ML Aggarwal Solution Class 10 Chapter 13 Similarity Exercise 13.2

 Exercise 13.2

Question 1

(a) In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find (i) AE : EC (ii) DE.

(b) In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.

(c) In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.































Sol :

(a) In the figure (i)

Given : DE || BC, AD = 3 cm, BD = 4 cm and BC = 5 cm.

To find (i) AE : EC and

(ii) DE Since DE || BC of ∆ABC

ADBD=AEEC

AEEC=ADBD

=34[AD=3 cmBD=4 cm]

AE : EC = 3 : 4


(ii) InΔADE and ΔABC

D=B and E=C[DEBC given ]

ΔADEΔABC

DEBC=ADAB

DE5=33+4

[AB=AD+BD=3cm+4cm]


DE5=37

DE=3×57

DE=157=217 cm


(b) In the figure (ii)

Given : PQ||AC, AP=4 cm

To find CQ and BQ

Now PQ||AC (Given)

∠BQP=∠BCA (Alternate angles)

Also, ∠B=∠B (common)

ΔABCΔBPQ

BQBC=BPAB=PQAC

BQBC=66+4=PQAC

BQBC=610=PQAC

BQ8=610=PQAC

[BC=8 cm given ]


Now, BQ8=610

BQ=610×8=4810=4.8 cm

Also, CQ=BCBQ

CQ=(84.8)cm

CQ=3.2 cm

Hence, CQ=3.2 cm and BQ=4.8 cm

Ans (c) In the figure (iii)


Given : XY||QR,

PX=1 cm, QX=3 cm, YR=4.5 cm and QR=9 cm

To find : PY=XY

Now , XY||QR (Given)

Figure to be added

PXQX=PYYR

13=PY4.5

4.5×13=PY

1.5=PY

PY=1.5 cm Ans

Also, X=Q

and Y=R (XY || QR given)

ΔPXYΔPQR

XYQR=PXPQ

XY9=11+3[PQ=1+3=4 cm]

XY9=14XY=94=2.25 cm


Question 2

In the given figure, DE || BC.

(i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.

(ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.












Sol :
In the given figure, DE || BC
(i) AD = x, DB = x – 2, AE = x + 2, EC = x – 1
In ∆ABC,
∵DE || BC
ADDB=AEEC

xx2=x+2x1 (By cross multiplication)

x(x1)=(x2)(x+2)

x2x=x24

x=4x=4








(ii) DB=x-3, AB=2x

EC=x-2, AC=2x+3

In ΔABC

∴DE||BC

ABDB=ACEC

2xx3=2x+3x2

By cross multiplication

2x(x-2)=(2 x+3)(x-3)

2x24x=2x26x+3x9

2x24x2x2+6x3x=9

x=9x=9

x=9

Figure to be added

PEEQ=3.93=3930=1310

PFFR=89

PEEQPFFR


Question 3

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:

(i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

Sol :

(i) In ∆PQR, E and F are the points on the sides PQ and PR respectively

PE = 3.9 cm, EQ = 3 cm, PF = 8 cm,

RF = 9 cm

Is EF || QR ?

EF is not parallel to QR


(ii) PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm, PF=0.36 cm

PQPE=1.280.18=12818=649

PRPF=2.560.36=25636=649

PQPE=PRPF

∴EF||QR


Question 4

A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR? Give reasons for your answer.

Sol :

In ∆PQR, A and B are points on the sides PQ and PR such that

PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm












Now, PQPA=12.55=2.51

and PRPB=PB+BRPB=4+64=104=2.51

PQPA=PRPB

ABQR


Question 5

(a) In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.

(b) In figure (ii) given below, AB || DE and BD || EF. Prove that DC² = CF x AC.

Sol :

(a) Given: In the figure,

DE || BC and BD = CE

To prove: ∆ABC is an isosceles triangle

Proof: In ∆ABC, DE || BC












ADDB=AEEC

But DB=EC (given )..(i)

∴AD=AE...(ii)

Adding (i) and (ii)

AD+DB=AE+EC

⇒AB=AC

∴ΔABC is an isosceles triangle 


(b) Given : In the given figure 

AB || DE, BD || EF

To prove : DC2=CF×AC

Proof : In ΔABC,DEAB

DCCA=CECB..(i)

In ΔCDE

EF ||| DB

CFCD=CECB...(ii)









From (i) and (ii)

DCCA=CFCD

DCAC=CFDC

By cross multiplication

DC2=CF×AC


Question 6

(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.

