = (p + tan ) ......(i) 1 + tan2 = sec2 1 + tan2 = (p + tan )2 [From (i)] 1 + tan2 = p2 + 2p . tan + tan2 1 = p2 + 2p . tan 1 – p2 = 2p . tan 1 – p2 = tan 2p 1 – p2 tan = 2p 1 – p2 Substituting tan = in (i) 2p
S C H O O L S E C TI O N
213
MT
GEOMETRY
EDUCARE LTD.
1 – p2 2p 2 2p 1– p2 sec = 2p 2 p 1 sec = 2p sec = p
1 p2 sec = 2p 1 cos = sec 2p cos = 1 p2 sin cos = tan sin = tan × cos 1 – p2 2p sin = 2p 1 p2
sin =
32.
1 – p2 1 p2
1 + sin x – cos x 1 + sin x + cos x Prove that 1 + sin x + cos x + 1 + sin x – cos x = 2 cosec x .
Proof :
(4 marks)
1 + sin x – cos x 1 + sin x + cos x L.H.S. = 1 + sin x + cos x + 1 + sin x – cos x = = =
1 + sin x – cos x 2 + 1 + sin x + cos x 2 1 + sin x + cos x 21 + sin x – cos x 2 1 + sin x – cos x + 1 + sin x + cos x 1 + sin x + cos x 1 + sin x – cos x 1 sin x 2
– 2 1 sin x . cos x cos2 x 1 sin x 2 1 sin x . cos x cos2 x 2
1 sin x 2 2 1 sin x 2 cos 2 x
2
2
– cos x
= 1 2 sin x sin2 x – cos 2 x =
2 1 + 2 sin x + sin2 x + 2 cos2 x 2
2
1 – cos x + sin x + 2 sin x sin2 x + cos2 x 1 2 + 4 sin x + 2 sin2 + 2cos 2 x = 2 2 sin2 x + sin2 x + 2 sin x sin x 1 – cos 2 + 4 sin x + 2 sin2 x +cos 2 x = 2 sin2 x + 2 sin x 2 + 4sin x + 2 (1) = 2 sin x (sin x + 1) [ sin2 x + cos2 x = 1] 4 + 4sin x = 2 sin x (sin x + 1) 4 (1 + sin x) = 2 sin x (sin x + 1) 2 = sin x
214
x
S C H O O L S E C TI O N
MT
33.
Sol.
GEOMETRY
EDUCARE LTD.
= = 1 + sin x 1 + sin x
2 cosec x R.H.S. – cos x 1 + sin x + cos x + + cos x 1 + sin x – cos x = 2 cosec x
The angle of elevation of a jet plane from a point on the ground is 60º. After a flight of 30 secs., the angle of elevation changes to 30º. If the jet plane is flying at a constant height of 3600 3 m, find the speed of the (5 marks) jet plane. A B Let A and B be the first and second position of a jet plane. P be the point of observer 3600 3 APD and BPC are the angles 60º of elevation 30º C m APD = 60º, mBPC = 30º P D AD and BC represent constant height at which the jet plane is flying.
AD = BC = 3600 3 m In right angled ADP, AD tan 60º = PD 3600 3 3 = PD PD = 3600 m In right angled BCP, BC tan 30º = PC 1 3600 3 = 3 PC PC = 3600 × 3 PC = 10800 m
[By definition]
[By definition]
PD + DC = PC [ P - D - C] 3600 + DC = 10800 DC = 10800 – 3600 DC = 7200 m ABCD is a rectangle [By definition] AB = DC = 7200 m [Opposite sides of rectangle] Distance covered in 30 seconds = 7200 m Dis tance Speed = Time AB = 30 sec s 7200 m = 30 sec s 7200 30 = [1 km = 1000 m and 1 hr = 3600 seconds] 1000 3600 7200 3600 = 1000 30 = 864 km /hr. Speed of the jet plane is 864 km/hr.
S C H O O L S E C TI O N
215
MT
GEOMETRY
34.
Sol.
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A straight highway leads to the foot of tower. A man standing at the top of tower observes a car at an angle of depression of 30º, which is approaching the foot of the tower with uniform speed. Six seconds later, the angle of depression of the car is found to be 60º. Find the time taken by the car to reach the foot of the tower from this point. (5 marks) Seg AB represents a tower. Let A be a position of an observer. D and C are the initial and final position of the car. EAD and EAC are angles of depression [Converse of alternate angles m EAD = m ADB = 30º test] m EAC = m ACB = 60º The car took 6 sec. to travel from D to C Let the speed of car be x units/seconds E A 30º Distance = speed × time 60º CD = x × 6 CD = 6x units ......(i) In right angled ABC
tan 60 =
AB BC
=
AB BC
3
B
60º
C
30º D
AB =
.....(ii) 3 BC In right angled ABD, tan 30 = 1
AB BD
=
AB BD
AB =
BD 3
3 BC =
[By definition]
3
3BC = 3BC = 3BC – BC = 2BC = 2BC =
BC =
BC = Time = =
.....(iii)
BD 3 BD BC + CD CD CD 6x 6x 2 3x
[From (ii) and (iii)] [ B - C - D]
[From (i)]
Dis tan ce Speed
BC x
3x x = 3 seconds =
The time taken by the car to reach the foot of the tower is 3 seconds. 216
S C H O O L S E C TI O N
MT 35.
Sol.
GEOMETRY
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A pilot in an aeroplane observes that Vashi bridge is on one side of the plane and Worli sea-link is just on the opposite side. The angle of depression of Vashi bridge and Worli sea-link are 60º and 30º respectively. If the aeroplane is at a height of 5500 3 m at that time, what is the distance between Vashi bridge and Worli sea-link ? (5 marks) Let A be the point of observer Let B and C represent the positions of Vashi bridge and Worli sealink respectively. AD represents the height of a plane from the ground AD = 5500 3 m EAB and FAC are angles of depression m EAB = m ABD = 60º [Converse of alternate angle test] m FAC = m ACD = 30º E F A In right angled ADB, 30º 60º AD [By definition] tan 60º = BD 5500 3 550 3 m = 3 BD BD = 5500 m 30º 60º C B D In right angled ADC, AD [By definition] tan 30 = DC 1 5000 3 = 3 DC DC = 5500 × 3 DC = 16500 m BC = BD + DC [B - D - C] BC = 5500 + 16500 BC = 22000 m
Distance between Vashi bridge and Worli sea-link is 22 km. 46. Sol.
5 sin 2 B + 7 cos 2 C + 4 7 In a right angled ABC, A = 90º and . If 2 27 3 + 8 tan 60 AC = 3 find the perimeter of ABC. (5 marks) In ABC, C m A = 90º [Given] AC 3 ......(i) sin B = BC AC .....(ii) cos C = A BC B 2 2 5 sin B 7 cos C 4 7 = 3 8 tan2 60 27 5 sin2 B 7 cos2 C 4 7 2 = 38 3 27
2
5 sin B 7 cos 2 C 4 7 = 3 8 3 27
7 5 sin2 B 7 cos 2 C 4 = 27 27
5 sin2 B 7 cos2 C 4 = 7 = 3 5 sin2 B 7 cos 2 C
S C H O O L S E C TI O N
[Multiplying throughout by 27] 217
MT
GEOMETRY 2
AC AC 5× +7× BC BC 2
AC 12 BC
2
3 12 × BC
12 ×
EDUCARE LTD.
2
= 3
[From (i) and (ii)]
= 3 = 3
9 BC2
BC2
BC2 BC
= 3
12 × 9 3 = 36 = 6 units =
[Taking square roots]
In ABC,
m A = 90º BC2 (6) 2 36 AB 2 AB 2
= = = = =
AB
=
AB Perimeter of ABC
AC2 + AB2 (3)2 + AB2 9 + AB2 36 – 9 27
[Given) [By Pythagoras theorem]
9×3
= 3 3 units = AB + BC +AC = 3 3+6+3 = 3 3+9
60.
Perimeter of ABC
1 Prove that 1 + tan2 A
Proof :
= 3
1 1 1 + cot2 A = sin2 A – sin4 A .
1 L.H.S. = 1 tan2 A
1 1 × 2 sin A cos2 A 1
= sin2 A 1 – sin2 A
1 = 2 sin A – sin4 A = R.H.S. 218
1 1 + tan2 A
(4 marks)
1 1 cot2 A
= (1 + cot2 A) (1 + tan2 A) = cosec2 A × sec2 A =
3 3 units
[ 1 + cot2 A = cosec2 A 1 + tan2 A = sec2 A]
[sin2 A + cos2 A = 1 cos2 A = 1 – sin2 A]
1 1 1 + cot2 A = sin2 A – sin4 A S C H O O L S E C TI O N
MT
GEOMETRY
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MCQ’s 1.
For the angle in standard position if the initial arm rotates 300º in anticlockwise direction, then in which quadrant will the terminal arm be (a) I (b) II (c) III (d) IV
2.
For the angle in standard position, if the terminal arm passes through the point (8, – 15) then what is the value of sin : 15 –15 (b) (a) 17 17 8 –8 (c) (d) 17 17
3.
If 5 sin = 12 cos and ‘’ is an acute angle. What is the value of cosec ? 12 13 (b) (a) 13 12 5 13 (c) (d) 13 5
4.
5.
1 sin 1 – sin 8 and is in III quadrant then what is the value of 1 cos 1 – cos 7 7 8 (a) (b) 8 7 64 49 (c) (d) 49 64 If tan =
If tan =
3 , what is the value of cos2 – sin2 ? 4
(a)
–4 25
(b)
4 25
(c)
–7 25
(d)
7 25
6.
What is the value of (sin2 17º – cos2 73º) ? (a) 1 (b) – 1 1 (c) 0 (d) 3
7.
If x sin (90º – ) cot (90º – ) = cos (90º – ) then what is the value of ‘x’ ? (a) – 1 (b) 1 (c) – 2 (d) 2
8.
If and (2 – 45º) are acute angles such that sin = cos (2 – 45º) then what is the value of tan ? (a) 0
(b)
1 3
(c) 1
(d)
3
S C H O O L S E C TI O N
219
MT
GEOMETRY
9.
10.
The ratio of the length of a rod and its shadow is 1 : of elevation of the sun ? (a) 30º (b) 45º (c) 60º (d) 98º
EDUCARE LTD.
3 , what is the angle
If the angle of elevation of the top of a tower from a distance of 100 m from its foot is 60º, then what is height of the tower ? (a) 50 3 m (b) 100 3 m (c)
100 3
m
(d)
50 3
m
11.
The tops of two poles of height 20 m and 16 m are connected by a wire. If the wire makes an angle of 30º with the horizontal then what is the length of the wire ? (a) 6 m (b) 8 m (c) 10 m (d) 12 m
12.
If ‘5’ and ‘4’ are acute angles such that sin 4 = cos 5 then what is the value of (2 sin 3 – 3 tan 3) ? (a) 1 – 3 (b) 1 (c) 0 (d) – 1
13.
sec2 + cosec2 = ............... . (a) sin2 . cos2 (c) sec . cosec
14.
The value of
sin (a) 1 – cos 1 – cos (c) sin
(b) (d)
sin + cos sec2 . cosec2
1 – cos 1 cos is ................. . (b) (d)
sin2 1 cos 1 cos sin
15.
Which of the following is not a measure of quadrantal angle ? (a) 180º (b) 270º (c) 450º (d) 420º
16.
If cot a = 2 and ‘a’ is in IIIrd quadrant, then cosec a is .............. . (a) (c) 5
5
(b)
– 5
(d)
1 5
17.
(sec + tan ) (sec – tan ) = ............. . (a) 1 (b) – 1 (c) 2 sec (d) 2 tan
18.
If the terminal arm of an angle passes through the point (0, 2) then the terminal arm lies ........... . (a) on X axis (b) on Y axis rd (c) in 3 quadrant (d) in 4th quadrant
220
S C H O O L S E C TI O N
MT
GEOMETRY
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19.
If the possible angles are 90 and – 270, the terminal arm is on the ......... . (a) negative X axis (b) positive X axis (c) negative Y axis (d) positive Y axis
20.
If and are complementary angles, then which of following is true. (b) cos = cos (a) sin = sin (c) tan = tan (d) sec = cosec
21.
If 3 sec – 4 cosec = 0 then tan = 3 (b) (a) 4 3 (c) (d) 5
22.
The slope of a hill makes an angle of 60º with the horizontal. If one has to walk 500 m to reach the top of the hill then the height of the hill is ........... . (a) 500 3 m (c) 500 m
(b) (d)
............. . 5 4 4 3
250 3 m 250 m
23.
In the IInd quadrant ................. is positive. (a) sin (b) cos (c) tan (d) cot
24.
If tan = (a)
2 3
(c)
25.
3
1 3
what is the value of sec ? (b) (d)
3 2 4 3
The terminal arm passes through the point
3 2
(a) (c) 2
(b) (d)
–1, 3
then cos = .......... .
–1 2 1 2
: ANSWERS : 1. 3.
(d) IV
2.
(b)
–15 17
13 12
4.
(d)
49 64
6.
(c) 0
8.
(c) 1
(b)
7.
7 25 (b) 1
9. 11.
(a) 30º (b) 8 m
5.
(d)
S C H O O L S E C TI O N
10. 12.
(b) 100 3 m (c) 0 221
MT
GEOMETRY
13.
(d) sec2 . cosec2
14.
15. 17. 19.
(d) 420º (a) 1 (d) positive Y axis 4 (d) 3
16. 18. 20.
1 – cos sin (b) – 5 (b) on Y axis (d) sec = cosec
22.
(b) 250 3 m
24.
(a)
21. 23.
(a) sin
25.
(b)
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(c)
2 3
–1 2
1 Mark Sums 1. Sol. 2. Sol.
For the angle in standard position, if the initial arm rotates 260º in clockwise direction, then state the quadrant in which the terminal arm lies. For an angle in standard position, if the initial arm rotates 260º in clockwise direction, then the terminal arm lies in the first quadrant. If the angle is in quadrant I and initial arm rotates in clockwise direction, write the possible values of angles. If the angle is in I quadrant and initial arm rotates in clockwise direction, then its lies between – 360º and – 270º.
3. Sol.
Find the value of 3 sin2 3 sin2 + 3 cos2 = = =
4.
If x--coordinate of point A is negative and y-coordinate is positive, then in which quadrant point A lie ? If x-co-ordinate of point A is negative and y-co-ordinate is positive. Then, point A lies in the II quadrant.
Sol.
+ 3 cos2 . 3 (sin2 + cos2 ) 3 (1) 3
[ sin2 + cos2 = 1]
5. Sol.
If = – 30º, find the value of sin . = – 30º [Given] sin = sin (– 30) = – sin 30 1 = – 2 1 sin (– 30) = – 2
6.
What is the directed angle, whose terminal arm lies along the coordinate axes, called ? If the terminal arm of a directed angle lies along the co-ordinate axes, then it is called a quadrantal angle.
Sol. 7. Sol.
222
If the initial arm rotates 70º in clockwise direction, then in which quadrant will the terminal arm lie ? If the initial arm rotates 70º in clockwise direction then the terminal arm lies in the IV quadrant. S C H O O L S E C TI O N
MT 8. Sol.
9.
GEOMETRY
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If sin = 1, sin But, sin 90 sin
what is the value of ? = 1 [Given] = 1 = sin 90
= 90º
1 , what is value of cosec2 ? 3 sin2 + cos2 = 1 2 1 sin2 + = 1 3 1 sin2 = 1 – 9 9 – 1 sin2 = 9 8 sin2 = 9 1 cosec2 = sin2 1 2 cosec = 8 9 9 cosec2 = 8
If cos =
Sol.
10. Sol.
If the terminal arm lies on the positive y-axis, what are the possible angles ? If the terminal arm lies on the positive Y-axis, then angle made is 90º or – 270º.
11.
If + = 90º and tan = + = 90º 3 tan = 4 cot = tan
Sol.
12. Sol.
[Given] [ cot = tan (90 – )]
3 4
If sec = 2, what is the value of tan2 ? sec = 2 [Given] But, sec 60 = 2 sec = sec 60
13. Sol.
cot =
3 then what is the value of cot ? 4 [Given]
= 60º
What is the value of (1 + cot2 ) (1 + cos ) (1 – cos ) ? (1 + cot2 ) (1 + cos ) (1 – cos ) = cosec2 (1 – cos2 ) [ 1 + cot2 = cosec2 ] 2 2 = cosec × sin [sin2 + cos2 = 1, sin2 = 1 – cos2 ] 1 sin2 = 2 sin =
1
S C H O O L S E C TI O N
223
MT
GEOMETRY
14.
If sin =
4 , what is the value of cot2 ? 5
sin
Sol.
EDUCARE LTD.
cosec
=
4 5
[Given]
1 = sin =
1 4 5
5 4 = cosec 2 = cosec2 – 1
cosec
1 + cot2 cot2
cot
cot2
cot2
=
25 – 16 16
cot2
=
9 16
=
2
15.
17.
=
1 cot2 – sin2 cot2 – cosec2
=
–1
What is (1 + = sec2 = sec2
[ 1 + cot2 = cosec2 cot2 – cosec2 = – 1] the value of (1 + tan2 ) (1 – sin ) (1 + sin ) ? tan2 ) (1 – sin ) (1 + sin ) (1 – sin2 ) [ 1 + tan2 = sec2 ] 2 × cos [sin2 + cos2 = 1 cos2 = 1 – sin2 ]
=
1 cos2 cos2
=
1
If tan = tan =
Sol.
224
5 = –1 4 25 –1 = 16
1 What is the value of cot2 – sin2 ?
Sol.
16. Sol.
2
3 , then = ? 3
But, tan 60 = 3 tan = tan 60
= 60
[Given]
S C H O O L S E C TI O N
MT 18.
1 , what is the value of sin4 – cos4 ? 2 1 sin2 – cos2 = [Given] 2 4 4 sin – cos (sin2 )2 – (cos2 )2 (sin2 + cos2 ) (sin2 – cos2 ) 1 (sin2 – cos2 ) [ sin2 + cos2 = 1] 2 2 sin – cos
If sin2 – cos2 =
Sol. = = = = = 19. Sol.
GEOMETRY
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1 2
If 3 sin – 4 cos = 0, what is the value of tan ? 3 sin – 4 cos = 0 3 sin = 4 cos sin 4 cos = 3
tan =
4 3
20. Sol.
What are the possible angles, if the terminal arm is in the I quadrant ? If the terminal arm is in the I quadrant and it in the anticlockwise are 0º and 90º. If it moves in the clockwise direction, the possible angles are – 270º and – 360º.
21.
If tan =
15 , what is the value sec ? tan = 15 1 + tan2 = sec2
1+
Sol.
2
= sec2
sec = 4
[Taking square roots]
If + = 90º and cosec = 2 , then find the value of sec ? + = 90º [Given] [Given] cosec = 2 sec = cosec [ sec = cosec (90 – )]
23. Sol.
1 + 15 = sec2 sec2 = 16
22. Sol.
15
sec =
2
Find tan , for the angle , whose terminal arm passes through (3, 4). The terminal arm passes through (3, 4) x=3 y=4 y tan = x
tan =
S C H O O L S E C TI O N
4 3 225
MT
GEOMETRY
24. Sol. 25. Sol.
26. Sol.
For the angle in standard position, if the initial arm rotates 340º in the anticlockwise direction, state the quadrant in which the terminal arm lies. For the standard angle, if the initial arm rotates 340º in the anticlockwise direction then the terminal arm lies in the IV quadrant. State the value of tan (– 60). tan (– 60) = – tan 60 =
– 3
tan (– 60) = – 3
If the terminal arm passes through the point (1, 1) making an angle , find the value of sec . The terminal arm passes through point (1, 1) x = 1 and y = 1 r
=
x2 y2
r
=
(1)2 (1)2
r
=
11
r
=
2 r x
sec =
27.
EDUCARE LTD.
sec =
2 1
sec =
2
If sec = sec =
Sol.
2 3
, find the value of ?
2
[Given]
3
But, sec 30º =
sec = sec 30º
= 30º
2 3
226
S C H O O L S E C TI O N
5. `
Co-ordinate Geometry
Co-ordinate Geometry : Co-ordinate geometry is a branch of mathematics where we follow a algebraic approach to geometry. Revision of concepts and formulae studied in Std. IX
(x 2 – x1 )2 + (y 2 – y1 )2
1.
Distance formula : AB =
2.
mx 2 + nx1 my 2 + ny1 , Section formula for internal division : P m+n m+n
3.
mx 2 – nx1 my 2 – ny1 , Section formula for external division : P m– n m– n
4.
x1 + x 2 y1 + y 2 , Midpoint formula : P 2 2
5.
Area of a triangle : A (ABC) =
`
Inclination of a line :
1 [x (y – y3) + x2 (y3 – y1) + x3 (y1 – y2)] 2 1 2
In the adjoining figure, Let l be the line in the XY- plane. Let be the angle made by the line l with positive direction of X-axis. The inclination of a line l is the X smallest positive angle made by it with positive direction of X-axis (anticlockwise). Thus is called the inclination of a line where 0 < < 180.
Y B O
Inclination X
A Y
Inclination of X-axis is zero degree and inclination of Y-axis is 90º.
`
Slope of a line : In mathematics, slope or gradient of a line describes its steepness or incline. A higher slope value indicates a steeper incline. The slope of inclined plane is the ratio of vertical (rise) height and horizontal distance (run). Vertical height i.e. slope = Horizontal dis tance If the inclination of the line is , then tangent of is called the slope of the line. It is denoted by m.
S C H O O L S E C TI O N
227
MT
GEOMETRY
Slope of a line = tan = m Slope of X-axis is 0. Slope of Y-axis is not defined. Slopes of parallel lines are equal.
EDUCARE LTD.
Y A X
O
X
B Y
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 1. (i) Sol.
Find the slope of line with inclinations : 45º Inclination of the line = 45º Slope of the line = tan = tan 45º = 1 Slope of the line is 1
(ii) Sol.
30º
(iii) Sol.
(1 mark)
(1 mark)
Inclination of the line = 30º Slope of the line = tan = tan 30º 1 = 3 1 Slope of the line is 3 (1 mark)
90º Inclination of the line = Slope of the line = = =
90º tan tan 90º Not defined
Slope of the line is not defined PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 1. (i) Sol.
Find the slope of the inclination of the line of the following : = 30º (1 mark) Inclination of the line = 30º Slope of the line = tan = tan 30º 1 = 3 1 Slope of the line is 3
(ii) Sol.
= 60º Inclination of the line = 60º Slope of the line = tan = tan 60º = 3 Slope of the line is
228
(1 mark)
3 S C H O O L S E C TI O N
MT
`
GEOMETRY
EDUCARE LTD.
Slope of a line passing through two points : Slope of a line passing through two points A (x1, y1) and B (x2, y2) is y 2 – y1 m= x – x 2 1 EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133)
2. (i) Sol.
Find the slope of the line passing through the following points : A (– 2, 1) and B (0, 3) (1 mark) (ii) P (1, – 1) and Q (– 2, 5) (1 mark) A (– 2, 1) (x1, y1) Sol. P (1, – 1) (x1, y1) B (0, 3) (x2, y2) Q (– 2, 5) (x2, y2) y 2 – y1 y 2 – y1 Slope of line PQ = x – x Slope of line AB = x – x 2 1 2 1 5 – (–1) = – 2 –1 3 –1 = 0 – (– 2) 5 1 = – 2 –1 2 6 = 2 = –3 = 1 = –2 Slope of line AB is 1 Slope of line PQ is – 2.
(iii) Sol.
C (3, 5) and D (– 2, – 3) (1 mark) (iv) C (3, 5) (x1, y1) Sol. D (– 2, – 3) (x2, y2) y 2 – y1 Slope of line CD = x – x 2 1 –3–5 = –2–3 –8 = –5 8 = 5 Slope of line CD is
(v) Sol.
8 5
M (4, 0) and N (– 3, – 2) (1 mark) (vi) M (4, 0) (x1, y1) Sol. N (– 3, – 2) (x2, y2) y 2 – y1 Slope of line MN = x – x 2 1
G (– 4, – 5) and H (– 1, – 2) (1 mark) G (– 4, – 5) (x1, y1) H (– 1, – 2) (x2, y2) y 2 – y1 Slope of line GH = x – x 2 1
(– 2) – (– 5) = (–1) – (– 4)
–2 5 –1 4 3 = 3 = 1 =
Slope of line GH is 1. B (0, – 5) and D (1, 2) (1 mark) B (0, – 5) (x1, y1) D (1, 2) (x2, y2) y 2 – y1 Slope of line BD = x – x 2 1
=
–2 – 0 –3 – 4
=
2 – (– 5) 1–0
=
–2 –7
=
25 1
=
2 7
=
Slope of line MN is
2 . 7
S C H O O L S E C TI O N
7 1 = 7
Slope of line BD is 7. 229
MT
GEOMETRY
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PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 2. (i) Sol.
Find the slope of the line passing through the following pairs : (– 1, 3) and (3, 5) (2 marks) Let, A (– 1, 3) (x1, y1) B (3, 5) (x2, y2) y 2 – y1 Slope of line AB = x – x 2 1 = = = =
5–3 3 – (–1) 2 3 1 2 4 1 2
Slope of line passing through points (– 1, 3) and (3, 5) is (ii) Sol.
(– 4, 5) and (2, 3) Let, A (– 4, 5) (x1, B (2, 3) (x2, y2 Slope of line AB = x 2
(2 marks) y1 ) y2 ) – y1 – x1
=
3–5 2 – (– 4)
=
3–5 24
=
–2 6
=
–1 3
Slope of line passing through points (– 4, 5) and (2, 3) is (iii) Sol.
–1 3 (2 marks)
(7, 8) and (3, 4) Let, A (7, 8) (x1, y1) B (3, 4) (x2, y2) Slope of line AB =
1 . 2
y 2 – y1 x 2 – x1
4–8 3–7 –4 = –4 = 1 Slope of line passing through points (7, 8) and (3, 4) is 1. =
230
S C H O O L S E C TI O N
MT (iv) Sol.
GEOMETRY
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(2 marks)
(3, 6) and (– 6, – 7) Let, A (3, 6) (x1, y1) B (– 6, – 7) (x2, y2) y 2 – y1 Slope of line AB = x – x 2 1 = = =
–7 – 6 –6–3 –13 –9
13 9 13 . 9
Slope of line passing through points (3, 6) and (– 6, – 7) is (v) Sol.
–1, 2 3 and – 2, 3 Let, A –1, 2 3 (x , y ) B – 2, 3 (x , y ) Slope of line AB = = = = =
1
1
2
2
(2 marks)
y 2 – y1 x 2 – x1 3 –2 3 – 2 – (–1)
– 3 –2 1 – 3 –1
3
Slope of line passing through points –1, 2 3 and – 2, 3 (vi) Sol.
0, – 3 and (3, 0) Let, A 0, – 3
=
is
3.
(2 marks)
(x1, (x2, B (3, 0) y2 – Slope of line AB = x – 2 =
y1 ) y2 ) y1 x1
0– – 3
3–0 3 3
Slope of line passing through points 0, – 3 and (3, 0) is
3 . 3
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 6. Sol.
A (0, 0), B (7, 2), C (7, 7) and D (2, 7) are the vertices of a quadrilateral. Find the slope of each diagonal. (2 marks) For ABCD, A (0, 0), B (7, 2), C (7, 7), D (2, 7) seg AC and seg BD are the diagonals of ABCD
S C H O O L S E C TI O N
231
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GEOMETRY
Slope of a line Slope of diagonal
Slope of diagonal Slope of diagonal
Slope of diagonal
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y 2 – y1 = x – x 2 1 7–0 AC = 7–0 7 = 7 = 1 AC is 1 7–2 BD = 2 – 7 5 = –5 = –1 BD is –1
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 7. Sol.
The vertices of a triangle are A (3, – 4), B (5, 7) and C (– 4, 5). Find the slope of each side of the triangle ABC. (3 marks) For ABC, A (3, – 4), B (5, 7), C (– 4, 5) y 2 – y1 Slope of a line = x – x 2 1 7 – (– 4) Slope of side AB = 5 –3
74 2 11 2 11 2 5–7 –4 – 5 –2 –9 2 9
= = Slope of side AB
=
Slope of side BC
= = =
2 9
Slope of side BC is Slope of side AC
= = = =
Slope of side AC is 232
5 – (– 4) –4 – 3 54 –7 9 –7 –9 7
–9 7 S C H O O L S E C TI O N
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GEOMETRY
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EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 3. (i) Sol.
Using slope concept. Check whether the following points are collinear : A (7, 8), B (– 5, 2) and C (3, 6) (3 marks) A (7, 8) (x1, y1) B (– 5, 2) (x2, y2) (x3, y3) C (3, 6) y2 – y 1 Slope of line AB = x – x 2 1 2–8 = –5 – 7 –6 = –12 1 = 2 y3 – y 2 Slope of line BC = x3 – x2 6–2 = 3 – (– 5) 4 = 35 4 = 8 1 = 2 Slope of line AB and slope of line BC are equal and point B is a common point for both the lines Points A, B and C are collinear.
