Written according to the New Text book (2011-2012) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.
Std. X
Geometry Mr. Biju Babu (B.E Prod.)
Mr. Kannan G. Iyer
Mr. Pankaj Pankaj Dubey
(B.Sc. Maths)
(B.E. Mech.)
Salient Features:
Written as per the new textbook. Exhaustive coverage of entire syllabus. Precise theory for every topic. Covers answers to all textual exercises and problem set. Includes additional problems for practice. Comprehensive solution to Question Bank. Constructions drawn with accurate measurements. Attractive layout of the content. Self evaluative in nature. Includes Board Question Paper of March 2013.
Tar
get
PUBLICATIONS PVT. LTD.
Mumbai, Maharashtra Tel: 022 – 6551 6551 Website : www.targetpublications.in www.targetpublications.in email : mail@targetpublications
[email protected] .in
Written according to the New Text book (2011-2012) published by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.
Std. X
Geometry Mr. Biju Babu (B.E Prod.)
Mr. Kannan G. Iyer
Mr. Pankaj Pankaj Dubey
(B.Sc. Maths)
(B.E. Mech.)
Salient Features:
Written as per the new textbook. Exhaustive coverage of entire syllabus. Precise theory for every topic. Covers answers to all textual exercises and problem set. Includes additional problems for practice. Comprehensive solution to Question Bank. Constructions drawn with accurate measurements. Attractive layout of the content. Self evaluative in nature. Includes Board Question Paper of March 2013.
Tar
get
PUBLICATIONS PVT. LTD.
Mumbai, Maharashtra Tel: 022 – 6551 6551 Website : www.targetpublications.in www.targetpublications.in email : mail@targetpublications
[email protected] .in
Std. X
Target Target Publication s Pvt Ltd.
Third Edition : March 2013 2013
Price:
190/-
Printed at: Spark Offset Nerul Navi Mumabi
Published by
Target PUBLICATIONS PVT. LTD. Shiv Mandir Sabhagriha, Mhatre Nagar, Near LIC Colony, Mithagar Road, Mulund (E), Mumbai - 400 081 Off.Tel: 022 – 6551 6551 email:
[email protected]
PREFACE
Geometry is the mathematics of properties, measurement and relationships of points, lines, angles, surfaces and solids. It is widely used in the fields of science, engineering, computers, architecture etc. It is a vast subject dealing with the study of properties, definitions, theorems, areas, perimeter, angles, triangles, mensuration, co-ordinates, constructions etc. The study of Geometry requires a deep and intrinsic understanding of concepts. Hence to ease this task we bring to you “Std. X: Geometry” a complete and thorough guide critically analysed and extensively drafted to boost the students confidence. The question answer format of this book helps the student to understand and grasp each and every concept thoroughly. The book is based on the new text book and covers the entire syllabus. It contains answers to textual exercises, problems sets and Question bank. It also includes additional questions for practice. All the diagrams are neat and have proper labelling. The book has a unique feature that all the constructions are as per the scale. Another feature of the book is its layout which is attractive and inspires the student to read. There is always room for improvement and hence we welcome all suggestions and regret any errors that may have occurred in the making of this book. A book affects eternity; one can never tell where its influence stops.
Best of luck to all the aspirants! Your’s faithfully Publisher
No.
Topi c Name
Page No.
1
Similarity
1
2
Circle
65
3
Geometric Constructions
122
4
Trigonometry
182
5
Co-ordinate Geometry
224
6
Mensuration
272
7
Question Bank (Hot Problems)
320
8
Board Paper March 2013
364
9
Marking Scheme
367
Std. X : Geometry
TARGET Publications
01
SIMILARITY
Concepts of Std. IX Similarity of triangles
For a given one-to-one correspondence between the vertices of two triangles if, i. their corresponding angles are congruent and ii. their corresponding sides are in proportion then the correspondence is known as similarity and the two triangles are said to be similar.
P
A 2 B
°
°
9
3
6
×
CQ
4
×
R
6
In the figure, for correspondence ABC ↔ PQR. i. ii.
∠A ≅ ∠P, ∠B ≅ ∠Q, ∠C ≅ ∠R AB PQ i.e.
=
2 3
AB PQ
,
=
BC QR
=
BC QR
4 6
=
2 3
,
AC PR
=
6 9
=
2 3
AC
=
PR
Hence, ∆ABC and ∆PQR are similar triangles and are symbolically written as ∆ABC
∆PQR.
Test of similarity of triangles
1.
S S S test of similarity: For a given one-to-one correspondence between the vertices of two triangles, the two triangles are similar, if the sides of one triangle are proportional to the corresponding sides of the other triangle.
P
A
B
2
6
2
3
1
C
4
Q
R
In the figure, AB PQ
∴
AB PQ
= =
1 2
,
BC QR
BC QR
=
=
2 4
=
1 2
,
AC PR
=
3 6
=
1 2
AC PR
then ∆ABC ↔ ∆PQR 2.
A A A test of similarity [A A test]: For a given one to one correspondence between the vertices of two triangles, the two triangles are similar if the angles of one triangle are congruent to the corresponding angles of the other triangle. In the figure,
if ∠A ≅ ∠P, ∠B ≅ ∠Q, ∠C ≅ ∠R
----- [By S−S−S test of similarity] P
×
A
×
B °
C
Q °
R
∆PQR then ∆ABC ----- [By A−A−A test of similarity] Note: A−A−A test is verified same as A −A test of similarity. Similarity
1
Std. X: Geometry 3.
TARGET Publications
S A S test of similarity: For a given one to one correspondence between the vertices of two triangles, the two triangles are similar if two sides of a triangle are proportional to the two corresponding sides of the other triangle and the corresponding included angles are also congruent. In the figure, 1 BC 2 1 AB = , = = 3 QR 6 3 PQ
AB
∴
PQ
=
BC QR
P
A 3
1 B °
C Q °
2
R
6
and ∠B ≅ ∠Q
then ∆ABC
∆PQR
----- [By S−A−S test of similarity]
Converse of the test for similarity: i. Converse of S S S test: If two triangles are similar, then the corresponding sides are in proportion. If ∆ABC ∆PQR then,
AB PQ ii.
