ML Aggarwal Solution Class 10 Chapter 13 Similarity Test

 Test

Question 1

In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.

Sol :

Given: In the given figure,

∠1 = ∠1 and ∠3 = ∠4












To prove : PT × QR = PR × ST
Proof: ∠1 = ∠2
Adding ∠6 to both sides
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR

In ∆PQR and ∆PST

$\angle \mathrm{SPT}=\angle \mathrm{QPR}$ (proved)

$\angle 3=\angle 4$ (Given)

$\therefore \Delta P Q R \sim \Delta P S T$

$\therefore \frac{\mathrm{PT}}{\mathrm{PR}}=\frac{\mathrm{ST}}{\mathrm{QR}} $

$\Rightarrow \mathrm{PT} \times \mathrm{QR}=\mathrm{PR} \times \mathrm{ST}$

(By cross multiplication)

$\therefore \mathrm{PT} \times \mathrm{QR}=\mathrm{PR} \times \mathrm{ST}$


Question 2

In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.












Sol :
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM × PC = PN × PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM, (Each 90°)
$\angle \mathrm{N}=\angle \mathrm{M}$ (Proved)
$\angle \mathrm{C}=\angle \mathrm{B}$ (AA axiom)
$\therefore \Delta \mathrm{CPN} \sim \Delta \mathrm{BPM}$
$\therefore \frac{\mathrm{PC}}{\mathrm{PB}}=\frac{\mathrm{PN}}{\mathrm{PM}}$
$\Rightarrow \mathrm{PC} \times \mathrm{PM}=\mathrm{PN} \times \mathrm{PB}$
Hence $\mathrm{PM} \times \mathrm{PC}=\mathrm{PN} \times \mathrm{PB}$


Question 3

(a) In the figure (1) given below. ∠AED = ∠ABC. Find the values of x and y.

(b) In the fig. (2) given below, $C D=\frac{1}{2} A C$ , B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :

(i) CE || AG

(ii) 3 ED = GD.

Sol :










(a) Given : In following figure, ∠AED = ∠ABC
Required: The values of x and y.
Now, in ∆ABC and ∆ADE
∠AED = ∠ABC (given)
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE
(By A.A. axiom of similarity)

$\therefore \frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}$

$ \Rightarrow \frac{3}{4+2}=\frac{y}{10}$

$\Rightarrow \quad \frac{3}{6}=\frac{y}{10}$

$ \Rightarrow \frac{10 \times 3}{6}=y$

$ \Rightarrow y=\frac{30}{6}$

$\Rightarrow \quad y=5$ ...(1)

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

$ \Rightarrow \frac{3+x}{4}=\frac{10}{5}$

(From 1, y=5 cm)

$\Rightarrow  5(3+x)=40$

$ \Rightarrow 15+5 x=40$

$\Rightarrow 5 x=40-15$

$ \Rightarrow 5 x=25$

$ \Rightarrow x=\frac{25}{5}$

$\Rightarrow x=5$

Hence, the value of x=5 cm and y=5 cm


(b) Given : $\mathrm{CD}=\frac{1}{2} \mathrm{AC}$, B is the mid point of AC and E is the mid-point of DF

Also, BF||AG

To prove : 

(i) CE||AG

(ii) 3ED=GD

Proof : (i) $\mathrm{CD}=\frac{1}{2} \mathrm{AC}$ (given)

$C D=\frac{1}{2}(2 B C)$

(B is the mid-point of BC ∴ A C=2 B C) 

∴ C is mid-point of BD also given E is mid-points of DF.

CD=BC

∴CE||BF (mid-point theorem)...(i)

∵BF||AG (given)...(2)

From (i) and (ii) we get

CE||AG (proved)


(ii) In ΔAGD,

CE||AG (above proved)

∴$\frac{E D}{G D}=\frac{D C}{A D} $

$\Rightarrow \frac{E D}{G D}=\frac{D C}{3 D C}

(∵AD=DC+CB+AB=DC+DC+DC=3DC)

$\Rightarrow  \frac{E D}{G D}=\frac{1}{3}$

∴3ED=GD (proved)


Question 4

In the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:

(i) DF || BH

(ii) AH = 3 AF.












Sol :
Given: E is the mid-point of BD
and F is mid-point of AC
also 2 AD = BD and EC || BH
To Prove : (i) DF || BH

(ii) AH = 3 AF

Proof : (i) $\mathrm{E}$ is the mid-point of $\mathrm{BD}$ (given)

$\therefore  2 \mathrm{DE}=\mathrm{BD}$...(1)

Also, $2 \mathrm{AD}=\mathrm{BD}$...(2)

From (1) and (2)

2DE=2ABD

⇒DE=AD

i.e. D is the mid-point of AE

Also F is the mid-point of AC 

∴DF||EC(given)..(3)

Also, EC||BH (given)...(4)

From (3) and (4)

∴DF||BH (proved)