(ii) Sol.
P (– 2, 3), Q (7, – 4) and R (2, 1) P (– 2, 3) (x1, y1) Q (7, – 4) (x2, y2) (x3, y3) R (2, 1) y2 – y 1 Slope of line PQ = x 2 – x1 –4 – 3 = 7 – (– 2) –7 = 72 –7 = 9 y3 – y 2 Slope of line QR = x3 – x2 1 – (– 4) = 2–7 1 4 = –5 5 = –5 = –1 Slope of line PQ and slope of line QR are not equal. Points P, Q and R are not collinear.
S C H O O L S E C TI O N
(3 marks)
233
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GEOMETRY
(iii) Sol.
X (– 1, 3), Y (8, – 3) and Z (2, 1) X (– 1, 3) (x1, y1) Y (8, – 3) (x2, y2) Z (2, 1) (x3, y3) y2 – y 1 Slope of line XY = x 2 – x1 –3 – 3 = 8 – (–1) =
EDUCARE LTD.
(3 marks)
–6 8 1
–6 9 –2 = 3 y3 – y 2 Slope of line YZ = x3 – x2 1 – (– 3) = 2–8 1 3 = –6 4 = –6 –2 = 3 Slope of line XY and slope of line YZ are equal and point Y is a common point for both the lines. Points X, Y and Z are collinear. =
(iv) Sol.
234
– 1) and T (3, 0) (3 marks) (x1, y1) (x2, y2) (x3, y3) y2 – y 1 Slope of line MN = x 2 – x1 –1 – (– 2) = 2 –1 –1 2 = 1 = 1 y3 – y 2 Slope of line NT = x3 – x2 0 – (–1) = 3–2 0 1 = 1 = 1 Slope of line MN and slope of line NT are equal and point N is a common point for both the lines. Points M, N and T are collinear. M (1, M N T
– 2), N (2, (1, – 2) (2, – 1) (3, 0)
S C H O O L S E C TI O N
MT (v) Sol.
GEOMETRY
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A (– 2, – A B D
2), B (1, 1) and D (3, 3) (– 2, – 2) (x1, y1) (1, 1) (x2, y2) (3, 3) (x3, y3) y2 – y 1 Slope of line AB = x 2 – x1
(3 marks)
1 – (– 2) 1 – (– 2) 1 2 = 1 2 3 = 3 = 1 y3 – y 2 Slope of line BD = x3 – x2 3 –1 = 3 –1 2 = 2 = 1 Slope of line AB and slope of line BD are equal and point B is a common point for both the lines. Points A, B and D are collinear. =
(vi) Sol.
V (– 7, 8), W (– 5, 2) and U (3, 6) V (– 7, 8) (x1, y1) W (– 5, 2) (x2, y2) U (3, 6) (x3, y3) y2 – y 1 Slope of line VW = x 2 – x1 2–8 = – 5 – (– 7)
(3 marks)
–6 –5 7 –6 = 2 = –3 y3 – y 2 Slope of line WU = x3 – x2 6–2 = 3 – (– 5) 4 = 35 4 = 8 1 = 2 Slope of line VW and slope of line WU are not equal. Points V, W and U are not collinear. =
S C H O O L S E C TI O N
235
MT
GEOMETRY
EDUCARE LTD.
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 3. (i) Sol.
Check whether points are collinear or not : (– 1, 2), (3, 4), (5, – 6) Let, A (– 1, 2) (x1, y1) B (3, 4) (x2, y2) C (5, – 6) (x3, y3) y2 – y 1 Slope of line AB = x 2 – x1 4–2 = 3 – (–1) 2 = 3 1 2 = 4 1 = 2 y3 – y 2 Slope of line BC = x – x 3 2 =
(3 marks)
–6 – 4 5–3
–10 2 = –5 Slope of line AB and slope of line BC are not equal. The points (1, – 2), (3, 4) and (5, – 6) are not collinear. =
(ii) Sol.
(3 marks)
(4, – 5), (7, 8), (– 2, – 3) Let, A (4, – 5) (x1, y1) B (7, 8) (x2, y2) C (– 2, – 3) (x3, y3) y2 – y 1 Slope of line AB = x 2 – x1
8 – (– 5) 7–4 85 = 3 13 = 3 y3 – y 2 Slope of line BC = x – x 3 2 –3 – 8 = –2 – 7 –11 = –9 11 = 9 Slope of line AB and slope of line BC are not equal. The points (4, – 5), (7, 8) and (– 2, – 3) are not collinear. =
236
S C H O O L S E C TI O N
MT (iii) Sol.
GEOMETRY
EDUCARE LTD.
(7, – 1), (0, 3), (4, 0) Let, A (7, – 1) (x1, y1) B (0, 3) (x2, y2) C (4, 0) (x3, y3) y2 – y 1 Slope of line AB = x 2 – x1
(3 marks)
3 – (–1) 0–7 3 1 = –7 4 = –7 –4 = 7 y3 – y 2 Slope of line BC = x3 – x2 0–3 = 4–0 –3 = 4 Slope of line AB and slope of line BC are not equal. The points (7, – 1), (0, 3) and (4, 0) are not collinear. =
(iv) Sol.
(– 1, 8), (9, – 2), (3, 4) Let, A (– 1, 8) (x1, y1) B (9, – 2) (x2, y2) C (3, 4) (x3, y3) y2 – y 1 Slope of line AB = x 2 – x1 =
–2 – 8 9 – (–1)
=
–10 9 1
= = Slope of line BC
=
(3 marks)
–10 10 –1 y3 – y 2 x3 – x2
4 – (– 2) 3–9 42 = –6 6 = –6 = –1 Slope of line AB and slope of line BC are equal and point B is a common point for both the lines.. The points (– 1, 8), (9, – 2) and (3, 4) are collinear. =
S C H O O L S E C TI O N
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GEOMETRY
(v) Sol.
(0, 6), (3, 0), (+ Let, A (0, B (3, C (2,
2, + 4) 6) (x1, y1) 0) (x2, y2) 4) (x3, y3) y2 – y 1 Slope of line AB = x 2 – x1 = = =
Slope of line BC
EDUCARE LTD.
(3 marks)
0–6 3–0 –6 3 –2
=
y3 – y 2 x3 – x2
=
4–0 2–3
4 –1 = –4 Slope of line AB and slope of line BC are not equal. The points (0, 6), (3, 0) and (2, 4) are not collinear. =
(vi) Sol.
–1 1 2 1 0 , , , 1 , , 4 3 3 4 –1 Let, A 0, (x1, y1) 4 1 B , 1 3
(3 marks)
(x2, y2)
2 1 C , (x3, y3) 3 4 y2 – y 1 Slope of line AB = x 2 – x1
=
=
=
= = 238
–1 1– 4 1 –0 3 1 1 4 1 3 5 4 1 3 5 3 × 4 1 15 4 S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
Slope of line BC
=
y3 – y 2 x3 – x2
=
1 –1 4 2 1 – 3 3
=
–3 4 1 3
=
–3 3 × 4 1
–9 4 Slope of line AB and slope of line BC are not equal. =
–1 1 2 1 The points 0, , , 1 and , are not collinear. 4 3 3 4 EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 9.
If the slope of the line joining points (k, – 3) and (4, 5) is
Sol.
value of k. Let A (k, – 3) (x1, y1) B (4, 5) (x2, y2) Slope of line AB
=
Slope of line AB
=
1 2 1 2 1 2 4–k –k –k k
= = = = = = =
1 2 y 2 – y1 x 2 – x1
1 then find the 2 (3 marks)
[Given]
5 – (– 3) 4–k 53 4–k 8 4–k 16 16 – 4 12 – 12
The value of k is – 12. PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 8. Sol.
–3 Find x if the slope of line joining (x, – 2) and (8, – 11) is . (3 marks) 4 Let, A (x, – 2) (x1, y1) B (8, – 11) (x2, y2) –3 [Given] Slope of line AB = 4
S C H O O L S E C TI O N
239
MT
GEOMETRY
Slope of line AB
=
–3 4 –3 4 –3 4 3 (8 – x) 24 – 3x 3x 3x
= = = =
x
=
x
=
= = =
EDUCARE LTD.
y 2 – y1 x 2 – x1 –11 – (– 2) 8–x –11 2 8–x –9 8– x 9×4 36 24 – 36 – 12 –12 3 –4
The value of x is – 4. EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 4. Sol.
Find the value of k if (– 3, 11), (6, 2) and (k, 4) are collinear points.(3 marks) Let, A (– 3, 11) (x1, y1) B (6, 2) (x2, y2) C (k, 4) (x3, y3) Points A, B and C are collinear Slope of line AB = Slope of line BC y 2 – y1 y3 – y2 = x 2 – x1 x3 – x2 2 – 11 4–2 = 6 – (– 3) k – 6 –9 2 = 63 k –6 2 –9 = k –6 9 2 –1 = k –6 – (k – 6) = 2 –k+6 = 2 –k = 2–6 –k = –4 k = 4 The value of k is 4 PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200)
4. Sol.
1 If (1, 2), , 3 and (0, k) are collinear points find the value of k.(3 marks) 2 Let, A (1, 2) (x1, y1) 1 B , 3 (x2, y2) 2 C (0, k) (x3, y3)
240
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
Points A, B and C are collinear Slope of line AB = Slope of line BC y 2 – y1 y3 – y2 = x 2 – x1 x3 – x2 3–2 1 –1 2 1 1 – 2 1 k k
=
= = = =
k –3 1 0– 2 k –3 –1 2 k–3 1+3 4
The value of k is 4 PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 5.
2 1 1 4 If the points , , , k and , 5 3 2 5
0 are collinear then find the value (3 marks)
of k. Sol.
2 1 Let, A , (x1, y1) 5 3 1 B , k (x2, y2) 2
4 C , 0 (x3, y3) 5 Points A, B and C are collinear Slope of line AB = Slope of line BC y 2 – y1 y3 – y2 = x 2 – x1 x3 – x2 1 k – 0–k 3 = 4 1 1 2 – – 5 2 2 5 3k – 1 –k 3 = 3 1 10 10 3k – 1 10 × 10 = – k × 3 3 3k – 1 = – k 3k + k = 1 4k = 1 1 k = 4 1 The value of k is 4 S C H O O L S E C TI O N
241
MT
GEOMETRY
EDUCARE LTD.
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 9. Sol.
The points (k, 3), (2, – 4) and (– k + 1, – 2) are collinear, find k. (3 marks) Let, A (k, 3), B (2, – 4), C (– k + 1, – 2) Points A, B and C are collinear Slope of line AB = Slope of line BC
–4 – 3 2–k
=
– 2 – (– 4) (– k 1) – 2
–7 2–k
=
–2 4 –k 1 – 2
–7 2–k – 7 (– k – 1) 7k + 7 7k + 2k 9k
= = = =
k
=
–3 9
k
=
–1 3
The value of k is
2 –k – 1 2 (2 – k) 4 – 2k 4–7 –3
=
–1 . 3
EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 8. Sol.
Show that the line joining joining (– 2, 14) and (6, 9). Let, A (– 1, 1), B (– 9, y2 Slope of a line = x 2 Slope of side AB
(– 1, 1) and (– 9, 6) is parallel to the line (3 marks) 6), C (– 2, 14), D (6, 9) – y1 – x1
6 –1 = – 9 – (–1) 5 = –9 1 5 = –8
Slope of line AB Slope of line CD
=
–5 8
9 – 14 = 6 – (– 2) –5 = 62
Slope of line CD 242
=
–5 8 S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
Slope of line AB and slope of line CD are equal. line AB || line CD The line joining (– 1, 1) and (– 9, 6) is parallel to the line joining (– 2, 14) and (6, 9). EXERCISE - 5.1 (TEXT BOOK PAGE NO. 133) 5. Sol.
Show that (– 2, 1), (0, 3), (2, 1) and (0, – 1) are the vertices of a parallelogram. (4 marks) Let, P (– 2, 1), Q (0, 3), R (2, 1), S (0, – 1) S (0, –1) y 2 – y1 P (– 2, 1) Slope of a line = x – x 2 1 3 –1 Slope of line PQ = 0 – (– 2) 2 Q (0, 3) R (2, 1) 02 2 2 Slope of line PQ = 1 –1 – 1 Slope of line RS = 0–2
Slope of line RS Slope of line PQ line PQ || line RS Slope of line QR
–2 = –2 = 1 = Slope of line RS .......(i)
1–3 = 2–0
–2 2 = –1 =
Slope of line QR Slope of line PS
–1 – 1 = 0 – (– 2)
–2 = 02 –2 2 = –1 = Slope of line PS =
Slope of line PS Slope of line QR line QR || line PS
......(ii)
In PQRS, side PQ || side RS
[From (i)]
side QR || side PS
[From (ii)]
PQRS is a parallelogram
[By definition]
The points (–2, 1), (0, 3), (2, 1) and (0, – 1) are the vertices of parallelogram. S C H O O L S E C TI O N
243
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PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 6. Sol.
If P (– 2, 4), Q (4, 8), R (10, 5) and S (4, 1) are the vertices of a quadrilateral (4 marks) show that it is a parallelogram. P (– 2, 4), Q (4, 8), R (10, 5), S (4, 1) y 2 – y1 S (4, 1) P (– 2, 4) Slope of a line = x – x 2 1 8–4 Slope of line PQ = 4 – (– 2) 4 = 42 Q (4, 8) R (10, 5) 4 = 6 2 Slope of line PQ = 3 1–5 Slope of line RS = 4 – 10 –4 = –6 2 Slope of line RS = 3 Slope of line PQ = Slope of line RS line PQ || line RS ........(i) 5–8 Slope of line QR = 10 – 4 –3 = 6 –1 Slope of line QR = 2 1– 4 Slope of line PS = 4 – (– 2)
–3 = 42 –3 = 6 –1 Slope of line PS = 2 Slope of line QR = Slope of line PS line QR || line PS In PQRS, side PQ || side RS side QR || side PS PQRS is a parallelogram
........(ii) [From (i)] [From (ii)] [By definition]
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 7. Sol. 244
Find the value of k if line PQ will be parallel to line RS where P (2, 4), Q (3, 6), R (8, 1) and S (10, k). (4 marks) P (2, 4), Q (3, 6), R (8, 1), S (10, k) line PQ || line RS [Given] S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
Slope of line PQ 6–4 3–2 2 1 k–1 k k
= = = = = =
Slope of line RS k –1 10 – 8 k –1 2 4 4+1 5
The value of k is 5
`
Equation of a line : An equation of a line essentially defines the conditions which must be satisfied by every point on the line. EXERCISE - 5.2 (TEXT BOOK PAGE NO. 137)
4. Sol.
If (4, – 3) is a point on the line 5x + 8y = c, then find c. The equation of the line is 5x + 8y = c Let P (4, – 3) Point P lies on the line 5x + 8y = c Co-ordinates of P satisfies the equation of the line 5 (4) + 8 (– 3) = c 20 – 24 = x c = –4
(2 marks)
The value of c is – 4 EXERCISE - 5.2 (TEXT BOOK PAGE NO. 137) 5. Sol.
`
If (– 2, – 3) is a point on the line 2y = mx + 5, find m. The equation of the line is 2y = mx + 5 Let P (– 2, – 3) Point P lies on the line 2y = mx + 5 Co-ordinates of point P satisfies the equation of the line 2 (– 3) = m (– 2) + 5 – 6 = – 2m + 5 – 2m = – 6 – 5 – 2m = – 11 11 m = 2 11 The value of m is 2
(2 marks)
Intercept of a line : A line can have two intercepts viz., x intercept and y intercept.
•
The x intercept of a line is the point at which the line intersects the X-axis x intercept (x, 0) Hence, the x intercept of a line is the value of the x-co-ordinate of the point at which the line intersects the X-axis.
•
The y intercept of a line is the point at which the line intersects the Y-axis. y intercept (0, y) Hence, the y intercept of a line is the value of the y-co-ordinate of the point at which the line intersects the Y-axis.
S C H O O L S E C TI O N
245
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GEOMETRY
In the adjoining figure, Line l intersects the X-axis The x intercept of line l Line l intersects the Y-axis The y intercept of line l
Y
at point (– 3, 0) is – 3 at point (0, 4) is 4.
EDUCARE LTD.
l
(0, 4) X
(–3, 0) O
(I) Equation of a line in slope intercept form : If the slope of a line is m and its y intercept is c then the equation of the line is y = mx + c. The equation of a line passing through the origin and having slope ‘m’ is y = mx. EXERCISE - 5.2 (TEXT BOOK PAGE NO. 137) 2. (i) Sol.
Write the equation of the lines if m and c are given as follows : m = 5, c = – 1 (1 mark) Slope of the line (m) = 5 y intercept of the line (c) = – 1 By slope intercept form, The equation of the line is y = mx + c y = 5 (x) + (–1) y = 5x – 1 The equation of the given line is y = 5x – 1
(ii) Sol.
m = – 4, c = – 3 Slope of the line (m) = – 4 y intercept of the line (c) = – 3 By slope intercept form, The equation of the line is y = mx + c y = (– 4)x + (– 3) y = – 4x – 3
(1 mark)
The equation of the given line is y = – 4x – 3 (iii) Sol.
m = – 2, c = 3 Slope of the line (m) = – 2 y intercept of the line (c) = 3 By slope intercept form, The equation of the line is y = mx + c y = (– 2)x + 3 y = – 2x + 3
(1 mark)
The equation of the given line is y = – 2x + 3 (iv) Sol.
m = 4, c = 0 Slope of the line (m) = 4 y intercept of the line (c) = 0 By slope intercept form, The equation of the line is y = mx + c y = (4)x + (0) y = 4x
(1 mark)
The equation of the given line is y = 4x 246
S C H O O L S E C TI O N
MT (v) Sol.
GEOMETRY
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m = 0, c = 2 Slope of the line (m) = 0 y intercept of the line (c) = 2 By slope intercept form, The equation of the line is y = mx + c y = (0)x + 2 y=2
(1 mark)
The equation of the given line is y = 2 (vi) Sol.
m = 0, c = – 3 Slope of the line (m) = 0 y intercept of the line (c) = – 3 By slope intercept form, The equation of the line is y = mx + c y = (0)x + (– 3) y=0–3 y=–3
(1 mark)
The equation of the given line is y = – 3 EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141) 1. Sol.
Slope of line is 3 and y intercept is – 4. Write the equation of that line. (1 mark) Slope of the line (m) = 3 y intercept of the line (c) = – 4 By slope intercept form, The equation of the line is y = mx + c y = (3)x + (– 4) y = 3x – 4 The equation of the given line is y = 3x – 4.
2. Sol.
EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141) 1 Line PQ intersects Y-axis in (0, 3) with slope – write the equation of 2 (2 marks) line PQ. 1 Slope of the line PQ (m) = – 2 line PQ intersects Y-axis in (0, 3) Its y intercept (c) is 3 By slope intercept form, The equation of the line y = mx + c 1 y= – x+3 2 1 The equation of the given line is y = – x + 3 2 EXERCISE - 5.2 (TEXT BOOK PAGE NO. 137)
3. (i) Sol.
Write the equation of line in slope intercept form : 2y – 3x + 5 = 0 (1 mark) 2y – 3x + 5 = 0 2y = 3x – 5 3 5 y = x – [Dividing throughout by 2] 2 2 3 –5 and c = Where m = 2 2
S C H O O L S E C TI O N
247
MT
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(ii) Sol.
(iii) Sol.
(iv) Sol.
3y – x = 1 3y – x = 1 3x = x + 1 1 1 y = x 3 3 1 1 and c = Where m = 3 3 x + 2y – 4 = 0 x + 2y – 4 = 0 2y = – x + 4 –1 4 x y = 2 2 –1 x2 y = 2 –1 Where m = and c = 2 2 3x – 2y = 5 3x – 2y = 5 – 2y = – 3x + 5 –3 5 y = –2 x –2 3 5 y = x – 2 2 3 –5 Where m = and c = 2 2
EDUCARE LTD.
(1 mark)
[Dividing throughout by 3]
(1 mark)
[Dividing throughout by 2]
(1 mark)
[Dividing throughout by – 2]
EXERCISE - 5.2 (TEXT BOOK PAGE NO. 137) 1. (i) Sol.
(ii) Sol.
(iii) Sol.
248
Find the slope and y-intercept of the lines given below : y = 3x – 5 (1 mark) y = 3x – 5 y = (3) x + (– 5) Comparing with the equation of a line in slope intercept form y = mx + c m = 3 and c = – 5 Slope of the line is 3 The y intercept of the line is – 5.
2 x+4 (1 mark) 3 2 y= x+4 3 Comparing with the equation of a line in slope intercept form y = mx + c 2 and c = 4 m= 3 2 Slope of the line is 3 The y intercept of the line is 4. y=
y = – 2x + 3 (1 mark) y = – 2x + 3 Comparing with the equation of a line in slope intercept form y = mx + c m = – 2 and c = 3 Slope of the line is – 2 The y intercept of the line is 3. S C H O O L S E C TI O N
MT (iv) Sol.
GEOMETRY
EDUCARE LTD.
y = – 3x – 5 (1 mark) y = – 3x – 5 y = (– 3) x + (– 5) Comparing with the equation of a line in slope intercept form y = mx + c m = – 3 and c = – 5 Slope of the line is – 3 The y intercept of the line is – 5. EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141)
3. (i) Sol.
Write the slope of each of the line stated below : y – 5 = 2 (x – 7) y – 5 = 2 (x – 7) Comparing with the equation of a line in slope point form, y – y1 = m (x – x1) m=2
(1 mark)
Slope of the line y – 5 = 2 (x – 7) is 2 (ii) Sol.
y – 2 = 5 (x – 2) y – 2 = 5 (x – 2) Comparing with the equation of a line in slope point form, y – y1 = m (x – x1) m=5
(1 mark)
Slope of the line y – 2 = 5 (x – 2) is 5 (iii) Sol.
1 (x – 5) 2 1 y+3 = (x – 5) 2 Comparing with the equation of a line in slope point form, y – y1 = m (x – x1)
y+3=
m=
1 2
Slope of the line y + 3 = (iv) Sol.
(1 mark)
1 1 (x – 5) is 2 2
y = – 5 (x + 3) y = – 5 (x + 3) Comparing with the equation of a line in slope point form, y – y1 = m (x – x1) m=–5
(1 mark)
Slope of the line y = – 5 (x + 3) is – 5 (v) Sol.
3 (x + 3) = y – 1 3 (x + 3) = y – 1 y – 1 = 3 (x + 3) Comparing with the equation of a line in slope point form, y – y1 = m (x – x1) m=3
(1 mark)
Slope of the line 3 (x + 3) = y – 1 is 3 S C H O O L S E C TI O N
249
MT
GEOMETRY
EDUCARE LTD.
(II) Equation of a line in slope point form : If a line l having slope m and passes through point A (x1, y1) then, the equation of line l is y – y1 = m (x – x1). EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141) 4. (i) Sol.
Write the equation of the line passing through the point P and having slope m : P (3, 5) and m = 2 (2 marks) P (3, 5) (x1, y1) m=2 The equation of the line passing through P (3, 5) and having slope m = 2 is given by slope point form (y – y1) = m (x – x1) (y – 5) = 2 (x – 3) y – 5 = 2x – 6 2x – y – 6 + 5 = 0 2x – y – 1 = 0 The required equation of the line is 2x – y – 1 = 0
(ii)
1 (2 marks) 2 P (– 3, 7) (x1, y1) 1 m= 2 The equation of the line passing through P (– 3, 7) and having slope 1 m= is given by slope point form 2 (y – y1) = m (x – x1) 1 (y – 7) = [x – (– 3)] 2 1 y–7 = (x + 3) 2 2 (y – 7) = x + 3 2y – 14 = x + 3 x – 2y + 3 + 14 = 0 x – 2y + 17 = 0
P (– 3, 7) and m =
Sol.
The required equation of the line is x – 2y + 17 = 0 (iii) Sol.
250
3 P (– 2, – 3) and m = (2 marks) 5 P (– 2, – 3) (x1, y1) 3 m= 5 The equation of the line passing through P (– 2, – 3) and having slope 3 is given by slope point form m= 5 (y – y1) = m (x – x1) 3 y – (– 3) = [x – (– 2)] 5 3 y+3 = (x + 2) 5 5 (y + 3) = 3 (x + 2) S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
5y + 15 = 3x + 6 3x – 5y + 6 – 15 = 0 3x – 5y – 9 = 0 The required equation of the line is 3x – 5y – 9 = 0 (iv)
6 (2 marks) 7 P (0, 6) (x1, y1) 6 m= 7 The equation of the line passing through P (0, 6) and having slope 6 is given by slope point form m= 7 (y – y1) = m (x – x1) 6 y–6 = (x – 0) 7 7 (y – 6) = 6x 7y – 42 = 6x 6x – 7y + 42 = 0
P (0, 6) and m =
Sol.
The required equation of the line is 6x – 7y + 42 = 0 PROBLEM SET - 5 (TEXT BOOK PAGE NO. 201) 12.
Sol.
3 and which 2 passes through P where P divides the line segment joining A (– 2, 6) and B (3, – 4) in the ratio 2 : 3. (3 marks) A (– 2, 6), B (3, – 4) Point P divides seg AB internally in the ratio 2 : 3 Let, P (x, y) By section formula for internal division, mx 2 nx1 my 2 + ny1 y = x = m+n m+n 2 (3) + 3 (– 2) y –7 = = 2+3 7–3 6–6 – 8 + 18 = = 5 5 0 10 = = 5 5 = 0 = 2 P (0, 2) Write down the equation of a the line whose slope is
3 passes through the point P (0, 2) 2 of the line by slope point form is, m (x – x1) 3 (x – 0) 2 3x 3x 0
The line having slope The equation (y – y1) =
(y – 2) =
y–2 = 2y – 4 = 3x – 2y + 4 =
The equation of the required line is 3x – 2y + 4 = 0 S C H O O L S E C TI O N
251
MT
GEOMETRY
EDUCARE LTD.
EXERCISE - 5.4 (TEXT BOOK PAGE NO. 147) 10.
Sol.
In the adjoining figure, two lines are intersecting at point (3, 4). Find the equation of line PA and line PB. (3 marks) Inclination of line PA is 45º Slope of line PA = tan = tan 45º = 1 A Line PA passes through point P (3, 4) X The equation of the line by slope point form is, (y – y1) = m (x – x1) (y – 4) = 1 (x – 3) y–4 = x–3 x–y–3+4 = 0 x–y+1 = 0
Y P (3, 4)
60º
45º
X
B Y
The equation of line PA is x – y + 1 = 0 Inclination of line PB is 60º Slope of line PB = tan = tan 60º = 3 Line PB passes through point P (3, 4) The equation of the line by slope point form is, (y – y1) = m (x – x1)
(y – 4) =
y–4 =
3 (x – 3) 3x – 3 3
3x – y 4 – 3 3 = 0
The equation of PB is
3x – y 4 – 3 3 = 0
(III) Equation of a line in two point form : If a line l passes through point A (x1, y1) and B (x2, y2) then, x – x1 y – y1 the equation of line l is x – x = y – y 1 2 1 2 EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141) 7. (i) Sol.
Two points of each line are given below write the equation of these lines : A (– 3, 4), B (4, 5) (2 marks) A (– 3, 4) (x1, y1) B (4, 5) (x2, y2) The equation of line AB by two point form is, x – x1 y – y1 = x1 – x 2 y1 – y 2 x – (– 3) y –4 = (– 3) – 4 4–5 y –4 x3 = –1 –7 x + 3 = 7 (y – 4) x + 3 = 7y – 28 x – 7y + 3 + 28 = 0 x – 7y + 31 = 0 The equation of line AB is x – 7y + 31 = 0
252
S C H O O L S E C TI O N
MT (ii) Sol.
GEOMETRY
EDUCARE LTD.
C (4, – 5), D (– 1, – 2) C (4, – 5) (x1, y1) D (– 1, – 2) (x2, y2) The equation of line CD by two point form is, x – x1 y – y1 x1 – x 2 = y1 – y 2
x –4 y – (– 5) 4 – (–1) = – 5 – (– 2)
x –4 y5 4 1 = –5 2
x –4 5 – 3 (x – 4) – 3x + 12 – 3x – 5y + 12 – 25 – 3x – 5y – 13 3x + 5y + 13
= = = = = =
(2 marks)
y5 –3 5 (y + 5) 5y + 25 0 0 0
The equation of line CD is 3x + 5y + 13 = 0 (iii) Sol.
D (5, 6), E (– 5, – 6) (2 marks) D (5, 6) (x1, y1) E (– 5, – 6) (x2, y2) The equation of line CD by two point form is, x – x1 y – y1 = x1 – x 2 y1 – y 2 x –5 y –6 = 5 – (– 5) 6 – (– 6) x –5 y –6 = 55 66 x –5 y –6 = 10 12 12 (x – 5) = 10 (y – 6) 12x – 60 = 10y – 60 12x – 10y – 60 + 60 = 0 12x – 10y = 0 6x – 5y = 0 [Dividing throughout by 2] The equation of line DE is 6x – 5y = 0
(iv) Sol.
