GEOMETRY
MT EDUCARE LTD.
EXTRA HOTS SUMS CHAPTER : 1 - SIMILARITY
1.
Bisectors of
B and
C in
ABC meet meet each each other at P. Line AP AP cuts the AP A B + AC side BC at Q. Then prove that : = (5 marks) PQ BC Proof : In ABQ, A ray BP bisects ABQ [ G i ve n ] proper erty ty of an ang angle By prop AP AB P = ........(i) bisect cto or of a tria triang ngle le PQ BQ bise x x In ACQ, C B Q ray CP bisects ACQ [ G i ve n ] proper erty ty of an angl angle e By prop AP AC = ........(ii) bise ctor or of a tria triang ngle le PQ CQ bisect AP AB AC = = [From (i) and (ii)] PQ BQ CQ AB + AC AP = [By Theorem on equal ratios] BQ+CQ PQ AP AB + AC = [ B - Q - C] PQ BC
•
•
In PQR, PQR = 900, As shown in figure, seg QS side PR. seg QM is angle bisector of PQR. PM² PS Prove that : (5 marks) MR² SR Proof : In PQR, seg QM bisects PQR [ G i ve n ] PQ PM = QR MR 2 PQ² PM = ...... ......... ...(i) (i) QR ² MR2 In PQR, m PQR = 90º seg QS hy hypotenuse PR PQR ~ PSQ ~ QSR .... ...... .... ...(i .(ii) i) 2.
P S M
=
PSQ ~ PQR PQ PS = QR PQ PQ² = PR × PS Also, QSR ~ PQR SR QR = QR PR Q R ² = PR × SR PR × PS PM2 = PR × SR MR2 PS PM2 = SR MR2
Q
R
[Prope roperty of an angle bisec isec tor of a triangle] [Squ [Squar arin ing g both both sides sides]] [ G i v e n] [ G i v e n] [Th [Theore eorem m on simi simila lari rity ty of right angled triangles] [F ro m (ii )] [Correspon [Correspondin ding g sides
of simil similar ar tria triang ngle les] s] ... ....(i i i ) [F ro m (ii )] [Correspon [Correspondin ding g sides
of simil similar ar tria triang ngle les] s] .. .... .. (iv) [From (i), (iii) and (iv)]
341
GEOMETRY
3.
MT EDUCARE LTD.
Find Find the the radiu radius s of of a circle circle draw drawn n by a comp compas ass s when when angle angle bet betwe ween en two two 0 arms of compass is 120 and length of each arm is 24cm. (5 marks)
Sol.
B
2 In the adjoining figure, 4 120 m c m c seg AB and seg BC represents the 2 4 arms of compass. In ABC, A C D side AB side BC [ G i v e n] BAC BCA ........(i ) [ Isosc ele s tr tr iangle th th e o r e m ] 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 = 18 180º [F ro m (i )] 2 m BAC BAC = 180 180ºº – 120º 120º 2 m BAC = 60 60º m BAC = 30 30º ...... .(i i ) Draw seg BD side AC, A - D - C. In ABD, m BAD = 30º [From (ii) and A - D - C] m ADB = 90 90º [ G i ve n ] m ABD = 60º [ R e m ai n i n g a n g l e ] ABD is a 30º - 60º - 90º triangle. triangle. By 30º - 60º - 90º triangle theorem. 0
AD =
3 AB 2
3 2 AD = 12 Similarly, DC = 12 AD =
[Side opposite to 60º]
× 24
3 cm we can get 3 cm
AC =
AD + DC
AC =
12 3 + 12 3 24 3 cm.
AC =
[ A - D - C]
The radius of the circle is 24 3 cm. 4.
In BAC, BAC = 90º, segments AD, seg BE and seg CF are medians. B Prove : 2 (AD² + BE² + CF² ) = 3BC². (5 marks)
Proof : F
In BAC, seg AD is median on side BC. AB² AB² + AC² AC² = 2AD² 2AD ² + 2BD² 2BD² AB² AB ² + AC² AC² – 2BD² 2BD² = 2AD² 2AD²
D
[ Gi Gi ve ve n] n] A E [By Appollonius theorem]
C
2
1 AB² AB ² + AC² AC ² – 2 BC = 2AD² 2AD² 2 1 AB² AB² + AC² – 2 BC2 = 2AD² 2AD² 4 1 AB² AB² + AC² AC² – BC² BC ² = 2AD² 2AD² 2 342
[ D is the midpoint of seg BC]
.........(i)
GEOMETRY
MT EDUCARE LTD.