=
BC
=
AC
----- [Corresponding sides of similar triangles]
QR PR
Converse of A A A test: If two triangles are similar, then the corresponding angles are congruent. ∆PQR If ∆ABC then ∠A ≅ ∠P, ∠B ≅ ∠Q, and ∠C ≅ ∠R ----- [Corresponding angles of similar triangles]
Properties of ratios of areas of two triangles Property – I The ratio of areas of two triangles is equal to the ratio of the product of their base and corresponding height.
P A
Given: In ∆ABC and ∆PQR, seg AD ⊥ seg BC, B −D−C, seg PS ⊥ ray QR, S−Q−R To prove that:
A(∆ABC) A(∆PQR)
=
BC × AD
B
QR × PS
D
C
S
Q
Proof:
A(∆ABC) = A(∆PQR) =
1 2 1
BC × AD
QR × PS 2 Dividing (i) by (ii), we get 1 × BC × AD A(∆ABC) 2 = 1 A(∆PQR) × QR × PS 2 A( ∆ABC) BC × AD = A( ∆PQR) QR × PS 2
----- (i) [Area of a triangle =
1 2
× base × height]
----- (ii)
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R
Std. X : Geometry
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For Understanding When do you say the triangles have equal heights? We discuss this in three cases.
A
P
D
C S
Case – I In the adjoining figure,
segments AD and PS are the corresponding heights of ∆ABC and ∆PQR respectively. If AD = PS, then ∆ABC and ∆PQR are said to have equal height.
B
Case – II
A
In the adjoining figure, ∆ABC and ∆XYZ have their one vertex on one of the parallel lines and the other two vertices lie on the other parallel line, hence the two triangles are said to lie between the same parallel lines and are said to have equal heights.
B
Q
X
C
Case – III
Z
R
P
Q
Y
A
In the adjoining figure, ∆ABC, ∆ACD and ∆ABD have a common vertex A and the opposite sides BC, CD and BD of the triangles lie on the same line, hence ∆ABC, ∆ACD and ∆ABD are said to have equal heights and BC, CD and BD are their respective bases.
C
B
D
Property – II The ratio of areas of two triangles having equal height is equal to the ratio of their corresponding bases.
A
Example:
∆ABC, ∆ACD and ∆ABD have a common vertex A and their opposite sides BC, CD, BD lie on the same line, hence they have equal heights.
∴
A(∆ABC) A(∆ACD)
=
BC CD
,
A(∆ABC) A(∆ABD)
=
BC BD
,
A(∆ACD) A(∆ABD)
=
CD
B
D
C
BD
Property – III The ratio of areas of two triangles having equal base is equal to the ratio of their corresponding heights.
D
A
Example:
∆ABC and ∆DCB have a common base BC. A( ∆ABC) AP = ∴ A( ∆DCB) DQ
P
C
B
Q
Property IV Areas of two triangles having equal bases and equal heights are equal.
A
Example:
∆ABD and ∆ACD have a common vertex A and their opposite sides BD and DC lie on the same line, hence the triangles have equal heights. Also their bases BD and DC are equal.
∴
A(∆ABD) = A(∆ACD)
Similarity
B
D
C 3
Std. X: Geometry
TARGET Publications
Exercise 1.1 1.
In the adjoining figure, seg BE seg AB and seg BA If BE = 6 and AD = 9, find
A ( ∆ABE ) A(∆BAD)
seg AD.
E
6
B
.
Solution: A ( ∆ABE ) A( ∆BAD)
∴ ∴
A ( ∆ABE ) A( ∆BAD) A ( ∆ABE ) A(∆BAD)
2.
A = = =
BE AD
----- [Ratio of areas of two triangles having equal base is equal to the ratio of their corresponding heights.]
6 9 2 3
In the adjoining figure, seg SP side YK and seg YT If SP = 6, YK = 13, YT = 5 and TK = 12, then find A( SYK):A( YTK).
seg SK.
S
6
T 5
Solution: A( ∆SYK) A( ∆YTK)
∴ ∴
A( ∆SYK) A( ∆YTK) A(∆SYK) A(∆YTK)
= = =
D
9
YK × SP TK × YT 13×6
12
K 13 P Y ----- [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights.]
12×5 13 10
T 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) Solution: K R P RP:PK = 3:2 ----- [Given] Let the common multiple be x. RP = 3 x, PK = 2 x ∴ RK = RP + PK ----- [R −P−K] RK = 3 x + 2 x ∴ RK = 5 x ∴ A( ∆TRP) RP i. = ----- [Ratio of the areas of two triangles having equal heights A( ∆TPK) PK is equal to the ratio of their corresponding bases.] A( ∆TRP) 3 x = ∴ A( ∆TPK) 2 x 3.
A(∆TRP) A(∆TPK) 4
=
3 2
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ii.
A(∆TRK) A(∆TPK)
∴
A(∆TRK)
∴
A(∆TRK)
A(∆TPK)
=
=
A(∆TPK)
iii.
∴ ∴
A( ∆TRP) A( ∆TRK) A( ∆TRP) A( ∆TRK) A(∆TRP) A(∆TRK)
4.
=
=
=
=
RK
----- [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.]
PK 5x
2 x 5 2
RP RK
----- [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.]
3 x 5 x 3 5
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.
Solution: Let A1 and A2 be the areas of larger triangle and smaller triangle respectively. A1 A2
=
6
----- [Given]
5
Let the corresponding heights be h 1 and h2 respectively.
∴
A1 A2
=
h1 h2
∴
6
∴
h2
∴
h2 =
∴
h2 = 7.5 cm
5
=
=
----- [Ratio of the areas of two triangles having equal base is equal to the ratio of their corresponding heights.]
9
----- [∵ h1 = 9, given]
h2 5× 9 6 15 2
Height of smaller triangle is 7.5 cm.
5.
In the adjoining figure, seg PR seg BC, seg AS
seg BC and seg QT
seg BC.
A
Find the following ratios: i.
iii.
Similarity
A(∆ABC) A(∆PBC) A(∆PRC) A(∆BQT)
ii.
iv.
A(∆ABS) A(∆ASC) A(∆BPR) A(∆CQT)
P Q B
R
S
T
C 5
Std. X: Geometry
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Solution: A( ∆ABC)
i.
A( ∆PBC)
ii.
iii. iv.
