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

$\therefore \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

$ \Rightarrow \frac{\mathrm{AE}}{\mathrm{EC}}=\frac{\mathrm{AD}}{\mathrm{BD}}$

$=\frac{3}{4} \quad[\because \mathrm{AD}=3 \mathrm{~cm} \mathrm{BD}=4 \mathrm{~cm}]$

AE : EC = 3 : 4


(ii) $\operatorname{In} \Delta \mathrm{ADE}$ and $\Delta \mathrm{ABC}$

$\angle \mathrm{D}=\angle \mathrm{B}$ and $\angle \mathrm{E}=\angle \mathrm{C} \quad[\because \mathrm{DE} \| \mathrm{BC}$ given $]$

$\Rightarrow \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$

$\therefore \quad \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{AB}} $

$\Rightarrow \frac{\mathrm{DE}}{5}=\frac{3}{3+4}$

$[\because A B=A D+B D=3 c m+4 c m]$


$\Rightarrow  \frac{\mathrm{DE}}{5}=\frac{3}{7}$

$ \Rightarrow \mathrm{DE}=\frac{3 \times 5}{7}$

$\mathrm{DE}=\frac{15}{7}=2 \frac{1}{7} \mathrm{~cm}$


(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)

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{BPQ}$

$\therefore \frac{\mathrm{BQ}}{\mathrm{BC}}=\frac{\mathrm{BP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{AC}}$

$\Rightarrow \quad \frac{B Q}{B C}=\frac{6}{6+4}=\frac{P Q}{A C}$

$\Rightarrow \quad \frac{B Q}{B C}=\frac{6}{10}=\frac{P Q}{A C}$

$\Rightarrow \quad \frac{B Q}{8}=\frac{6}{10}=\frac{P Q}{A C}$

$[\because \mathrm{BC}=8 \mathrm{~cm}$ given ]


Now, $\frac{\mathrm{BQ}}{8}=\frac{6}{10}$

$\Rightarrow  B Q=\frac{6}{10} \times 8=\frac{48}{10}=4.8 \mathrm{~cm}$

Also, $C Q=B C-B Q$

$ \Rightarrow C Q=(8-4.8) \mathrm{cm}$

$\Rightarrow \mathrm{CQ}=3.2 \mathrm{~cm}$

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

$\therefore  \frac{P X}{Q X}=\frac{P Y}{Y R}$

$\Rightarrow   \frac{1}{3}=\frac{\mathrm{PY}}{4.5}$

$\Rightarrow \quad \frac{4.5 \times 1}{3}=\mathrm{PY} $

$\Rightarrow 1.5=\mathrm{PY}$

$\Rightarrow  \mathrm{PY}=1.5 \mathrm{~cm}$ Ans

Also, $\angle \mathrm{X}=\angle \mathrm{Q}$

and $\angle Y=\angle R$ (XY || QR given)

$\because \Delta \mathrm{PXY} \sim \Delta \mathrm{PQR}$

$\therefore \frac{\mathrm{XY}}{\mathrm{QR}}=\frac{\mathrm{PX}}{\mathrm{PQ}}$

$\Rightarrow  \frac{X Y}{9}=\frac{1}{1+3} \quad[P Q=1+3=4 \mathrm{~cm}]$

$\Rightarrow  \frac{X Y}{9}=\frac{1}{4} \Rightarrow X Y=\frac{9}{4}=2.25 \mathrm{~cm}$


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
$\therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

$\Rightarrow \frac{x}{x-2}=\frac{x+2}{x-1}$ (By cross multiplication)

$x(x-1)=(x-2) \cdot(x+2)$

$x^{2}-x=x^{2}-4$

$-x=-4 \Rightarrow x=4$








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

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

In ΔABC

∴DE||BC

$\therefore \frac{\mathrm{AB}}{\mathrm{DB}}=\frac{\mathrm{AC}}{\mathrm{EC}}$

$\Rightarrow \frac{2 x}{x-3}=\frac{2 x+3}{x-2}$

By cross multiplication

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

$\Rightarrow 2 x^{2}-4 x=2 x^{2}-6 x+3 x-9$

$\Rightarrow 2 x^{2}-4 x-2 x^{2}+6 x-3 x=-9$

$\Rightarrow-x=-9 \Rightarrow x=9$

$\therefore x=9$

Figure to be added

$\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3}=\frac{39}{30}=\frac{13}{10}$

$\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9}$

$\because \frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}$


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 ?