Multiplying throughout by 2, we get 2AB2 + 2AC 2 – BC2 = 4AD2 Similarly, we can prove that 2AB² + 2BC² – AC² = 4BE² .. .. . . . . . ( i i ) 2AC² + 2BC² – AB² = 4CF² ........( ii i ) Adding (i), (ii) and (iii) we get 2AB 2AB2 + 2AC2 – BC2 + 2AB² + 2BC² 2BC² – AC² AC² + 2AC² + 2BC² 2BC² – AB² = 4AD 4AD 2 + 4BE2 + 4CF2 3AB2 + 3AC2 + 3BC2 = 4AD2 + 4BE2 + 4CF2 3 (AB2 + AC2 + BC2) = 4 (AD 2 + BE2 + CF2) ..... ...... ...( .(iv iv)) In BAC, m BAC = 90º [Given] 2 2 2 BC = AB + AC ........(v) [By Pythagoras theorem] 2 2 2 2 2 3 (BC + BC ) = 4 (AD + BE + CF ) [From (iv) and (v)] 2 2 2 2 3 × 2 BC = 4 (AD + BE + CF ) 6BC2 = 4 (AD 2 + BE2 + CF2) 3BC2 = 2 (AD 2 + BE2 + CF2) [Dividing throughout by 2] 2 (AD² + BE² + CF²) = 3BC² 5.
In ABCD ABCD points points P, Q, R and S lies lies on sides sides AB, BC, BC, CD and AD respec respecti tivel vely y such that seg PS || seg BD || seg QR and seg PQ || seg SR. Then prove D that seg PQ || seg AC. (5 marks) S A
Proof :
seg PS || seg BD On transversal AB, APS AB D . .. . .. . ( i ) of corresponding In APS and ABD, PAS BAD APS AB D APS ABD AP PS AS = = . .....( i i ) AB BD AD seg QR || seg BD On transversal BC, CQR CB D . ......( i ii ) In CQR and CBD, QCR BC D CQR CB D CQR CB D CQ QR CR = = .. .. .. .( iv) CB BD CD In PQRS, seg PQ || seg RS seg PS || seg QR PQRS is a parall elog ram PS = QR .. .... .. (v)
CQ = CB AP = AB AB = AP
PS CR = BD CD CQ CB CB CQ
.. .. ..( vi)
[G i ve n] P
R
[C o nve r se angles test] B
Q
C
[Com mon angl es] [F ro m (i )] [By A-A test of similarity] [ c . s .s .t .] [G i ve n] [Converse of corresponding angles test] [Com mon angl es] [Fro m (iii )] [By AA test of similarity] [ c . s .s .t .] [G i ve n] [G i ve n] [B y de fi nit io n]
[From (iv) and (v)] [From (ii) and (vi) [By Invertendo] 343
GEOMETRY
MT EDUCARE LTD.
CB – CQ A B – AP = CQ AP BQ BP = ... .. .(vii) CQ AP In ABC, BQ BP = CQ AP seg PQ || seg AC 6.
In ABC, m BAC = 90º. seg DE side AB, seg DF
[By Dividendo] [ A - P - B and B - Q - C]
[From (vii)] [By converse of B.P.T.] A
side AC,
F
prove A ( AEDF) = AE × EB × AF × FC (5 marks) E Proof : In ADB, m ADB = 90º [ G i ve n ] B C seg DC si s ide AB [ G i ve n ] D DE2 = AE × EB ......( i) [By property of geometric mean] In ADC, m ADC = 90º [ G i ve n ] seg DF si side AC [ G i ve n ] 2 DF = AF × FC .......( ii ) [By property of geometric m e a n ] 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 [Given]] m EDF = 90º [ R e m ai n i n g a n g l e ] AEDF is a recta ngle [B y de fi nit io n] A ( AEDF) = DE ×DF......(iv) From (iii) and (iv), A ( AEDF) = AE × EB × AF × FC [From (iii) and (iv)] CHAPTER : 2 - CIRCLE
1.
From From the the end end poi point nts s of a diame diamete terr of circle circle perpen perpendi dicul cular ars s are are drawn 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 cir circl cle e wit with h cen centr tre e O. O. (ii) seg AB is is the diameter diameter of the the circle. circle. D (iii) Line Line l is tangent to the circle at point C. A (iv) (iv) seg AD line l . (v ) seg BE BE line l . C To Prove : OD : OD = OE. O Construction : Draw : 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 [Perp Perpe endic ndicul ular ars s drawn awn to the the sam ame e line are parallel to each other] On transversal AB and DE, AO DC = ...... ......... ...(i) (i) [By [By pro prope pert rty y of of int inter erce cept pts s mad made e by by OB CE three parallel lines]
344
GEOMETRY
MT EDUCARE LTD.
But, AO = OB AO = 1 OB DC = 1 CE D C = CE In OCD and OCE , seg OC seg OC OCD OC D OC E seg DC seg CE OCD OC D OC E seg OD seg OE OD = O E
[Radii of t he same circle] . . . . . . .. .( i i ) [F rom (i ) and (i i)] .........( ii i) [C om mo n si de] [Each is a righ t angl e] [F rom (ii i)] [B y SAS te st of co ngruen ce ] [ c . s .c .t .] C
2.