E (– 2, – 3), F (– 4, 7) E (– 2, – 3) (x1, y1) F (– 4, 7) (x2, y2) The equation of line EF by two point form is, x – x1 y – y1 = x1 – x 2 y1 – y 2
S C H O O L S E C TI O N
(2 marks)
x – (– 2) y – (– 3) = – 2 – (– 4) –3 – 7 253
MT
GEOMETRY
x2 –2 4 = x2 = 2 – 10 (x + 2) = – 10x – 20 = – 10x – 2y – 20 – 6 = – 10x – 2y – 26 = 5x + y + 13 =
EDUCARE LTD.
y3 –10 y3 –10 2 (y + 3) 2y + 6 0 0 0 [Dividing throughout by – 2]
The equation of line EF the 5x + y + 13 = 0 (v) Sol.
G (0, 2), H (– 3, – 1) G (0, 2) (x1, y1) H (– 3, – 1) (x2, y2) The equation of line GH by two point form is, x – x1 y – y1 = x1 – x 2 y1 – y 2 x –0 y –2 0 – (– 3) = 2 – (–1) y –2 x = 21 3 x y –2 = 3 3 x = y–2 x–y+2 = 0
(2 marks)
The equation of line GH is x – y + 2 = 0 (vi) Sol.
R (4, – 5), T (– 3, 0) R (4, – 5) (x1, y1) T (– 3, 0) (x2, y2) The equation of line RT by two point form is, x – x1 y – y1 x1 – x 2 = y1 – y 2
x –4 y – (– 5) 4 – (– 3) = – 5 – 0
x –4 y5 = 43 –5
x –4 = 7 – 5 (x – 4) = – 5x + 20 = – 5x – 7y + 20 – 35 = – 5x – 7y – 15 = 5x + 7y + 15 =
(2 marks)
y5 –5 7 (y + 5) 7y + 35 0 0 0
The equation of line RT is 5x + 7y + 15 = 0 254
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141) 5. (i) Sol.
Write the equation of the line passing through each of the pair of points given below in the form y = mx + c (2, 3) and (4, 7) (2 marks) Let, A (2, 3) (x1, y1) B (4, 7) (x2, y2) The line passes through points A and B The equation of the line by two point form is x – x1 y – y1 = x1 – x 2 y1 – y 2 x –2 y –3 = 2 –4 3–7 x –2 y –3 = –2 –4 4 (x – 2) = 2 (y – 3) 4x – 8 = 2y – 6 2y = 4x – 8 + 6 2y = 4x – 2 y = 2x – 1 [Dividing throughout by 2] y = 2x – 1 is the equation of the line passing through (2, 3) and (4,7)
(ii) Sol.
(0, 5) and (5, 6) (2 marks) Let, A (0, 5) (x1, y1) B (5, 6) (x2, y2) The line passes through points A and B The equation of the line by two point form is x – x1 y – y1 x1 – x 2 = y1 – y 2 x –0 y –5 = 0–5 5–6 x y –5 = –5 –1 x = 5 (y – 5) x = 5y – 25 5y = x + 25 1 y = x+5 [Dividing throughout by 5] 5 1 y = x + 5 is the equation of the line passing through (0, 5) and (5, 6) 5
(iii) Sol.
(– 3, 5) and (4, – 7) Let, A (– 3, 5) (x1, y1) B (4, – 7) (x2, y2) The line passes through points A and B The equation of the line by two point form is x – x1 y – y1 = x1 – x 2 y1 – y 2 x – (– 3) y –5 = –3 – 4 5 – (– 7)
S C H O O L S E C TI O N
(2 marks)
255
MT
GEOMETRY
x3 –7 12 (x + 3) 12x + 36 7y 7y
= = = =
y =
y= (iv) Sol.
=
y –5 12 – 7 (y – 5) – 7y + 35 – 12x + 35 – 36 – 12x – 1 –12 1 x – 7 7
EDUCARE LTD.
[Dividing throughout by 7]
–12 1 x – is the equation of the line passing through (– 3, 5) and (4, – 7) 7 7
(– 2, – 5) and (– 4, – 3) (2 marks) Let, A (– 2, – 5) (x1, y1) B (– 4, – 3) (x2, y2) The line passes through points A and B The equation of the line by two point form is x – x1 y – y1 = x1 – x 2 y1 – y 2 x – (– 2) y – (– 5) = – 2 – (– 4) – 5 – (– 3) x2 y5 –2 4 = – 5 3 y5 x2 = –2 2 – 2 (x + 2) = 2 (y + 5) – 2x – 4 = 2y + 10 2y = – 2x – 4 – 10 2y = – 2x – 14 y = –x–7 [Dividing throughout by 2] y = – x – 7 is the equation of the line passing through (– 2, – 5) and (– 4, – 3). EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146)
8. Sol.
A (3, 7), B (5, 11), C (– 2, 8) are the vertices of ABC. AD is one of the medians of the triangle. Find the equation of the median AD. (3 marks) A (3, 7) A (3, 7), B (5, 11), C (– 2, 8) Seg AD is median of ABC D is the midpoint of seg BC By midpoint theorem,
x1 + x 2 y1 + y 2 , D 2 2
5 (– 2) 11 8 , 2 2
B (5, 11)
D
C (– 2, 8)
5 – 2 9 , 2 2 3 19 , 2 2 The equation of median AD by two point form is 256
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
x – x1 x1 – x 2 = x –3 3 = 3– 2 x –3 = (6 – 3) 2 x –3 = –3 2 2 (x – 3) = 3 5 (x – 3) = 5x – 15 = 5x – 3y – 15 + 21 = 5x – 3y + 6 =
y – y1 y1 – y 2 y –7 19 7– 2 y –7 (14 – 19) y –7 –5 2 2 (y – 7) 5 3 (y – 7) 3y – 21 0 0
2
The equation of median AD is 5x – 3y + 6 = 0. EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146) 9. Sol.
P (– 3, 4), Q (2, 3), R (– 2, – 5) are the vertices of the PQR. Find the equations of all the medians of PQR. (5 marks) P (– 3, 4), Q (2, 3), R (– 2, – 5) Let seg PA, seg QB and seg RC be the medians on the sides QR, PR and PQ respectively. P (– 3, 4) A, B and C are the midpoints of sides QR, PR and PQ respectively By midpoint formula,
A
B
C
x1 + x 2 y1 + y 2 , 2 2 2 (– 2) 3 (– 5) , 2 2 2 – 2 3 – 5 , 2 2 0 –2 , 2 2 (0, – 1)
Q (– 3, 4)
A
R (– 2, – 5)
– 3 (– 2) 4 (– 5) , 2 2 –3 – 2 4 – 5 , 2 2
– 5 –1 , 2 2
–3 2 4 3 , 2 2
–1 7 , 2 2
S C H O O L S E C TI O N
B
C
257
MT
GEOMETRY
EDUCARE LTD.
By two point form, Equation of line PA is x – x1 y – y1 x1 – x 2 = y1 – y 2
x – (– 3) y –4 (– 3) – 0 = 4 – (–1)
x3 –3 5 (x + 3) 5x + 15 5x + 3y + 15 – 12 5x + 3y + 3
= = = = =
Equation of line QB x – x1 x1 – x 2 = x –2 = –5 2– 2 x –2 = 5 2 2 x –2 = (4 5) 2 2 (x – 2) = 9 7 (x – 2) = 7x – 14 = 7x – 9y – 14 + 27 = 7x – 9y + 13 =
y –4 5 – 3 (y – 4) – 3y + 12 0 0 is y – y1 y1 – y 2 y –3 –1 3– 2 y –3 1 3 2 y –3 (6 1) 2 2 (y – 3) 7 9 (y – 3) 9y – 27 0 0
Equation of line RC, x – x1 y – y1 x1 – x 2 = y1 – y 2 x – (– 2) y – (– 5) = –1 (– 2) – (– 5) – 7 2 2 x 2 y5 = –5 – 7 –2 1 2 2 x2 y 5 (– 4 1) = (–10 – 7) 2 2 2 (x 2) 2 (y 5) = –3 –17 17 (x + 2) = 3 (y + 5) 17x + 34 = 3y + 15 17x – 3y + 34 – 15 = 0 17x – 3y + 19 = 0
The equation of the median of PQR are 5x + 3y + 3 = 0, 7x – 9y + 13 = 0 and 17x – 3y + 19 = 0. 258
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 201) 13. Sol.
In triangle ABC the co-ordinates of vertices A, B and C are (4, 7), (– 2, 3) and (0, 1) respectively. Find the equation of medians passing through vertices A, B and C. (3 marks) A (4, 7), B (– 2, 3), C (0, 1) Let, seg AD, seg BE and seg CF be the medians on sides BC, AC and AB respectively. D, E and F are the midpoints of sides BC, AC and AB respectively. By midpoint formula, A (4, 7) D
x1 + x 2 y1 + y 2 , 2 2
– 2 + 0 3 + 1 , 2 2 – 2 4 , 2 2 (– 1, 2) E
E
F
B (– 2, 3)
D
C (0, 1)
x1 + x 2 y1 + y 2 , 2 2 4 0 7 1 , 2 2
4 8 , 2 2 (2, 4) F
x1 + x 2 y1 + y 2 , 2 2
4 (– 2) 7 3 , 2 2 4 – 2 10 , 2 2 2 , 5 2 (1, 5) By two point form, The equation of median AD, x – x1 y – y1 x1 – x 2 = y1 – y 2
x –4 y –7 4 – (– 1) = 7 – 2
x –4 y –7 4+ 1 = 5
x –4 5 x–4 x–y–4+7 x–y+3
S C H O O L S E C TI O N
y –7 5 = y–7 = 0 = 0 =
259
MT
GEOMETRY
EDUCARE LTD.
The equation of the median BE x – x1 y – y1 x1 – x 2 = y1 – y 2 x – (– 2) y –3 = –2 – 2 3–4
x2 –4 x+2 x+2 x – 4y + 2 + 12 x – 4y + 14 The equation of x – x1 x1 – x 2 x –0 0 –1 x –1 4x 4x – y + 1
y –3 –1 = 4 (y – 3) = 4y – 12 = 0 = 0 the median CF y – y1 = y – y 1 2 y –1 = 1–5 y –1 = –4 = y–1 = 0 =
The equation of the medians of ABC are x – y + 3 = 0, x – 4y + 14 = 0 and 4x – y + 1 = 0 PROBLEM SET - 5 (TEXT BOOK PAGE NO. 201) 14. Sol.
A (5, 4), B (– 3, – 2) and C (1, – 8) are the vertices of a triangle ABC. Find the equation of median AD and line parallel to AC passing through point B. (3 marks) A (5, 4), B (– 3, – 2), C (1, – 8) seg AD is the median of seg BC D is midpoint of seg BC
x1 + x 2 y1 + y 2 , D 2 2 – 3 + 1 – 2 + (– 8) , 2 2 – 2 – 2 – 8 , 2 2 –10 –1 , 2 (– 1, – 5) By two point form, The equation of median AD x – x1 y – y1 = x1 – x 2 y1 – y 2 x –5 y –4 = 5 – (–1) 4 – (– 5) x –5 y –4 = 5 +1 4+5 x –5 y –4 = 6 9 260
S C H O O L S E C TI O N
MT
GEOMETRY
EDUCARE LTD.
9 (x – 5) 9x – 45 9x – 6y – 45 + 24 9x – 6y – 21 3x – 2y – 7
= = = = =
6 (y – 4) 6y – 24 0 0 0
[Dividing throughout by 3]
The equation of median AD is 3x – 2y – 7 = 0
y 2 – y1 = x – x 2 1 –8 – 4 = 1–5 –12 = –4 = 3 Slope of parallel lines are equal Slope of the line parallel to line AC is 3 The line passes through B (– 3, – 2) The equation of the line parallel to line AC passing through point B by the slope point form is y – y1 = m (x – x1) y – (– 2) = 3 [x – (– 3)] y + 2 = 3 (x + 3) y + 2 = 3x + 9 3x – y + 9 – 2 = 0 3x – y + 7 = 0 Slope of line AC
The equation of the line parallel to AC passing through point B is 3x – y + 7 = 0 PROBLEM SET - 5 (TEXT BOOK PAGE NO. 201) 11. Sol.
Find the equation of the line which passes through (2, 7) and whose y-intercept is 3. (2 marks) Let A (2, 7) The y intercept of the line is 3 The line intersects the y-axis at point (0, 3) Let B (0, 3) The line passes through point A and B The equation of the line AB By two point form x – x1 y – y1 = x1 – x 2 y1 – y 2 x –2 y –7 = 2–0 7–3 x –2 y –7 = 2 4 4 (x – 2) = 2 (y – 7) 4x – 8 = 2y – 14 4x – 2y – 8 + 14 = 0 4x – 2y + 6 = 0 2x – y + 3 = 0 [Dividing throughout by 2] The required equation of the line is 2x – y + 3 = 0
S C H O O L S E C TI O N
261
MT
GEOMETRY
EDUCARE LTD.
(III) Equation of a line in double intercept form : If a line l has x intercept a and y intercept b then, the equation of line l is
x y + =1 a b
EXERCISE - 5.3 (TEXT BOOK PAGE NO. 141) 6. (i) Sol.
(iii) Sol.
Convert the equation in y = mx + c form :
x y + =1 (1 mark) (ii) 2 3 x y =1 Sol. 2 3 Multiplying throughout by 3, 3x y =3 2 – 3x 3 y= 2
x y – =1 (1 mark) 3 4 x y – =1 3 4 Multiplying throughout by 4, 4x – y =4 3 4x –4 y= 3
x y + =1 – 4 –1 x y – 4 –1 = 1 –x – y =1 4 –1 x –1 y= 4
x y + =2 (1 mark) 4 3 x y =2 4 3 Multiplying throughout by 3, 3x y =6 4
(1 mark) (iv) Sol.
y=
– 3x 6 4
EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146) 3. (i) Sol.
Write the equation of each of the following line in double intercept form and write x intercept and y intercept. (2 marks) x+y=2 x+y=2 Dividing throughout by 2,
x y =1 2 2 x intercept of line x + y = 2 is 2 y intercept of line x + y = 2 is 2
(ii) Sol.
2x – y = 3 2x – y = 3 Dividing throughout by 3,
(2 marks)
2x y – =1 3 3 x y =1 3 –3 2 3 2 y intercept of line 2x – y = 3 is – 3
x intercept of line 2x – y = 3 is
262
S C H O O L S E C TI O N
MT (iii) Sol.
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(2 marks)
3x + y = 4 3x + y = 4 Dividing throughout by 4,
3x y 4 = 4 4 4 x y =1 4 4 3 4 3 y intercept of line 3x + y = 4 is 4
x intercept of line 3x + y = 4 is
(iv) Sol.
(2 marks)
4x – y – 7 = 0 4x – y – 7 = 0 4x – y = 7 Dividing throughout by 7, 4x y 7 – = 7 7 7 x y =1 7 –7 4
7 4 y intercept of line 4x – y – 7 = 0 is – 7
x intercept of line 4x – y – 7 = 0 is
(v) Sol.
(vi) Sol.
(2 marks)
2x + 3y – 7 = 0 2x + 3y – 7 = 0 2x + 3y = 7 Dividing throughout by 7,
2x 3y 7 = 7 7 7 x y 7 7 =1 2 3
x intercept of line 2x + 3y – 7 = 0 is
7 2
y intercept of line 2x + 3y – 7 = 0 is
7 3
2x – y = 11 2x – y = 11 Dividing throughout by 11, 2x y 11 – = 11 11 11 x y 11 =1 –11 2
(2 marks)
11 2 y intercept of line 2x – y = 11 is – 11
x intercept of line 2x – y = 11 is
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EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146) 1. (i) Sol.
Find x and y intercepts of each of the line : y = 2x – 3 y = 2x – 3 ......(i) Substituting y = 0 in equation (i), 0 = 2x – 3 2x = 3 3 x = 2 3 The x intercept of the line y = 2x – 3 is 2 Substituting x = 0 in equation (i), y = 2 (0) – 3 y = 0–3 y = –3
(2 marks)
The y intercept of the line y = 2x – 3 is – 3 (ii) Sol.
y–x=–5 y–x = –5 ......(i) Substituting y = 0 in equation (i), 0–x = –5 –x = –5 x = 5 The x intercept of the line y – x = – 5 is 5 Substituting x = 0 in equation (i), y–0 = –5 y = –5
(2 marks)
The y intercept of the line y – x = – 5 is – 5 (iii)
y=
Sol.
2 x–4 3
(2 marks)
2 x–4 ......(i) 3 Substituting y = 0 in equation (i), 2 x–4 0 = 3 2 x = 4 3 3 x = 4× 2 x = 2×3 x = 6 y =
The x intercept of the line y =
2 x – 4 is 6 3
Substituting x = 0 in equation (i), 2 (0) – 4 y = 3 y = –4 The y intercept of the line y = 264
2 x – 4 is – 4 3 S C H O O L S E C TI O N
MT (iv) Sol.
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2 =0 7 2 5x – y – 7 Substituting 2 5x – 0 – 7 2 5x – 7
(2 marks)
5x – y –
= 0
......(i)
y = 0 in equation (i), = 0 = 0
2 7 2 1 x = × 7 5 2 x = 35
5x =
The x intercept of the line 5x – y –
2 2 = 0 is 7 35
Substituting x = 0 in equation (i), 2 = 0 5 (0) – y – 7 2 –y– = 0 7 –2 y = 7 The y intercept of the line 5x – y – (v) Sol.
2 –2 = 0 is 7 7
x + 2y – 6 = 0 x + 2y – 6 = 0 .......(i) Substituting y = 0 in equation (i), x + 2 (0) – 6 = 0 x–6 = 0 x = 6
(2 marks)
The x intercept of the line x + 2y – 6 = 0 is 6 Substituting 0 + 2y – 6 2y y
x = 0 in equation (i), = 0 = 6 = 3
The y intercept of the line x + 2y – 6 = 0 is 3 (vi) Sol.
5 =0 3 5 3x + y – 3 Substituting 5 3x + 0 – 3 5 3x – 3
(2 marks)
3x + y –
S C H O O L S E C TI O N
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.......(i)
y = 0 in equation (i), = 0 = 0 265
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5 3
3x =
x =
5 1 × 3 3
x =
5 9
The x intercept of the line 3x + y –
5 5 = 0 is 3 9
Substituting x = 0 in equation (i), 5 = 0 3 (0) + y – 3 5 0+y– = 0 3 5 y = 3 The y intercept of the line 3x + y – (vii) Sol.
5 5 = 0 is 3 3 (2 marks)
3y – 2x + 5 = 0 3y – 2x + 5 = 0 .......(i) Substituting y = 0 in equation (i), 3 (0) – 2x + 5 = 0 0 – 2x + 5 = 0 – 2x + 5 = 0 – 2x = – 5 5 x = 2 The x intercept of the line 3y – 2x + 5 = 0 is
5 2
Substituting 3y – 2(0) + 5 3y + 5 3y
x = 0 in equation (i), = 0 = 0 = –5 –5 y = 3
The y intercept of the line 3y – 2x + 5 = 0 is (viii) Sol.
7x + 6y – 1 = 0 7x + 6y – 1 = 0 .......(i) Substituting y = 0 in equation (i), 7x + 6 (0) – 1 = 0 7x – 1 = 0 7x = 1 1 x = 7 The x intercept of the line 7x + 6y – 1 = 0 is
–5 3 (2 marks)
1 7
Substituting x = 0 in equation (i), 266
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7 (0) + 6y – 1 = 0 6y – 1 = 0 6y = 1 1 y = 6 The y intercept of the line 7x + 6y – 1 = 0 is (ix) Sol.
1 6
y – 3x = 0 y – 3x = 0 .......(i) Substituting y = 0 in equation (i), 0 – 3x = 0 – 3x = 0 x = 0 The x intercept of the line y – 3x = 0 is 0 Substituting x = 0 in equation (i), y – 3 (0) = 0 y–0 = 0 y = 0
(2 marks)
The y intercept of the line y – 3x = 0 is 0 (x) Sol.
4x – 7y = 0 4x – 7y = 0 .......(i) Substituting y = 0 in equation (i), 4x – 7(0) = 0 4x = 0 x = 0 The x intercept of the line 4x – 7y = 0 is 0 Substituting x = 0 in equation (i), 4 (0) – 7y = 0 – 7y = 0 y = 0
(2 marks)
The y intercept of the line 4x – 7y = 0 is 0
`
General equation of a line : The general equation of line is ax + by + c = 0 – coefficient of x Slope of the line = m = coefficient of y – co ns tant term x intercept of the line = coefficient of x – co ns tant term y intercept of the line = coefficient of y EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146)
2. (i) Sol.
Find the slope and y intercept for each of the following line : 2x – 3y = 7 (2 marks) 2x – 3y = 7 2x – 3y – 7 = 0 – coefficient of x –2 2 Slope of line 2x – 3y = 7 is coefficient of y = – 3 = 3
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Slope of line 2x – 3y = 7 is
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2 3
– cons tan t term – (– 7) –7 y intercept of line 2x – 3y = 7 is coefficient of y = – 3 = 3 –7 y intercept of line 2x – 3y = 7 is 3 (ii) Sol.
(2 marks)
y – 3x – 6 = 0 y – 3x – 6 = 0 – 3x + y – 6 = 0
– coefficient of x – (– 3) =3 Slope of line y – 3x – 6 = 0 is coefficient of y = 1 Slope of line y – 3x – 6 = 0 is 3. – cons tan t term – (– 6) y intercept of line y – 3x – 6 = 0 is coefficient of y = =6 1 y intercept of line y – 3x – 6 = 0 is 6. (iii) Sol.
(2 marks)
2y + 2x – 5 = 0 2y + 2x – 5 = 0 2x + 2y – 5 = 0
– coefficient of x –2 =–1 Slope of line 2x + 2y – 5 = 0 is coefficient of y = 2 Slope of line 2x + 2y – 5 = 0 is – 1. – cons tan t term – (– 5) 5 y intercept of line 2x + 2y – 5 = 0 is coefficient of y = = 2 2 5 y intercept of line 2x + 2y – 5 = 0 is . 2 (iv) Sol.
(v) Sol.
(vi) Sol.
7x – y + 3 = 0 7x – y + 3 = 0
(2 marks)
4x – y = 0 4x – y = 0
(2 marks)
8x – 4y – 1 = 0 8x – 4y – 1 = 0
(2 marks)
– coefficient of x –7 Slope of line 7x – y + 3 = 0 is coefficient of y = –1 = 7 Slope of line 7x – y + 3 = 0 is 7. – cons tan t term –3 y intercept of line 7x – y + 3 = 0 is coefficient of y = –1 = 3 y intercept of line7x – y + 3 = 0 is 3.
– coefficient of x –4 Slope of line 4x – y = 0 is coefficient of y = –1 = 4 Slope of line 4x – y = 0 is 4. – cons tan t term –0 y intercept of line 4x – y = 0 is coefficient of y = –1 = 0 y intercept of line 4x – y = 0 is 0.
– coefficient of x –8 Slope of line 8x – 4y – 1 = 0 is coefficient of y = – 4 = 2 Slope of line 8x – 4y – 1 = 0 is 2. 268
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– cons tan t term – (–1) 1 –1 y intercept of line 8x – 4y – 1 = 0 is coefficient of y = – 4 = – 4 = 4 y intercept of line 8x – 4y – 1 = 0 is (vii) Sol.
–1 . 4 (2 marks)
5x – 2y = 3 5x – 2y = 3 5x – 2y – 3 = 0
– coefficient of x –5 5 Slope of line 5x – 2y = 3 is coefficient of y = – 2 = 2 Slope of line 5x – 2y = 3 is
5 . 2
– cons tan t term – (– 3) 3 –3 y intercept of line 5x – 2y = 3 is coefficient of y = – 2 = – 2 = 2 y intercept of line 5x – 2y = 3 is (viii) Sol.
–3 2 (2 marks)
5x – 8y = – 2 5x – 8y = – 2 5x – 8y + 2 = 0
– coefficient of x –5 5 Slope of line 5x – 8y = – 2 is coefficient of y = – 8 = 8 5 Slope of line 5x – 8y = – 2 is . 8 – cons tan t term –2 1 y intercept of line 5x – 8y = – 2 is coefficient of y = – 8 = 4 y intercept of line 5x – 8y = – 2 is
1 . 4
PROBLEM SET - 5 (TEXT BOOK PAGE NO. 200) 10.
Sol.
The equation of a line is 3x – 4y + 12 = 0. It intersects X-axis in point A and y-axis in point B, find the co-ordinates of points A and B, find the length of AB. (3 marks) The equation of line is 3x – 4y + 12 = 0 y intercept of the line =
– co ns tant term –12 = coefficient of y –4 = 3
– cons tan t term –12 x intercept of the line = coefficient of x = = – 4 3 The line intersects the X-axis at point A Its y-co-ordinate is 0 A (– 4, 0) The line intersects the Y-axis at point B Its x-co-ordinate is 0 B (0, 3)
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By Distance formula, AB
=
(x 2 – x1 )2 (y 2 – y1 )2
=
[0 – (– 4)]2 (3 – 0)2
=
(0 4)2 32
=
42 32
=
16 9
= 25 = 5 units l (AB) = 5 units EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146) 4. Sol.
Find the equation of the line passing through (2, –1) and parallel to 3x + 4y = 10. (3 marks) The equation of the given line is 3x + 4y = 10 i.e., 3x + 4y – 10 = 0 – coefficient of x –3 Slope of the given line is coefficient of y = 4 The required line is parallel to the given line –3 Slope of the required line is 4 –3 The required line whose slope is passes through point (2, – 1) 4 The equation of the line by slope point form is (y – y1) = m (x – x1) –3 y – (– 1) = (x – 2) 4 –3 y+1 = (x – 2) 4 4 (y + 1) = – 3 (x – 2) 4y + 4 = – 3x + 6 3x + 4y + 4 – 6 = 0 3x + 4y – 2 = 0 The equation of the line passing through (2, –1) and parallel to 3x + 4y = 10 is 3x + 4y – 2 = 0. EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146)
5. Sol.
Find the equation of the line passing through (– 3, – 5) and parallel to x – 2y – 7 = 0. (3 marks) The equation of the given line is x – 2y – 7 = 0
– coefficient of x –1 1 = coefficient of y –2 = 2 The required line is parallel to the given line 1 Slope of the required line is 2 1 The required line whose slope is passes through (– 3, – 5) 2 Slope of the given line is
270
S C H O O L S E C TI O N
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The equation of the line by slope point form is, (y – y1) = (x – x1) 1 y – (– 5) = [ x – (– 3)] 2 1 y+5 = (x + 3) 2 2 (y + 5) = x + 3 2y + 10 = x + 3 x – 2y + 3 – 10 = 0 x – 2y – 7 = 0 The equation of the line passing through (– 3, – 5) and parallel to x – 2y – 7 = 0 is x – 2y – 7 = 0. EXERCISE - 5.4 (TEXT BOOK PAGE NO. 146) 6. Sol.
Find the equation of the line parallel to 4x + 3y = 5 and having x-intercept (– 3). (3 marks) The equation of the given line is 4x + 3y = 5 i.e., 4x + 3y – 5 = 0 – coefficient of x –4 Slope of the given line 4x + 3y = 5 is coefficient of y = 3 The required line is parallel to the given line –4 Slope of the required line is 3 The required line has x intercept – 3 The line intersects the X-axis at point (– 3, 0) Let A (– 3, 0) –4 and passes through point A (– 3, 0) The required line has slope 3 The equation of the line by slope point form is, y – y1 = m (x – x1) –4 x – (– 3) y–0 = 3 3y = – 4 (x + 3) 3y = – 4x – 12 4x + 3y + 12 = 0 The equation of required line is 4x + 3y + 12 = 0 EXERCISE - 5.4 (TEXT BOOK PAGE NO. 141)
7. (i) Sol.
Write the equation of each of the following line : The X-axis and Y-axis The equation of X-axis is y = 0 The equation of Y-axis is x = 0
(ii) Sol.
The line passing through the origin and the point (– 3, 5) Let O (0, 0) (x1, y1) A (– 3, 5) (x2, y2) The equation of line OA by two point form is, x – x1 y – y1 x1 – x 2 = y1 – y 2 x –0 y –0 = 0 – (– 3) 0 – 5
S C H O O L S E C TI O N
(1 mark)
(2 marks)
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y x = –5 3 – 5x = 3y 5x + 3y = 0
The equation of the line passing through the origin and the point (– 3, 5) is 5x + 3y = 0. (iii) Sol.
(2 marks)
The line passing through the points (2, 3) and (4, 5). Let A (2, 3) (x1, y1), B (4, 5) (x2, y2) The equation of line AB by two point form is, x – x1 y – y1 = x1 – x 2 y1 – y 2 x –2 y –3 2–4 = 3–5 x –2 y –3 = –2 –2 x–2 = y–3 x–y–2+3 = 0 x–y+1 = 0
The equation of the line passing through the points (2, 3) and (4, 5) is x – y + 1 = 0 (iv) Sol.