$\therefore$ EF is not parallel to QR


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

$\frac{P Q}{P E}=\frac{1.28}{0.18}=\frac{128}{18}=\frac{64}{9}$

$\frac{\mathrm{PR}}{\mathrm{PF}}=\frac{2.56}{0.36}=\frac{256}{36}=\frac{64}{9}$

$\because \frac{\mathrm{PQ}}{\mathrm{PE}}=\frac{\mathrm{PR}}{\mathrm{PF}}$

∴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, $\frac{P Q}{P A}=\frac{12.5}{5}=\frac{2.5}{1}$

and $\frac{P R}{P B}=\frac{P B+B R}{P B}=\frac{4+6}{4}=\frac{10}{4}=\frac{2.5}{1}$

$\because \frac{\mathrm{PQ}}{\mathrm{PA}}=\frac{\mathrm{PR}}{\mathrm{PB}}$

$\therefore \mathrm{AB} \| \mathrm{QR}$


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












$\therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

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 : $\mathrm{DC}^{2}=\mathrm{CF} \times \mathrm{AC}$

Proof : In $\Delta \mathrm{ABC}, \mathrm{DE} \| \mathrm{AB}$

$\therefore \frac{\mathrm{DC}}{\mathrm{CA}}=\frac{\mathrm{CE}}{\mathrm{CB}}$..(i)

In ΔCDE

EF ||| DB

$\frac{CF}{C D}=\frac{C E}{C B}$...(ii)









From (i) and (ii)

$\frac{\mathrm{DC}}{\mathrm{CA}}=\frac{\mathrm{CF}}{\mathrm{CD}}$

$ \Rightarrow \frac{\mathrm{DC}}{\mathrm{AC}}=\frac{\mathrm{CF}}{\mathrm{DC}}$

By cross multiplication

$\mathrm{DC}^{2}=\mathrm{CF} \times \mathrm{AC}$


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 $\frac{A D}{B D}=\frac{A E}{E C}$. 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

$\frac{B E}{B C}=\frac{B D}{B A}$

(Corallary of basic proportionality theorem)

$\Rightarrow \frac{B E}{B E+E C}=\frac{B D}{A B}$
$\Rightarrow \frac{4}{(4+2)}=\frac{B D}{A B}$...(i)

Now, In ΔBLA

CD || LA

$\therefore \frac{\mathrm{BC}}{\mathrm{BL}}=\frac{\mathrm{BD}}{\mathrm{AB}}$

(Corallary of basic proportionality theorem)

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{BC}+\mathrm{CL}}=\frac{\mathrm{BD}}{\mathrm{AB}}$

$\frac{6}{6+\mathrm{CL}}=\frac{\mathrm{BD}}{\mathrm{AB}}$...(ii)

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

$\frac{6}{6+\mathrm{CL}}=\frac{4}{6}$

$\Rightarrow 6 \times 6=4 \times(6+\mathrm{CL})$

$\Rightarrow 24+4 \mathrm{CL}=36$

$\Rightarrow 4 \mathrm{CL}=36-24$

$\Rightarrow \mathrm{CL}=\frac{12}{4}=3 \mathrm{~cm}$

∴The length of CL=3 cm


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

$\frac{A D}{D B}=\frac{A E}{E C}$

To prove : ΔBAC is an isosceles triangle

Proof : In ΔADE

∠D=∠E (Given)

∴AD=AE (sides opposite to equal angles)

In ΔABC,

$\because \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ ...(i)

$\therefore \mathrm{DE} \| \mathrm{BC}$

$\because \mathrm{AD}=\mathrm{AE}$

$\therefore \mathrm{DB}=\mathrm{EC}$  [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