(ii) Now E is mid-point of BD and EC||BH (given)

∴C is the mid-point of AH

i.e. FC=CH...(5)

But F is mid-point of AC

i.e. AF=FC...(6)

From (5) and (6), we get
FC=AF=CH

i.e. $\mathrm{AF}=\frac{1}{3} \mathrm{AH}$ 
i.e. 3AF=AH

Hence, AH=3AF

Proved

Question 5

In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Sol :











Given : In ∆ABC, D and E are the points
on the sides AB and AC respectively
DE || BC
AD = 2.4 cm, AE = 3.2 cm,
DE = 2 cm, BC = 5 cm

To find : BD and CE
Proof : In ΔABC

∵DE||BC

∴$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}$

$\frac{2.4}{\mathrm{AB}}=\frac{3.2}{\mathrm{AC}}=\frac{2}{5}$

Now, $\frac{2.4}{\mathrm{AB}}=\frac{2}{5} $

$\Rightarrow \mathrm{AB}=\frac{2.4 \times 5}{2}=6 \mathrm{~cm}$

and $\frac{3.2}{\mathrm{AC}}=\frac{2}{5} $

$\Rightarrow \mathrm{AC}=\frac{3.2 \times 5}{2}=8 \mathrm{~cm}$

∴ BD=AB-AD=6-2.4=3.6cm

and CE=AC-AE=8-3.2=4.8 cm


Question 6

In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.

Sol :

In ∆ABC, D and E are points on the sides AB and AC respectively

AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm


To prove : Whether $\mathrm{DE} \| \mathrm{BC}$

Proof : In $\triangle \mathrm{ABC}$

AD=5.7 cm, B D=9.5 cm, A E=3.3 cm,AC=8.8 cm

$\therefore \mathrm{EC}=\mathrm{AC}-\mathrm{AE}=8.8 \mathrm{~cm}-3.3 \mathrm{~cm}=5.5 \mathrm{~cm}$

Now, $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{5.7}{9.5}=\frac{3}{5}$

and $\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{3.3}{5.5}=\frac{3}{5}$

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

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


Question 7

In a ∆ABC, DE is parallel to the base BC, with D on AB and E on AC. If 

$\frac{A D}{D B}=\frac{2}{3}, \frac{B C}{D E}$

Sol :

In ∆ABC, DE || BC

D is on AB and E is on AC












$\frac{A D}{D B}=\frac{2}{3}$

To find $\frac{\mathrm{BC}}{\mathrm{DE}}$

In ΔABC

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

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

$\Rightarrow \frac{2}{2+3}=\frac{D E}{B C}$

$\Rightarrow \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{2}{5} $

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{DE}}=\frac{5}{2}$


Question 8

If the area of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.

Sol :

Let ∆ABC and ∆DEF are similar and area of

∆ABC = 360 cm²











and area of $\Delta \mathrm{DEF}=250 \mathrm{~cm}^{2}$

BC=8cm

Let EF=x cm

Now $\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

$\Rightarrow \frac{360}{250}=\frac{(8)^{2}}{x^{2}} $

$\Rightarrow \frac{64}{x^{2}}=\frac{360}{250}$

$\Rightarrow x^{2}=\frac{250 \times 64}{360}=\frac{400}{9}$

$\therefore x=\sqrt{\frac{400}{9}}=\frac{20}{3}=6 \frac{2}{3}$

$\therefore \mathrm{EF}=6 \frac{2}{3}$ cm


Question 9

In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find

(i) BC

(ii) DC

(iii) area of ∆ACD : area of ∆BCA.










Sol :

In ∆ABC and ∆ACD

∠C = ∠C (Common)

∠ABC = ∠CAD (given)

∴ ∆ABC ~ ∆ACD

$\therefore \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{AC}}{\mathrm{DC}}$

$\Rightarrow \frac{5}{4}=\frac{B C}{3}=\frac{3}{D C}$

(i) $\frac{B C}{3}=\frac{5}{4}$

$\Rightarrow \mathrm{BC}=\frac{3 \times 5}{4}=\frac{15}{4}=3.75 \mathrm{~cm}$


(ii) $\frac{3}{\mathrm{DC}}=\frac{5}{4}$

$\Rightarrow \mathrm{DC}=\frac{3 \times 4}{5}=\frac{12}{5}=2 \cdot 4 \mathrm{~cm}$


(iii) In ΔACD and ΔABC

$\angle \mathrm{CAD}=\angle \mathrm{ABC}$ (given)

$\angle \mathrm{ACD}=\angle \mathrm{ACB}$ (common)

$\therefore \Delta A C D \sim \Delta A B C$ (AA axiom)

Figure to be added

$\frac{\operatorname{area}(\Delta \mathrm{ACD})}{\text { area }(\Delta \mathrm{ABC})}$

$=\frac{\mathrm{AD}^{2}}{\mathrm{AB}^{2}}=\frac{(4)^{2}}{(5)^{2}}=\frac{16}{25}$

$\operatorname{ar}(\Delta \mathrm{ACD}): \operatorname{ar}(\Delta \mathrm{ABC})=16: 25$


Question 10

In the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.