The line passing through the points (3, 4) and having slope 5.(2 marks) Let A (3, 4) (x1, y1) and m = 5 The equation of the line passing through A and having slope 5 by slope point form is, y – y1 = m (x – x1) y – 4 = 5 (x – 3) y – 4 = 5x – 15 5x – y – 15 + 4 = 0 5x – y – 11 = 0 The equation of the line passing through the points (3, 4) and having slope 5 is 5x – y – 11 = 0.
(v) Sol.
The line parallel to X-axis and passing through the point (– 3, 4). (1 mark) The equation of the line parallel to X-axis and passing through the point (– 3, 4) is y = 4.
(vi) Sol.
The line parallel to Y-axis and passing through the point (– 3, 5). (1 mark) The equation of the line parallel to Y-axis and passing through the point (– 3, 5) is x = – 3.
HOTS PROBLEM (Problems for developing Higher Order Thinking Skill) 36. Sol.
Point (m, 2m – 1) lies on the line Let, A (m, 2m – 1)
3x 2y – = –1 find m. 5 3
(2 marks)
3x 2y – = – 1 5 3 Co-ordinates of point A satisfies the equation, 3m 2 (2m – 1) – = –1 5 3 Point A lies on the line
272
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Multiplying throughout by 15,
3m 2 (2m – 1) 15 – 15 5 3 3 (3m) – 5 (4m – 2) 9m – 20m + 10 – 11m – 11m
The value of m is 37. Sol.
25 11
= 15 (– 1) = = = =
– 15 – 15 – 15 – 10 – 25 25 m = 11
The line x – 6y + 11 = 0 bisects the segment joining (8, – 1) and (0, k) find the value of k. (3 marks) Let, A (8, – 1), B (0, k) Let, line x – 6y + 11 = 0 bisect seg AB at point P Point P is the midpoint of seg AB By midpoint formula, x1 x 2 y1 y 2 , P 2 2 8 0 –1 k , P 2 2 –1 k P 4 , 2 –1 k Point P 4 , lies on line x – 6y + 11 = 0 2 Co-ordinates of point P satisfies the equation,
–1 k 4–6 2 4 – 3 (– 1 + k) 4 + 3 – 3k 7 – 3k – 3k
+ 11 = 0 + + + + –
11 11 11 18 3k k
= = = = = =
0 0 0 0 – 18 6
The value of k is 6. 38.
The point P divides the segment AB joining points A (2, 1) and B (– 3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0. Justify. (4 marks) Proof : Let A (2, 1), B (– 3, 6) Point P divides seg AB internally in the ratio 2 : 3 By section formula for internal division, mx 2 nx1 my 2 ny1 , P m n mn 2 (– 3) 3 (2) 2 (6) 3 (1) , P 23 23 6 6 12 3 , P 5 5 0 15 P , 5 5 P (0, 3) S C H O O L S E C TI O N
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Given equation of the line is x – 5y + 15 = 0 Substituting x = 0 and y = 3 on the L.H.S. of the equation L.H.S. = x – 5y + 15 = 0 – 5 (3) + 15 = 0 – 15 + 15 = 0 = R.H.S. Point P (0, 3) lies on the x – 5y + 15 = 0 39.
The side AB of an equilateral Y triangle ABC is parallel to X-axis. Find the slopes of all sides. (4 marks)
C
120º A 60º 60º • B F X
O Y
D
E
• M
X
Construction : Extend ray CA and ray CB intersecting the X-axis at points D and E respectively. Extend ray AB upto point F such that A - B - F. Take point M as shown in the figure. Sol. ABC is an equilateral triangle side AB || X axis [Given] Slope of X axis is 0 Slope of side AB is 0 [Slopes of parallel lines are equal] line AB || X axis [Given] On transversal CD, CAB CDE [Converse of corresponding angles test] But, m CAB = 60º [Angle of equialteral triangle] m CDE = 60º The inclination of line AC is 60º
Slope of side AC = tan 60º = 3 m CBF + m CBA = 180 [Linear pair axiom] m CBF + 60 = 180 m CBF = 180 – 60 m CBF = 120º ......(i) line AB || X axis On transversal CE, CBF CEM [Converse of corresponding angles test] m CEM = 120º Inclination of line CB is 120º Slope of line CB = tan 120º ......(ii) Slope of line CB = cot (90 – 120) [tan = cot (90 – )] Slope of line CB = cot – 30 Slope of line CB = – cot 30
Slope of line CB = – 3
40. Sol.
274
Find the value of k so that PQ will be parallel to RS where P (2, 4), Q (3, 6), R (8, 1) and S (10, k). (2 marks) P (2, 4), Q (3, 6), R (8, 1), S (10, k) Line PQ is parallel to line RS Slope of line PQ = Slope of line RS S C H O O L S E C TI O N
MT
41. Sol.
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6–4 3–2 2 1 4 k k Value of k is 5.
k –1 = 10 – 8 k –1 = 2 = k–1 = 4+1 = 5
Find the equation of the straight line passing through the origin and the point of intersection of the lines x + 2y = 7 and x – y = 4. (3 marks) Let line x + 2y = 7 and x – y = 4 intersect at point A x + 2y = 7 .......(i) x–y=4 ......(ii) Subtracting (ii) from (i), x + 2y = 7 x – y = 4 (–) (+) (–) 3y = 3 y = 1 Substituting y = 1 in equation (ii), x–1 = 4 x = 4+1 x = 5 A (5, 1) The straight line passes through A (5, 1) and O (0, 0) The equation of the line by two point form, y – y1 x – x1 = y1 – y 2 x1 – x 2 x –5 y –1 5 – 0 = 1–0 x –5 y –1 = 5 1 x – 5 = 5 (y – 1) x – 5 = 5y – 5 x – 5y – 5 + 5 = 0 x – 5y = 0 The equation of the line passing through the origin and the point of intersection of the lines x + 2y = 7 and x – y = 4 is x – 5y = 0.
42.
x=5
Sol.
Lines x = 5 and y = 4 form a rectangle with coordinate axes. Find the equation of the diagonals. (4 marks) Let line y = 4 intersect the Y-axis at point A A (0, 4) Y Let line x = 5 intersect the X-axis at point C y=4 B A C (5, 0) Let line y = 4 and x = 5 intersect at point B B (5, 4) X X The origin O (0, 0) O C ABCO is a rectangle seg AC and seg BO are the diagonals Y Equation of line AC by two point form,
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x – x1 y – y1 = x1 – x 2 y1 – y 2 x –0 y –4 = 0–5 4–0 x y –4 = –5 4 4x = – 5 (y – 4) – 5y + 20 = 4x 4x + 5y – 20 = 0 Equation of diagonal AC is 4x + 5y – 20 = 0. Equation of line BD by two point from ix, x – x1 y – y1 = x1 – x 2 y1 – y 2 x –5 y –4 = 5–0 4–0 x –5 y –4 = 5 4 4 (x – 5) = 5 (y – 4) 5y – 20 = 4x – 20 4x – 5y = 0
Equation of diagonal BD is 4x – 5y = 0. 43. Sol.
276
If the points A (1, 2), B (4, 0), C (3, 5) are the vertices of a triangle ABC. Find the equation of the line passing through the mid points of AB and (4 marks) AC. A (1, 2), B (4, 6), C (3, 5) Let D and E be the midpoints of sides AB and AC respectively Point D is the midpoint of side AB x1 x 2 y1 y 2 , D 2 2 1 4 2 6 , 2 2 5 8 , 2 2 5 , 4 2 Point E is the midpoint of side AC x1 x 2 y1 y 2 , E 2 2 1 + 3 2 5 , 2 2 4 7 , 2 2 7 2 , 2 The required line passes through points D and E The equation of the line by two point form, x – x1 y – y1 = x1 – x 2 y1 – y 2 x – 52 y –4 5 –2 = 4–7 2 2 S C H O O L S E C TI O N
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2x
– 5
5 – 4
2 =
2 2x – 5 = 5–4 2x – 5 = 2x – 5 = 2x – 2y – 5 + 8 = 2x – 2y + 3 =
y –4
8 – 7
2
(y – 4) 1 2 2 (y – 4) 2y – 8 0 0
The equation of the required line is 2x – 2y + 3 = 0 44. Sol.
Find the equation of the line making an angle of 45º with positive X-axis and at a distance 2 2 from the origin. (3 marks) Let line l be the line making an angle of 45º with positive X axis at point A Slope of line l = tan 45 = 1 Let seg OP line l as shown in the figure,
45. Sol.
OP = 2 2 units OAP = 45 In OPA, m OPA = 90º m OAP = 45º m PAO = 45º OPA is a 45º - 45º - 90º triangle By 45º - 45º - 90º triangle theorem, 1 OP = OA 2 2 2 = 2 2 OA
[Given] [Vertically opposite angles] Y B•
X
A 45º
O
2 2 Y
• C
X
P
OA = 2 2 2 OA = 4 units A is at distance of 4 from the origin A (4, 0) line l has slope 1 and passes through point A (4, 0) Equation of l by slope point (y – y1) = m (x – x1) y – 0 = 1 (x – 4) y=x–4 x–y–4=0
Find the equation of the line passing through the point of intersection of 4x + 3y + 2 = 0 and 6x + 5y + 6 = 0 and the point of intersection of lines 4x – 3y – 17 = 0 and 2x + 3y + 5 = 0. (5 marks) Let point P be the point of intersection of lines 4x + 3y + 2 = 0 and 6x + 5y + 6 = 0 4x + 3y + 2 = 0 4x + 3y = –2 ......(i) Multiplying throughout by 3 we get, 12x + 9y = –6 .....(ii) 6x + 5y + 6 = 0 6x + 5y = –6 .....(iii)
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Multiplying throughout by – 2 we get, – 12x – 10y = 12 ......(iv) Adding (ii) and (iv), 12x + 9y = – 6 – 12x – 10y = 12 –y = 6 y = –6 Substituting y = – 6 in equation (i), 4x + 3 (– 6) = – 2 4x – 18 = – 2 4x = – 2 + 18 4x = 16 16 x = 4 x = 4 P (4, – 6) Let Q be the point of intersection of lines 4x – 3y – 17 = 0 and 2x + 3y + 5 = 0 4x – 3y – 17 = 0 4x – 3y = 17 .....(v) 2x + 3y + 5 = 0 2x + 3y = – 5 .....(vi) Multiplying throughout by – 2 we get, – 4x – 6y = 10 .....(vii) Adding (v) and (vii), 4x – 3y = 17 – 4x – 6y = 10 – 9y = 27 y = –3 Substituting y = – 3 in equation (v), 4x – 3 (– 3) = 17 4x + 9 = 17 4x = 17 – 9 4x = 8 x = 2 Q (2, – 3) The equation of line PQ by two point from, y – y1 x – x1 = y1 – y 2 x1 – x 2
y – (–6) x –4 = –6 – (–3) 4–2 y6 x –4 = –6 3 2 y6 x –4 = –3 2 2 (y + 6) = – 3 (x – 4) 2y + 12 = – 3x + 12 3x + 2y + 12 – 12 = 0 3x + 2y = 0 The required equation of line is 3x + 2y = 0. S C H O O L S E C TI O N
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MCQ’s 1.
What is the slope of a line having inclination 45º ? 1 (b) (a) 3 3 (c) 2 (d) 1
2.
If the slope of a line joining the points (2, k) and (–3, 0) is the value of k ? (a) 2 (c) – 5
3.
(b) (d)
2 then what is 5
–2 5
If the slope of a line joining the points (k, – 5) and (– 2, – 4) is is the value of k ? (a) – 1 (c) – 4
(b) (d)
1 then what 2
1 4
4.
For what value of x will be points (x, – 1) (2, 1) and (4, 5) lie on a line. (a) 1 (b) – 1 (c) 4 (d) 2
5.
A (1, 2), B (– 2, – 1) and C (2, – 2) are the vertices of ABC. What is the slope of line AC ? (a) 4 (b) – 4 (c)
6.
1 4
(d)
–1 4
If (– 4, 3) is a point on the line 3y = mx – 1 then what is the value of m ?
–5 2 (c) – 2 (a)
(b) (d)
5 2 2
7.
The slope of a line is 4 and y-intercept is – 3, then its equation is (a) y = 4x – 3 (b) y = 4x + 3 (c) y = – 4x – 3 (d) y = – 4x + 3
8.
The slope of a line is line ? (a) 2y = x – 10 (c) 2y = – x – 5
9.
–1 and its y-intercept is 5. What is the equation of 2 (b) (d)
2y = – x + 10 2y = x + 5
What are the values of m and n, if D (m, – 2) is the midpoint of the segment joining (– 3, n) and (2, – 5). –1 1 (a) m = 1, n = (b) m = – 1 n = 2 2 –1 –1 (c) m = ,n=1 (d) m = ,n=–1 2 2
S C H O O L S E C TI O N
279
MT
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10.
If the points (k, 2k), (3k, 3k) and (3, 1) are collinear then what is the value of k ? –1 1 (b) (a) 3 3 (c) – 3 (d) 3
11.
What is the slope of line (a) (c)
–4 3 –3 4
x y – 2 ? 3 4 (b) (d)
4 3 3 4
12.
The vertices of ABC are A (3, – 4), B (5, 7) and C (– 4, 5). Then the slope of seg BC is : 2 9 (b) (a) 9 2 –2 –9 (c) (d) 9 2
13.
The equation of line 2y – 3x + 5 = 0, in slope intercept form is ......... . –3 5 –3 5 x – x (b) y = (a) y = 2 2 2 2 3 5 3 5 (c) y = x – (d) y = x 2 2 2 2
14.
The point (4, – 3) lies on the 5x + 8y = c, then the value of c is .......... . (a) 4 (b) – 4 (c)
1 4
(d)
–1 4
15.
The slope of a line is 3 and y intercept is – 4, then the equation of line is .............. . (a) y = 3x + 4 (b) y = 3x – 4 (c) y = – 3x + 4 (d) y = – 3x – 4
16.
Slope of the line 3 (x + 3) = y – 1, .......... . 1 (a) 3 (b) 3 –1 (c) – 3 (d) 3
17.
The equation of a line passing through origin with slope m = (a) 3x – 2y = 0 (c) 2x + 3y = 0
18.
280
(b) (d)
3x + 2y = 0 2x – 3y = 0
3 , is ........ . 2
y – 2 = 4 (x + 3) passes through the point ............ . (a) (– 3, 2) (b) (2, – 3) (c) (3, – 2) (d) (– 2,3) S C H O O L S E C TI O N
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19.
The slope of the line y – 3x – 6 = 0 is ............... . 1 –1 (b) (a) 3 3 (c) 3 (d) – 3
20.
The equation of line 4x – y – 7 = 0 in the double intercept form is ......... . x y x y 1 1 7 –7 (a) (b) 7 7 4 4 x y x y 1 1 (c) – 7 – 7 (d) – 7 7 4 4
21.
Equation of X-axis is ........... . (a) x = 0 (c) x = k
(b) (d)
y=0 y=k
22.
The equation of a line parallel to y = – 3x + 2 and passing through point (3, 5) is ............. . (a) y – 3 = – 3 (x – 5) (b) y – 5 = – 3 (x – 3) (c) y – 5 = 2 (x – 3) (d) y – 3 = 2 (x – 5)
23.
Equation of a line parallel to X-axis and passing through (– 3, 4) is ............ . (a) x = – 3 (b) x = 3 (c) y = – 4 (d) y = 4
24.
The equation of a line passing through the point (2, 7) and whose y-intercept is 3 is ............... . (a) 2x – y + 3 = 0 (b) 2x – y – 3 = 0 (c) 2x + y + 3 = 0 (d) 2x + y – 3 = 0
25.
The line y = – 3x lies in the ................. quadrant. (a) I and II (b) II and IV (c) III and IV (d) IV and III
: ANSWERS : 1. 3.
(d) 1 (c) – 4
2. 4.
5.
(b) – 4
6.
7.
(a) y = 4x – 3 –1 ,n=1 (c) m = 2 4 (b) 3
8.
9. 11.
15.
3 5 x – 2 2 (b) y = 3x – 4
17.
(a) 3x – 2y = 0
13.
(c) y =
S C H O O L S E C TI O N
10. 12.
(a) 2 (d) 2 –5 (a) 2 (b) 2y = – x + 10 –1 (a) 3 2 (a) 9
14.
(b) – 4
16.
(a) 3
18.
(a) (– 3, 2) 281
MT
GEOMETRY
19.
(c) 3
20.
21.
(b) y = 0
22.
x y 1 –7 (a) 7 4 (b) y – 5 = – 3 (x – 3)
23.
(d) y = 4
24.
(a) 2x – y + 3 = 0
25.
(b) II and IV
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1 Mark Sums 1. Sol.
Find the slope of a line whose Inclination of the line () = Slope of the line = = = Slope of the line is 1.
2. Sol.
Write the slope of line 2y = 3x – 5. 2y = 3x – 5 3 5 y= x – [Dividing throughout by 2] 2 2 Comparing the above equation with slope intercept form y = mx + c we get m =
3. Sol.
6. Sol. 282
7 2
y-intercept of line is
7 2
If m = 5 and c = – 3, then write the equation of the line. m = 5, c = – 3 By slope point form, the equation of line is y = mx + c y = 5x – 3
Sol.
3 2
Write the y-intercept of the line 3y = 2x + 7 3y = 2x + 7 2 7 y= x [Dividing throughout by 3] 3 2 Comparing the above equation with slope intercept form y = mx + c we
5.
3 2
Slope of line is
get c =
4. Sol.
inclination is 45º. 45º tan tan 45º 1
5x – y – 3 = 0
Write the equation of X-axis. Equation of X-axis is y = 0 Write the equation of a line parallel to Y-axis and passing through point(3, 0). Equation of a line parallel to Y-axis and passing through point (3, 0) is x = 3. S C H O O L S E C TI O N
MT 7. Sol.
GEOMETRY
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Find the slope of a line having Inclination of the line () = Slope of the line = = =
Slope of the line is
inclination 60º. 60º tan tan 60
3
3.
8. Sol.
State the slope of X-axis. Slope of X-axis = tan 0 = 0 Slope of X-axis = 0
9. Sol.
If (– 3, – 2) lies on the line 2y = mx + 5, find m. (– 3, – 2) lies on the line 2y = mx + 5 Its co-ordinates will satisfy the equation 2 (– 2) = m (– 3) + 5 – 4 = – 3m + 5 3m = 5 + 4 3m = 9 9 m = 3
10. Sol. 11. Sol.
Write the equation of a line parallel to Y-axis and passing through the point (3, 4). Equation of a line parallel to Y-axis and passing through the poitn (3, 4) is x = 3. If the slope of a line is 2 and its y-intercept is 5, write its equation. Slope of the line (m) = 2 Its y-intercepts (c) = 5 Equation of the line by slope-intercept form, y = mx + c
12. Sol.
13. Sol.
Sol.
y = 2x + 5
The equation of a line passing through the origin is y = 3x. Find the x-coordinate of a point on the line, if its y-coordinate is 3. Equation of the line passing through origin y = 3x y-co-ordinate of a point lying on this line is 3 [Given] 3 = 3x 3 x= 3 x=1 A line has the equation y = 3x – 2. State its y-intercept. Equation of the line is y = 3x – 2 Comparing the given equation with slope-intercept form y = mx + c, c=–2
14.
m = 3
y intercept of the line is – 2.
Write the equation of a line parallel to X-axis and passing through the point (– 2, 5). Equation of a line parallel to X-axis and passing through the point (– 2, 5) is y = 5.
S C H O O L S E C TI O N
283
GEOMETRY
15.
Sol.
16. Sol.
What are the co-ordinates of the midpoint of the line segment joining A (2, 5) and B (4, 1) ? A (2, 5) and B (4, 1) Co-ordinates of the midpoint of line segment AB by midpoint formula is,
=
2 4 5 1 , 2 2 6 6 , 2 2 (3, 3)
What is the x-intercept of line 3x – 4y = 12 ? Substituting y = 0 in the equation 3x + 4y = 12 3x – 4(0) = 12 3x – 0 = 12 3x = 12 12 x = 3 x = 4
18. Sol.
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Write the x-intercept and the y-intercept of the line represented by the x y + = 1. equation 2 3 x y 1 with the double intercept form, Comparing the equation 2 3 x y 1 we get a = 2 and b = 3 a b x intercept = 2 and y intercept = 3
=
17. Sol.
MT
x intercept is 4.
What is the equation of a line whose slope is – 2 and y-intercept is 3 ? Slope (m) = – 2 y intercept (c) = 3 Equation of the line by slope-intercept form is y = mx + c y = – 2x + 3
2x + y – 3 = 0
19. Sol.
What is the y-intercept of line 2x – 3y = 4 ? 2x – 3y = 4 3y = 2x – 4 Dividing throughout by 3, 2 4 y= x– 3 3 Comparing the above equation with slope intercept form y = mx + c, we get –4 y intercept = 3
20.
The slope of line PQ is – 1 and the slope of line QR is – 1. What can you say about the points P, Q and R ? Slope of line PQ = – 1 Slope of line QR = – 1
Sol. 284
S C H O O L S E C TI O N
MT
21. Sol.
GEOMETRY
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Slopes of line PQ and line QR are equal and point Q is a common point for both the lines Points P, Q and R are collinear
What is the equation of a line parallel to X-axis and passing through the point (5, – 7) ? Equation of a line parallel to X-axis and psssing through the point (5, –7) is y = – 7
22. Sol.
What is the slope of line whose inclination is 0º ? Inclination of line () = 0º Its slope = tan Its slope = tan 0 Its slope = 0
23. Sol.
What is the x-intercept of the line 3x – 4y = 7 ? Substituting y = 0 in the equation, 3x – 4y = 7 3x – 4 (0) = 7 3x – 0 = 7 3x = 7 7 x = 3 7 x intercept is 3
24.
The slope of line AB is
Sol.
25.
2 . What is the slope of line DE which is parallel 3
to line AB ? line DE || line AB [Given] Slope of line DE = slope of line AB 2 [Given] But, slope of line AB = 3 2 Slope of line DE = 3
What is the equation of a line whose slope is – 3 and the y-intercept
Sol.
Slope of the line (m) = – 3 5 It’s y intercept (c) = 2 Equation of the line by slope-intercept form, y = mx + c 5 y = – 3x + 2 2y = – 6x + 5 [Dividing throughout by 2]
6x + 2y – 5 = 0
5 ? 2
S C H O O L S E C TI O N
285
6. `
Mensuration
Introduction : Mensuration is a special branch of mathematics that deals with the measurement of geometric figures. In previous classes we have studied certain concepts related to areas of plane figures (shapes) such as triangles, quadrilaterals, polygons and circles. Now we will study how to find some measurements related to circle and the surface area and the volume of solid figures.
`
Circle : Arc, Sector, Segment : Area of sector : Sector of a circle is the part Y of the circle enclosed by two radii of the circle and their intercepted arc. (i.e. arc between the two ends of radii) × r 2 Area of the sector (A) = 360
O
Major arc
r
B
Central angle Minor arc
A
X
Length of an arc : Length of an arc of a circle (arc length) is the distance along the curved line making up the arc. × 2r Length of the arc (l) = 360
Relation between the area of the sector and the length of an arc : Area of the sector =
r × length of arc 2
EXERCISE - 6.1 (TEXT BOOK PAGE NO. 157) 1. (i) Sol.
The diameter of a circle is 10 cm. Find the length of the arc, when the corresponding central angle is as given below : ( = 3.14) (2 marks) 144º Diameter of a circle = 10 cm
10 2 = 5 cm
Its radius (r) =
l =
× 2r 360
l =
144 × 2 × 3.14 × 5 360
l =
144 × 3.14 × 10 360 l = 12.56 cm
The length of the arc is 12.56 cm. 286
S C H O O L S E C TI O N
MT (ii) Sol.
GEOMETRY
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(2 marks)
45º Diameter of a circle = 10 cm 10 Its radius (r) = 2 = 5 cm l =
× 2r 360
l =
45 × 2 × 3.14 × 5 360
l =
45 × 3.14 × 10 360
l =
5 314 × 4 100
l =
785 2 100
l =
392.5 100 l = 3.925 l = 3.93 cm
The length of the arc is 3.93 cm. (iii) Sol.
270º Diameter of a circle = 10 cm
10 2 = 5 cm
It radius (r) =
l =
× 2r 360
l =
270 × 2 × 3.14 × 5 360
l =
(2 marks)
3 × 3.14 × 10 4 l = 23.55 cm
The length of the arc is 23.55 cm. (iv) Sol.
180º Diameter of a circle = 10 cm
Its radius (r) = = l = = =
(2 marks)
10 2 5 cm × 2r 360 180 × 2 × 3.14 × 5 360 15.70 cm
The length of the arc is 15.70 cm. S C H O O L S E C TI O N
287
MT
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EXERCISE - 6.1 (TEXT BOOK PAGE NO. 157) 2. (i) Sol.
Find the angle subtended at the centre of a circle by an arc, given the following : 22 ) (2 marks) radius of circle = 5.5 m, length of arc = 6.05 m ( = 7 l = × 2r 360 22 6.05 = ×2× × 5.5 7 360 605 22 = × × 11 100 7 360 605 × 360 × 7 100 × 22 × 11 = = 63º Measure of the arc is 63º.
(ii) Sol.
(2 marks)
radius of circle = 20 m length of arc = 78.50 m ( = 3.14) l = × 2r 360 78.50 = × 2 × 3.14 × 20 360 785 314 = × × 40 10 360 100 785 × 9 × 100 = 10 × 314 = 225º Measure of an arc is 225º. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
3. (i) Sol. (ii) Sol.
The radius of the circle is 7 cm and m (arc RYS) = 60º, with the help of the figure, answer the following questions : (3 marks) Name the shaded portion. S P-RYS Find the area of the circle. Area of circle = r 2 22 = ×7×7 7 = 154 cm2
X
P
60º
Y R
Area of a circle is 154 cm2. (iii) Sol.
288
Find A (P-RYS)
× r2 360 60 22 = × ×7×7 360 7 77 = 3 = 25.67 Area of the sector P -RYS is 25.67 cm2
Area of the sector =
S C H O O L S E C TI O N
MT (iv) Sol.
GEOMETRY
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Find A (P-RXS) Area of sector P-RXY = Area of circle – Area of sector P-RYS = 154 – 25.67 = 128.33 cm2 Area of sector P-RXY is 128.33 cm2. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
4. (i) Sol.
The radius of a circle is 7 cm. Find area of the sector of this circle if the angle of the sector is : 30º (2 marks) Area of the sector = × r2 360 30 22 = × ×7×7 360 7 77 = 6 = 12.83 Area of the sector is 12.83 cm2.
(ii) Sol.
(2 marks)
210º Area of the sector = = = =
× r2 360 210 22 × ×7×7 360 7 539 6 89.83
Area of the sector is 89.83 cm2. (iii) Sol.
(2 marks)
3 rt. angles Area of the sector = = = =
× r2 360 270 22 × ×7×7 360 7 231 2 115.50
Area of the sector is 115.50 cm2. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 5. Sol.
An arc of a circle having measure 36 has length 176 m. Find the (2 marks) circumference of the circle. Length of arc (l) = 176 m measure of arc () = 36º l = × 2r 360 36 176 = × 2r 360
S C H O O L S E C TI O N
289
MT
GEOMETRY
176
176 × 10 2r But, circumference
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1 × 2r 10 = 2r = 1760 = 2r =
Circumference of the circle is 1760 m. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 6. Sol.
An arc of length 4 cm subtends an angle of measure 40º at the centre. Find the radius and the area of the sector formed by this arc.(2 marks) Length of arc (l) = 4 cm measure of arc () = 40º l = × 2r 360 40 4 = ×2××r 360 4×9 = r 2 r = 18 cm. l ×r Area of the sector = 2 4 × 18 = 2 = 36cm 2 Radius of the circle is 18 cm and Area of the sector is 36 cm2. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158)
7. Sol.
If the area of the minor sector is 392.5 sq. cm and the corresponding central angle is 72º, find the radius. ( = 3.14) (2 marks) Measure of arc () = 72º Area of the sector = 392.5 cm2 Area of the sector 392.5
3925 10 3925 × 360 × 100 10 × 72 × 314 r2 r
× r2 360 72 = × 3.14 × r2 360 72 314 = × × r2 360 100 =
= r2 = 625 = 25
[Taking square roots]
Radius of the circle is 25 cm. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 8. Sol. 290
Find the area of sector whose arc length and radius are 10 cm and 5 cm respectively. (2 marks) Length of arc (l) = 10 cm Radius of a circle (r) = 5 cm S C H O O L S E C TI O N
MT
GEOMETRY
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Area of the sector
=
r ×l 2
5 × 10 2 = 25 cm2 =
Area of the sector is 25 cm2. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 9. Sol.