$\therefore \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}$..(i)

Similarly in ΔOPR

AC || PR

$\therefore \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}}$ ...(ii)

From (i) and (ii)
$\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$

Now in ΔOQR
$\because \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$
∴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 $\frac{A O}{B O}=\frac{C O}{D O}$
Sol :
Given : ABCD is a trapezium in which AB || DC
Its diagonals AC and BD intersect each other at O











To prove : $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$

Proof : In $\Delta \mathrm{OAB}$ and $\Delta \mathrm{OCD}$ 
$\angle \mathrm{AOB}=\angle \mathrm{COD}$ (Vertically opposite angles)

$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (Alternate angles)
and $\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate angles)
$\therefore \Delta \mathrm{OAB} \sim \Delta \mathrm{OCD}$

$\therefore \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$
$\Rightarrow \frac{\mathrm{AO}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{DO}}$ (by alternendo)

Question 9

(a) In the figure (1) given below, AB || CR and LM || QR.
(i) Prove that $\frac{B M}{M C}=\frac{A L}{L Q}$
(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) $\frac{B M}{M C}=\frac{A L}{L Q}$

(ii) To calculate LM : QR

$\therefore  \frac{\mathrm{AM}}{\mathrm{MR}}=\frac{\mathrm{AL}}{\mathrm{LQ}}$..(i)

Now, in ΔAMB and ΔMCR

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

∴ΔAMB=ΔMCR
∴$\frac{A M}{M R}=\frac{B M}{M C}$...(ii)

From (i) and (ii), we get

$\frac{\mathrm{BM}}{\mathrm{MC}}=\frac{\mathrm{AL}}{\mathrm{LQ}}$

(ii) From (2),
$\frac{\mathrm{AM}}{\mathrm{MR}}=\frac{\mathrm{BM}}{\mathrm{MC}}$

$ \Rightarrow \frac{\mathrm{AM}}{\mathrm{MR}}=\frac{1}{2}$..(iii)

$\left[\because \mathrm{BM}: \mathrm{MC}=1: 2 \quad \therefore \frac{\mathrm{BM}}{\mathrm{MC}}=\frac{1}{2}\right]$

$\because \mathrm{LM} \| \mathrm{QR}$ (given)
$\because \frac{\mathrm{AM}}{\mathrm{MR}}=\frac{\mathrm{LM}}{\mathrm{QR}}$...(iv)

or, $\frac{A R}{A M}=\frac{Q R}{L M}$

or, $\frac{\mathrm{AM}+\mathrm{MR}}{\mathrm{AM}}=\frac{\mathrm{QR}}{\mathrm{LM}}$

or, $1+\frac{\mathrm{MR}}{\mathrm{AM}}=\frac{\mathrm{QR}}{\mathrm{LM}}$

or, $1+\frac{2}{1}=\frac{Q R}{L M}$

$\left[\because\right.$ From $\left.(i i i), \frac{\mathrm{AM}}{\mathrm{MR}}=\frac{1}{2} \therefore \frac{\mathrm{MR}}{\mathrm{AM}}=\frac{2}{1}\right]$

or, $\frac{3}{1}=\frac{Q R}{L M}$

or, $\frac{L M}{Q R}=\frac{3}{1}$

$\therefore \frac{\mathrm{LM}}{\mathrm{QR}}=1: 3$








(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 $\Delta \mathrm{BCE}, \mathrm{DA} \| \mathrm{CE}$

$\therefore \frac{B D}{D C}=\frac{A B}{A E}$

$\Rightarrow   \frac{B D}{D C}=\frac{A B}{A C}$ (∵From (4), AC=AE)

$\Rightarrow \frac{3}{\mathrm{DC}}=\frac{6}{4}$

$ \Rightarrow 3 \times 4=6 \times \mathrm{DC}$

$\Rightarrow  \mathrm{DC}=\frac{3 \times 4}{6}=2$

∴BC=BD+DC

=3cm+2cm=5cm

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