The The bise bisect cto ors of the the angle gles A,B of ABC ABC intersect in I, the bisectors of the corresponding exterior angles intersect in E. Prove that AIBE is cyclic. (5 marks) Proof : A P
)
I )
• • ×
B ×
Q
Take points P and Q as shown in the figure. E m CAB + m BAP = 180º [Linear pair ax 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 . .... (i i ) m IAE + m IBE = 90º + 90º [Adding (i (i) an and (i (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, m PAC = m ABC = y m PAE = m PAC + m CAE m PAE = (y + x) . .. . . . . . ( i i i ) PEA is a exterior angle of ABE, m PEA = m ABE + m BAE m PEA = m ABC + m BAE m PEA = (y + x) .. .. .. .. (i v) In PAE, PAE PEA seg AP seg PE
[ Ray AE bisects BAC] .. .. . . . . .( i ) [Angles in altern ate segme nt ] .. .. . . . . . ( i i ) [Angle addition propert y] y] [From (i (i) an and (i (ii)] [Remote in interi or or an angle th theo re re m] m] [ B - E - C] [From (i) and (ii)] [From (iii) and (iv)] [Converse of isosceles triangle theorem] 345
GEOMETRY
MT EDUCARE LTD.
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. seg AJ seg DJ .... .. ....( i) i) [The lengths of two tangent segments from an external point to a circle are equal] 0 m ODM = 90 ...... ... ....(i .(i i) [Rad [Radiu ius s is perp perpen endi dicu cula larr to the the tang tangen ent] t] In OBD, seg OB seg OD [Radii of the same circle] OBD OD B [ I s o s c e l e s t r i an g l e t h e o r e m ] Let, Let, m OBD = m ODB = xº ......(iii) m ODM = m ODB + m BDM BDM [Ang [Angle le addi additi tion on prop proper erty ty]] 90 = x + m BDM [From (ii) and (ii i)] m BDM = (90 – x)º .. .. ..( iv) But, BDM JDC ...... .(v) [Verticall [Ver tically y opposi te angles] angles ] m JDC = (90 – x)º ..... .(vi) .(vi ) [From (iv) and (v)] In BAC, m BAC = 90º [Radius is perpendicular to ta tangent ] m ABC = xº [From (iii) and A - O - B, B - D - C] m ACB = (90 – x)º [ R e m a i n i n g an g l e ] m JCD = (90 – x)º ..... ..(vii) ..(v ii) [B - D - C and A - J - C] In JDC, JCD JDC [From (vi) and (vii)] seg seg DJ seg CJ .... ...... ..(v (vii iii) i) [Converse of Isosceles triangle theorem] seg AJ seg CJ [From (i) and (viii)] 5.
If fou fourr tang tangen ents ts of of a cir circle cle dete determ rmin ine e a rec recta tang ngle le the then n show show that that it must be a square. square. (5 marks) A
P
B
Given : (i) Lines AB, BC, CD and and AD 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 D R Proof : AP = AS ..........( i) [The lengths of the two BP = B Q .........( i i) tangent segments from an CR = CQ . ........( i ii ) external point to a circle DR = DS .. ... .. ..(i v) 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 rect angl e [ G i ve n ] AB = C D .. ... .. .. .( vi) [ Opposite sides of a rectangle are AD = B C .. .. .. .. .( vi i) congruent] 346
GEOMETRY
MT EDUCARE LTD.
AB + AB = BC + BC 2 AB = 2 BC AB = B C ABCD is a square
[From (v), (vi) and (vii)]
[A rectangle in which adjacent sides are congruent, is a square]
6.
Two Two con conce cent ntri ric c cir circl cles es wit with h cen centr tre e O. Se Seg g AB, AB, seg seg BC BC a and nd seg seg AC are 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, seg OP se seg AB .......( i) seg OQ se s eg BC . . .. . . . ( i i ) With respect to bigger circle, seg OP ch chord AB AP = BP .. ... ( i i i ) seg OQ ch chord BC BQ = QC
.. .. .(iv)
In ABC, P and and Q are are midp midpoi oint nts s of side sides s AB and BC respectively. seg PQ || seg AC 1 AC PQ = 2
B
[Radius is perpendicular to the tangent ] [F ro m (i )] [Perpendicul ar draw n from ce nt re of circle to chord bisects the chord] [F ro m (i i) Perp endi cular dr aw n from ce ntre of circle to chord bisects the chord] [Fro [From m (iii (iii)) and and (iv) (iv)]]
[By Midpoint theorem]
CHAPTER : 3 - GEOMETRIC GEOMETRIC CONSTRUTION
1.
Point I is the the inc 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 [Ang [Angle le ad additi dition on pro prope pert rty] y] In BIC, m BIC + m IBC + m ICB = 180º 120 + x + y = 180 x + y = 180 – 120 x + y = 60 ...... ... ....(i .(i ) In m
ABC, 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 can be constructed median AP.
= 180 = 180 = 180 = 180 [F rom (i )] = 180 = 180 – 120 = 60º with base BC, vertical angle
BAC and
347
GEOMETRY
MT EDUCARE LTD.
(Rough Figure)
A
A
A 60º 60º 3
I
O
c m
120º
B
•
30º
30º B
120º
•
P 4 cm
P
× ×
C
C
2.