A( ∆ABS) A( ∆ASC) A( ∆PRC) A( ∆BQT) A( ∆BPR) A( ∆CQT)
=
=
AS PR BS SC
----- [Ratio of the areas of two triangles having equal bases is equal to the ratio of their corresponding heights.]
----- [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.]
=
RC × PR BT × QT
----- [Ratio of the areas of two triangles is equal to the ratio of product of their bases and corresponding heights.]
=
BR × PR CT × QT
----- [Ratio of the areas of two triangles is equal to the ratio of product of their bases and corresponding heights.]
6.
In the adjoining figure, seg DH seg EF and seg GK seg EF. 2 If DH = 12 cm, GK = 20 cm and A( DEF) = 300 cm , then find i. length of EF ii. A( GEF) iii. A( DFGE) Solution: 1 i. Area of triangle = × base × height 2 1 A(∆DEF) = ∴ × EF × DH 2 1 300 = ----- [∵ A(∆DEF) = 300 cm 2] ∴ × EF × 12 2 300 = EF × 6 ∴ 300 = EF ∴ 6 EF = 50 cm.
ii.
∴ ∴ ∴ ∴
A( ∆DEF) A( ∆GEF) 300 A( ∆GEF)
= =
DH GK
D 12 K
E
H
F
20 G
----- [Ratio of the areas of two triangles having equal bases is equal to the ratio of their corresponding heights.]
12 20
300 × 20 = 12 × A(∆GEF) 300 × 20 = A(∆GEF) 12 300 × 20 A(∆GEF) = 12 2
A( GEF) = 500 cm
iii.
∴
A(DFGE) = A( ∆DEF) + A(∆GEF) ----- [Area addition property] A(DFGE) = 300 + 500 = 800 2
A( DFGE) = 800 cm 7.
P
In the adjoining figure, seg ST || side QR. Find the following ratios. i.
A(∆PST) A(∆QST)
ii.
A(∆PST) A(∆RST)
iii.
A(∆QST)
S
A(∆RST)
Q 6
T R
Similarity
Std. X : Geometry
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Solution: A( ∆PST)
i.
A( ∆QST) A( ∆PST)
ii.
A( ∆RST)
=
PS QS
=
[Ratio of the areas of two triangles having equal heights is e ual to the ratio of their corres ondin bases.
PT TR
iii.
∆QST and ∆RST lie between the same parallels ST and QR
∴ ∴
Their heights are equal. A(∆QST) = A(∆RST) A( ∆QST) =1 A( ∆RST)
∴
----- [Areas of two triangles having common base i.e. ST and equal heights are equal.]
Basic Proportionality Theorem If a line parallel to a side of a triangle intersects the other sides in two distinct points, then the line divides these sides in proportion. P Given: In ∆PQR, line l || side QR. Line l intersects side PQ and side PR in points M and N N M l respectively such that P −M−Q and P− N−R. To Prove that:
PM MQ
=
PN Q
NR
R
Construction: Draw seg QN and seg RM. Proof: In ∆PMN and ∆QMN, where P−M−Q
A( ∆PMN) A( ∆QMN)
=
PM MQ
In ∆PMN and ∆RMN, where P− N−R A( ∆PMN) PN = NR A( ∆RMN) A(∆QMN) = A(∆RMN)
∴ ∴
A( ∆PMN) A( ∆QMN) PM MQ
=
=
A( ∆PMN) A( ∆RMN)
PN NR
----- (i) [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.]
----- (ii) [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.] ----- (iii) [Areas of two triangles having equal bases and equal heights are equal.] ----- (iv) [From (i), (ii) and (iii)] ----- [From (i), (ii) and (iv)]
Converse of Basic Proportionality Theorem If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. [March 2013] P Given:
Line l intersects the side PQ and side PR of ∆PQR in PM PN the points M and N respectively such that = . MQ NR
To Prove that: Line l || side QR.
Similarity
K M
Q
N
l
R 7
Std. X: Geometry
TARGET Publications
Proof: Let us consider that line l is not parallel to the side QR. Then, there must be another line passing through M which is parallel to the side QR Let line MK be that line. Line MK intersects the side PR at K,
----- [P−K −R]
In ∆PQR, line MK || side QR. PM
∴
MQ PM
But,
MQ PK
∴
KR
= =
PK
----- (i) [By B.P.T.]
KR PN
----- (ii) [Given]
NR PN
----- [From (i) and (ii)]
NR
PK + KR PN + NR = KR NR PR PR = KR NR KR = NR
∴ ∴ ∴ ∴ ∴ ∴ ∴
=
----- [By componendo] ----- [P−K −R, P− N−R]
Points K and N are not different. Line MK and line MN coincide. line MN || side QR line l || side QR
Applications of Basic Proportionality Theorem i.
Property of intercepts made by three parallel lines on a transversal: 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 three parallel lines.
Given:
Line l || line m || line n The transversals x and y intersect these parallel lines at points A, B, C and P, Q, R respectively.
To Prove that:
AB BC
=
PQ QR
Construction: Draw seg AR to intersect line m at point H. Proof:
AB
=
AH
BC HR In ∆ARP, seg HQ || side AP PQ AH = ∴ QR HR
∴ 8
AB BC
=
PQ QR
m ----- (i) [By B.P.T.]