If the area of minor sector of a circle with radius 11.2 cm is 49.28cm2, find the measure of the arc. (2 marks) Radius of a circle = 11.2 cm Area of the sector = 49.28 cm2 Area of the sector
49.28
4928 100 4928 × 360 × 7 × 10 × 10 100 × 22 × 112 × 112
× r2 360 22 = × × 11.2 × 11.2 7 360 22 112 112 = × × × 7 10 10 360 =
= = 45º
Measure of arc is 45º. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 10. Sol.
Find the length of the arc of a circle with radius 0.7 m and area of the sector is 0.49 m2. (2 marks) Radius of a circle = 0.7 cm Area of the sector = 0.49 m2 r Area of the sector = ×l 2 0.7 0.49 = ×l 2
49 100 49 20 100 7 l
=
7 ×l 20
= l = 1.4
The length of the arc is 1.4 m. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 11. Sol.
Two arcs of the same circle have their lengths in the ratio 4:5. Find the ratio of the areas of the corresponding sectors. (2 marks) Ratio of lengths of two arcs is 4 : 5. Let the common multiple be ‘x’ Lengths of two arcs are (4x) units and (5x) units respectively Let the lengths of two arcs be ‘l1’, and ‘l2’ and Areas of their corresponding sectors be A1 and A2.
S C H O O L S E C TI O N
291
MT
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l1 = (4x) units l2 = (5x) units Both Arcs are of the same circle. Their radii are equal Now, r × l1 .......(i) A1 = 2 r × l2 ......(ii) A2 = 2 Dividing (i) and (ii) we get, A1 l1 = A2 l2 A1 4x A2 = 5x A1 : A 2 = 4 : 5 Ratio of the areas of sectors is 4 : 5. EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 12.
Sol.
Adjoining figure depicts a racing track 7m whose left and right ends are semicircular. 105 m The distance between two inner parallel 70 m line segments is 70 m and they are each 70 m 105 m long. If the track is 7 m wide, find 105 m the difference in the lengths of the inner 7m edge and outer edge of the track. (4 marks) Diameter of inner circular edge (d 1) = 70 m Width of the track = 7 m Diameter of outer circular edge (d2) = 70 + 7 + 7 = 84 m The inner and outer edges of the racing tracks comprises of two semicircles and parallel segments of length 105 m each
1 1 d2 + 105 + d2 + 105 2 2 = d2 + 210 = (84 + 210) m
Length of outer edge =
1 1 d1 + 105 + d1 + 105 2 2 = d1 + 210 = (70 + 210) m Difference in the lengths of = (84 + 210) – (70 + 210) inner and outer edge = 84 + 210 – 70 – 210 = 14 Length of inner edge =
= 14 ×
22 7
= 44 m The difference in the lengths of inner edge and outer edge of the track is 44 m. 292
S C H O O L S E C TI O N
MT
GEOMETRY
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EXERCISE - 6.1 (TEXT BOOK PAGE NO. 158) 13.
(i) Sol.
10 m In the adjoining figure, A horse is tied to a pole fixed 10 m at one corner of a 30 m × 30 m square field of grass by means of a 10 m long rope. ( = 3.14). (3 marks) Find the area of that part of the field in which the horse can graze. Side of a square = 30 m. 30 m Length of the rope = radius of the sector Radius of the sector (r) = 10 m Measure of arc () = 90º [Angle of a square] Area of field that can be grazed = Area of sector
=
× r2 360
=
90 × 3.14 × 10 × 10 360
=
1 × 314 4
Area of field that can be grazed = 78.5 m2 (ii)
Sol.
10 m 10 m In the adjoining figure, What will be the area of the part of the field in which the horse can graze, if the pole was fixed on a side exactly at the middle of the side? If the pole is fixed at the middle of the middle of the side of a square, then 30 m Area of field that can be grazed = 2 × Area of sector = 2 × 78.5
Area of field that can be grazed = 157 m2 PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202) 2. Sol.
The area of a circle is 314 sq.cm and area of its sector is 31.4 sq.cm. Find the area of its major sector. (2 marks) 2 Area of a circle = 314 cm Area of major sector = Area of a circle – Area of its minor sector = 314 – 31.4 = 282.6 Area of major sector is 282.6 cm2 PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
3. Sol.
1 C r, for a circle having radius, circumference and area r, C, 2 A respectively. (2 marks) L.H.S. = A L.H.S. = r 2 ........(i) Prove A =
S C H O O L S E C TI O N
293
MT
GEOMETRY
R.H.S. =
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1 Cr 2
1 × 2r × r 2 R.H.S. = r 2 L.H.S. = R.H.S 1 A = Cr 2 R.H.S =
........(ii) [From (i) and (ii)]
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202) 4. Sol.
The radius of a circle is 3.5 cm and area of the sector is 3.85 cm2. Find the length of the corresponding arc and the measure of arc. (3 marks) Radius of a circle (r) = 3.5 cm Area of the sector = 3.85 cm2 Area of sector =
3.85 =
r ×l 2 3.5 ×l 2
3.85 × 2 = l 3.5 3.85 × 2 × 10 = l 100 × 35 22 10 l = 2.2 cm l =
Area of sector =
r 2 360
3.85 =
22 3.5 3.5 360 7
3.85 =
22 35 35 360 7 10 10
385 35 11 = 100 360 10 385 × 360 × 10 100 × 11 × 35 = = 36º
Length of arc is 2.2 cm and measure of an arc is 36º. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202) 6.
Sol.
294
Find the perimeter of each of these sectors. (Give your answers in terms of ) (3 marks) Radius of the sector (r) = 8 cm Measure of arc () = 40º
40º 8 cm
10 cm
120º
S C H O O L S E C TI O N
MT
GEOMETRY
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2r 360 40 = ×2××8 360 16 = cm 9 Perimeter of the sector = r + r + l Length of arc (l) =
= 8+8+
16 9
16 9 = 16 1 + 9 = 16
Perimeter of the sector = (b)
16 (9 + ) cm 9
Radius of a sector (r) Measure of arc ()
= 10 cm = 126º 2r Length of arc (l) = 360 126 = × 2 × × 10 360 = 7 cm Perimeter of the sector = r + r + l = 10 + 10 + 7 Perimeter of the sector = (20 + 7 ) cm. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202)
7.
Find the area of the shaded part. (Give your answers in terms of ) 8 cm (b) 50º (a) m 3c
9c m
70º
30º
(3 marks)
4 cm
Sol.
(a) Area of shaded part
=
Area of sector I + Area of sector II
=
30 90 360 3 3 360 4 4
Area of shaded part
=
3 4 sq.units 4
(b) Area of shaded part
=
Area of sector I + Area of sector II
=
70 50 360 9 9 360 8 8
=
63 80 sq. units 4 9
Area of shaded part S C H O O L S E C TI O N
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203) In the adjoining figure, seg QR is a tangent to the circle with centre O. Point Q is the point of contact. O Radius of the circle is 10 cm. T OR = 20 cm. Find the area of the shaded region. ( = 3.14, 3 = 1.73 ) R (4 marks) Q In OQR, [Radius is perpendicular to the tangent] m OQR = 90º 2 2 OQ + QR = OR 2 [By Pythagoras theorem] 2 2 2 10 + QR = 20 QR 2 = 400 – 100 QR 2 = 300 QR = 300 10 cm
10.
Sol.
QR
=
QR QR QR
= 10 3 = 10 (1.73) = 17.3 cm
Area of OQR
=
1 × Product of Perpendicular sides 2
=
1 × OQ × QR 2
= In OQR,
m OQR OQ OR
100 × 3
= = = =
1 10 17.3 2 86.5 cm2 90º 10 cm 20 cm
1 OR 2 By converse of 30º - 60º - 90º triangle theorem. m ORQ = 30º m QOR = 60º [Remaining angle] Now, For sector O-QXT Measure of arc () = 60º Radius (r) = 10 cm × r2 Area of Sector O-QXT = 360 60 = × 3.14 × 10 × 10 360 157 = 3 Area of sector O-QXT = 52.33 cm2 Area of shaded region = Area of OQR – Area of sector O-QXT = 86.5 – 52.33 = 34.17 cm2
OQ
=
Area of the shaded region is 34.17 cm2 296
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203) 11.
Sol.
In the adjoining figure, PR = 6 units and PQ = 8 units. P Semicircles are draw taking sides PR, RQ and PQ as diameters as shown in the figure. Find out the area of the shaded portion. ( = 3.14) (5 marks) Diameter PR = 6 units Q R Its radius (r1) = 3 units Diameter PQ = 8 units Its radius (r2) = 4 units In PQR, m RPQ = 90º ......(i) [Angle subtended by a semicircle] 2 2 2 [By Pythagoras theorem] QR = PR + PQ 2 QR = 62 + 82 QR 2 = 36 + 64 QR = 100 QR = 10 units [Taking square roots] Diameter QR = 10 units Its radius (r3) = 5 units PQR is a right angled triangle [From (i)]
1 × product of perpendicular sides 2 1 = × PR × PQ 2 1 ×6×8 = 2 = 24 sq. units. Area of shaded portion = Area of semicircle with diameter PR + Area of semicircle with diameter PQ + Area of PQR – Area of semicircle with diameter QR A (PQR) =
=
1 1 1 r12 + r22 + 24 – r 2 2 2 2 3
1 1 2 1 2 2 = r1 r2 – r3 24 2 2 2 =
1 (r12 + r22 – r32) + 24 2
=
1 × 3.14 (32 + 42 – 52) + 24 2
=
1 × 3.14 × (9 + 16 – 25) + 24 2
1 × 3.14 (0) + 24 2 = 0 + 24 = 24 sq. units =
Area of shaded portion = 24 sq.units S C H O O L S E C TI O N
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A
Area of segment of a circle : A segment of a circle is the region bounded by a chord and an arc.
O
sin 2 – Area of segment = r 2 360
P
Q R
EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162) 1.
Sol.
In the adjoining figure, A (O-AXB) = 75.36 cm2 and radius = 12 cm, find the area O of the segment AXB. ( = 3.14). (2 marks) Radius of a circle (r) = 12 cm A (O-AXB) = 75.36 cm2 1 2 Area of OAB = r sin 2 1 = × 12 × 12 × sin 60º 2 3 = 72 × 2 = 36 3 = 36 × 1.73 = 62.28 cm2 Area of the segment AXB = A (O-AXB) – A (OAB) = 75.36 – 62.28 = 13.08 cm2
A 60º
X B
Area of the segment AXB is 13.08 cm2. EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162) 2. Sol.
A Calculate the area of the shaded region in the adjoining figure where X ABCD is a square with side 8 cm each. Mark point X as shown in the figure ABCD is a square [Given] side = 8 cm D 8 cm Radius (r) = side of a square r = 8 cm Measure of arc () = 90º [Angle of a square]
B
(3 marks)
C
sin Area of the segment AXC = r2 360 2 3.14 90 sin 90 = 82 2 360 1.57 1 = 64 2 2
1.57 1 = 64 2 298
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= = Area of shaded region = = =
64 0.57 2 36.48 cm2 2 2 × Area of segment AXC 36.48 2× 2 36.48 cm2
Area of shaded region is 36.48 cm2. EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162) P In the adjoining figure, P is the centre of the circle with radius 18 cm. If the area of the PQR is 100 cm2 and area of the segment QXR is 13.04 cm2. Find the central angle . ( = 3.14) (3 marks) Q R Radius of a circle (r) = 18 cm Area of PQR = 100 cm2 X Area of the segment QXR = 13.04 cm2 Area of sector P-QXR = Area of PQR + Area of segment QXR = 100 + 13.04 Area of sector P-QXR = 113.04 cm2 Area of sector = × r2 360 113.04 = × 3.14 × 18 × 18 360 11304 = × 314 × 18 × 18 360 11304 × 360 314 × 18 × 18 = = 40 18
cm
3.
Sol.
Central angle is 40º. EXERCISE - 6.2 (TEXT BOOK PAGE NO. 162) 4.
Sol.
In the adjoining figure, the centre of the circle is A and ABCDEF is a regular hexagon of side 6 cm. Find the following : ( 3 = 1.73, = 3.14) B (i) Area of segment BPF (5 marks) (ii) Area of the shaded portion C Side of hexagon = radius of a circle Radius (r) = 6 cm Measure of arc () = Angle of a regular hexagon = 120º Area of sector A-BPF = × r2 360 120 = × 3.14 × 6 × 6 360 = 3.14 × 12 = 37.68 cm2
S C H O O L S E C TI O N
A M P P
F E
D
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In ABF, seg AB seg AF ABF AFB ......(i) mBAF + mABF + mAFB = 180º
120 + mABF + mABF = 2m ABF = 2m ABF =
m ABF =
m ABF = m ABM =
[Radii of the same circle] [Isosceles triangle theorem] [Sum of measures of angles of a triangle is 180º] 180 [Given, from (i)] 180 – 120 60 60 2 30º 30º .......(ii) [B - M - F]
In AMB, m AMB = 90º m ABM = 30º m BAM = 60º AMB is 30º - 60º - 90º triangle, By 30º - 60º - 90º triangle theorem, AM =
1 AB 2
[By construction] [From (ii)] [Remaining angle]
[Side opposite to 30º]
1 ×6 2 AM = 3 cm AM =
BM = BM =
3 × AB 2
[Side opposite to 60º]
3 ×6 2
BM = 3 3 cm seg AM chord BF 1 BM = BF 2 3 3 =
[The perpendicular drawn from the centre of circle to a chord, bi sec ts the chord]
1 BF 2
BF = 6 3 BF = 6 (1.73) BF = 10.38 cm. Area of ABF = =
1 × base × height 2
1 × BF × AM 2
1 × 10.38 × 3 2 = 15.57 cm2 Area of segment BPF = Area of sector A-BPF – Area of ABF = 37.68 – 15.57 =
Area of segment BPF = 22.11 cm2 300
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(ii)
side
= 6 cm
Area of regular hexagon ABCDEF = =
3 3 × (side)2 2 3 3 ×6×6 2
= 54
3 = 54 × 1.73 = 93.42 cm2
Area of the shaded portion
= Area of regular hexagon ABCDEF – Area of ABF = 93.42 – 15.57 = 77.85 cm2
Area of segment BPF is 22.11 cm2 and Area of shaded portion is 77.85 cm2. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202) 8.
Find the area of the shaded region. ( = 3.14,
3 = 1.73)
(3 marks)
12
cm
60º
Sol.
Radius of the sector (r) Measure of arc () Area of segment
= 12 cm = 60º
sin – = r2 2 360 sin 60 2 3.14 × 60 – = 12 2 360 3.14 3 1 – = 144 2 2 6 3.14 3 – = 144 4 6
6.28 – 3(1.73) = 144 12
6.28 – 5.19 = 144 12
144 × 1.09 12 = 12 × 1.09 = 13.08 cm2 Area of shaded region is 13.08 cm2. =
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203) 9.
Sol.
In the adjoining figure, m POQ = 30º and radius OP = 12 cm. Find the following (Given = 3.14) O (i) Area of sector O-PRQ (ii) Area of OPQ M (3 marks) (iii) Area of segment PRQ 12 cm 30º Radius of the circle (r) = 12 cm Q • Measure of arc () = 30º R P Area of sector O - PRQ = × r2 360 30 = × 3.14 × 12 × 12 360 = 37.68 cm2 In OMP, m OMP = 90º [Given] m POM = 30º [Given and O - M - Q] m OPM = 60º [Remaining angle] OMP is 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem 1 PM = OP 2 1 × 12 = 2 PM = 6 cm. OP = OQ = 12 cm [Radii of same circle] 1 × base × height Area of OPQ = 2 1 × OQ × PM = 2 1 × 12 × 6 = 2 = 36 cm2 Area of segment PRQ = Area of sector O-PRQ – Area of OPQ = 37.68 – 36 = 1.68 cm2 (i) Area of sector O-PRQ is 37.68 cm2 (ii) Area of OPQ is 36 cm2 (iii)Area of segment PRQ is 1.68 cm2 PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
12.
Sol.
302
In the adjoining figure, P PR and QS are two diameters of the circle. If PR = 28 cm and PS = 14 3 cm, find 120º (i) Area of triangle OPS O (ii) The total area of two shaded segments. Q ( 3 = 1.73) (4 marks) Draw seg OM side PS 1 PR OP = [Radius is half of diameter] 2 1 28 OP = 2 OP = 14 cm
S
R
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seg OM chord PS 1 PS PM = 2
1 14 3 2 = 7 3 cm
PM
PM In OMP, OMP = 90º OM2 + PM2 = OP 2
=
OM2 + 7 3 OM2 OM2 OM
[By construction] [The perpendicular drawn from the centre of a circle to a chord bi sec ts the chord]
2
= = = =
Area of OPS = Area of OPS = = = =
[By construction] [By Pythagoras theorem]
142 196 – 147 49 7 cm [Taking square roots] 1 × base × height 2 1 × PS × OM 2 1 × 14 3 × 7 2 49 3 49 (1.73)
Area of OPS = 84.77 cm2 Area of sector OPS
= = =
× r2 360 120 22 14 14 360 7 616 3 205.33 cm2 Area of sector OPS – Area of OPS 205.33 – 84.77 120.56 cm2
= Area of segment PS = = = Similarly we can prove, Area of segment QR = 120.56 cm2 Total area of two shaded segments = 120.56 + 120.56 = 241.12 cm2 Area of OPS is 84.77 cm2 and total area of two shaded segments is 241.12 cm2. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 203)
S C H O O L S E C TI O N
cm
60º
cm
(Given = 3.14, 3 = 1.73) (5 marks) For a segment PMQ, radius (r) = 10 cm measure of arc () = 60º
O 10
Sol.
In the adjoining figure, seg PQ is a diameter of semicircle PNQ. The centre of arc PMQ is O. OP = OQ = 10 cm and m POQ = 60º. Find the area of the shaded portion
10
13.
P
Q M • Q
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sin 2 – Area of segment PMQ = r 2 360 sin 60 2 3.14 × 60 – = 10 2 360 3.14 3 1 – = 100 2 2 6 3.14 3 = 100 6 – 4
6.28 – 3(1.73) = 100 12
6.28 – 5.19 = 100 12 100 1.09 = 12 109 = 12 Area of segment PMQ = 9.08 cm2 In OPQ, seg OP seg OQ [Radii of same circle] OPQ OQP [Isosceles triangle theorem] Let, m OPQ = m OQP = x m OPQ + m OQP + m POQ = 180º [Sum of the measures of angles of a triangle is 180º] x + x + 60 = 180 2x = 180 – 60 2x = 120
120 2 x = 60 m POQ = m OPQ = m OQP = 60º OPQ is an equilateral triangle
x =
[An equiangular triangle is an equilateral triangle] OP = OQ = PQ = 10 cm [Sides of an equilateral triangle] Diameter PQ = 10 cm
10 2 = 5 cm
Radius (r) =
Area of semicircle = = = Area of the shaded portion = = =
1 2 r 2 1 3.14 5 5 2 39.25 cm2 Area of semicircle – Area of segment PMQ 39.25 – 9.08 30.17 cm2
The area of shaded portion is 30.17 cm2. 304
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CUBOID [RECTANGULAR PARALLELOPIPED] A cuboid is a solid figure bounded by six rectangular faces, where the opposite faces are equal. A cuboid has a length, breadth and height denoted as ‘l’, ‘b’ and ‘h’ respectively as shown in the figure,
b
l
h
In our day to day life we come across cuboids such as rectangular room, rectangular box, brick, rectangular fish tank, etc. The following are the formulae for the surface area of cuboid :
FORMULAE 1. 2. 3.
Total surface area of a cuboid = 2 (lb + bh + lh) Vertical surface area of a cuboid = 2 (l + b) × h Volume of a cuboid = l × b × h EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
1. Sol.
The dimensions of a cuboid in cm are 16 × 14 × 20. Find its total surface area. (2 marks) Length of a cuboid (l) = 16 cm its breadth (b) = 14 cm its height (h) = 20 cm Total surface area of a cuboid = 2 (lb + bh + lh) = 2 (16 × 14 + 14 × 20 + 16 × 20) = 2 (224 + 280 + 320) = 2 × 824 = 1648 cm2 Total surface area of a cuboid is 1648 cm2. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
4. Sol.
The cuboid water tank has length 2 m, breadth 1.6 m and height 1.8m. Find the capacity of the tank in litres. (2 marks) Length of the cuboidal water tank (l) = 2 m its breadth (b) = 1.6 m and its height (h) = 1.8 m. Volume of cuboidal water tank = l × b × h = 2 × 1.6 × 1.8 = 5.76 m3 = 5.76 × 1000 litres [l m3 = 1000 litres] Volume of cuboidal water tank is 5760 litres. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
6. Sol.
A fish tank is in the form of a cuboid whose external measures are 80cm × 40cm × 30cm. The base, side faces and back faces are to be covered (2 marks) with a coloured paper. Find the area of the paper needed. Length of cuboidal fish tank (l) = 80 cm its breadth (b) = 40 cm its height (h) = 30 cm
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Cuboid is make up of 6 rectangular faces Area of the base of the fish tank = l × b = 80 × 40 = 3200 cm2 Area of two side faces = 2 × b × h = 2 × 40 × 30 = 2400 cm2 Area of back face = l × h = 80 × 30 = 2400 cm2 Area of the paper needed = 3200 + 2400 + 2400 = 8000 cm2 The area of the paper needed is 8000 cm2. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169) 7.
Sol.
Find the total cost of white washing the 4 walls of a cuboidal room at the rate Rs. 15 per m2. The internal measures of the cuboidal room are (3 marks) length 10 m, breadth 4 m and height 4 m. Length of the cuboidal room (l) = 10m Its breadth (b) = 14m Its height (h) = 4m Vertical Surface area of the room = 2 (l + b) × h = 2 (10 + 4) × 4 = 2 × 14 × 4 = 112m 2 Area of white washing = 112m 2 Rate of white washing = Rs 15 per m2 Total cost = Area of white washing × rate of white washing = 112 × 15 = 1680 Total cost of white washing is Rs. 1680. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
9. Sol.
A beam 4m long, 50cm wide and 20cm deep is made of wood which (3 marks) weighs 25kg per m3. Find the weight of the beam. Length of the beam (l) = 4m its breadth (b) = 50cm =
50 m 100
5 m 10 = 20 cm =
its height (h)
306
=
20 m 100
=
2 m 10 S C H O O L S E C TI O N
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Volume of the beam
= l×b×h 5 2 = 4× × 10 10 40 3 = m 100 Weight of the beam = 25kg per m3 40 Total weight of the beam = 25 × 100 = 10 kg The weight of the beam is 10 kg. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 202) 5. Sol.
The length, breadth and height of a cuboid are in the ratio 5:4:2. If the total surface area is 1216 cm2, find the dimensions of the solid. (3 marks) Ratio of the length, breadth and height of a cuboid is 5 : 4 : 2 Let the common multiple be ‘x’ Length of a cuboid = 5x cm its breadth = 4x cm and its height = 2x cm Total surface area of a cuboid = 1216 cm2 Total surface area of a cuboid = 2 (lb + bh + lh) 1216 = 2 [(5x) (4x) + (4x) (2x) + (5x) (2x)] 1216 = 20x2 + 8x2 + 10x2 2 608 = 38x2 608 = x2 38 x2 = 16 x = 4 [Taking square roots] Length of a cuboid = 5x = 5 (4) = 20 cm its Breadth = 4x = 4 (4) = 16cm and its height = 2x = 2 (4) = 8 cm Dimensions of a cuboid are 20 cm, 16 cm and 8 cm.
CUBE A cube is a cuboid bounded by six equal square faces. Hence its length, breadth and height are equal. The edge of the cube = length = breadth = height The edge of the cube is denoted as ‘l’ l A dice is an example of cube. The following are the formulae for the surface area of the cube :
FORMULAE 1. Total surface area of a cube = 6l2 2. Vertical surface area of a cube = 4l2 3. Volume of cube = l3 S C H O O L S E C TI O N
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EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169) 2. Sol.
The side of a cube is 60 cm. Find the total surface area of the cube. Side of a cube (l) = 60 cm Total surface area of a cube = 6l2 = 6 (60)2 = 6 × 60 × 60 = 21600 cm2 Total surface area of a cube is 21600 cm2. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
3. Sol.
The perimeter of one face of a cube Find (i) the total area of the 6 faces Perimeter of one face of a cube = Perimeter of one face of a cube = 4l =
l =
l = Total surface area of a cube = = = = Volume of the cube = = =
is 24 cm. (ii) the volume of the cube. 24 cm 4l 24
24 4 6 cm. 6l2 6 (6)2 6×6×6 216 cm2 l3 63 216 cm3.
Total area of the 6 faces is 216 cm2 and volume of the cube is 216 cm3. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169) 5. Sol.
The volume of a cube is 1000 cm3. Find its total surface area. Volume of a cube = 1000 cm3 Volume of a cube = l3 l3 = 1000 l = 10 cm [Taking cube roots] Total surface area of a cube = 6l2 = 6 × 102 = 6 × 10 × 10 = 600 cm2 Total surface area of a cube is 600 cm2. EXERCISE - 6.4 (TEXT BOOK PAGE NO. 169)
8. Sol.
A solid cube is cut into two cuboids exactly at middle. Find the ratio of the total surface area of the given cube and that of the cuboid. Side of a cube = l Total Surface of a cube = 6l2 l Length of cuboid (l1) = side of a cube l1 = l
l 2 Its height (h1) = l
Its Breadth (b1) =
308
l
l l 2 2
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Total surface area of a cuboid = 2 (l1b1 + b1h1 + l1h1)
l l + × l + l × l = 2 l × 2 2 l2 l2 + + l2 = 2 2 2
2l 2 + l2 = 2 2 = 2 × 2l2 Total surface area of a cuboid = 4l2
T.S.A of a cube Ratio of the total surface area of a cube and cuboid = T.S.A of a cuboid =
6l 2 4l 2
3 2 = 3:2 =
The ratio of the total surface area of the given cube and that of the cuboid is 3 : 2.
RIGHT CIRCULAR CYLINDER A right circular cylinder (Cylinder) is a solid figure bounded by two flat circular surfaces and a curved surface. h The perpendicular distance between the two base faces is called height of the cylinder and is denoted by ‘h’. The radius of the base of the cylinder is denoted by ‘r’. r The cylinders which we see regularly are drum, pipe, road roller, coins, test tube, refill of a ball pen, syringe etc. The following are the formulae for the surface area of a right circular cylinder :
FORMULAE 1. Curved surface area of a right circular cylinder = 2rh 2. Total surface area of a right circular cylinder = 2r (r + h) 3. Volume of a right circular cylinder = r2h EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173) 1.
The radius of the base of a right circular cylinder is 3 cm height is 7cm. Find (i) curved surface area (ii) total surface area (iii) volume of the
22 7 Radius of a right circular cylinder = 3cm its height (h) = 7cm (i) Curved surface area of a cylinder = 2 rh 22 = 2× ×3×7 7 Curved surface area of a cylinder = 132 cm2 closed righ t circu lar cylin der. Given =
Sol.
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(ii) Total Surface area of a cylinder = 2r (r + h) 22 × 3 (3 + 7) = 2× 7 22 × 3 × 10 = 2× 7 1320 = 7 = 188.57 cm2 (iii) Volume of the cylinder = r 2 h 22 ×3×3×7 = 7 = 198 cm3 Curved surface area is 132 cm2 Total surface area is 188.57cm2 and volume of the cylinder is 198 cm3 EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173) 2. Sol.
The volume of a cylinder is 504 cm3 and height is 14 cm. Find its curved surface area and total surface area. Express answer in terms of .(3 marks) Volume of a cylinder = 504 cm3 Its height (h) = 14 cm Volume of a cylinder = r 2 h 504 = × r2 × 14 504 = r2 14 r 2 = 36 r = 6 cm [Taking square roots] Curved surface area of a cylinder = 2rh = 2 × × 6 × 14 = 168 cm2 Total surface area of a cylinder = 2r (r + h) = 2 × × 6 (6 + 14) = 2 × × 6 × 20 = 240 cm2 Curved surface area of a cylinder is 168 cm2 and Total surface area of a cylinder is 240 cm2 EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173)
3. Sol.
310
The radius and height of a cylinder are n same ratio 3:7 and its volume is 1584 cm3. Find its radius. (3 marks) The ratio of radius and height of a cylinder is 3 :7 Let the common multiple be ‘x’ Radius of cylinder (r) = ‘3x’ cm and its height (h) = ‘7x’ cm Volume of a cylinder = 1584 cm3 Volume of a cylinder = r 2 h 22 1584 = × (3x) × (3x) × (7x) 7 1584 = 22 × 9 × x3 1584 = x3 22 × 9 x3 = 8 x = 2 [Taking cube roots] S C H O O L S E C TI O N
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Radius (r) = 3x = 3(2) = 6cm Radius of the cylinder is 6 cm. EXERCISE - 6.5 (TEXT BOOK PAGE NO. 173) 4. Sol.