Pointt O is the Poin the orth rthoc ocen entr tre e of ABC ABC, m BOC = 105º, seg AD seg BC. BC = 6.5 cm and AD = 3.5 cm. Draw Draw ABC. (5 marks) Analysis : BOC EOF [V er ti call y opps oi te angl es] 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. B
A
A
D 6.5 cm
Y
75º 3.5 cm
3.5 cm P 150º B
348
15º 6.5 cm
15º C
X
C
GEOMETRY
MT EDUCARE LTD.
3.
Cons Consttruct ruct ABC such such that that BC = 8.8 cm, cm, ABC is 2.2 cm.
B = 50º, radius of incircle of (5 marks)
(Rough Figure) A
I
A
B
Q
5
50º
2.2 cm M 8.8 cm
C
P
4 . 8
c m
B
c m 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 Draw a lin line e par paral alle lell to to sid side e BC BC at at a dist distan ance ce of 2.2 2.2 cm cm fro from m BC. BC. Point Point of interse intersecti ction on of line line para parallel llel to BC and angle angle bisect bisector or is incent incentre. re. Let incentre be I. From From I, draw draw seg seg IM side side BC BC.. seg seg IM is in-r in-rad adiu ius. s. Draw Draw inci incirc rcle le with with IM as radi radius us to touc touch h sid sides es AB an and d BC. BC. From From C dra draw w the the tang tangen entt to to the the circ circle le to meet meet ray ray BA BA at at A. A. Draw Draw a sect sector or O-AX O-AXB B wit with h radi radius us 7 cm cm and and m (arc (arc AXB) AXB) = 50º 50º.. Draw Draw a circle touching the sides OA and OB and also the arc. (5 marks)
(Rough Figure)
A
X
c m 7
I
50º O
7 cm
B
349
GEOMETRY
MT EDUCARE LTD.
P A
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 draw the the PQ ray ray OX to cut cut ray ray OA at P an and d ray ray OB at Q. Draw bisector of Q. It intersects ray OX at I. Draw Draw inci incirc rcle le with with I as as cen centr tre e and and IX as radi radius us.. This circle touches the ray OA, ray OB and also arc AXB.
5.
In ABC, ABC, BC = 5.8 5.8 cm cm,, seg seg BP CQ = 4.8 cm. Construct ABC.
seg seg AC, AC, seg seg CQ
seg seg AB, AB, BP = 5 cm , (5 marks)
(Rough Figure) A
Q
A
5
B
I
B
350
c m
M 8.8 cm
2.2 cm
C
4 . 8
5.8 cm
Y
2.2 cm • 50º •
P
X
c m
C
GEOMETRY
MT EDUCARE LTD.
Steps of construction : 1. 2. 3. 4. 5.
6.
Dra Draw seg seg BC of leng length th 5.8 cm. Draw Draw a semi semici circ rcle le with with seg seg BC BC as as the the dia diame meter ter.. Taking Taking B as the the cent centre re and and radi radius us 5 cm cut cut an an arc arc on the semi semicir circle cle to to get get point P and draw seg BP. Taking Taking C as the centre centre and radius radius 4.8 cm cut cut an an arc arc on the semicir semicircle cle to get point Q and draw seg CQ. Exte Extend nd seg seg BQ and and seg seg CP to inte inters rsec ectt at at poi point nt 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 through point P. (5 marks)
(Rough Figure) P
3 c m
P m c N 5
3 c m
O 3 cm
m c 5 N
M
O
T
3 cm
l
M
T
Steps of construction : 1. 2. 3. 4. 5. 6. 7. 8.
Draw line l . Take a po point M on li line l and and draw a perpendicular to line l at at point M. With point point M as as the the centr centre, e, cut cut an arc of radi radius us 5 cm on the perp perpend endicu icular lar to get point P. With With point point M as the the cent centre re and and rad radiu ius s 3 cm cut cut an an arc on seg seg PM PM to get get point point N. Draw Draw a line line m per perpe pend ndic icul ular ar to lin line e PM PM at at poi point nt N. With point point P as as the the centre centre cut an arc arc of of radiu radius s 3 cm cm on line line m to get get poin pointt O. With point point O as the the centre centre and and seg OP as the the radius radius,, draw draw the requir required ed circle circle.. l Draw Draw a per perpe pend ndic icul ular ar from from poin pointt O to to lin line e to to get point T. CHAPTER : 4 - TRIGONOMETRY
1.
If
Proof :
1 + x 2 sin = x, prove that tan2 + cot2 = x2 +
1 + x 2 sin
=
1 x 2
.
(5 marks)
x
x
sin
=
sin 2
=
1 + x2 x 2 1 + x2
[Squaring both sides] 351
GEOMETRY
MT EDUCARE LTD.
sin2 cos2
+ cos2
= =
cos2
=
cos2
=
cos2
=
tan
1 1 – sin2
2
1–
1 + x2 1
1 + x2 sin2 cos 2 x 2
=
1 x x 2
=
=
1
1 x2 1 x2 1
2
1 x2 x2 1
=
tan2 1
= L .H . S.
