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.
(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
=34[∵AD=3 cmBD=4 cm]
AE : EC = 3 : 4
(ii) InΔADE and ΔABC
∠D=∠B and ∠E=∠C[∵DE‖BC given ]
⇒ΔADE∼ΔABC
∴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)
∴ΔABC∼ΔBPQ
∴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=BC−BQ
⇒CQ=(8−4.8)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)
∵ΔPXY∼ΔPQR
∴XYQR=PXPQ
⇒XY9=11+3[PQ=1+3=4 cm]
⇒XY9=14⇒XY=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.
⇒xx−2=x+2x−1 (By cross multiplication)
x(x−1)=(x−2)⋅(x+2)
x2−x=x2−4
−x=−4⇒x=4
(ii) DB=x-3, AB=2x
EC=x-2, AC=2x+3
In ΔABC
∴DE||BC
∴ABDB=ACEC
⇒2xx−3=2x+3x−2
By cross multiplication
2x(x-2)=(2 x+3)(x-3)
⇒2x2−4x=2x2−6x+3x−9
⇒2x2−4x−2x2+6x−3x=−9
⇒−x=−9⇒x=9
∴x=9
Figure to be added
PEEQ=3.93=3930=1310
PFFR=89
∵PEEQ≠PFFR
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 ?
(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
and PRPB=PB+BRPB=4+64=104=2.51
∵PQPA=PRPB
∴AB‖QR
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
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,DE‖AB
∴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
(Corallary of basic proportionality theorem)
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=36−24
⇒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)
∴DE‖BC
∵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.
Question 8
Question 9
or, AM+MRAM=QRLM
or, 1+MRAM=QRLM
or, 1+21=QRLM
[∵ From (iii),AMMR=12∴MRAM=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,DA‖CE
∴BDDC=ABAE
⇒BDDC=ABAC (∵From (4), AC=AE)
⇒3DC=64
⇒3×4=6×DC
⇒DC=3×46=2
∴BC=BD+DC
=3cm+2cm=5cm
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