(b) In the given figure, ∠D = ∠E and ADBD=AEEC. Prove that BAC is an isosceles triangle













Sol :

(a) Given : CD || LA and DE || AC

Length of BE = 4 cm

Length of EC = 2 cm

Now, in ∆BCA

DE || AC

BEBC=BDBA

(Corallary of basic proportionality theorem)

BEBE+EC=BDAB
4(4+2)=BDAB...(i)

Now, In ΔBLA

CD || LA

BCBL=BDAB

(Corallary of basic proportionality theorem)

BCBC+CL=BDAB

66+CL=BDAB...(ii)

Combining equation (i) and (ii), we get

66+CL=46

6×6=4×(6+CL)

24+4CL=36

4CL=3624

CL=124=3 cm

∴The length of CL=3 cm


(b) Given : In the given figure , ∠D=∠E

ADDB=AEEC

To prove : ΔBAC is an isosceles triangle

Proof : In ΔADE

∠D=∠E (Given)

∴AD=AE (sides opposite to equal angles)

In ΔABC,

ADDB=AEEC ...(i)

DEBC

AD=AE

DB=EC  [From (i)]

Adding, we get

AD+DB=AE+EC

⇒AB=AC

∴ΔABC is an isosceles triangle


Question 7

In the figure given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.












Sol :
In the given figure, A, B, C are points on
OP, OQ and OR respectively
and AB || PQ and AC || PR
To prove: BC || QR
Proof: In POQ,
AB || PQ

OAAP=OBBQ..(i)

Similarly in ΔOPR

AC || PR

OAAP=OCCR ...(ii)

From (i) and (ii)
OBBQ=OCCR

Now in ΔOQR
OBBQ=OCCR
∴BC || QR


Question 8

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that AOBO=CODO
Sol :
Given : ABCD is a trapezium in which AB || DC
Its diagonals AC and BD intersect each other at O











To prove : AOBO=CODO

Proof : In ΔOAB and ΔOCD 
AOB=COD (Vertically opposite angles)

OAB=OCD (Alternate angles)
and OBA=ODC (Alternate angles)
ΔOABΔOCD

OAOC=OBOD
AOOB=CODO (by alternendo)

Question 9

(a) In the figure (1) given below, AB || CR and LM || QR.
(i) Prove that BMMC=ALLQ
(ii) Calculate LM : QR, given that BM : MC = 1 : 2.
(b) In the figure (2) given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC























Sol :
(a) Given: AB || CR and LM II QR.
Also BM : MC = 1:2
To Prove:
(i) BMMC=ALLQ

(ii) To calculate LM : QR

AMMR=ALLQ..(i)

Now, in ΔAMB and ΔMCR

∠AMB=∠CMR (vertically opposite angles)
∠MBA=∠MCR (alternate angles)
[∵AB||CR given]

∴ΔAMB=ΔMCR
AMMR=BMMC...(ii)

From (i) and (ii), we get

BMMC=ALLQ

(ii) From (2),
AMMR=BMMC

AMMR=12..(iii)

[BM:MC=1:2BMMC=12]

LMQR (given)
AMMR=LMQR...(iv)

or, ARAM=QRLM

or, AM+MRAM=QRLM

or, 1+MRAM=QRLM

or, 1+21=QRLM

[ From (iii),AMMR=12MRAM=21]

or, 31=QRLM

or, LMQR=31

LMQR=1:3








(b) Given : ΔABC, AD is (internal) bisector of ∠BAC.

AB=6 cm. AC=4cm and 3D=3cm

To calculate : The value of BC

Construction : Through C, draw straight line CE || DA meeting BA produced in E

Now, in ΔABC

As AD is bisector of ∠A,

Figure to be added

∠1=∠2...(i)

∵CE || DE (By construction ) and AC cuts them,

∠2=∠4 (Alternate angles)...(ii)

Again CE||DA and BE cuts them,

∠1=∠3 (corresponding angles)..(iii)

From (i), (ii) and (iii), we get

∠3=∠4

⇒AC=AE...(iv)

In ΔBCE,DACE

BDDC=ABAE

BDDC=ABAC (∵From (4), AC=AE)

3DC=64

3×4=6×DC

DC=3×46=2

∴BC=BD+DC

=3cm+2cm=5cm

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2