1 + x2
1 x2 – x2
=
cot2
x 2
=
x 2 tan2
=
x2 +
+ cot2 1
=
x 2 R.H .S .S.
=
x2 +
1 tan2
+ cot2
tan 2.
Prove :
Proof :
1 – co cot
L.H.S. = = = = =
cot
+
sin cos cos
2
= 1 + tan
+
= = = 352
cos
(5 marks)
cot 1 – tan
cos 1 – sin sin – cos sin
cos si n 1 – s i n c o s cos cos – cos sin
cos si n
si n sin – cos sin2 (sin – cos )
sin (sin
– cos )
–
sin cos cos – sin
sin
cos 2 (cos – sin )
sin
cos2 (sin – cos )
2
=
+ cot
1 – ta tan
tan 1 – cot
sin cos sin cos
x
(sin
sin2 cos2 – – cos ) cos si n sin3 – cos3 1 – cos ) cos × sin
(sin
1 (sin – cos )
1 (sin
– cos ) (sin2 sin . cos cos sin
cos 2 )
GEOMETRY
MT EDUCARE LTD.
sin2
=
+ si sin . co cos + co cos 2 cos . si s in
sin2 cos . sin
=
sin cos
si n cos 1 cos si n = tan + 1 + cot = 1 + tan + cot = R.H .S. tan cot + = 1 + tan 1 – co cot 1 – ta tan
. cos . sin
cos 2 cos . sin
=
+ cot
3. Prove : sin8 – cos8 = (sin2 – cos2 ) (1 – 2 sin2 cos2 ). (5 marks) 8 8 Proof : L . H . S . = si n – cos 4 = (sin )2 – (cos 4 )2 = (sin4 – cos4 ) (sin4 + cos 4 ) = (sin2 – cos2 ) (sin2 + cos2 ) (sin4 + cos4 ) = (sin2 – cos2 ) (sin4 + cos 4 ) [ sin2 + cos 2 = 1] = (sin2 – cos2 ) (sin4 + cos4 + 2sin2 cos2 – 2sin2 cos cos2 ) = (sin2 – cos 2 ) [(sin2 + cos 2 )2 – 2 sin 2 cos2 ] = (sin2 – cos 2 ) (1 – 2sin 2 cos2 ) [ sin2 + cos 2 = 1] = R.H .S. 8 sin – cos8 = (sin 2 – cos2 ) (1 – 2 sin2 cos2 ). 4.
Sol.
A 1.5 1.5 m tall tall boy boy is is sta standi nding ng at at som some e dist distanc ance e from from a 30 m tall tall buil buildi ding. ng. 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
[By definition]
28.5 GF
3
=
GF
=
GF
=
GF
=
GF
= 9.5 3 m
28.5 3 28.5
3
28.5 3
3 3 3
353
GEOMETRY
MT EDUCARE LTD.
In right angled BGD, BG tan 30 30º = GD 28.5 1 = 9.5 3 x 3
[By definition]
9.5 3 x = 28.5 3 x
= 28.5 3 – 9.5 3
x
= 19 3
Distance he walked towards the building is 19 3 m. 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 + cosec 2 A + cos2A + 2cos A . sec A + sec2 A = (sin2 A + cos 2 A) + (cosec 2 A) + (sec 2 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 + cot 2 A + 1 + tan 2 A + 2 + 2 = 7 + tan2 A + cot 2 A = R .H . S . (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A.
6.
1 – sin Prove that : cos (sec
Proof :
L .H . S.
cos
=
cos
1 cos sin sin
cos
=
1 sin – cos sin sin
– cos
=
sin
=
R .H . S . 1 – sin cos (sec
tan A 7.
sec A - 1
–
354
sin3
+ co cos 3
– cos . co cos
×
×
×
(s i n (s i n
si n
– cos
cos ) (sin – si n
= sin
(5 marks)
– cos ) . cos
cos 2 )
– cos )
(1 – sin
– cos
– cos 2
cos ) (sin2
(s i n
× si sin
. cos sin 2 × – cosec ) sin 3
. co cos )
– cos 2 + co cos 3
[ sin2 + cos 2
= 1]
= sin
tan A +
s ec A + 1
= 2 cosec A
tan A Proof :
– cos 2
– cosec )
1 si n
. cos )
=
si n
sin2
. cos
(sec
. cos
( 1 – si n
=
+ co cos 3
1 – sin
1 – si n =
. cos sin 2 × – cosec ) sin 3
L . H . S. =
sec A - 1
t an A +
sec A + 1
(5 marks)
GEOMETRY
MT EDUCARE LTD.
=
=
= = = =
sinA sinA cosA cosA 1 1 + -1 +1 cos A cos A sinA sinA cosA cosA (1 – cos A ) + (1+ cosA ) cos A cos A sinA sinA + 1 – cos A 1+cosA
1 1 + sin A cos A 1 + co cos A 1 – co 1 + cos A + 1 – cos A sin A (1–cos A) (1+co (1+cos s A) (1–cos 2 sin A 1 – cos² A
=
sin² A
sin A
c os² A = 1 sin² A + co 1 - co cos² A = si sin² A
2
2 sinA = 2 cosec A = R.H.S. tan A tan A + = 2 cosec A sec A - 1 s ec A + 1 =
6.