Keeping the height same, how many times the rod of the given cylinder should be made to get the cylinder of double the volume of given cylinder ? (3 marks) Let the radius and the volume of the given cylinder be r1 and v1 respectively. The radius and the volume of the required cylinder be r2 and v2 respectively. Let the heights of the cylinder be h [ their heights are same] From the given condition, v2 = 2v1 2 r2 h = r12 h r22 = 2r12 r 2 = 2 r1 [Taking square roots] The radius of the required cylinder should be the given cylinder
2 times the radius
EXERCISE - 6.8 (TEXT BOOK PAGE NO. 184) 1.
Sol.
A cylindrical hole of diameter 30 cm is bored through a cuboid wooden block with side 1 meter. Find the volume of the object so formed ( = 3.14) (4 marks) side of cubical wooden block = 1 m = 100 cm Volume of cubical wooden block = l3 = (100) 3 = 1000000 cm3 A cylindrical hole is bored through the cubical wooden block Height of cylindrical hole (h) = 1m = 100 cm Diameter of cylindrical hole = 30 cm
Its radius (r)
=
= Volume of cylindrical hole = = = Volume of the object so formed = = =
30 2 15 cm r 2 h 3.14 × 15 × 15 × 100 70650 cm3 Volume of cubical wooden block – Volume of cylindrical hole 1000000 – 70650 929350 cm3
Volume of the object so formed is 929350 cm3. S C H O O L S E C TI O N
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Sol.
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EXERCISE - 6.8 (TEXT BOOK PAGE NO. 185) An ink container of cylindrical shape is filled with ink upto 91%. Ball pen refills of length 12 cm and inner diameter 2 m are filled upto 84%. If the height and radius of the ink container are 14 cm and 6 cm respectively, find the number of refills that can be filled with this ink. (4 marks) Height of the cylindrical container (h) = 14cm Its radius (r) = 6 cm Volume of cylindrical container = r2h = × 6 × 6 × 14 = 504 cm3 But, volume of ink filled = 91% of 504 in the cylindrical container 91 × 504 cm 3 = 100 Length of ball pen refill (h1) = 12m its inner diameter = 2 mm Its radius (r1) = 1 mm 1 = cm 10 Volume of the refill = r 1 2 h 1 1 1 = × × × 12 10 10 12 = cm3 100 12 But, volume of ink filled = 84% of 100 84 12 × = cm3 100 100 Number of refills that can be filled with ink Volume of ink filled in the cylindrical container = Volume of ink filled in each refill 91 × 504 100 84 ×12 = 100 ×100 91 × 504 100 ×100 × = 100 84 ×12 = 4550 Number of refills that can be filled with this ink is 4550. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
15. Sol.
A building has 8 right cylindrical pillars whose cross sectional diameter is 1 m and whose height is 4.2 m. Find the expenditure to paint those pillars at the rate of Rs. 24 per m2. (3 marks) Diameter of a pillar = 1 m
312
1 = m 2 Its height (h) = 4.2 m Curved surface area of a pillars = 2rh 22 1 = 2× × × 4.2 7 2 = 13.2 m2 Its radius (r)
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Curved surface area of 8 pillars = = Rate of painting = Total expenditure = = =
8 × 13.2 105.6 m2 Rs. 24 per m2 Area to be painted × Rate of painting 105.6 × 24 2534.40
Total expenditure to paint the pillars is Rs. 2534.40. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
16.
Sol.
A 10 m deep well of diameter 1.4 m us dug up in a field and the earth from digging is spread evenly on the adjoining cuboid field. The length and breadth of that filled are 55m and 14 m respectively. Find the thickness of the earth layer spread. (4 marks) Diameter of well = 1.4 m 1.4 Its radius (r) = 2 = 0.7 m Its depth (h) = 10 m Volume of cylindrical well = r 2 h 22 0.7 0.7 10 = 7 22 7 7 10 = 7 10 10 154 = 10 = 15.4 m3 Volume of earth dug is 15.4 m3 Now, Earth dug from the well is spread evenly on the adjoining cuboid field Volume of cuboid = Volume of earth dug = 15.4 m3 Length of a cuboid (l) = 55 m Its breadth (b) = 14 m Volume of cuboid = l × b × h 15.4 = 55 × 14 × h 154 10 × 55 × 14 = h 1 h = m 50 h = 0.02 m
The thickness of the earth layer spread is 0.02 cm. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
18. Sol.
A roller of diameter 0.9 m and length 1.8 m is used to press the ground. Find the area of ground pressed by it in 500 revolutions. (Given = 3.14) (3 marks) Diameter of the roller = 0.9 m 0.9 its radius (r) = 2 = 0.45 m its length (h) = 1.8 m
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Curved surface area of the roller = 2rh = 2 × 3.14 × 0.45 × 1.8 = 6.28 × 0.81 = 5.0868 m2 Area of the ground pressed by the roller in 1 revolution = curved surface area of roller Area of the ground pressed in one revolution = 5.0868 m2 Area of the ground pressed in 500 revolution = 500 × 5.0868 50868 = 500 10000 = 2543.4 m2 Area of the ground pressed by the roller = 2543.4 m2.
O
RIGHT CIRCULAR CONE
An ice-cream cone, a clown’s hat, a funnel are examples of cones. A cone has one circular flat surface and one l h curved surface. In the diagram alongside, seg OA is the height of the cone denoted by ‘h’. P A r seg AP is the radius of the base denoted by ‘r’. seg OP is the slant height of the cone denoted by ‘l’. The h, r and l of a cone represents the sides of a right angled triangle where l is the hypotenuse. l2 = r2 + h2.
FORMULA Volume of a right circular cone =
1 × r2h 3
EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178) 1. Sol.
The curved surface area of a cone is 4070 cm 2 and its diameter is 70 cm. What is its slant height ? (2 marks) Diameter of a cone = 70 cm. 70 Its radius (r) = 2 = 35 cm. Curved surface area of a cone = 4070 cm2 Curved surface area of a cone = rl 22 35 l 4070 = 7 4070 22 × 5 = l l = 37 Slant height of a cone is 37 cm. EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
2. Sol. 314
The base radii of two right circular cones of the same height are in the ratio 2:3. Find the ratio of their volumes. (3 marks) Let the radii of two right circular cone be r1 and r2 and their volumes be v1 and v2 respectively S C H O O L S E C TI O N
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r1 r2
=
v1 v2
=
v1 v2
r12 = 2 r2
v1 v2
r1 = r2
v1 v2 = v1 v2 = v1 : v2 =
2 3
.......(i)
(Given)
1 2 r1 h 3 1 2 r2 h 3
2
2
2 3 4 9 4:9
[From (i)]
Ratio of volumes of two right circular cone is 4 : 9 EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178) 3. Sol.
A cone of height 24 cm has a plane base of surface area 154 cm2. Find its volume. (2 marks) Height of a cone (h) = 24 cm Surface area of base = 154 cm2 1 Volume of a cone = × Surface area of base × height 3 1 × r2 × h = 3 1 × 154 × 24 = 3 = 1232 cm3 3 Volume of the cone is 1232 cm . EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178)
4. Sol.
Curved surface area of a cone with Find the height of the cone. Curved surface area of a cone = its radius (r) = Curved surface area of a cone = 1640 =
base radius 40 cm is 1640 sq.cm. (3 marks) 1640cm2 40 cm. rl × 40 × l
1640 = l 40 l = 41 cm
Now,
r2 + h2 402 + h2 h2 h2 h2 h Height of a cone is 9 cm.
S C H O O L S E C TI O N
= = = = = =
l2 412 412 – 402 1681 – 1600 81 9 cm [Taking square roots]
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204) 17. Sol.
The total surface area of cone is 71.28 cm2. Find the volume of this cone if the diameter of the base is 5.6 cm. (3 marks) Diameter of the base = 5.6 cm 5.6 Its radius (r) = 2 = 2.8 cm Total surface area of a cone = 71.28 cm2 Total surface area of cone = r (r + l) 22 71.28 = × 2.8 (2.8 + l) 7 7128 22 28 = × (2.8 + l) 100 7 10 7128 10 100 22 4 = 2.8 + l 81 = 2.8 + l 10 8.1 – 2.8 = l l = 5.3 cm r2 + h2 = l2 (2.8)2 + h2 = (5.3)2 h 2 = (5.3)2 – (2.8)2 h 2 = (5.3 + 2.8) (5.3 – 2.8) h 2 = 8.1 × 2.5 81 × 25 h 2 = 10 × 10 9×5 h = [Taking square roots] 10 45 h = 10 h = 4.5 cm 1 2 r h Volume of a cone = 3 1 22 2.8 2.8 4.5 = 3 7 1 22 28 28 45 = 3 7 10 10 10 36960 = 1000
Volume of a cone is 36.96 cm3. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
19.
Sol.
316
The diameter of the base of metallic cone is 2 cm and height is 10 cm. 900 such cones are molten to form 1 right circular cylinder whose radius is 10 cm. Find total surface area of the right circular cylinder so formed. (Given = 3.14) (4 marks) Diameter of the base of metallic cone = 2 cm 2 Its radius (r) = = 1 cm 2 Its height (h) = 10 cm S C H O O L S E C TI O N
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Volume of a metallic cone
=
1 2 r h 3
=
1 × × 1 × 1 × 10 3
=
10 cm3 3 10 3 3000 cm3 right circular cylinder 3000 10 cm and height be h2 r12h1 × 10 × 10 h2 30 cm 2r1 (r1 + h1) 2 × 3.14 × 10 (10 + 30) 6.28 × 40 2512 cm2
Volume of 900 metallic cones = 900 = 900 cones are melted to form a Volume of a cylinder = For a cylinder, Radius (r2) = Volume of a cylinder = 3000 = h1 = Total surface area of cylinder = = = =
Total surface area of the right circular cylinder is 2512 cm2. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
20. Sol.
The volume of a cone of height 5 cm is 753.6 cm3. This cone and a cylinder have equal radii and height. Find the total surface area of cylinder. (Given = 3.14) (3 marks) Height of a cone (h) = 5 cm Volume of a cone = 753.6 cm3 Volume of a cone
=
1 2 r h 3
753.6
=
1 × 3.14 × r2 × 5 3
7536 10
=
1 314 × × r2 × 5 3 100
7536 × 3 × 100 10 × 314 × 5
= r2
r 2 = 144 r = 12 cm [Taking square roots] Cone and cylinder have equal radii and height Radius of a cylinder = 12 cm and its heights = 5 cm. Total surface area of cylinder = 2r (r + h) = 2 × 3.14 × 12 (12 + 5) = 75.36 × 17 = 1281.12 cm2 Total surface area of a cylinder is 1281.12 cm2.
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FRUSTUM OF THE CONE : If the cone is cut off by a plane parallel to the base not passing through the vertex, two parts are formed as (i) cone (a part towards the vertex) (ii) frustum of cone (the part left over on the other side i.e. towards base of the original cone)
r2
r1
FORMULAE 1. 2. 3. 4.
l
h
Slant height (l) of the frustum = h 2 + r1 – r2 Curved surface area = p (r1 + r2) l
2
Total surface area of the frustum = r1 + r2 l + r12 + r22 1 r12 + r22 + r1 × r2 h Volume of the frustum = 3
EXERCISE - 6.6 (TEXT BOOK PAGE NO. 178) 5. Sol.
The curved surface area of the frustum of a cone is 180 sq. cm and the perimeters of its circular bases are 18 cm and 6 cm respectively. Find the slant height of the frustum of a cone. (3 marks) Curved surface area of the frustum of a cone = 180 cm2 Perimeters of circular bases are 18 cm and 6 cm 2r 1 = 18 ........(i) 2r 2 = 6 ........(ii) Adding (i) and (ii), we get 2r1 + 2r2 = 18 + 6 2 (r1 + r2) = 24 24 (r1 + r2) = 2 (r1 + r2) = 12 .......(iii) Curved surface area of the frustum of a cone = (r1 + r2) l 180 = (r1 + r2) l 180 = 12 × l [From (iii)] l = 15 cm Slant height of the frustum of a cone is 15 cm. PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
21.
The height of a cone is 40 cm. A small cone is cut off at the top of a 1 of the volume of the given plane parallel to its base. If its volume be 64 (5 marks) cone, at what height above the base is the section cut ?
Sol.
318
Let the radius, height and volume of the smaller cone be r1, h1 and v1 respectively. The radius height and volume of the given figure cone be r2, h2 and v2 respectively. h2 = 40 m [Given] Consider points A, B, C, E and F as shown in the figure, In AEF and ABC, A A [Common angle] AEF ABC [Each is 90º] AEF ~ ABC [By AA test of similarity]
A
E
F
r1
l
h B
r2
C
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AE EF = AB BC h1 r1 = h2 r2 1 v2 V1 = 64 V1 1 V2 = 64 1 r2 h V1 3 1 1 = 1 r2 h V2 3 2 2
r1 h1 = × r h 2 2
1 64
h1 h1 = × h2 h2
1 64 h1 h2 h1 40
h1 = h
......(i) [Given] ......(ii)
2
1 64
[c.s.s.t.]
2
[From (i)]
3
2
1 = 4 1 = 4 40 h1 = 4 h 1 = 10 cm The height above the base in the section cut = h2 – h1 = 40 – 10 = 30 cm The height above the base in the section cut is 30 cm.
SPHERE The set of all points of space which are at a fixed distance from a fixed point is called a sphere. The fixed point is called the centre and the fixed distance is called the Radius of the sphere. A Oh• r In the adjoining figure, point O is the centre of the sphere and seg OA is the radius of the sphere which is denoted as ‘r’. Since the entire surface of the sphere is curved, its area is called as curved surface area or simply surface area of the sphere. Some common examples of a sphere are cricket ball, football, globe, spherical soap bubble etc. The following are the formulae for surface area of the sphere :
FORMULAE Surface area (curved surface area) of a sphere = 4r2 4 Volume of a sphere = × r3 3 S C H O O L S E C TI O N
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EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
1. Sol.
22 Find the volume and surface area of a sphere of radius 4.2 cm. = 7 (3 marks) Radius of a sphere (r) = 4.2 cm 4 Volume of a sphere = r3 3 4 22 = × × 4.2 × 4.2 × 4.2 3 7 4 22 42 42 42 = × × × × 3 7 10 10 10 310464 = 1000 = 310.464 = 310.46 cm3 Surface area of a sphere = 4r 2 22 = 4× × 4.2 × 4.2 7 22 42 42 = 4× × × 7 10 10 22176 = 100 = 221.76 cm2 Volume of sphere is 310.46 cm3 and surface area of a sphere is 221.76 cm2. EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
2. Sol.
320
The volumes of two spheres are in the ratio 27 : 64. Find their radii if the sum of their radii is 28 cm. (3 marks) Let the radii of two spheres be r1 and r2 and their volumes be v1 and v2. v1 27 ........(i) [Given] v 2 = 64 4 3 r1 3 v1 v 2 = 4 r23 3 4 3 r1 3 27 = 4 3 64 r2 3 r13 27 = 3 r2 64 r1 3 [Taking cube roots] r2 = 4 Let the common multiple be x r1 = 3x and r2 = 4x [Given] r1 + r2 = 28 3x + 4x = 28 7x = 28 28 x = 7 x = 4 S C H O O L S E C TI O N
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r 1 = 3x r 1 = 3 (4) r 1 = 12 cm
r 2 = 4x r 2 = 4 (4) r 2 = 16 cm
Radii of two spheres are 12 cm and 16 cm. EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204) 3. Sol.
22 The surface area of a sphere is 616cm2. What is its volume ? = 7 (3 marks) Surface area of sphere = 616 cm2 Surface area of a sphere = 4r 2 22 616 = 4 × × r2 7 616 7 = r2 4 22 r 2 = 49 r = 7 cm [Taking square roots] 4 3 Volume of a sphere = r 3 4 22 777 = 3 7 4312 = 3 = 1437.33 cm3 The volume is 1437.33 cm3. EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204)
4. Sol.
If the radius of a sphere is doubled, what will be the ratio of its surface area and volume as to that of the first sphere. Let the original radius be ‘r1’ Original surface Area (A 1)= 4r 1 2 New radius (r2) = 2r 1 New surface Area (A 2) = 4r 2 2 = 4 × × (2r1)2 = 4 × × 4r12 = 16r 1 2 A2 Ratio of New surface area to the original surface area = A 1 16r12 = 4r12 4 = 1 Ratio of New surface area to the original surface area is 4 : 1 4 r 3 Now, original volume (v1) = 3 1 4 New volume (v2) = r 3 3 2 4 = (2r 1) 3 3 4 = × × 8r13 3 32 v2 = 3 r 1 3
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v2 Ratio of New volume to the original volume = v 1 32 3 r1 3 = 4 3 r1 3 32 = 4 8 = 1 Ratio of New volume to the original volume is 8 : 1
HEMISPHERE Half of a sphere is called as hemisphere. Any hemisphere is made up of a curved surface and a plane circular surface. The following are the formulae for the surface area of a hemisphere :
FORMULAE 1. Curved surface area of a hemisphere = 2r2 2. Total surface area of a hemisphere = 3r2 2 × r 3 3. Volume of a hemisphere = 3 EXERCISE - 6.7 (TEXT BOOK PAGE NO. 204) 5.
The curved surface area of a hemisphere is 905
Sol.
1 cm2, what is its volume? 7 (3 marks)
Curved surface area of a hemisphere = 905
Curved surface area of a hemisphere 1 905 7 6336 7 6336 7 7 2 22 r2 r
1 cm2 7
= 2r 2 22 = 2× × r2 7 22 = 2× × r2 7 = r2
= 144 = 12 cm [Taking square roots] 2 3 r Volume of a hemisphere = 3 2 22 = × × 12 × 12 × 12 3 7 25344 = 7 = 3620.57 cm3 Volume of a hemisphere is 3620.57 cm3. 322
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PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204) 14. Sol.
The following shapes are made up of cones and hemispheres. Find their volume. (3 marks) (a) For a hemisphere, Height = radius (r) = 3.5 cm Volume of the figure = Volume of cone + Volume of hemisphere =
1 2 3 r1h + r 3 3
=
1 2 r [h + 2r] 3
=
1 22 3.5 3.5 [12.3 2(3.5)] 3 7
=
1 22 35 35 (12.3 7) 3 7 10 10
12.3 cm
3.5 cm
1 385 19.3 3 10 385 193 = 3 10 10 =
74305 = 3 100
24768.33 100 = 247.68 cm3 =
Volume of the given figure is 247.68 cm3. (b) Diameter of smaller hemisphere = 3 cm 3 Its radius r1 = = 1.5 cm 2 Diameter of bigger figure = 10 cm 10 Its radius r2 = = 5 cm 2 Volume of the figure = Volume of smaller hemisphere + volume of bigger hemisphere 2 3 2 3 r1 r2 = 3 3 2 3 r1 r23 = 3 2 22 (1.5)3 53 = 3 7 2 22 (3.375 125) = 3 7 2 22 128.375 = 3 7 5648.5 = 21 = 268.98 cm3
3 cm
10 cm
Volume of the given figure is 268.98 cm3. S C H O O L S E C TI O N
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EXERCISE - 6.8 (TEXT BOOK PAGE NO. 184) 2.
Sol.
A toy is a combination of a cylinder, hemisphere and a cone, each with radius 10cm. Height of the conical part is 10 cm and total height is 60cm. Find the total surface area of the toy. ( = 3.14, 2 = 1.41) (5 marks) 10 cm 10 cm
10 cm
60 cm A toy is a combination of cylinder, radius 10 cm r = 10 cm Height of the conical part (h) = Height of the hemispherical part = Total height of the toy = Height of the cylindrical part (h1) = l2 = l2 = l2 = l2 = l = l = Slant height of the conical part (l) = = = Total surface area of the toy =
= = = = = =
hemisphere and cone, each with 10 cm its radius = 10cm 60cm 60 – 10 – 10 = 60 – 20 = 40 cm r2 + h2 102 + 102 100 + 100 200 [Taking square roots] 200 10 2 cm
10 2 10 × 1.41 14.1 cm Curved surface area of the conical part + Curved surface area of the cylindrical part + Curved surface area of the hemispherical part rl + 2rh1 + 2r2 r (l + 2h1 + 2r) 3.14 × 10 (14.1 + 2 × 40 + 2 × 10) 31.4 (14.1 + 80 + 20) 31.4 × 114.1 3582.74 cm2
Total surface area of the toy is 3582.74 cm2. EXERCISE - 6.8 (TEXT BOOK PAGE NO. 184) 3.
Sol.
324
A test tube has diameter 20 mm and height is 15 cm. The lower portion is a hemisphere in the adjoining figure. Find the capacity of the test tube. ( = 3.14) (5 marks) Diameter of a test tube = 20 mm 15 cm 20 its radius (r) = 2 = 10 mm = 1 cm Its height (h) = 15 cm Height of hemispherical part (h1) = radius of hemisphere = 1 cm Height of cylindrical part (h2) = h – h1 = 15 – 1 = 14 cm S C H O O L S E C TI O N
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Volume of test tube = Volume of cylindrical part + Volume of hemispherical part 2 3 = r2h2 + r 3 2 r = r2 h2 + 3 2 = 3.14 (1) 14 + 3 44 = 3.14 × 3 138.16 = 3 Volume of test tube = 46.05 cm3 Capacity of a test tube is 46.05 cm3 EXERCISE - 6.8 (TEXT BOOK PAGE NO. 185) 5. Sol.
A cylinder of radius 12 cm contains water upto depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus water level is (5 marks) raised by 6.75 cm. what is the radius of the ball ? Radius of the cylinder (r) = 12 cm A spherical iron ball is dropped into the cylinder and the water level rises by 6.75 cm Volume of water displaced = volume of the iron ball Height of the raised water level (h) = 6.75 m Volume of water displaced = r 2 h = × 12 × 12 × 6.75 cm3 Volume of iron ball = × 12 × 12 × 6.75 cm3 4 3 r But, Volume of iron ball = 3 4 × 12 × 12 × 6.75 = × × r3 6.75 cm 3 12 × 12 × 6.75 × 3 = r3 4 20 cm r 3 = 3 × 12 × 6.75 × 3 r 3 = 3 × 3 × 3 × 4 × 6.75 r 3 = 3 × 3 × 3 × 27 r = 3 3 × 3 × 3 × 3 × 3 × 3 [Taking cube roots] r = 3×3 r = 9
Radius of the iron ball is 9 cm.
22.
Sol.
A piece of cheese is cut in the shape of the sector of a circle of radius 6 cm. The thickness of the cheese is 7 cm. Find (i) The curved surface area of the cheese. (ii) The volume of the cheese piece. (4 marks) For a sector, Measure of arc () = 60º Radius (r) = 6 cm
S C H O O L S E C TI O N
6 cm 60º
7 cm
PROBLEM SET - 6 (TEXT BOOK PAGE NO. 204)
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(i) Curved surface area of the cheese = Length of arc × height 2r h = 360 60 22 2 67 = 360 7 = 44 cm2 The curved surface area of the cheese is 44 cm2. (ii)
Volume of the cheese piece
= A (sector) × height r 2 h = 360 60 22 667 = 360 7 = 132 cm3
The volume of the cheese piece 132 cm3. EXERCISE - 6.3 (TEXT BOOK PAGE NO. 164) 1.
Which are polyhedrons from the following ?
Nail (i)
Sol.
(i) No
Unsharpened Pencil (ii)
Tile
Diamond
Test tube
(iii)
(iv)
(v)
(ii) Yes
(iii) Yes
(iv) Yes
(v) No
EXERCISE - 6.3 (TEXT BOOK PAGE NO. 164) 2. Sol.
Using Euler’s formula, find V, if E = 30, F = 12. If the solid figure is a prism, how many sides the base polygon has. (1 mark) F+V = E+2 12 + V = 30 + 2 V = 32 – 12 V = 20 1 20 = 10 Number of sides of base polygon = 2 EXERCISE - 6.3 (TEXT BOOK PAGE NO. 164)
3. (i) Sol.
Verify Euler’s formula for these solids : (2 marks)
326
F = 8, V = 12, E = 18 L.H.S. = F + V = 8 + 12 L.H.S. = 20 R.H.S. = E + 2 = 18 + 2 R.H.S. = 20 L.H.S. = R.H.S. F+V = E+2
......(i) ......(ii) [From (i) and (ii)]
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(2 marks)
(ii) Sol.
F = 8, V = 6, E = 12 L.H.S. = F + V = 8+6 L.H.S. = 14 R.H.S. = E + 2 = 12 + 2 R.H.S. = 14 L.H.S. = R.H.S. F+V = E+2
......(i) ......(ii) [From (i) and (ii)] (2 marks)
(iii) Sol.
F = 8, V = 12, E = 18 L.H.S. = F + V = 8 + 12 L.H.S. = 20 R.H.S. = E + 2 = 18 + 2 R.H.S. = 20 L.H.S. = R.H.S. F+V = E+2
......(i) ......(ii) [From (i) and (ii)] (2 marks)
(iv) Sol.
F = 6, V = 6, E = 10 L.H.S. = F + V = 6+6 L.H.S. = 12 R.H.S. = E + 2 = 10 + 2 R.H.S. = 12 L.H.S. = R.H.S. F+V = E+2
......(i) ......(ii) [From (i) and (ii)]
HOTS PROBLEM (Problems for developing Higher Order Thinking Skill) 47.
A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is
6: .
(4 marks)
Proof : Surface are of sphere = surface area of cube 4r 2 = 6l2 ......(i)
r2 l2
=
6 4
r2 l2
=
3 2
r l
=
Volume of sphere Volume of cube
4 = =
S C H O O L S E C TI O N
3 ......(ii) 2×
r3 3 l3 4 r 3 3l 3 327
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= = = = = =
Volume of sphere Volume of cube
=
4 r 2 × r 3l 2 × l 4 r 2 r × 3l 2 l 6l 2 r × [From (i)] 3l 2 l r 2× l 3 2× 2× 2× 2× 3 2× 6
Ratio of the volume of the sphere to that of the cube is 48.
Sol.
49. Sol.
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6
Marbles of diameter 1.4 cm are dropped into a beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm. (5 marks) Diameter of marble = 1.4 cm 1.4 its radius (r) = 2 = 0.7 cm 4 3 r Volume of a marble = 3 4 7 7 7 ×× × × = cm3 3 10 10 10 Marbles are submerged fully in the water, water level rises by 5.6 cm Height of water displaced (h) = 5.6 cm Diameter of beaker = 7 cm 7 Its radius (r1) = cm 2 Volume of water displaced = r12 h 7 7 56 cm3 = × × × 2 2 10 Volume of water displaced Number of marbles = Volume of marble 7 7 56 4 7 7 7 = × 2 2 10 3 10 10 10 7 7 56 3 1 10 10 10 × × × × × = × × × 2 2 10 4 7 7 7 = 150 Number of marbles is 150. Water flows at the rate of 10 m per minute through a cylindrical pipe having its diameter is 20 mm. How much time will it take to fill a conical vessel of base diameter 40 cm and depth 24 cm ? (5 marks) Diameter of conical vessel = 40 cm 40 Its radius (r) = = 20cm 2 S C H O O L S E C TI O N
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Its depth (h) = 24 cm Volume of conical vessel
= =
= = Diameter of cylindrical pipe= 20 Its radius (r1)
= =
1 2 r h 3 1 × × 20 × 20 × 24 3 20 × 20 × 8 × 3200 cm3 mm 20 2 10 mm
10 cm [ 1 cm = 10 mm] 10 = 1 cm Water flowing in 1 minute (h) = 10 m = 10 × 100 cm [ 1 m = 100 cm] = 1000 cm Volume of water flowing in 1 minute through a cylindrical pipe r12 h × 1 × 1 × 1000 1000 cm3 Volume of conical vessel Time taken to fill conical vessel = Volume of water flowing in 1 minute =
= = =
=
3200 1000
=
32 mins 10
32 × 60 secs [1 minute = 60 seconds] 10 = 192 seconds = 3 minutes and 12 seconds
=
The time taken to fill the conical vessel is 3 minutes and 12 seconds. 50. Sol.
Find the length of 13.2 kg. copper wire of diameter 4 mm, when 1 cubic cm of copper weighs 8.4 gm. (4 marks) 3 Volume of 8.4 gm of copper = 1 cm
13200 8.4 13200 × 10 = 84 11000 cm3 = 7 Diameter of copper wire = 4 mm
Volume of 13.2 kg i.e. 13200 gm of copper =
4 2 = 2 mm
Its radius (r) =
= S C H O O L S E C TI O N
2 cm 10 329
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Volume of copper wire = r 2 h
11000 7 11000 × 7 × 10 × 10 7 × 22 × 2 × 2 h h
=
22 2 2 × × ×h 7 10 10
= h = 12500 cm = 125 m
[ 1 metre = 100 cm]
Length of wire is 125 m. 51. Sol.
A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 (4 marks) cm thick iron. Find the volume of the iron. Girth of garden roller = 440 cm Girth of garden roller = 2 r 2r = 440
22 × r = 440 7 440 × 7 r = 2 × 22 r = 70 cm Radius of outer cylinder (r) = 70 cm Width of the roller (h) = 63 cm Thickness of the roller = 4 cm Radius of inner cylinder (r1) = 70 – 4 = 66 cm Volume of iron = Volume of outer cylinder – Volume of inner cylinder = r2h – r12 h 2×
= h r2 – r12 =
22 × 63 × 702 – 662 7
= 198 (70 66) (70 – 66) = 198 × 136 × 4 = 107712 cm3 The volume of the iron is 107712 cm3. 52. Sol.