n
y
A
l
In ∆ACR, seg BH || side CR
∴
x
B C
P H
Q R
----- (ii) [By B.P.T.] ----- [From (i) and (ii)]
Similarity
Std. X: Geometry
∴ ∴
∴ ∴
AC2 = AD2 + DC2 AC2 = AD2 + (BD + BC)2 AC2 = AD2 + BD2 + 2BD⋅BC + BC2 In ∆ADB, ∠ADB = 90° AB2 = AD2 + BD2 AC2 = AB2 + 2BD⋅BC + BC2 2 2 2 AC = AB + BC + 2BD BC
TARGET Publications
----- [By Pythagoras theorem] ----- [D−B−C] ----- (i) ----- [Given] ----- (ii) [By Pythagoras theorem] ----- [From (i) and (ii)]
Appollonius theorem It is a theorem relating to the length of the median of a triangle to the length of the sides. Here, we shall prove this theorem for an acute angled triangle using application of Pythagoras theorem. A Given: In ∆ABC, seg AD is the median. 2 2 2 2 To prove that: AB + AC = 2AD + 2CD Construction: Draw seg AE ⊥ seg BC such that B −E−D. Proof: In ∆ABD, B E D ----- [Given] ∠ADB < 90° AB2 = AD2 + BD2 − 2BD⋅DE ----- (i) [By application of Pythagoras theorem] ∴ In ∆ADC, ----- [Given] ∠ADC > 90° 2 2 2 AC = AD + CD + 2CD⋅DE ----- (ii) [By application of Pythagoras theorem] ∴ Adding (i) and (ii) AB2 + AC2 = 2AD2 − 2BD⋅DE + 2CD⋅DE + BD2 + CD2 ----- (iii) But, BD = CD ----- (iv) [D is the midpoint of seg BC.] 2 2 2 2 AB + AC = 2AD − 2CD⋅DE + 2CD⋅DE + CD + CD2 ----- [From (iii) and (iv)] 2 2 2 2 AB + AC = 2AD + 2CD
C
Exercise 1.7 2
2
1. In ABC, AP is a median. If AP = 7, AB + AC = 260, then find BC. Solution: In ∆ABC, seg AP is the median ----- [Given] 2 2 2 2 AB + AC = 2AP + 2PC ----- [By Appollonius Principle] 2 2 260 = 2(7) + 2PC ∴ 260 = 2 × 49 + 2PC 2 ∴ 260 = 98 + 2PC2 ∴ ∴ 260 − 98 = 2PC2 162 = 2PC2 ∴ 162 = PC2 ∴ 2 81 = PC2 ∴ PC = 9 units ----- [Taking square roots] ∴ 38
A
B
P
C
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Std. X : Geometry
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∴
BC = 2 × PC BC = 2 × 9 BC
2.
----- [P is the midpoint of seg BC.]
= 18 units 2
2
A
In the adjoining figure, AB + AC = 122, BC = 10. Find the length of the median on side BC.
Solution: seg AQ is the median on side BC. 1 BQ = BC ∴ 2 1 BQ = ∴ × 10 2 BQ = 5 units ∴ In ∆ABC, seg AQ is the median AB2 + AC2 = 2AQ2 + 2BQ2 ∴ 122 = 2AQ2 + 2(5)2 ∴ 122 = 2AQ2 + 50 ∴ ∴ 122 − 50 = 2AQ2 72 = 2AQ2 ∴ 72 AQ2 = ∴ 2 2 AQ = 36 ∴ AQ = 6 units ∴
----- [Q is the midpoint of side BC.]
B
Q
C
----- [By Appollonius theorem]
----- [Taking square roots]
Length of the median on side BC = 6 units. 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 diagonal. D 17 cm A Solution: Let ABCD be the parallelogram and its 11 cm diagonals AC and BD intersect each other at O. O
AB = 11 cm, AD = 17 cm, BD = 26cm 1
∴
BO =
∴
× 26 2 BO = 13cm In ∆ABD, O is the midpoint of seg BD seg AO is the median AB2 + AD2 = 2AO2 + 2BO2 112 + 172 = 2AO2 + 2(13)2 121 + 289 = 2AO2 + 2 × 169 410 = 2(AO)2 + 338
∴
∴ ∴ ∴ ∴ ∴
BO =
Similarity
2 1
BD
C B ----- [Diagonals of a parallelogram bisect each other.]
----- [Diagonals of a parallelogram bisect each other.] ----- [By definition] ----- [By Appollonius theorem]
39
Std. X: Geometry
∴ ∴
∴ ∴ ∴
∴ ∴
TARGET Publications
410 − 338 = 2AO2 72 = 2AO2 72 = AO2 2 AO2 = 36 AO = 6 units 1 AO = AC 2 1 6 = AC 2 AC = 12 units
----- [Taking square roots] ----- [Diagonals of a parallelogram bisect each other.]
Length of the other diagonal = 12 units. 4.
In the adjoining figure, LMN = 90 and LKN = 90 , seg MK LN. Prove that R is the midpoint of seg MK. Proof: In ∆LMN, L m∠LMN = 90 ° ----- [Given] seg MR ⊥ hypotenuse LN ----- [Given] 2 MR = LR × RN ----- (i) [By property of geometric mean] ∴ In ∆LKN, ----- [Given] ∠LKN = 90° seg KR ⊥ hypotenuse LN ----- [Given] 2 KR = LR × RN ----- (ii) [By property of geometric mean] ∴ 2 2 MR = KR ----- [From (i) and (ii)] ∴ MR = KR ∴ R is midpoint of seg MK. 5.
ii.
2
2
AC = AD + BC DM + 2
2
AB = AD
BC DM +
Proof: i. In ∆ADC, ∠ADC > 90° AC2 = AD 2 + CD2 +2 CD⋅DM ∴ 1 CD = BC 2
R
N
K
A
seg AD is the median of ABC and AM BC. Prove that: i.
M
BC
2
2 BC
2
2
B
M
D
C
----- [Given] ----- (i) [By application of theorem of Pythagoras] ----- (ii) [D is the midpoint of seg BC.]
2
∴
⎛1 ⎞ ⎛1 ⎞ AC = AD + ⎜ BC ⎟ + 2 ⎜ BC ⎟ ⋅DM ⎝2 ⎠ ⎝2 ⎠
∴
⎛ BC ⎞ AC = AD + ⎜ ⎟ + BC⋅DM ⎝ 2 ⎠
∴
⎛ BC ⎞ AC = AD + BC DM + ⎜ ⎟ ⎝ 2 ⎠
2
2
----- [From (i) and (ii)]
2
40
2
2
2
2
2
Similarity
Std. X : Geometry
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ii.
∴
In acute angled ∆ABD, seg AM ⊥ side BD AB2 = AD 2 + BD2 − 2 BD⋅DM 1 BD = BC 2
----- (i) [By application of theorem of Pythagoras] ----- (ii) [D is the midpoint of seg BC.]
2
∴
⎛1 ⎞ ⎛1 ⎞ AB = AD + ⎜ BC ⎟ − 2 ⎜ BC ⎟ ⋅DM ⎝2 ⎠ ⎝2 ⎠
∴
⎛ BC ⎞ AB = AD + ⎜ ⎟ − BC⋅DM ⎝ 2 ⎠
2
2
----- [From (i) and (ii)]
2
2
2
2
2
⎛ BC ⎞ BC DM + ⎜ ⎟ ⎝ 2 ⎠
2
∴
AB = AD
6.