Sol.
From From the the top top of a light light hous house, e, 80 80 metr metres es high, high, two two ship ships s on on same same side side of 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 the lighthouse are in a line). (5 marks) E A In the adjoining figure, 0 30 45 0 seg AB represents the lighthouse. A is the position of the observer m D and C are the position of the ships. 0 8 Draw ray AE || seg BD. EAD and EAC are the angles of depression. 45 0 m EAD = 30º 30º and m EAC = 45º 30 0 B C D On transversal AD m EAD = m ADB = 30º [C onve rs e of al tern at e angl es test ] On transversal AC m EAC = m ACB = 45º [C onve rs e of altern at e angl es test ] In right angled ABD, AB tan 30 = [By definition] DB 1 80 = 3 DB DB = 80 3 m
......( i ) 355
GEOMETRY
MT EDUCARE LTD.
In right angled tan 45 = 1
=
CB BD
= =
80 3 = CD
=
CD
=
ABC,
AB CB 80 CB 80 m BC + CD
[By definition] .......(ii) [ D - C - B]
80 + CD
[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-ORDINA CO-ORDI NATE TE GEOMETRY
1. Sol.
A (8, (8, 5), 5), B (9, (9, – 7), 7), C (– (– 4, 4, 2) and D (2, (2, 6) 6) are the the ver verti tices ces of a ABCD. 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, P Q R S
x1 x 2 y1 y 2 8 9 5 – 7 , 2 , 2 2 2 9 – 4 –7 2 5 –5 2 , 2 , 2 2 – 4 2 2 6 , 2 (– 1, 4) 2 11 8 2 5 6 , 5 , 2 2 2
y 2 – y 1 Slope of PQ = x – x = 2 1
–
5 4 – – 2 Slope of QR = 5 = –1 – 2 11 – 4 2 Slope of RS = = 5 – (– 1) 1)
17 2 , –1
5 – 3 – (–1) 2 2 1 5 17 = –12 = 4 – 2 2 2 13
2
– 7 2
=
–13 7
3 1 2 = 4 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 . . .. . . . ( i i ) PQRS is a parallelogram. [From (i), (ii) and by definition]
356
MT EDUCARE LTD.
2. Sol.
GEOMETRY
Find Find the the equ equati ation ons s of the the lines lines which which thr throu ough gh the the poi point nt (3, (3, 4) 4) and and the 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 a b Since the line passes through the point (3, 4) 3 4 1 a b 3b + 4a = ab .. .. ... .. ..(i i) i) From (i), a = 14 – b Substituting a = 14 – b in (ii) we get, 3b + 4 (14 – b) = (14 – b)b 3b + 56 – 4b = 14b – b 2 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
The equation of the line is
4x + 3y – 24 = 0 and x + y – 7 = 0 3.
Sol.
Find Find the the equ equati ation on of of a lin line e whic which h pass passes es thr throu ough gh the the poi point nt (– (– 3, 7) and makes intercepts on the co-ordinate axes which are equal in magnitude but opposite in sign. Let the intercepts intercepts made by the line line on the co-ordinate co-ordinate axes axes be a and b. a=–b .......( i ) x y 1 The equation of the line is a b x y 1 – b b – 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 x – y + 10 = 0
4.
Find Find the the equ equat atio ion n of of a lin line e which which conta contain ins s the the poi point nt (4, (4, 1) and whose whose x-intercept is twice its y-intercept. Sol. 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 357
GEOMETRY
MT EDUCARE LTD.
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 5. Sol.
Find Find the equati equation on of side side AC of an isosc isoscele eles s ABC, ABC, if the equati equation on of side 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 A
17 5 3 , 3
(k 4 – 6)2 (k – 4)2 (k – 2)2 + (k – 4) 2 k2 – 4k + 4 + k 2 – 8k + 16 – 12k + 20 20
20 12 5 = 3 5 17 4 = = 3 3 =
Equation of side AC by two point form is
x – x1 x1 – x 2
=
y – y 1 y1 – y 2
x – 6 y – 4 17 = 5 6 – 4 – 3 3
x – 6 y – 4
=
x – 6 y – 4
=
17 6 – 3 5 4 – 3 1 3 7 3
x – 6 1 = y – 4 7 7 (x – 6) = y – 4 7x – 42 = y – 4 7x – y – 38 = 0 358
GEOMETRY
MT EDUCARE LTD.
6. Sol.
Find Find the the equa equati tion ons s of of the the line line which which cut cut off off inte interce rcept pts s on on the the axe axes s whos whose e 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 .. .. ... .. ..(i i) i) From (ii) – 6 b= a Substituting this in (i) we get, 6 a– = 1 a a2 – 6 = a a2 – a – 6 = 0 a2 – 3a + 2a – 6 = 0 a (a – 3) + 2 (a (a – 3) 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 1 and 3 –2 – 2 3 2x – 3y = 6 and – 3x + 2y = 6 2x – 3y – 6 = 0 and 3x – 2y + 6 = 0 CHAPTER : 6 - MENSURATION MENSURATION
1.