Sol :

(a) In the figure

Diagonals of parallelogram ABCD are

AC and BD which intersect each other at O.

OE is drawn parallel to CB to meet AB in E.

To find : 
Area of ΔAOE : area of parallelogram ABCD
∵Diagonals of a ||gm bisect each other at O
∴Four triangles in equal area are formed

∴ area of $\Delta \mathrm{OAB}=\frac{1}{4}$ area of $\| \mathrm{gm} \mathrm{ABCD}$

∵ O is the mid-point of AC of ΔABC and DE||CB

∴E is also mid-point of AB

∴OE is the median of ΔAOB

But median of a triangle bisect the triangle into two equal triangles in areas

∴Area of $\Delta \mathrm{AOE}=\frac{1}{2}$ area of $\Delta \mathrm{AOB}$

$=\frac{1}{2} \times \frac{1}{4}$ area of $\| \mathrm{gm} \mathrm{ABCD}$

$=\frac{1}{8}$ area of $\| \mathrm{gm} \mathrm{ ABCD}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{AOE}}{\text { area of } \| \mathrm{gm} \mathrm{ABCD}}=\frac{1}{8}$

$\Rightarrow$ area of $\Delta \mathrm{AOE}$ : area of $\| \mathrm{gm} \mathrm{ABCD}$

=1 : 8


Question 11

In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.










Sol :
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{DC}}=\frac{3}{2}$

To find : area $\triangle \mathrm{AOB}$ : area of $\Delta \mathrm{COD}$

$\angle \mathrm{AOB}=\angle \mathrm{COD}$

(vertically opposite angles)

$\angle \mathrm{OAB}=\angle \mathrm{OCD}$ (alternate angles)

$\therefore \Delta \mathrm{AOB} \sim \Delta \mathrm{COD}$ (AA axiom)

$\frac{\text { area } \Delta \mathrm{AOB}}{\text { area } \Delta \mathrm{COD}}=\frac{\mathrm{AB}^{2}}{\mathrm{DC}^{2}}=\frac{(3)^{2}}{(2)^{2}}=\frac{9}{4}$

Ratio in their areas =9 : 4


Question 12

In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .








(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Sol :
Given : In || gm ABCD,
E is the midpoint of BC and DE meets the diagonal AC at O
and meet AB produced at F.
To prove : (i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4
Proof: In ∆AOD and ∆EDC
∠AOD = ∠EOC (vertically opposite angle)
∠OAD = ∠OCB (alt. angles)
∆AOD ~ ∆EOC (AA postulate)

$\therefore \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{DO}}{\mathrm{OE}}=\frac{\mathrm{AD}}{\mathrm{EC}}=\frac{2 \mathrm{EC}}{\mathrm{EC}}$

$\left(\because \mathrm{EC}=\frac{1}{2} \mathrm{BC}\right.$ or $\left.=\frac{1}{2} \mathrm{AD}\right)$

$\therefore \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{DO}}{\mathrm{OE}}=\frac{2}{1}$

$\therefore \mathrm{DO}: \mathrm{OE}=2: 1$


(ii) $\because \Delta \mathrm{AOD} \sim \Delta \mathrm{OEC}$ [proved in (i)]

∴$\frac{\text { area of } \Delta \text { OEC }}{\text { area of } \Delta \mathrm{AOD}}=\frac{(\mathrm{OE})^{2}}{(\mathrm{DO})^{2}}$

$=\frac{(1)^{2}}{(2)^{2}}=\frac{1}{4}$

⇒Area of ΔOEC : Area of ΔAOD= 1 : 4


Question 13

A model of a ship is made to a scale of 1 : 250. Calculate :

(i) the length of the ship, if the length of model is 1.6 m.

(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m².

(iii) the volume of the model, if the volume of the ship is 1 km³.

Sol :

Scale factor (k) of the model of the ship $=\frac{1}{250}$

(i) Length of model = 1.6 m

$\therefore$ Length of ship $=\frac{1}{k}$ (length of model)

$=250(1 \cdot 6 \mathrm{~m})=400~m$


(ii) Area of deck of ship on the model=2.4m²

$\therefore$ Area of deck on ship $=\frac{1}{k^{2}}$ (Area of deck on model )$=(250)^{2} \times 2 \cdot 4 \mathrm{~m}^{2}=150000 \mathrm{~m}^{2}$


(iii) Volume of ship $=1 \mathrm{~km}^{3}$

$\therefore$ Volume of ship on the model $=(k)^{3}$

(Volume of actual ship)

$=\frac{1}{(250)^{3}} \times 1 \mathrm{~km}^{3}$

$=\frac{1}{(250)^{3}} \times \frac{(1000)^{3}}{1}$

$=(4)^{3}=64 m^{3}$

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