In the adjoining figure, PQR = 90 , T is the mid point of side QR. 2 2 2 Prove that: (PR) = 4(PT) 3(PQ)
Proof : In ∆PQR, seg PT is the median PQ2 + PR 2 = 2PT2 + 2QT2 ∴ PR 2 = 2PT2 + 2QT2 − PQ2 ∴ In ∆PQT, ∠PQT = 90° PT2 = PQ2 + QT2 ∴ QT2 = PT2 − PQ2 ∴ PR 2 = 2PT2 + 2[PT2 − PQ2] − PQ2 ∴ PR 2 = 2PT2 + 2PT2 − 2PQ2 − PQ2 ∴ 2
PR = 4PT
2
----- [By definition] ----- [By Appollonius theorem] ----- (i)
P
Q
T
R
----- [Given] ----- [By Pythagoras theorem] ----- (ii) ----- [From (i) and (ii)]
2
3PQ
Problem Set ‐ I 1.
In each of the following figures, you find two triangles. Indicate whether the triangles are similar. Give reasons in support of your answer. (i)
(iii)
(ii)
4.6 2.3
5
8
60°
5
4 3
5
3
60°
10 2.5
(iv)
(v)
24
25 13
12
(vi)
52°
2.5
2.5
42° 7
Similarity
5
2.5
41
Std. X: Geometry Solution: i. The given two triangles are similar. Reason: AB 4.6 2 = = 2.3 1 PQ BC QR
=
AC
10 5
2
=
8
1 2
= = PR 4 1 In ∆ABC and ∆PQR, AB BC AC = = PQ QR PR
∴
TARGET Publications
A ----- (i)
P 4.6
----- (ii)
2.3
4
Q
----- [From (i), (ii) and (iii)]
∆ABC ∼ ∆PQR
----- [By S−S−S test of similarity] Hence, the given two triangles are similar. A
D 5
iv. The given two triangles are not similar. Reason: AB 24 2 = = PQ 12 1 BC QR AC PR AB PQ
= =
≠
7
3
25 13 BC QR
AC PR
B
A
P
60°
Q
60°
B
C
A P
----- (i) 24
----- (iii)
≠
3
C
25 13
12
----- (ii)
5
5 O
iii. The given two triangles are similar. Reason: In ∆APQ and ∆ABC, ----- [Each is 60°] ∠APQ ≅ ∠ABC ----- [Common angle] ∠PAQ ≅ ∠BAC ----- [By A−A test of similarity] ∴ ∆APQ ∼ ∆ABC Hence, given two triangles are similar.
∴
C
10
----- (iii)
ii. The given two triangles are similar. Reason: OA 5 = ----- (i) OB 3 5 OD = ----- (ii) OC 3 In ∆AOD and ∆BOC, OA OD = ----- [From (i) and (ii)] OB OC ----- [Vertically opposite angles] ∠AOD ≅ ∠BOC ----- [By S−A−S test of similarity] ∴ ∆AOD ∼ ∆BOC Hence, the given two triangles are similar.
∴
R
5
B
8
B
7
C Q
5
R
----- [From (i), (ii) and (iii)]
∆ABC is not similar to ∆PQR. Hence, the given two triangles are not similar.
42
Similarity
Std. X : Geometry
TARGET Publications
2.5 v. The given two triangles are similar. A D Reason: In ∆ABD and ∆CBD, 2.5 2.5 seg AB ≅ seg BC ----- [Each is 2.5 units] seg AD ≅ seg CD ----- [Each is 2.5 units] seg BD ≅ seg BD ----- [Common side] B C 2.5 ----- [By S−S−S test of congruency] ∴ ∆ABD ≅ ∆CBD ----- [Two congruent triangles are similar to each other] ∴ ∆ABD ∼ ∆CBD Hence, the given two triangles are similar. D vi. The given two triangles are not similar. Reason: 52° m∠ADO = 52° O B A m∠BCO = 42°
∠ADO ≠ ∠ BCO ∴
42°
Corresponding angles of ∆ADO and ∆BCO are not congruent. ADO is not similar to BCO.
C
2.
A triangle ABC with sides AB = 6 cm, BC = 12 cm and AC = 8 cm is enlarged to PQR such that its largest side is 18 cm. Find the ratio and hence, find the lengths of the remaining sides of PQR. Solution: P
A 6cm
B
8cm
12cm
∆ABC ∼ ∆PQR ∴
AB PQ 6 PQ
∴
6 PQ
= = =
BC QR 12 18
=
=
8 PR
=
AC PR
C
Q
18 cm
R
----- [A figure and its enlarged figure are similar] ----- [Corresponding sides of similar triangles]
8 PR 2 3
----- (i)
Ratio of the sides is 2:3
6 PQ
∴
=
PQ =
2
----- [From (i)]
3 3× 6 2
PQ = 9 cm
8 PR
∴
=
PR =
2 3 8×3
----- [From (i)]
2
PR = 12 cm
Similarity
43
Std. X : Geometry
TARGET Publications
In ∆ABC and ∆DEF,
∴
AB
∴
∆ABC ∼ ∆DEF ∠C ≅ ∠F In ∆ABC, ∠A + ∠B + ∠C = 180° 80° + 60° + ∠C = 180° ∠C = 180° − 140° ∠C = 40°
∴ ∴ ∴
DE
=
BC EF
=
CA FD
----- [By S−S−S test of similarity] ----- (iv) [Corresponding angles of similar triangles] ----- [Sum of the measures of all angles of a triangle is 180 °.]
----- (v) ----- [From (iv) and (v)]
F = 40 10.
----- [From (i), (ii) and (iii)]
A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the shadow of length 40 m on the ground. Determine the height of the tower.
Solution:
P
Let AB represent the vertical stick, AB = 12 m.
A
BC represents the shadow of the stick, BC = 8 m. PQ represents the height of the tower.
12 m
QR represents the shadow of the tower. QR = 40 m
B
∆ABC ∼ ∆PQR ∴
AB
∴
12
∴
PQ PQ
= =
BC QR
=
AC PR
8m
C
Q
40 m
R
----- [c.s.s.t.]