A tin tinmak maker er con conve vert rts s a cub cubic ical al met metall allic ic box box int into o 10 10 cyli cylindr ndric ical al tins. tins. Side Side of the cube is 50 cm and radius of the cylinder is 7 cm. Find the height of
Sol.
each cylinder so made if the wastage wastage of 12% is incurred incurred in the process. process. 22 (Given = ). 7 Side of the cubical metallic box ( l ) = 50 cm Total surface area of cubical box = 6 l 2 = 6 × (50)2 = 6 × 2500 = 15000 cm 2 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 Total surf surface ace area area of cubi cubical cal box – Wast Wastage age incu incurre rred d in the the proces process s = 15000 - 1800 = 13200 cm2 Area of metal sheet used to make each cylindrical tin 13200 = 10 = 1320 cm2 359
GEOMETRY
MT EDUCARE LTD.
Radius (r) = 7 cm Area of metal sheet used to = Total Total surfa surface ce area of cylinder cylinder make each cylindrical tin Total surface area of cylinder = 2 r (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.
The The thr three ee face faces s A, A, B, B, C of a cub cuboi oid d hav have e sur surfa face ce area area 45 450 0 cm cm2, 600 cm2 and 300 cm2 respectively. Find the volume of the cuboid. Surface area rea of of fa face A = 450 cm 2 Surf Surface ace area area of face face A = l × × h l × × h = 450 .... .( i) 2 Surface area rea of face B = 600cm 00cm Surf Surface ace area area of face face B = l × × b l × × b = 600 .... .( i i ) 2 Surface area rea of of fa face C = 300 cm Surface area rea of of fa face C = b × h b × h = 300 .. ... ( ii i) Multiplying (i), (ii) and (iii), l × × h × l × × b × b × h = 450 × 600 × 300 l 2 × b 2 × h 2 = 450 × 2 × 30 300 × 30 300 l × × l × ×
b×h =
b×h l × × b × h But, Volume Volume of the cuboid cuboid Volume of of th the cuboid
900
300
300 [Taking square roots]
= 30 × 300 = 9000cm 3 = l × × b × h = 9000cm 0cm 3
Volume of the cuboid is 9000 cm 3. 3.
Sol.
360
Oil tins tins of of cubo cuboida idall shape shape are made made from from a metall metallic ic sheet sheet with with lengt length h 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 rea of of met metal alli lic c she sheet et = l × × b l = 800 × 40 400 = 320000 cm 2 Length of the oil tin ( l 1) = 60 cm its breadth (b 1) = 40 cm its height (h 1) = 20 cm Area of metallic sheet required for each tin = surfac surface e area area of vertica verticall face faces s + surface surface area area of of the the base base = [2 (l 1 + b1) × h1] + [l 1 b 1]
GEOMETRY
MT EDUCARE LTD.
= = = =
=
Sol.
Area of metal required for each tin
=
320000 6400 50
50 Oil tins can be made.
=
4.
[2 (60 (60 + 40) 40) × 20] 20] + [60 [60 × 40] 40] (2 × 100 × 20) + (6 (60 × 40) 4000 + 2400 6400 cm2 Number of tins that can be made Area Area of met metalli llic shee sheett
Plasti Plastic c drum drum of cyli cylindr ndric ical al sha shape pe is is made made by melt melting ing spheri spherica call solid solid plasti plastic c balls of radius 1 cm. Find the number of balls required to make a drum of thickness 2 cm, height 90 cm and outer radius 30 cm. (5 marks) Outer radius of the drum (r 1) = 30cm It s th thic kn es s = 2 cm inner radius of the drum (r 2) = 30 – 2 = 28 cm Outer height of cylindrical plastic drum (h 1) = 90cm Inner height of cylindrical plastic drum (h 2) = Oute Outerr hei heigh ghtt – thic thickn knes ess s of base base = 90 – 2 = 88 c m Volume of plastic required for the cylindrical drum = Volum Volume e of outer outer cylin cylinde derr – Volum Volume e of inne innerr cylin cylinde derr 2 2 = r1 h1 – r 2 h2 = [(30)2 × 90 – (28) 2 × 88] = × (900 × 90 – 784 × 88] = × (81000 – 68992) = 12008 cm 3 Radius of spherical solid plastic ball (r) = 1cm Volume of each each plastic plastic ball ball = =
4 × 3
=
4 3
Num Number ber of ball balls s requ requir ired ed to make the drum
4 3
r3 ×r×1×1×1 cm3
Vol Volume ume of plas plasti tic c requ requir ired ed for for the the drum drum =
=
=
Volu Volum me of each each plas plasti tic c ball ball 12008 4 3
12008
3 4
= 9006 Number of plastic balls required to make the cylindrical drum is 9006. 361
GEOMETRY
5.