A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. Find the depth and capacity of cup. ( 3 = 1.73) (5 marks) Diameter of semicircle sheet = 28 cm 28 Its radius (r) = 2 = 14 cm A semicircular sheet is bent to form a open cone Slant height of a cone (l) = radius of a semicircular sheet l = 14 cm Circumference of a base of a cone = length of semicircle = r
22 × 14 7 = 44 cm
=
330
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Let the radius of a cone be r1 Circumference of a base of a cone = 2r 1 44 = 2r 1 22 × r1 44 = 2 × 7 44 × 7 2 × 22 = r 1 r 1 = 7 cm
l2 142 h2 h2 h2
= = = = =
r12 h2 72 + h2 142 – 72 196 – 49 147
h
=
h
=
147 49 × 3
h h h
= 7 3 = 7 × 1.73 = 12.11 cm 1 2 r1 h = 3 1 22 × × 7 × 7 × 12.11 = 3 7 1864.94 = 3 = 621.646 = 621.65 cm3
Volume of conical cup
Depth of a conical cup is 12.11 cm and volume of conical cup is 621.65 cm3. 53. Sol.
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights. (3 marks) A cone and a hemisphere have equal bases their radii are equal Height of hemisphere = radius of hemisphere Volume of a cone = Volume of hemisphere
1 2 2 3 r h = r 3 3 h = 2r h 2 = r 1 Ratio of heights of a cone and a hemisphere is 2 : 1
54.
Sol.
A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28 cm and 21 cm respectively. Find the height of the bucket. (4 marks) Radii of circular ends are 25 cm and 21 cm r1 = 28 cm and r2 = 21 cm Volume of bucket = 28.490 litres [ 1 litre = 1000 cm3] = 28.490 × 1000 cm3 3 = 28490 cm
S C H O O L S E C TI O N
331
MT
GEOMETRY
Volume of bucket =
28490 =
28490 =
28490 =
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1 r12 r22 r1 r2 × h 3 1 22 × 282 212 28 × 21 × h 3 7 22 784 441 588 × h 21 22 × 1813 × h 21
28490 × 21 22 × 1813 = h h = 15 cm
The height of the bucket is 15 cm. 55.
Sol.
A oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If diameter of the top and bottom of the frustum is 18 cm and 8 cm respectively and the slant height of the frustum of cone is 13 cm. Find the surface area of the tin required to make the funnel. (Express your answer in terms of ) (4 marks) Diameters of circular ends of frustum are 18 cm and 8 cm 18 8 r1 = = 9 cm and r2 = = 4 cm 2 2 Slant height (l) = 13 cm Curved surface area of frustum of frustum = (r1 + r2) l = (9 + 4) × 13 = × 13 × 13 = 169 cm2 Radius of a cylinder (r2) = 4 cm Its height (h) = 10 cm Curved surface area of a cylinder = 2r 2 h = 2 × × 4 × 10 = 80 cm2 Surface area of tin required to make the funnel = Curved surface area of frustum + curved surface area of cylinder = 169 + 80 = 249 cm2 The surface area of the tin required to make the funnel is 249 cm2.
56.
Sol.
332
There are 3 stair-steps as shown in the figure. Each stair-step has width 25 cm, height 12 cm and length 50 cm. How many bricks have been (5 marks) used in it if each brick is 12.5 cm × 6.25 cm × 4 cm. Length of a stair-step (l) = 50 cm its breadth (b) = 25 cm its height (h) = 12 cm Volume of a stair-step = l × b × h = 50 × 25 × 12 = 15000 cm3 Volume of 3 stair-step = 6 × 15000 = 90000 cm3 Length of a brick (l1) = 12.5 cm its breadth (b2) = 6.25 cm its height (h1) = 4 cm S C H O O L S E C TI O N
MT
GEOMETRY
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Volume of a brick = l1 × b1 × h1 = 12.5 × 6.25 × 4 = 312.5 cm3 Volume of 3 stair-steps Number of bricks required = Volume of each brick 90000 = 312.5 6 × 50 × 25 × 12 = 12.5 × 6.25 × 4 = 288 Number of bricks required is 288. 57.
If V is the volume of a cuboid of dimensions a × b × c and S its surface area, then prove that
1 2 1 1 1 = + + . V S a b c
(3 marks)
1 V 1 L.H.S. = .......(i) abc 2 1 1 1 R.H.S. = S a b c 2 bc ac ab = 2 (ab bc ac) abc 1 (ab + bc + ac) = ab + bc + ac × abc 1 R.H.S. = ......(ii) abc L.H.S. = R.H.S. [From (i) and (ii)]
Proof :
L.H.S. =
2 1 1 1 1 + + = S a b c V
MCQ’s 1.
An arc of a circle having measure 45º has length 25 cm. What is the circumference of the circle ? (a) 200 cm (b) 100 cm (c) 50 cm (d) 45 cm
2.
The measure of arc of circle is 90º. If the radius of the circle is 7 cm. What is the area of sector ? (b) 77 cm2 (a) 78.5 cm2 2 (c) 35.8 cm (d) 38.5 cm2
3.
P is the centre of a circle, m (arc RYS) = 60º. If the radius of the circle is 4.2 cm, what is the perimeter of P-RYS ? (a) 4.4 m (b) 8.4 m (c) 12.8 m (d) 17.2 m
S C H O O L S E C TI O N
333
MT
GEOMETRY
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4.
The radius of a circle is 3.5 cm. What is the measure of arc whose length is 5.5 cm ? (a) 30º (b) 45º (c) 60º (d) 90º
5.
A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. What is the radius of the cone ? (a) 2 cm (b) 4 cm (c) 8 cm (d) 16 cm
6.
Which of the following represents Euler’s formula ? (a) F – V = E + 2 (b) F + V = E + 2 (c) F + E = V + 2 (d) F – V = E – 2
7.
What is the curved surface area of a cone of height 15 cm and base radius 8 cm ? (a) 60 cm2 (b) 68 cm2 2 (c) 120 cm (d) 136 cm2
8.
How many solid metallic spheres each of diameter 6 cm are required to be melted to form a solid metallic cylinder of height 45 cm and diameter 4 cm. (a) 3 (b) 4 (c) 15 (d) 6
9.
The radius and slant height of a cone are 5 cm and 10 cm respectively. What is its curved surface area ? (a) 314 cm2 (b) 157 cm2 2 (c) 78.5 cm (d) 100 cm2
10.
The radii of the circular ends of a bucket which is 24 cm high are 14 cm and 7 cm respectively. What will be its slant height ? (a) 12 cm (b) 21 cm (c) 45 cm (d) 25 cm
11.
The diameter of a sphere 6 cm is melted and drawn into a wire of diameter 2 mm. What will be the length of the wire ? (a) 12 m (b) 18 m (c) 24 m (d) 36 m
12.
Two cubes each with 12 m edge are joined end to end. What is the difference in surface area of the resulting cuboid and the surface area of two cubes ? (a) 288 m2 (b) 144 m2 2 (c) 1440 m (d) 770 m2
13.
Area of a sector with central angle 60º will be ............ of area of a circle.
14.
334
(a)
2 3
(c)
1 2
rd
1 6
th
(b)
1 4
th
(d)
If area of semicircle is 77 cm2 its perimeter is ................. . (a) 72 cm (b) 120 cm (c) 36 cm (d) 40 cm S C H O O L S E C TI O N
MT 15.
GEOMETRY
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Area of minor segment AXB is ................ cm2. (a) 38.5 cm2 (b) 24.5 cm2 2 (c) 14.0 cm (d) 154 cm2
O
7
B x
16.
The capacity of a bowl is 144 cm . Find the radius. (a) 8 cm (b) 4 cm (c) 7 cm (d) 6 cm
17.
A solid metallic ball of radius 14 cm is melted and recasted into small balls of radius 2 cm. Find how many such balls can be made ? (a) 434 (b) 343 (c) 433 (d) 344
18.
Find the capacity of swimming pool of length 20 m breadth 5 m and depth 4 m ? (a) 40000 l (b) 400000 l (c) 4000 l (d) 4000000 l
19.
A cube of side 40 cm is divided into 8 equal cubes. Then its surface area will increase .................. times. (a) 4 (b) 8 (c) 2 (d) 5
20.
Find the number of coins 2.4 cm in diameter and 2 mm thick to be melted to form a right circular cylinder of height 12 cm and diameter 6 cm ? (a) 350 (b) 370 (c) 400 (d) 375
21.
The curved surface area of a right cone is double that of another right cone. If the ratio of their slant heights is 1 : 2, find the ratio of their radii ? (a) 1 : 4 (b) 2 : 3 (c) 3 : 2 (d) 4 : 1
22.
The area swept out by a horse tied in a rectangular grass field with a rope 8 m long is ............... . (a) 16 cm2 (b) 64 cm2 2 (c) 48 cm (d) 32 cm2
23.
The angle swept by the minute hand of a clock of length 9 cm in 15 mins is ................. . (a) 90 (b) 45 (c) 30 (d) 60
24.
A (sector) =
25.
A cylinder and a cone have equal radii and equal heights. If the volume of the cylinder is 300 cm3, then what is the volume of the cone ? (a) 100 cm3 (b) 10 cm3 3 (c) 110 cm (d) 300 cm3
3
A
1 A (circle). Hence, measure of the corresponding central 12 angle will be ................ . (a) 30 (b) 45 (c) 60 (d) 90
S C H O O L S E C TI O N
335
MT
GEOMETRY
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: ANSWERS : 1. 3. 5. 7. 9. 11.
(a) (c) (c) (d) (b) (d)
13.
(b)
15. 17. 19. 21. 23. 25.
(c) (b) (c) (d) (a) (a)
200 cm 12.8 m 8 cm 136 cm2 157 cm2 36 m
2. 4. 6. 8. 10. 12.
(d) (d) (b) (c) (d) (a)
38.5 cm2 90º F+V=E+2 15 25 cm 288 m2
14.
(c) 36 cm
16. 18. 20. 22. 24.
(d) (b) (d) (a) (a)
th
1 6 14.0 cm2 343 2 4:1 90 100 cm2
6 cm 400000 l 375 16 cm2 30
1 Mark Sums 1. Sol.
Find the length of the arc when the corresponding central angle is 270º and circumference is 31.4 cm. Measure of central angle () = 270º Circumfernce (2r) = 31.4 cm Length of the arc =
2. Sol.
2r 360
l =
270 × 31.4 360
l =
3 31.4 4
l =
94.2 4
l = 23.55 cm
If length of an arc is 7 cm, 2r = 36, then find the angle subtended at the centre by the arc. Length of the arc (l) = 7 cm 2r = 36
2r 360 36 7 = 360 7 360 = 36 = 7 × 10
= 70º
l =
336
The angle subtended at the centre by the arc is 70º. S C H O O L S E C TI O N
MT 3. Sol.
Find the area of a circle with radius 7 cm. Radius of circle (r) = 7 cm Area of the circle = r 2 22 77 = 7 = 154 cm2
4. Sol.
The area of a circle is 154 cm2.
Using Euler’s formula, write the value of V, if E = 30 and F = 12. F+ V = E+2 12 + V = 30 + 2 12 + V = 32 V = 32 – 12
5. Sol.
Sol.
Sol.
area of its minor sector is 31.4 cm2. Area of circle – Area of minor sector 314 – 3.14 282.6 cm2
The area of the major sector is 282.6 cm2.
The length of the side of a cube is 6 cm.
A cubical tank has each side of length in cubic metres. Side of a cubical tank (l) = Volume of cubical tank = = =
8.
= 20
Perimeter of one face of a cube is 24 cm. Find the length of its side. Perimeter of one face of a cube = 24 cm Perimeter of one face of a cube = 4l 24 = 4l 24 l = 4 l =6
7.
V
The area of a circle is 314 cm2 and the Find the area of its major sector. Area of major sector = = =
6. Sol.
GEOMETRY
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2 m. Find the capacity of the tank 2m l3 23 8 m3
Capacity of the cubical tank is 8 m3.
If the radius is 2 cm and length of corresponding arc is 3.14 cm, find the area of a sector. Radius (r) = 2 cm Length of arc (l) = 3.14 cm r Area of sector = l 2 2 = 3.14 × 2 = 3.14 cm2
The area of a sector is 3.14 cm2.
S C H O O L S E C TI O N
337
MT
GEOMETRY
9. Sol.
The dimensions of a cuboid are Length of a cuboid (l) Its breadth (b) Its height (h) Volume of a cuboid
10. Sol.
Sol.
Sol.
side 5 cm ? 5 cm l3 (5) 3 125 cm3
Volume of the cube is 125 cm3.
The value of R is 10.
The radius of the base of a cone its slant height ? Radius of base of cone (r) Its height (h) l2 l2 l2 l2 l
338
Length of the side of cube is 10 cm.
The area of a circle with radius R is equal to the sum of the areas of circles with radii 6 cm and 8 cm. What is the value of R ? According to given condition, R 2 = × (6)2 + × (8)2 R 2 = [62 + 82] R 2 = 36 + 64 R 2 = 100 R = 10 [Taking square roots]
14.
The total surface area of cube is 6 cm2.
What is the volume of a cube with Side of a cube (l) = Volume of cube = = =
13.
Volume of cuboid is 60 cm3.
Volume of a cube is 1000 cm3, find the length of its side. Volume of a cube = 1000 cm3 Volume of a cube = l3 l3 = 100 l = 10 [Taking cube roots]
12. Sol.
5 cm, 4 cm and 3 cm. Find its volume. = 5 cm = 4 cm = 3 cm = l×b×h = 5×4×3 = 60 cm3
Find the total surface area of a cube with side 1 cm. Length of side of cube (i) = 1 cm Total surface area of a cube = 6l2 = 6 (l)2 = 6 (1) = 6 cm2
11. Sol.
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is 7 cm and its height is 24 cm. What is = = = = = = =
7 cm 24 cm r2 + h2 72 + 242 49 + 576 625 25 [Taking square roots]
Slant height of cone is 25 cm. S C H O O L S E C TI O N
MT 15. Sol.
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Using Euler’s formula, find F, if V = 6 and E = 12. F+V=E+2 F + 6 = 12 + 2 F + 6 = 14 F = 14 – 6
16. Sol.
Sol.
Sol.
Sol.
20. Sol.
Length of the arc is 22 cm.
Radius of a circle is 10 cm. The length of an arc of this circle is 25 cm. What is the area of the sector ? Radius of circle (r) = 10 cm Length of arc (l) = 25 cm r Area of sector = l 2 10 = 25 × 2 = 25 × 2 = 125 cm2
19.
Area of the sector is 456 cm2.
The corresponding central angle of an arc is 90º. What is the length of this arc, if the radius of the circle is 14 cm ? Measure of central angle () = 90º Radius (r) = 14 cm 2r Length of the arc (l) = 360 90 22 2 14 l = 360 7 l = 22
18.
F=8
The area of a circle is 1368 cm2. What is the area of the sector of the circle whose corresponding central angle is 120º ? Area of circle = 1368 cm2 Area of sector of the circle whose corresponding central is 120º 1 1368 = 3 = 456 cm2
17.
GEOMETRY
The area of the sector is 125 cm2.
A cylinder and a cone have equal radii and equal heights ? If the volume of the cylinder is 300 cm3, what is the volume of the cone ? A cylinder and cone have equal height and equal radii 1 volume of cylinder Volume of cone = 3 1 300 = 3 = 100 cm3 Volume of the cone is 100 cm3. What is the corresponding angle of a sector whose area is one-forth of the area of the circle ? The corresponding angle of a sector whose area is one-forth of the area of the circle is 90º.
S C H O O L S E C TI O N
339
MT
GEOMETRY
21. Sol.
The area of a sector with corresponding angle 45º is 8cm2. What is the area of the circle ? ( = 3.14) Area of sector = 8cm 2 Measure of the arc ()= 45º r 2 Area of sector = 360 45 r 2 8 = 360 360 r 2 = 8 × 45 360 2 r = 8 3.14 45 r 2 = 3.14 × 64 r 2 = 200.96
22. Sol.
Sol.
Sol.
Sol.
r = 8
The radius of the base of a its slant height ? Radius of base of cone (r) its height (h) l2 l2 l2 l2 l
25.
x = 8
The area of a circle with radius 17 cm is equal to the sum of the areas of circles with radii r cm and 15 cm respectively. What is the value of r ? According to given condition, (17) 2 = r2 + (15)2 (17) 2 = (r2 + 152) 172 = r2 + 152 r 2 = 172 – 152 r 2 = 289 – 225 r 2 = 64
24.
Area of the circle is 200.96 cm2.
The dimensions of a cuboid are 3 cm × 9 cm × x cm. The volume of this cuboid is equal to the volume of a cube with side 6 cm. What is the value of x ? Volume of cuboid = Volume of cube [Given] 3 3 × 9 × x = (6) 3×9×x = 6×6×6 666 x = 39
23.
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[Taking square roots]
cone is 7 cm and its height is 24 cm. What is = = = = = = =
7 cm 24 cm r2 + h2 72 + 212 49 + 576 625 25
[Taking square roots]
Slant height of cone is 25 cm
What is the corresponding central angle of a sector whose area is onetenth the area of the circle ? The corresponding central angle of a sector whose area is one tenth the area of the circle is 36º.
340
S C H O O L S E C TI O N
MT
GEOMETRY
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EXTRA HOTS SUMS CHAPTER : 1 - SIMILARITY Bisectors of B and C in ABC meet each other at P. Line AP cuts the AP AB + AC = (5 marks) side BC at Q. Then prove that : PQ BC Proof : In ABQ, A ray BP bisects ABQ [Given] By property of an angle AP AB P = ........(i) bisector of a triangle PQ BQ • In ACQ, C B Q ray CP bisects ACQ [Given] By property of an angle AP AC = ........(ii) bisector of a triangle PQ CQ AP AB AC = = [From (i) and (ii)] PQ BQ CQ AB + AC AP = BQ + CQ [By Theorem on equal ratios] PQ AP AB + AC = [ B - Q - C] PQ BC 1.
•
In PQR, PQR = 900, As shown in figure, seg QS side PR. seg QM is angle bisector of PQR. PM² PS = (5 marks) Prove that : MR² SR Proof : In PQR, [Given] seg QM bisects PQR PQ PM = QR MR PQ² PM2 .........(i) 2 = QR ² MR In PQR, m PQR = 90º seg QS hypotenuse PR PQR ~ PSQ ~ QSR .........(ii) 2.
PSQ ~ PQR PQ PS = QR PQ PQ² = PR × PS Also, QSR ~ PQR SR QR = QR PR QR² = PR × SR PR × PS PM2 2 = PR × SR MR PS PM2 2 = SR MR
S C H O O L S E C TI O N
xx
P S M
Q
R
[Pr operty of an angle bi sec tor of a triangle] [Squaring both sides] [Given] [Given] [Theorem on similarity of right angled triangles] [From (ii)] [Corresponding sides
of similar triangles] .......(iii) [From (ii)] [Corresponding sides
of similar triangles] ........(iv) [From (i), (iii) and (iv)]
341
MT
GEOMETRY
3.
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Find the radius of a circle drawn by a compass when angle between two arms of compass is 1200 and length of each arm is 24cm. (5 marks)
Sol.
B
24
AD =
[Side opposite to 60º]
3 2 AD = 12 Similarly, DC = 12 AC = AD AC = 12 AC = 24
The radius of the circle is 24 3 cm.
4.
3 AB 2
AD =
cm
cm
24
In the adjoining figure, 120 0 seg AB and seg BC represents the arms of compass. In ABC, A C D side AB side BC [Given] BAC BCA ........(i) [Isosceles triangle theorem] In ABC, m ABC + m BAC + m BCA = 180º [ Sum of the measures of the angles of a triangles is 180º] 120º + m BAC + m BAC = 180º [From (i)] 2 m BAC = 180º – 120º 2 m BAC = 60º m BAC = 30º .......(ii) Draw seg BD side AC, A - D - C. In ABD, [From (ii) and A - D - C] m BAD = 30º m ADB = 90º [Given] m ABD = 60º [Remaining angle] ABD is a 30º - 60º - 90º triangle. By 30º - 60º - 90º triangle theorem.
× 24
3 cm we can get 3 cm + DC 3 + 12 3 3 cm.
[ A - D - C]
In BAC, BAC = 90º, segments AD, seg BE and B seg CF are medians. Prove : 2 (AD² + BE² + CF² ) = 3BC². (5 marks)
Proof : F
In BAC, seg AD is median on side BC. AB² + AC² = 2AD² + 2BD² AB² + AC² – 2BD² = 2AD²
D
[Given] A E [By Appollonius theorem]
C
2
342
1 AB² + AC² – 2 BC = 2AD² 2 1 AB² + AC² – 2 BC2 = 2AD² 4 1 AB² + AC² – BC² = 2AD² 2
[ D is the midpoint of seg BC]
.........(i) S C H O O L S E C TI O N
MT
5.
GEOMETRY
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Multiplying throughout by 2, we get 2AB2 + 2AC2 – BC2 = 4AD2 Similarly, we can prove that 2AB² + 2BC² – AC² = 4BE² .........(ii) 2AC² + 2BC² – AB² = 4CF² ........(iii) Adding (i), (ii) and (iii) we get 2AB2 + 2AC2 – BC2 + 2AB² + 2BC² – AC² + 2AC² + 2BC² – AB² = 4AD2 + 4BE2 + 4CF2 3AB2 + 3AC2 + 3BC2 = 4AD2 + 4BE2 + 4CF2 3 (AB2 + AC2 + BC2) = 4 (AD2 + BE2 + CF2) ........(iv) In BAC, [Given] m BAC = 90º BC2 = AB2 + AC2 ........(v) [By Pythagoras theorem] [From (iv) and (v)] 3 (BC2 + BC2) = 4 (AD2 + BE2 + CF2) 3 × 2 BC2 = 4 (AD2 + BE2 + CF2) 6BC2 = 4 (AD2 + BE2 + CF2) 3BC2 = 2 (AD2 + BE2 + CF2) [Dividing throughout by 2] 2 (AD² + BE² + CF²) = 3BC²
In ABCD points P, Q, R and S lies on sides AB, BC, CD and AD respectively such that seg PS || seg BD || seg QR and seg PQ || seg SR. Then prove D that seg PQ || seg AC. (5 marks) S
Proof :
seg PS || seg BD On transversal AB, APS ABD .......(i) of corresponding In APS and ABD, PAS BAD APS ABD APS ABD AP PS AS = = ......(ii) AB BD AD seg QR || seg BD On transversal BC, CQR CBD .......(iii) In CQR and CBD, QCR BCD CQR CBD CQR CBD CQ QR CR = = .......(iv) CB BD CD In PQRS, seg PQ || seg RS seg PS || seg QR PQRS is a parallelogram PS = QR ........(v) CQ PS CR = = ......(vi) CB BD CD AP CQ = AB CB CB AB = CQ AP
S C H O O L S E C TI O N
A
[Given]
P
[Converse angles test] B
R
Q
[Common angles] [From (i)] [By A-A test of similarity]
C
[c.s.s.t.] [Given] [Converse of corresponding angles test] [Common angles] [From (iii)] [By AA test of similarity] [c.s.s.t.] [Given] [Given] [By definition] [From (iv) and (v)] [From (ii) and (vi) [By Invertendo] 343
MT
GEOMETRY
6.
CB – CQ AB – AP = CQ AP BQ BP = ......(vii) CQ AP In ABC, BQ BP = CQ AP seg PQ || seg AC
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[By Dividendo] [ A - P - B and B - Q - C]
[From (vii)] [By converse of B.P.T.]
In ABC, m BAC = 90º. seg DE side AB, seg DF side AC,
A F
(5 marks) prove A (AEDF) = AE × EB × AF × FC E Proof : In ADB, [Given] m ADB = 90º B C seg DC side AB [Given] D DE2 = AE × EB ......(i) [By property of geometric mean] In ADC, [Given] m ADC = 90º seg DF side AC [Given] DF2 = AF × FC .......(ii) [By property of geometric mean] Multiplying (i) and (ii), DE2 × DF2 = AE × EB × AF × FC DE × DF = AE × EB × AF × FC ......(iii) In AEDF, m EAF = m AED = m AFD = 90º [Given] m EDF = 90º [Remaining angle] AEDF is a rectangle [By definition] A (AEDF) = DE ×DF ......(iv) From (iii) and (iv), A (AEDF) = AE × EB × AF × FC [From (iii) and (iv)]
CHAPTER : 2 - CIRCLE 1.
From the end points of a diameter of circle perpendiculars are drawn to a tangent of the same circle. Show that their feet on the tangent are equidistant from the centre of the circle. (5 marks) Given : (i) A circle with centre O. (ii) seg AB is the diameter of the circle. D (iii) Line l is tangent to the circle at point C. A (iv) seg AD line l . (v) seg BE line l . C To Prove : OD = OE. O Construction : Draw seg OC. Proof : seg AD line l [Given] E seg OC line l [Radius is perpendicular to the tangent] l B seg BE line l [Given] seg AD || seg OC || seg BE [Perpendiculars drawn to the same line are parallel to each other] On transversal AB and DE, AO DC = .........(i) [By property of intercepts made by OB CE three parallel lines]
344
S C H O O L S E C TI O N
MT
GEOMETRY
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But, AO = OB AO = 1 OB DC = 1 CE DC = CE In OCD and OCE , seg OC seg OC OCD OCE seg DC seg CE OCD OCE seg OD seg OE OD = OE
[Radii of the same circle] .........(ii) [From (i) and (ii)] .........(iii) [Common side] [Each is a right angle] [From (iii)] [By SAS test of congruence] [c.s.c.t.]
The bisectors of the angles A,B of ABC intersect in I, the bisectors of the corresponding exterior angles intersect in E. Prove that AIBE is cyclic. (5 marks) Proof : A P
C
2.
)
I
)
B
• • ×
×
Q
Take points P and Q as shown in the figure. E m CAB + m BAP = 180º [Linear pair axiom] 1 1 1 1 m CAB + m BAP = × 180º[Multiplying throughout by ] 2 2 2 2 m IAB + m BAE = 90º [ Ray AI and ray AE bisects CAB and BAP respectively] m IAE = 90º ......(i) [Angle addition property] Similarly, m IBE = 90º .....(ii) m IAE + m IBE = 90º + 90º [Adding (i) and (ii)] m IAE + m IBE = 180º AIBE is cyclic [If opposite angles of a quadrilateral are supplementary then quadrilateral is cyclic]
In the adjoining figure, line AP is a tangent to a circle with centre O at point A. seg AF is angle bisector of BAC. Prove that : seg AP seg PE. B Proof :
A
3.
P
0 0
O E
C
(5 marks)
F
BAE CAE Let, m BAE = m CAE = x PAC PAC Let, mPAC = mABC = y m PAE = m PAC + m CAE m PAE = (y + x) ........(iii) PEA is a exterior angle of ABE, m PEA = m ABE + m BAE m PEA = m ABC + m BAE m PEA = (y + x) ........(iv) In PAE, PAE PEA seg AP seg PE
S C H O O L S E C TI O N
[ Ray AE bisects BAC] .........(i) [Angles in alternate segment] .........(ii) [Angle addition property] [From (i) and (ii)] [Remote interior angle theorem] [ B - E - C] [From (i) and (ii)] [From (iii) and (iv)] [Converse of isosceles triangle theorem] 345
MT
GEOMETRY
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4.
In the adjoining figure, B AB is diameter of a circle with centre O, •M seg AC is tangent to the circle at point A. D line JD touches circle at point D, O and intersects segment AC in point J. Prove that : seg AJ seg CJ. (5 marks) Proof : C J A Take a point M on line DJ such that M - D - J. ........(i) [The lengths of two tangent segments seg AJ seg DJ from an external point to a circle are equal] .......(ii) [Radius is perpendicular to the tangent] m ODM = 900 In OBD, [Radii of the same circle] seg OB seg OD OBD ODB [Isosceles triangle theorem] Let, m OBD = m ODB = xº ......(iii) [Angle addition property] m ODM = m ODB + m BDM 90 = x + m BDM [From (ii) and (iii)] mBDM = (90 – x)º ......(iv) .......(v) [Vertically opposite angles] But, BDM JDC m JDC = (90 – x)º ......(vi) [From (iv) and (v)] In BAC, [Radius is perpendicular to tangent] m BAC = 90º [From (iii) and A - O - B, B - D - C] m ABC = xº m ACB = (90 – x)º [Remaining angle] .......(vii) [B - D - C and A - J - C] m JCD = (90 – x)º In JDC, JCD JDC [From (vi) and (vii)] seg DJ seg CJ ......(viii) [Converse of Isosceles triangle theorem] seg AJ seg CJ [From (i) and (viii)] 5.