8 40
PQ = 12 × 5 = 60 Height of the tower = 60 m
11.
In each of the figure an altitude is drawn to the hypotenuse. The lengths of different segments are marked in each figure. Determine the value of x, y, z in each case.
A
Solution: i.
In ∆ABC, m∠ABC = 90°
----- [Given]
seg BD ⊥ hypotenuse AC
----- [Given]
∴ ∴
y = 4 × 5
∴
y =
∴
y = 2 5
2
BD
= AD × DC
D
x
5 y
----- [By property of geometric mean]
2
4×5
4
B
z
C
----- [Taking square roots]
In ∆ADB,
∴
m∠ADB = 90° AB2 = AD2 + BD2 AB2 = 42 + y2
Similarity
----- [Given] ----- [By Pythagoras theorem] 49
Std. X: Geometry
∴ ∴ ∴
x = 4 + (2
∴
x = 6
∴ ∴ ∴ ∴ ∴ ∴ ∴ ii.
∴ ∴ ∴ ∴ ∴ ∴ ∴
∴ ∴
2
2
----- [By Pythagoras theorem]
z2 = (2 5 )2 + 52 z2 = 20 + 25 z2 = 45
9× 5
z=
----- [Taking square roots]
z= 3 5 In ∆PSQ, m ∠PSQ = 90° PQ2 = PS2 + QS2 62 = 42 + y2 36 = 16 + y2 2 y = 36 − 16 2 y = 20
4× 5
y =
----- [Given] ----- [By Pythagoras theorem]
P 4 S
6
x y
Q ----- [Taking square roots]
z
R
y = 2 5
In ∆PSQ, m ∠PSQ = 90° seg QS ⊥ hypotenuse PR QS2 = PS × SR 2 y = 4 × x 2
∴ ∴
20 = 4 x 20 x = 4
∴
x = 5
50
----- [Taking square roots]
In ∆BDC, m∠BDC = 90° BC2 = BD2 + CD2 z2 = y2 + 52
(2 5 )
∴
5 )2
2 x =16 + 20 2 x = 36
∴
∴ ∴ ∴ ∴ ∴ ∴
TARGET Publications
----- [Given] ----- [Given]
= 4 x
In ∆QSR, m ∠QSR = 90° QR 2 = QS2 + SR 2 z2 = y2 + x2
----- [Given] ----- [By Pythagoras theorem]
z2 = (2 5 )2 + (5)2 z2 = 20 + 25 z2 = 45 z=
9× 5
----- [Taking square roots]
z=3 5
Similarity
Std. X: Geometry 128
TARGET Publications
∴
MR 2 =
∴
MR = 8 units
----- [Taking square roots]
QR
= 2MR
----- [M is the midpoint of seg QR.]
QR
= 2 × 8 = 16 units
∴
∴
2
MR 2 = 64
QR = 16 units 14.
P
From the information given in the adjoining figure, Prove that: PM = PN = 3 a, where QR = a.
a
a
Proof: In ∆PQR, M
a
QM = QR = a
----- [Given]
∴
Q is midpoint of seg MR.
----- [By definition]
∴
seg PQ is the median
----- [By definition]
∴
PM2 + PR 2 = 2PQ2 + 2QM2
----- [By Appollonius theorem]
∴
PM2 + a2 = 2a2 + 2a2
∴
PM2 + a2 = 4a2
∴
PM2 = 4a2 − a2
∴
PM2 = 3a2
∴
PM = 3 a
Q
S a
a
R
N
Similarly, we can prove PN =
3a
PM = PN = 15.
3 a.
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 || BC.
D
Proof: DB = AB − AD
∴
DB = 5.6 − 1.4
∴
DB = 4.2 units
∴
AD DB
=
1.4 4.2
=
1 3
EC = AC − AE
∴
EC = 7.2 − 1.8
∴
EC = 5.4 units
∴
AE EC
=
1.8 5.4
=
1 3
----- [A−D−B]
A
1.8
E
C
7.2 ----- (i) ----- [A−E−C]
----- (ii)
In ∆ABC, AD DB
∴ 52
=
AE EC
seg DE || seg BC
----- [From (i) and (ii)] ----- [By converse of B.P.T.]
Similarity
Std. X: Geometry 18.
TARGET Publications
A
Let X be any point on side BC of ABC, XM and XN
M
are drawn parallel to BA and CA. MN meets
N
2
produced BC in T. Prove that TX = TB TC.
Proof:
T
In ∆TXM, seg BN || seg XM
∴
TN NM
=
B
C
X
----- [Given]
TB
----- (i) [By B.P.T.]
BX
In ∆TMC, seg XN || seg CM
∴
TN NM TB
∴
BX BX
∴
TB
= =
TX TX
TX TB 2
TX
=
----- [From (i) and (ii)]
CX CX
----- [By invertendo]
TX
TB
∴
----- (ii)
CX
BX + TB
∴
19.
=
----- [Given]
=
CX + TX TX
TC
----- [By componendo] ----- [T−B−X, T−X−C]
TX
= TB TC
Two triangles,
ABC and
D
DBC, lie on the same side of the
base BC. From a point P on BC, PQ || AB and PR || BD are
A
drawn. They intersect AC at Q and DC at R.
R Q
Prove that QR || AD.
B
Proof:
P
C
In ∆ABC, seg PQ || side AB
∴
CP PB
=
CQ AQ
----- [Given] ----- (i) [By B.P.T.]
In ∆BCD, seg PR || side BD CP PB
=
CR RD
----- [Given] ----- (ii) [By B.P.T.]
In ∆ACD, CQ AQ
=
CR RD
seg QR || seg AD 54
----- [From (i) and (ii)] ----- [By converse of B.P.T.]
Similarity
Std. X: Geometry 22.
TARGET Publications
The bisector of interior of exterior
A of ABC meets BC in D. The bisector
A meets BC produced in E. Prove that
BD
BE Construction: Draw seg CP || seg AE meeting AB at P. Proof:
=
CD CE
.