Sol.
MT EDUCARE LTD.
Wate Wa terr dri drips ps fro from m a tap tap at at the the rate rate of of 4 dro drops ps in in ever every y 3 sec secon onds ds.. Volu Volume me of one drop is 0.4 cm3. If dripped water is collected in a cylinder vessel of height height 7 cm and diameter diameter is 8 cm. In what time will the vessel be complet completely ely filled ? What is the volume of water collected ? How many such vessels will be completely completely filled in 3 hours hours and 40 minutes minutes ? (5 marks) Diame iamete terr of of the the cylin ylindr dric ical al vesse essell = 8cm Its radius (r) = 4 cm it s height (h) = 7 cm Volum olume e of th the cylin ylind drica ical vessel sel = r 2h 22 4 4 7 = 7 = 22 × 16 = 352 cm3 Vo lume of wate r coll ec ted = 352 cm 3 Volume of one drop of water = 0.4 cm 3 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 cm 3 1.6 Volume of water collected in 1 seconds = cm3 3 Time required to fill the cylindri cal vessel Volu Volume me of cylin cylindr dric ical al vess vessel el = Volu Volume me of wat water coll collec ecte ted d in each each sec sec ond ond =
352 1.6 3
=
352
= = = =
= =
3 1.6 352 3 10 1.6 10
352 3 10 16 660 seconds 11 11 minut es 3 hour hours s and and 40 min minut utes es = = = Number of vessels that can 220 11 20
[ 1 minutes = 60 seconds] 3 × 60 60 min min + 40 40 min min 180 + 40 220 mi minute utes be completely filled in 220 minutes
20 vessels can be filled in 3 hours and 40 minutes. 6.
Sol.
362
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 [ G i v e n] i.e. PB = BQ = 7 cm ABCDEF is a regular hexagon......
E
D
C X• N
A
P
Q B
MT EDUCARE LTD.
In BP m m
GEOMETRY
BPQ, = BQ BPQ = m BQP BPQ + m BQP + m
. .. . .. . ( i ) [ I s o sc e le s t r i a n g l e t h eo r em ] PBQ = 180º 180º [Sum of the measures of angles of a triangle is 180º] m BPQ + m BPQ BPQ + 120 120 = 180 180 [Fro [From m (i) (i) and and angl angle e of reg regul ular ar hexagon] 2 m BPQ = 180 – 120 2 m BPQ = 60 m BPQ = 30º ...... (i i ) In BNP, BNP = 90º [ C o ns t r uc t i o n] BPN = 30º [From (ii) and P - N - Q] PBN = 60º [ R e m ai n i n g a n g l e ] 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 3 BP 2 3 7 2
PN
=
PN
=
PN PQ
= 3.5 3 cm = 2PN
PQ
=
PQ
= 7 3 cm 1 PQ BN = 2 1 7 3 3.5 = 2
A ( BPQ)
= = = = = A (B-PXQ)
= = =
A (s (segme gment PX P X Q)
= = = =
[Side opposite to 60º]
[Perpendicular drawn from the centre of the circle to the chord bisects the chord]
2 3.5 3
3.5 3 3.5 3.5 × 3. 3.5 × 1.73 20.25 × 1.73 21.1925 21.19 cm2
r2
360 120 22 7 7 360 7 154 3 51.33 cm2 A (B – PXQ) – A ( BPQ) 51.33 – 21 21.19 30.14 cm2 363
GEOMETRY
MT EDUCARE LTD.
A (sh (shad aded ed port portio ion) n) = 6 × A (seg (segme ment nt PXQ) PXQ) = 6 × 30.14 = 180.84 cm2 The area of shaded portion is 180.84 cm 2. 7.
Sol.
A meta metalli llic c righ rightt circ circula ularr cylin cylindri drica call disc disc is is of hei height ght 30 cm and the diame diamete terr 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) Heigh Heightt of cylin cylindr drica icall disc disc (h) = 30 cm 1 it s di diam et er = 1 times the height 2 3 30 = 2 = 45 cm
45 cm 2 Volum lume of of cyl cyliindri drical cal di disc = r 2h 45 45 30 = 2 2 60750 cm2 = 4 Let the radius of the sphere be ‘r 1’ cm The metallic disc is melted and moulded into the sphere [Given] Volu Volume me meta metalli llic c sphe sphere re = Volu Volume me cylin ylindr dric ical al disc isc 4 3 60750 r1 cm3 = 3 4 4 45 45 3 r 30 = 1 3 2 2 45 45 30 3 3 = r1 4 2 2 45 45 45 3 = r1 222 45 r1 = [Taking cube roots] 2 r 1 = 22.5 cm Total surface area of sphere = 4 r 2 22 2 22.5 = 4 7 4 22 506.25 = 7 88 506.25 = 7 4455000 = 7 = 6364.28 cm2 (Approximately) Radi Radius us of cyli cylind ndri rica call dis disc c (r) (r)
=
Radius of the sphere is 22.5 cm and total surface area of the sphere is 6364.28 cm 2.
364