If four tangents of a circle determine a rectangle then show that it must be a square. (5 marks) A
P
B
Given : (i) Lines AB, BC, CD and AD are the tangents to the circle at points Q S P, Q, R and S respectively (ii) ABCD is a rectangle. To Prove : ABCD is a square C R D Proof : AP = AS ..........(i) [The lengths of the two BP = BQ .........(ii) tangent segments from an CR = CQ .........(iii) external point to a circle DR = DS .........(iv) are equal] Adding (i), (ii), (iii) and (iv), we get AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB + CD = AD + BC ..........(v) [ A - P - B, B - Q - C, C - R - D, A - S - D] ABCD is a rectangle [Given] AB = CD ..........(vi) [ Opposite sides of a rectangle are AD = BC .........(vii) congruent] 346
S C H O O L S E C TI O N
MT
GEOMETRY
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AB + AB = BC + BC 2 AB = 2 BC AB = BC ABCD is a square
[From (v), (vi) and (vii)] [A rectangle in which adjacent sides are congruent, is a square]
6.
Two concentric circles with centre O. Seg AB, seg BC and seg AC are the tangents to the smaller circle at points P, Q and R respectively and also they are chords of the bigger circle. 1 A AC . Prove that seg PQ || seg AC , PQ = (5 marks) 2 Proof : R P
O
C
Q
With respect to smaller circle, .......(i) seg OP seg AB seg OQ seg BC .......(ii) With respect to bigger circle, seg OP chord AB AP = BP .....(iii)
seg OQ chord BC BQ = QC
.....(iv)
In ABC, P and Q are midpoints of sides AB and BC respectively. seg PQ || seg AC 1 PQ = AC 2
B
[Radius is perpendicular to the tangent] [From (i)] [Perpendicular drawn from centre of circle to chord bisects the chord] [From (ii) Perpendicular drawn from centre of circle to chord bisects the chord] [From (iii) and (iv)] [By Midpoint theorem]
CHAPTER : 3 - GEOMETRIC CONSTRUTION 1.
Point I is the incentre of ABC, BIC = 120º, BC = 4 cm, median AP = 3 cm. Draw ABC. (5 marks) Analysis : Let m ABI = m IBC = x and m ACI = m ICB = y m ABC = 2x and m ACB = 2y [Angle addition property] In BIC, m BIC + m IBC + m ICB = 180º 120 + x + y = 180 x + y = 180 – 120 x + y = 60 .......(i)
In ABC, m BAC + m ABC + m ACB m BAC + 2x + 2y m BAC + 2 (x + y) m BAC + 2 (60) m BAC + 120 m BAC m BAC Now, ABC can be constructed median AP.
S C H O O L S E C TI O N
= 180 = 180 = 180 = 180 [From (i)] = 180 = 180 – 120 = 60º with base BC, vertical angle BAC and 347
MT
GEOMETRY
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(Rough Figure) A A
A 60º 3
I
O
cm
120º B
30º
B
30º P 4 cm
120º
•
•
P
× ×
C
C
Point O is the orthocentre of ABC, m BOC = 105º, seg AD seg BC. BC = 6.5 cm and AD = 3.5 cm. Draw ABC. (5 marks) Analysis : BOC EOF [Vertically oppsoite angles] m BOC = m EOF = 105º In AFOE, m FAE + m AFO + m EOF + m AEO = 360º [Angle sum property of a quadrilateral] m FAE + 90 + 105 + 90 = 360 (Rough Figure) A m FAE + 285 = 360 m FAE = 360 – 285 m FAE = 75º E m BAC = 75º [A - F - B, A - E - C] F O Now, ABC can be constructed with base BC, 105º vertical angle BAC and altitude AD. 2.
B
A
A
D 6.5 cm
C
Y
75º 3.5 cm
3.5 cm P B
348
15º
150º 6.5 cm
15º
C
X
S C H O O L S E C TI O N
MT 3.
GEOMETRY
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Construct ABC such that BC = 8.8 cm, B = 50º, radius of incircle of ABC is 2.2 cm. (5 marks)
(Rough Figure) A
I
A
B
Q
5
50º
2.2 cm M 8.8 cm
C
P 4.
cm
B
8
cm
C
O 5.8 cm
Steps of construction : 1. 2. 3. 4. 5. 6. 7. 8. 4.
Draw seg BC = 8.8 cm. At B, draw m ABC = 50º. Draw bisector of B, as incentre lies on angle bisector. Draw a line parallel to side BC at a distance of 2.2 cm from BC. Point of intersection of line parallel to BC and angle bisector is incentre. Let incentre be I. From I, draw seg IM side BC. seg IM is in-radius. Draw incircle with IM as radius to touch sides AB and BC. From C draw the tangent to the circle to meet ray BA at A. Draw a sector O-AXB with radius 7 cm and m (arc AXB) = 50º. Draw a circle touching the sides OA and OB and also the arc. (5 marks) A
X
cm
(Rough Figure)
7
I
O S C H O O L S E C TI O N
50º 7 cm
B
349
MT
GEOMETRY
A
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P
X
I
O
• 50º •
B
7 cm
×
×
Q
Steps of construction : 1. 2. 3. 4. 5.
Draw m O = 50º and arc AB of radius 7 cm. Draw bisector of O. It intersects arc AB at X. At X, draw the PQ ray OX to cut ray OA at P and ray OB at Q. Draw bisector of Q. It intersects ray OX at I. Draw incircle with I as centre and IX as radius. This circle touches the ray OA, ray OB and also arc AXB.
5.
In ABC, BC = 5.8 cm, seg BP seg AC, seg CQ seg AB, BP = 5 cm , CQ = 4.8 cm. Construct ABC. (5 marks)
(Rough Figure) A Q
A
5
B
I
350
• 50º •
cm
4.
5.8 cm
8
cm
C
Y
2.2 cm B
P
M 8.8 cm
2.2 cm C
X
S C H O O L S E C TI O N
MT
GEOMETRY
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Steps of construction : 1. 2. 3. 4. 5.
6.
Draw seg BC of length 5.8 cm. Draw a semicircle with seg BC as the diameter. Taking B as the centre and radius 5 cm cut an arc on the semicircle to get point P and draw seg BP. Taking C as the centre and radius 4.8 cm cut an arc on the semicircle to get point Q and draw seg CQ. Extend seg BQ and seg CP to intersect at point A. ABC is the required triangle. Draw a line l. Take a point P at a distance 5cm from line l. Draw a circle with radius 3cm such that the circle touches the line l and passes (5 marks) through point P.
(Rough Figure)
5 cm
cm
3 5 cm
3
P
P
cm
N
O 3 cm
M
O
N
T
3 cm l
M
T
Steps of construction : 1. 2. 3. 4. 5. 6. 7. 8.
Draw line l. Take a point M on line l and draw a perpendicular to line l at point M. With point M as the centre, cut an arc of radius 5 cm on the perpendicular to get point P. With point M as the centre and radius 3 cm cut an arc on seg PM to get point N. Draw a line m perpendicular to line PM at point N. With point P as the centre cut an arc of radius 3 cm on line m to get point O. With point O as the centre and seg OP as the radius, draw the required circle. Draw a perpendicular from point O to line l to get point T.
CHAPTER : 4 - TRIGONOMETRY 1.
If
Proof :
1 + x 2 sin = x, prove that tan2 + cot2 = x2 + 1 + x 2 sin
=
x
1 + x2 x2 1 + x2
sin
=
sin2
=
S C H O O L S E C TI O N
1 . x2
(5 marks)
x [Squaring both sides] 351
MT
GEOMETRY
sin2 + cos2 = cos2 =
1 1 – sin2
cos2
=
1–
cos2
=
cos2
=
tan2
= = = =
cot2
= =
L.H.S.
= = =
tan2 + cot2 =
x2 1 + x2
1 x2 – x2 1 + x2 1 1 + x2 sin2 cos 2 x2 1 2 1 x 1 x2 2 x 1 x2 1 1 x2 x2 1 tan2 1 x2 tan2 + cot2 1 x2 + 2 x R.H.S. 1 x2 + 2 x
tan cot Prove : 1 – cot + 1 – tan = 1 + tan + cot tan cot Proof : L.H.S. = 1 – cot + 1 – tan
(5 marks)
2.
= = = = = = = = 352
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sin cos 1 – sin cos sin sin – cos sin cos
cos sin cos sin
sin 1 – cos cos – sin cos sin cos sin cos cos sin – cos sin cos – sin sin2 cos 2 cos (sin – cos ) sin (cos – sin ) sin2 cos 2 – cos (sin – cos ) sin (sin – cos ) sin2 cos2 1 – (sin – cos ) cos sin sin3 – cos3 1 (sin – cos ) cos × sin 1 (sin – cos ) (sin2 sin . cos cos 2 ) (sin – cos ) cos sin S C H O O L S E C TI O N
MT
GEOMETRY
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=
sin2 + sin . cos + cos 2 cos . sin
=
sin2 sin . cos cos 2 cos . sin cos . sin cos . sin
=
= = = tan 1 – cot
sin cos 1 cos sin tan + 1 + cot 1 + tan + cot R.H.S. cot + = 1 + tan + cot 1 – tan
3. Prove : sin8 – cos8 = (sin2 – cos2 ) (1 – 2 sin2 cos2 ). (5 marks) 8 8 Proof : L.H.S. = sin – cos = (sin4 )2 – (cos4 )2 = (sin4 – cos4 ) (sin4 + cos4 ) = (sin2 – cos2 ) (sin2 + cos2 ) (sin4 + cos4 ) [ sin2 + cos2 = 1] = (sin2 – cos2 ) (sin4 + cos4 ) 2 2 4 4 2 = (sin – cos ) (sin + cos + 2sin cos2 – 2sin2 cos2 ) = (sin2 – cos2 ) [(sin2 + cos2 )2 – 2 sin2 cos2 ] = (sin2 – cos2 ) (1 – 2sin2 cos2 ) [ sin2 + cos2 = 1] = R.H.S. sin8 – cos8 = (sin2 – cos2 ) (1 – 2 sin2 cos2 ). 4.
Sol.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of building increases from 30º to 60º as he walks towards the building. Find the distance he walked towards the building. (5 marks) Let the distance he walked B towards the building x m Height of tower (AB) = 30 m 28.5 m Height of boy (CD) = 1.5 m 60º 30º But CD = EF = AG D G F AG = 1.5 m 1.5 m BG = AB – AG 1.5 m BG = 30 – 1.5 1.5 m A C BG = 28.5 m x E In right angled BGF, tan 60º =
BG GF
28.5 GF 28.5 = 3 28.5 3 = 3 3
3
GF
GF
GF
=
GF
= 9.5 3 m
S C H O O L S E C TI O N
[By definition]
=
28.5 3 3
353
MT
GEOMETRY
In right angled BGD, BG tan 30º = GD 28.5 1 = 9.5 3 x 3
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[By definition]
9.5 3 x = 28.5 3
x
x
Distance he walked towards the building is 19 3 m.
= 28.5 3 – 9.5 3 = 19 3
5. Prove : (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A. (5 marks) Proof : L.H.S = (sin A + cosec A)2 + (cos A + sec A)2 = sin2 A + 2sin A . cosec A + cosec2 A + cos2A + 2cos A . sec A + sec2 A = (sin2 A + cos2 A) + (cosec2 A) + (sec2 A) + 2 sin A . cosec A + 2 cos A . sec A 1 1 = 1 + (1 + cot2 A) + (1 + tan2 A) + 2 × sin A sin A + 2cos A cos A = 1 + 1 + cot2 A + 1 + tan2 A + 2 + 2 = 7 + tan2 A + cot2 A = R.H.S. (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A. 6.
Prove that :
Proof :
=
=
= = = =
7.
=
(5 marks)
1 – sin . cos sin2 – cos 2 × cos (sec – cosec ) sin3 + cos 3
1 – sin . cos (sin cos ) (sin – cos ) × 1 1 (sin cos ) (sin2 – sin . cos cos2 ) cos – cos sin (1 – sin . cos ) (sin – cos ) × sin – cos (1 – sin . cos ) [ sin2 + cos2 = 1] cos sin . cos 1 × sin – cos sin – cos sin sin sin – cos sin – cos sin R.H.S. 1 – sin . cos sin 2 – cos 2 × = sin cos (sec – cosec ) sin 3 + cos 3
tan A tan A + sec A - 1 sec A + 1 = 2 cosec A
Proof : 354
L.H.S.
1 – sin . cos sin 2 – cos 2 × = sin cos (sec – cosec ) sin 3 + cos 3
L.H.S. =
(5 marks)
tan A tan A + sec A - 1 sec A + 1 S C H O O L S E C TI O N
MT
GEOMETRY
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=
=
= = = = = =
6.
1 1 + sin A 1 – cos A 1 + cos A 1 + cos A + 1– cos A sin A (1– cos A) (1+cos A) 2 sin A 1 – cos ² A 2 sin A sin² A 2 sin A 2 cosec A R.H.S. tan A + sec A + 1 = 2 cosec A
sin² A + cos² A = 1 1 - cos ² A = sin² A
From the top of a light house, 80 metres high, two ships on same side of light house are observed . The angles of depression of the ships as seen from the lighthouse are found to be of 450 and 300. Find the distance between the two ships (Assume that the two ships and the bottom of (5 marks) the lighthouse are in a line). E A In the adjoining figure, 0 30 450 seg AB represents the lighthouse. A is the position of the observer D and C are the position of the ships. Draw ray AE || seg BD. EAD and EAC are the angles of depression. 450 m EAD = 30º and m EAC = 45º 300 B C D On transversal AD [Converse of alternate angles test] m EAD = m ADB = 30º On transversal AC [Converse of alternate angles test] m EAC = m ACB = 45º In right angled ABD, AB [By definition] tan 30 = DB 1 80 = 3 DB DB = 80 3 m ......(i) 80 m
Sol.
= = tan A sec A - 1
sinA sin A cos A cos A 1 1 + -1 +1 cos A cos A sinA sin A cos A cos A (1 – cos A ) + (1 + cos A ) cos A cos A sin A sin A + 1 – cos A 1 + cos A
S C H O O L S E C TI O N
355
MT
GEOMETRY
In right angled ABC, AB tan 45 = CB 80 1 = CB CB = 80 m BD = BC + CD
80 3 =
CD
=
CD
=
80 + CD
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[By definition] .......(ii) [ D - C - B] [From (i) and (ii)]
80 3 – 80 80 3 – 1 m
The distance between the two ships is 80
3 – 1 m.
CHAPTER : 5 - CO-ORDINATE GEOMETRY 1. Sol.
A (8, 5), B (9, – 7), C (– 4, 2) and D (2, 6) are the vertices of a ABCD. If P, Q, R and S are the midpoints of sides AB, BC, CD and DA respectively. Show that PQRS is a parallelogram, using the slopes. P, Q, R and S are the midpoints of side AB, BC, CD and AD of ABCD. A (8, 5), B (9, – 7), C (– 4, 2) and D (2, 6) By midpoint formula, x1 x 2 y1 y 2 8 9 5 – 7 17 , P 2 , 2 2 , –1 2 2
356
9 – 4 –7 2 5 –5 , Q , 2 2 2 2 –4 2 2 6 , R (– 1, 4) 2 2 11 8 2 5 6 , S 5 , 2 2 2 5 –3 – – (–1) 2 y 2 – y1 2 1 Slope of PQ = x – x = 5 17 = –12 = 4 – 2 1 2 2 2 5 4 – – 13 2 2 –13 Slope of QR = 5 = –7 = 7 –1 – 2 2 11 3 –4 1 Slope of RS = 2 = 2 = 4 5 – (– 1) 6 11 13 – (–1) 2 2 –13 Slope of PS = 17 = – 7 = 7 5– 2 2 Slope of PQ = Slope of RS seg PQ || seg RS .......(i) Slope of QR = slope of PS seg QR || seg PS .......(ii) PQRS is a parallelogram. [From (i), (ii) and by definition] S C H O O L S E C TI O N
MT 2. Sol.
3. Sol.
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Find the equations of the lines which through the point (3, 4) and the sum of whose intercepts on the axes is 14. Let the intercepts made by the lines on the co-ordinate axes be a and b respectively. a + b = 14 ........(i) x y 1 The equation of the line is a b Since the line passes through the point (3, 4) 3 4 1 a b 3b + 4a = ab .......(ii) From (i), a = 14 – b Substituting a = 14 – b in (ii) we get, 3b + 4 (14 – b) = (14 – b)b 3b + 56 – 4b = 14b – b2 b2 – 15b + 56 = 0 b2 – 8b – 7b + 56 = 0 b (b – 8) – 7 (b – 8) = 0 (b – 8) (b – 7) = 0 b = 8 OR b = 7 By (i) when b = 8, c = 14 – 8 = 6 and when b = 7, c = 14 – 7 = 7 Equations of the required lines are x y x y 1 and 1 6 8 7 7 4x + 3y = 24 and x + y = 7 4x + 3y – 24 = 0 and x + y – 7 = 0 Find the equation of a line which passes through the point (– 3, 7) and makes intercepts on the co-ordinate axes which are equal in magnitude but opposite in sign. Let the intercepts made by the line on the co-ordinate axes be a and b. a=–b .......(i) x y 1 The equation of the line is a b x y –b b 1 –x+y=b This line passes through the point (– 3, 7) – (– 3) + 7 = b b = 10 The equation of the line is – x + y = 10
4. Sol.
GEOMETRY
x – y + 10 = 0
Find the equation of a line which contains the point (4, 1) and whose x-intercept is twice its y-intercept. Let the intercepts made by the line on the co-ordinate axes be a and b respectively. a = 2b x y 1 The equation of the line is a b x y 1 2b b
S C H O O L S E C TI O N
357
MT
GEOMETRY
5. Sol.
x + 2y = 2b Since this line contains the point (4, 1) 4 + 2 (1) = 2b 6 = 2b b=3 Equation of the required line is x + 2y = 6
x + 2y – 6 = 0
Find the equation of side AC of an isosceles ABC, if the equation of side AB is x – y – 4 = 0 and B (4, 0) and C (6, 4) are the extremities of the base. Let A (h, k) Since A lies on side AB i.e. on x–y–4=0 h–k–4=0 h=k+4 A (k + 4, k) Since ABC is an isosceles triangle with BC as base, l (AB) = l (AC)
(k 4 – 4)2 (k – 0)2 = On squaring both sides, k2 + k2 = k2 + k2 = 0 = 12k =
k
k
k+4
(k 4 – 6)2 (k – 4)2
(k – 2)2 + (k – 4)2 k2 – 4k + 4 + k2 – 8k + 16 – 12k + 20 20 20 = 12 5 = 3 5 17 4 = = 3 3
17 5 , A 3 3 Equation of side AC by two point form is x – x1 y – y1 x1 – x 2 = y1 – y 2
x –6 y –4 17 = 5 6– 4– 3 3
17 3 5 4– 3 1 3 7 3 1 7 y–4 y–4 6–
x– 6 y –4
=
x– 6 y –4
=
x– 6 y –4 = 7 (x – 6) = 7x – 42 =
7x – y – 38 = 0
358
EDUCARE LTD.
S C H O O L S E C TI O N
MT 6. Sol.
GEOMETRY
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Find the equations of the line which cut off intercepts on the axes whose sum is 1 and product is – 6. Let the intercepts made by the line on the co-ordinates axes be a and b respectively. a+b=1 ........(i) and ab = – 6 .......(ii) From (ii) –6 b= a Substituting this in (i) we get, 6 = 1 a– a a2 – 6 = a 2 a –a–6 = 0 a2 – 3a + 2a – 6 = 0 a (a – 3) + 2 (a – 3) = 0 (a – 3) (a + 2) = 0 a = 3 or a = – 2 By (i) when a = 3, b = 1 – 3 = – 2 and when a = – 2, b = 1 – (– 2) = 3 Now, equation of the line making intercepts a and b is x y 1 a b Equations of the required lines are x y x y 1 and 1 3 –2 –2 3 2x – 3y = 6 and – 3x + 2y = 6
2x – 3y – 6 = 0 and 3x – 2y + 6 = 0
CHAPTER : 6 - MENSURATION 1.
Sol.
A tinmaker converts a cubical metallic box into 10 cylindrical tins. Side of the cube is 50 cm and radius of the cylinder is 7 cm. Find the height of each cylinder so made if the wastage of 12% is incurred in the process. 22 (Given = ). 7 Side of the cubical metallic box (l) = 50 cm Total surface area of cubical box = 6l2 = 6 × (50)2 = 6 × 2500 = 15000 cm2 Wastage incurred in the process of making 10 cylindrical tins = 12% of 15000 12 × 15000 = 100 = 1800cm2 Area of metal sheet used to make 10 cylindrical tins = Total surface area of cubical box – Wastage incurred in the process = 15000 - 1800 = 13200 cm2 Area of metal sheet used to make each cylindrical tin 13200 = 10 = 1320 cm2
S C H O O L S E C TI O N
359
MT
GEOMETRY
Radius (r) = 7 cm Area of metal sheet used to = Total surface area of cylinder make each cylindrical tin Total surface area of cylinder = 2r (r + h) 22 1320 = 2 × 7 (7 + h) 7 1320 = 2 22 × (7 + h) 1320 = 7+h 2 × 22 30 = 7+h h = 30 – 7 h = 23 cm
Height of each cylinder is 23 cm.
2. Sol.
3. Sol.
360
EDUCARE LTD.
The three faces A, B, C of a cuboid have surface area 450 cm2, 600 cm2 and 300 cm2 respectively. Find the volume of the cuboid. Surface area of face A = 450 cm2 Surface area of face A = l × h l × h = 450 .....(i) Surface area of face B = 600cm2 Surface area of face B = l × b l × b = 600 .....(ii) Surface area of face C = 300 cm2 Surface area of face C = b × h b × h = 300 .....(iii) Multiplying (i), (ii) and (iii), l × h × l × b × b × h = 450 × 600 × 300 l2 × b2 × h2 = 450 × 2 × 300 × 300
l×b×h l×b×h l×b×h But, Volume of the cuboid Volume of the cuboid
Volume of the cuboid is 9000 cm3.
= = = = =
900 300 300 [Taking square roots] 30 × 300 9000cm3 l×b×h 9000cm3
Oil tins of cuboidal shape are made from a metallic sheet with length 8 m and breadth 4 m . Each tin has dimensions 60 40 20 in cm and is open from the top. Find the number of such tins that can be made. Length of the metallic sheet (l) = 8 m = 8 × 100 B = 800 cm its breadth (b) = 4 m C = 4 × 100 A h b = 400 cm Area of metallic sheet = l × b l = 800 × 400 = 320000 cm2 Length of the oil tin (l1) = 60 cm its breadth (b1) = 40 cm its height (h1) = 20 cm Area of metallic sheet required for each tin = surface area of vertical faces + surface area of the base = [2 (l1 + b1) × h1] + [l1 b1] S C H O O L S E C TI O N
MT
= = = =
=
[2 (60 + 40) × 20] + [60 × 40] (2 × 100 × 20) + (60 × 40) 4000 + 2400 6400 cm2 Number of tins that can be made Area of metallic sheet Area of metal required for each tin 320000 6400 50
50 Oil tins can be made.
= =
4.
Sol.
GEOMETRY
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Plastic drum of cylindrical shape is made by melting spherical solid plastic balls of radius 1 cm. Find the number of balls required to make a drum (5 marks) of thickness 2 cm, height 90 cm and outer radius 30 cm. Outer radius of the drum (r1) = 30cm Its thickness = 2 cm inner radius of the drum (r2) = 30 – 2 = 28 cm Outer height of cylindrical plastic drum (h1) = 90cm Inner height of cylindrical plastic drum (h2) = Outer height – thickness of base = 90 – 2 = 88cm Volume of plastic required for the cylindrical drum = Volume of outer cylinder – Volume of inner cylinder = r12h1 – r22h2 = [(30)2 × 90 – (28)2 × 88] = × (900 × 90 – 784 × 88] = × (81000 – 68992) = 12008 cm3 Radius of spherical solid plastic ball (r) = 1cm Volume of each plastic ball = = =
Number of balls required to make the drum
=
=
4 r3 3 4 ××r×1×1×1 3 4 cm3 3 Volume of plastic required for the drum Volume of each plastic ball 12008 4 3
= 12008
3 4
= 9006
Number of plastic balls required to make the cylindrical drum is 9006.
S C H O O L S E C TI O N
361
MT
GEOMETRY
5.
Sol.
Water drips from a tap at the rate of 4 drops in every 3 seconds. Volume of one drop is 0.4 cm3. If dripped water is collected in a cylinder vessel of height 7 cm and diameter is 8 cm. In what time will the vessel be completely filled ? What is the volume of water collected ? How many such vessels (5 marks) will be completely filled in 3 hours and 40 minutes ? Diameter of the cylindrical vessel = 8cm Its radius (r) = 4 cm its height (h) = 7 cm Volume of the cylindrical vessel = r 2 h 22 447 = 7 = 22 × 16 = 352 cm3 Volume of water collected = 352 cm3 Volume of one drop of water = 0.4 cm3 Volume of 4 drops of water = 4 × 0.4 = 1.6 cm3 Water drips from the tap at the rate of 4 drops in every 3 seconds Volume of water collected in 3 seconds = 1.6 cm3 1.6 Volume of water collected in 1 seconds = cm3 3 Time required to fill the cylindrical vessel Volume of cylindrical vessel = Volume of water collected in each sec ond 352 = 1.6 3 3 352 = 1.6 352 3 10 = 1.6 10 352 3 10 = 16 = 660 seconds = 11 minutes [ 1 minutes = 60 seconds] 3 hours and 40 minutes = 3 × 60 min + 40 min = 180 + 40 = 220 minutes Number of vessels that can be completely filled in 220 minutes 220 = 11 = 20
6.
Sol.
362
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20 vessels can be filled in 3 hours and 40 minutes.
In the adjoining figure, ABCDEF is a regular hexagon with each side 14 cm. From each vertex, arcs with radius 7 cm are drawn. Find the area of the shaded portion. (5 marks) F Draw seg BN chord PQ Radii of each arc = 7 cm [Given] i.e. PB = BQ = 7 cm ABCDEF is a regular hexagon......
E
D
C Q
X• N
A
P
B
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GEOMETRY
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In BPQ, BP = BQ mBPQ = m BQP .......(i) [Isosceles triangle theorem] m BPQ + m BQP + m PBQ = 180º [Sum of the measures of angles of a triangle is 180º] m BPQ + m BPQ + 120 = 180 [From (i) and angle of regular hexagon] 2 mBPQ = 180 – 120 2 mBPQ = 60 m BPQ = 30º ......(ii) In BNP, BNP = 90º [Construction] BPN = 30º [From (ii) and P - N - Q] PBN = 60º [Remaining angle] BNP is 30º - 60º - 90º triangle By 30º - 60º - 90º triangle theorem, 1 BN = BP [Side opposite to 30º] 2 7 BN = 2 BN = 3.5 cm PN
3 BP 2 3 7 = 2
=
PN
PN PQ
= 3.5 3 cm = 2PN
PQ
=
PQ
A (BPQ)
A (B-PXQ)
A (segment PXQ)
S C H O O L S E C TI O N
[Side opposite to 60º]
[Perpendicular drawn from the centre of the circle to the chord bisects the chord]
2 3.5 3 7 3 cm = 1 PQ BN = 2 1 7 3 3.5 = 2 = 3.5 3 3.5 = 3.5 × 3.5 × 1.73 = 20.25 × 1.73 = 21.1925 = 21.19 cm2 r2 = 360 120 22 77 = 360 7 154 = 3 = 51.33 cm2 = A (B – PXQ) – A (BPQ) = 51.33 – 21.19 = 30.14 cm2 363
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GEOMETRY
7.
Sol.
A (shaded portion) = 6 × A (segment PXQ) = 6 × 30.14 = 180.84 cm2
The area of shaded portion is 180.84 cm2.
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A metallic right circular cylindrical disc is of height 30 cm and the diameter of the base is one half time the height. This metallic disc is melted and moulded into the sphere. Assuming that no metal is wasted during moulding, find the radius and total surface area of the sphere. (5 marks) Height of cylindrical disc (h) = 30 cm 1 its diameter = 1 times the height 2 3 30 = 2 = 45 cm 45 Radius of cylindrical disc (r) = cm 2 2 Volume of cylindrical disc = r h 45 45 30 = 2 2 60750 cm2 = 4 Let the radius of the sphere be ‘r1’ cm The metallic disc is melted and moulded into the sphere [Given] Volume metallic sphere = Volume cylindrical disc 4 3 60750 r1 cm3 = 3 4 4 45 45 r13 = 30 3 2 2 45 45 30 3 = r13 422 45 45 45 = r13 222 45 r1 = [Taking cube roots] 2 r 1 = 22.5 cm Total surface area of sphere = 4r 2 22 2 22.5 = 4 7 4 22 506.25 = 7 88 506.25 = 7 4455000 = 7 = 6364.28 cm2 (Approximately)
Radius of the sphere is 22.5 cm and total surface area of the sphere is 6364.28 cm2.
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