In ∆ABC, Ray AD is bisector of ∠BAC AB
∴
AC
=
BC CE
=
BD
BE
=
E
----- [Given]
BP
----- [B. P. T]
AP
CE
C
----- (i) [By property of angle bisector of triangle]
CD
BC + CE
D
----- [Given]
In ∆ABE seg CP || seg AE
∴
°
P B
F
A × ° ×
=
BP + AP
----- [componendo]
AP
AB
CE AP seg CP || seg AE, on transversal BF.
∠FAE ≅ ∠APC
----- (ii)
----- (iii) [corresponding angles]
seg CP || seg AE on transversal AC.
∠CAE ≅ ∠ACP Also, ∠FAE ≅ ∠CAE ∴ ∠APC ≅ ∠ACP In ∆APC, ∠APC ≅ ∠ACP AP = AC ∴ BE
∴
CE BD
∴
CD
= =
----- (iv) [alternate angles] ----- (v) [seg AE bisects ∠FAC] ----- (vi) [From (iii), (iv) and (v)] ----- [From (vi)] ----- (vii) [By converse of isosecles triangle theorem]
AB
----- (viii) [from (ii) and (vii)]
AC BE
----- [from (i) and (viii)
CE
∴
BD
23.
In the adjoining figure,
BE
=
CD
----- [alternendo]
CE
BC and
ABCD is a square. The BCE on side
F
ACF on the diagonal AC are similar to each other.
Then, show that A( BCE) =
1 2
D
C
A( ACF).
E
Proof:
ABCD is a square ∴
AC =
2 BC
----- (i) [∵ Diagonal of a square =
∆BCE ∼ ∆ACF ∴ ∴ 56
A(∆BCE) A(∆ACF) A(∆BCE) A(∆ACF)
= =
----- [Given]
A B 2 × side of square]
----- [Given]
(BC)
2
(AC) 2 (BC)2 ( 2.BC) 2
----- (ii) [By theorem on areas of similar triangles] ----- [From (i) and (ii)]
Similarity
Std. X : Geometry
TARGET Publications
A( ∆BCE)
∴
=
A( ∆ACF) A( ∆BCE)
∴
BC2 2BC 2 1
=
A( ∆ACF)
2 1
A( BCE) = 24.
2
A( ACF)
Two poles of height ‘a’ meters and ‘b’ metres are ‘p’ meters apart. Prove that the height ‘h’ drawn from the point of intersection N of the lines joining the top of each pole to the foot of the opposite pole is
ab a b
metres.
Proof:
∴ ∴ ∴ ∴
PQ
=
NT a h
=
a=
QR
P N b a
h R
In ∆PQR and ∆ NTR,
∠PQR ≅ ∠ NTR ∠PRQ ≅ ∠ NRT ∆PQR ∼ ∆ NTR
S
x
T p
y
Q
----- [Each is 90°] ----- [Common angle] ----- [By A.A. test of similarity] ----- [c.s.s.t.]
TR
p x
ph
----- (i)
x
In ∆SRQ and ∆ NTQ,
∠SRQ ≅ ∠ NTQ ∠SQR ≅ ∠ NQT ∆SRQ ~ ∆ NTQ ∴
SR
=
NT
∴
b
∴
b=
h
=
----- [Each is 90°] ----- [Common angle] ----- [By A−A test of similarity]
QR QT
p y
ph
----- (ii)
y
Consider, ab =
∴
ab =
ph
×
x
ph y
p2 h 2
----- (iii)
xy
Consider, a+b=
ph x
+
ph y
⎡1 1 ⎤ = ph ⎢ + ⎥ ⎣ x y ⎦ Similarity
57
Std. X: Geometry
∴
TARGET Publications
( y + x )
a + b = ph
----- (iv)
xy
Dividing (iii) by (iv) p 2 h 2 ab a+b
xy ph ( x + y )
=
xy
ab
∴
a+b ab
∴
a+b ab
∴
a+b
p2 h 2
=
xy
×
xy
ph( x + y )
ph
=
x + y
ph
=
----- [∵ x + y = p, R −T−Q]
p
ab
∴
h=
25.
In the adjoining figure, DEFG is a square and
a+b
Prove that: i.
AGF
DBG
ii.
AGF
EFC
iii.
DBG
EFC
iv.
2
BAC = 90 .
A G
DE = BD EC
B
Proof: i
∴
DEFG is a square
----- [Given]
seg GF || seg DE
----- [Opposite sides of a square]
seg GF || seg BC
----- (i) [B−D−E−C]
D
F
E
C
In ∆AGF and ∆DBG,
∴ ii
iii.
∠GAF ≅ ∠BDG
----- [Each is 90°]
∠AGF ≅ ∠DBG
----- [Corresponding angles of parallel lines GF and BC]
AGF
DBG
----- (ii) [By A−A test of similarity]
In ∆AGF and ∆EFC,
∠GAF ≅ ∠FEC
----- [Each is 90°]
∠AFG ≅ ∠ECF
----- [Corresponding angles on parallel lines GF and BC]
Since ∆AGF ∼ ∆EFC AGF EFC
----- (iii) [By A−A test of similarity]
DBG
EFC
∴
BD
∴
DG × FE = BD × EC
FE
=
DG EC
----- [From (ii) and (iii)] ----- [c.s.s.t.] ----- (iv)
But, DG = EF = DE
----- (v) [Sides of a square]
∴
----- [From (iv) and (v)]
DE × DE = DB × EC 2
DE = DB EC 58
Similarity
Std. X : Geometry
TARGET Publications
One‐Mark Questions
1.
In ABC and XYZ ,
AB YZ
=
BC ZX
=
AC XY
,
then state by which correspondence are ABC and XYZ similar.
Solution:
∆ ABC ∼ ∆XYZ by ABC ↔ YZX. 2.
T
In the figure, RP : PK = 3:2. Find
A
TRP
A
TPK
.
R
P
K
Solution: A ( ∆TRP ) 3 = A ( ∆TPK ) 2
3.
----- [Triangles with equal heights]
Write the statement of Basic Proportionality Theorem.
Solution: If a line parallel to a side of a triangle intersects the other sides in two distrinct points, then the line divides those sides in proportion.
4.
What is the ratio among the length of the sides of any triangle of angles 30 60
90 ?
Solution: The ratio is 1: 3 :2.
5.
What is the ratio among the length of the sides of any triangle of angles 45 45
90 ?
Solution: The ratio is 1:1: 2 .
6.
State the test by which the given triangles are similar.
A
C B Solution:
∆ABC ∼ ∆EDC by SAS test. Similarity
D
E 59
Std. X: Geometry 7.
TARGET Publications
In the adjoining figure, find
A
PQR
A
RSQ
P
.
R
Solution: A ( ∆PQR ) A ( ∆RSQ ) 8.
PQ
=
T
Q
----- [Triangles with common base]
ST
S
Draw a pair of triangles which are equal in areas.
A
Solution:
A(∆ABD) = A(∆ADC)
B
C
D
9. State the relation between diagonal of a square and its side. Solution:
Diagonal of a square =
2 × side.
10.
Adjacent sides of parallelogram are 11 cm and 17 cm respectively. If length of one diagonal is 26 cm, then using which theorem/property can we find the length of the other diagonal? Solution: We can find the length of the other diagonal by using Appollonius’ Theorem. 11.
A
In the adjoining figure,
60°
using given information, find BC.
Solution:
24
B
BC = =
3 2 3 2
× AC
30°
C
----- [side opposite to 60 °]
× 24
BC = 12 3 units 12.
Find the value of MN, so that A( ABC) = A( LMN). A P M N 4 cm
8 cm
B
D
C
L
5 cm
Solution: MN = 10 cm 60
Similarity
Std. X : Geometry
TARGET Publications
Additional Problems for Practice Based on Exercise 1.1 1.
P
In the adjoining figure, QR = 12 and SR = 4. Find values of i.
A ( ∆PSR ) A ( ∆PQR )
ii.
A ( ∆PQS) A ( ∆PQR )
iii.
A ( ∆PQS) A ( ∆PSR )
Q
R
4
S 12
2.
The ratio of the areas of two triangles with the equal heights is 3 : 4. Base of the smaller triangle is 15 cm. Find the corresponding base of the larger triangle.
3.
In the adjoining figure, seg AE ⊥ seg BC and seg DF ⊥ seg BC. Find i.
A ( ∆ABC ) A ( ∆DBC )
ii.
A ( ∆DBF ) A ( ∆DFC )
A D iii.
A ( ∆AEC ) A ( ∆DBF )
B
F
C
E
Based on Exercise 1.2 4.
5.
6.
7.
A
In the adjoining figure, seg EF | | side AC, AB = 18, AE = 10, BF = 4. Find BC.
E B
A
In the adjoining figure, seg DE | | side AC and seg DC | | side AP. BE BC Prove that = EC CP In the adjoining figure, PM = 10, MR = 8, QN = 5, NR = 4. State with reason whether line MN is parallel to side PQ or not ? In the adjoining figure, Ray AD is the angle bisector of ∠BAC of ∆ABC. From the given information find value of x.
D
B P
E
P
C
10 M 8 Q
N 4
5
R
B 10 D
15
x
A 8.
C
F
18
C
Bisectors of ∠B and ∠C in ∆ABC meet each other at P. Line AP cuts the side BC at Q. AP AB + AC Then prove that = . PQ BC
Similarity
61
Std. X: Geometry
TARGET Publications
Based on Exercise 1.3 9.
L
In the adjoining figure,
∆MPL ∼ ∆ NQL,
10.
P
Q
MP = 21, ML = 35, NQ = 18, QL = 24. Find PL and NL.
M
N
P
In the adjoining figure,
∆PQR and ∆RST are similar under PQR ↔ STR, PQ = 12, PR = 15, QR
3
T
Q
= . TR 2 Find ST and SR.
R S
11.
In the map of a triangular field, sides are shown by 8 cm, 7 cm and 6 cm. If the largest side of the triangular field is 400 m, find the remaining sides of the field.
12.
∆EFG ∼ ∆RST and EF = 8, FG = 10, EG = 6, RS = 4. Find ST and RT.
13.
In ABCD,
A
[Oct 09]
D
side BC | | side AD. P
Diagonals AC and BD intersect each other at P. If AP =
1 3
AC, then prove that DP =
1 2
BP.
B
C
Based on Exercise 1.4 QR
14.
If ∆PQR ∼ ∆PMN and 9A( ∆PQR) = 16A(∆PMN), then find
15.
∆LMN ∼ ∆RST and A(∆LMN) = 100 sq. cm, A( ∆RST) = 144 sq. cm, LM = 5 cm. Find RS.
16.
∆ABC and ∆DEF are equilateral triangles. A( ∆ABC):A(∆DEF) = 1:2 and AB = 4 cm. Find DE.
17.
If the areas of two similar triangles are equal, then prove that they are congruent.
18.
In the adjoining figure,
MN
.
A E
seg DE | | side AB, DC = 2BD, A(∆CDE) = 20 cm2. Find A(ABDE). 62
B
D
C
Similarity
Std. X : Geometry
TARGET Publications
Based on Exercise 1.5 19.
P
In the adjoining figure, ∠PQR = 90°, seg QS ⊥ side PR. Find values of x, y and z.
8
y
S 10
x
Q 20.
R
z
B
In the adjoining figure, ∠ABC = ∠ADC = 90°, seg BD ⊥ seg AC. Prove that: E is the midpoint of seg BD.
A
C
E D
21.
In the adjoining figure, ∠PRQ = 90°, seg RS ⊥ seg PQ. Prove that :
22.
23.
24.
PR 2 QR 2
=
R
PS QS
P P
In the adjoining figure, ∠PQR = 90°, ∠PSR = 90°. Find: i. PR and ii. RS
In the adjoining figure, ABCD is a trapezium, seg AB || seg DC, seg DE ⊥ side AB, seg CF ⊥ side AB. Find: i. DE and CF ii.
Q
S 48
S
30 Q
R
40
D
7
C 17
10 A BF
iii.
6
E
B
F
AB.
Starting from Anil’s house, Peter first goes 50m to south, then 75m to west, then 62 m to North and finally 40 m to east and reaches Salim’s house. Then find the distance between Anil’s house and Salim’s house.
Based on Exercise 1.6 25.
In the adjoining figure, ∠S = 90°, ∠T = x°, ∠R = ( x + 30)°, RT = 16. Find: i. RS ii.
R x+30°
ST
x°
S
Similarity
16
T 63
