ML Aggarwal Solution Class 10 Chapter 13 Similarity Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13

 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)

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

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

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

ML Aggarwal Solution Class 10 Chapter 13 Similarity MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22

 MCQs

Question 1

In the given figure, ∆ABC ~ ∆QPR. Then ∠R is

(a) 60°

(b) 50°

(c) 70°

(d) 80°

Sol :

In the given figure,

∆ABC ~ ∆QPR

∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R

But ∠A = 70°, ∠B = 50°

∴ ∠C = 180° – (70° + 500) = 180° – 120° = 60°

∠C = ∠R

∴ ∠R = ∠C = 60°

Question 2

In the given figure, ∆ABC ~ ∆QPR. The value of x is

(a) 2.25 cm

(b) 4 cm

(c) 4.5 cm

(d) 5.2 cm

Sol :

In the given figure

∆ABC ~ ∆QPR

$\therefore \frac{\mathrm{AC}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}$

$ \Rightarrow \frac{6}{3}=\frac{4.5}{x}$

$\Rightarrow x=\frac{3 \times 4.5}{6}=2.25$

$\therefore x=2.25 \mathrm{~cm}$

Ans (a)

Question 3

In the given figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to

(a) 50°

(b) 30°

(c) 60°

(d) 100°

Sol :

In the given figure two line segments AC and BD intersect each other at P and PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°

In ∆APB and ∆CPD

$\frac{A P}{P D}=\frac{6}{5}$ and $\frac{B P}{C P}=\frac{3}{2.5}=\frac{6}{5}$

$\because \frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BP}}{\mathrm{CP}}$ and

$\Delta \mathrm{APB}=\angle \mathrm{CPD}$

(Vertically opposite angles)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{CPD}$

$\therefore \angle \mathrm{A}=\angle \mathrm{D}=30$

and third $\angle \mathrm{PBA}=180^{\circ}-\left(50^{\circ}+30^{\circ}\right)$

=180°-80°=100°

Ans (d)

Question 4

If in two triangles ABC and PQR,

$\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$

then

(a) ∆PQR ~ ∆CAB

(b) ∆PQR ~ ∆ABC

(c) ∆CBA ~ ∆PQR

(d) ∆BCA ~ ∆PQR

Sol :

In two ∆ABC and ∆PQR

$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}$

$ \Rightarrow \frac{\mathrm{PQ}}{\mathrm{CA}}=\frac{\mathrm{QR}}{\mathrm{AB}}=\frac{\mathrm{RP}}{\mathrm{BC}}$

Ans (a)

Question 5

In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are

(a) congruent but not similar

(b) similar but not congruent

(c) neither congruent nor similar

(d) congruent as well as similar

Sol :

In ∆ABC and ∆DEF

∠B = ∠E, ∠F = ∠C

AB = 3DE

∵ Two angles of the one triangles are equal

to corresponding two angles of the other

But sides are not equal

∵ Triangles are similar but not congruent, (b)

Question 6

In the given figure, if D, E and F are midpoints of the sides BC, CA and AB respectively, then the two triangles ABC and DEF are

(a) similar

(b) congruent

(c) both similar and congruent

(d) neither similar nor congruent

Sol :

D, E and F are the mid-points of the sides BC, CA and AB of ∆ABC

then two triangles ABC and DEF are similar.

Ans (a)

Question 7

The given figure, AB || DE. The length of CD is

(a) 2.5 cm

(b) 2.7 cm

(c) $\frac{10}{3} \mathrm{~cm}$

(d) 3.5 cm

Sol :

In the given figure, AB || DE

∆ABC ~ ∆DCE

$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{CD}}$

$\Rightarrow \frac{5}{3}=\frac{4.5}{\mathrm{CD}} $

$\Rightarrow \mathrm{CD}=\frac{3 \times 4.5}{5}=2.7 \mathrm{~cm}$

Ans (b)

Question 8

If in triangles ABC and DEF, $\frac{A B}{D E}=\frac{B C}{F D}$  ,then they will be similar when

(a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠A = ∠F

Solution:

In two triangles ABC and DEF

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

They will be similar if their included angles are equal

∠B = ∠D 

Ans (c)

Question 9

If ∆PQR ~ ∆ABC, PQ = 6 cm, AB = 8 cm and perimeter of ∆ABC is 36 cm, then perimeter of ∆PQR is

(a) 20.25 cm

(b) 27 cm

(c) 48 cm

(d) 64 cm

Sol :

∆PQR ~ ∆ABC

PQ = 6 cm, AB = 8 cm

Perimeter of ∆ABC = 36 cm

Let perimeter of ∆PQR be x cm

$\therefore \frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{\text { Perimeter of } \Delta \mathrm{PQR}}{\text { Perimeter of } \Delta \mathrm{ABC}}$

$\Rightarrow \frac{6}{8}=\frac{x}{36}$

$ \Rightarrow x=\frac{6 \times 36}{8}=27 \mathrm{~cm}$

Perimeter of ΔPQR=27 cm

Ans (b)

Question 10

In the given figure, DE || BC and all measurements are in centimetres. The length of AE is

(a) 2 cm

(b) 2.25 cm

(c) 3.5 cm

(d) 4 cm

Sol :

In the given figure,

DE || BC

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

$ \Rightarrow \frac{3}{4}=\frac{x}{3}$

$\Rightarrow x=\frac{3 \times 3}{4}=\frac{9}{4}=2.25 \mathrm{~cm}$

Ans (b)

Question 11

In the given figure, PQ || CA and all lengths are given in centimetres. The length of BC is

(a) 6.4 cm

(b) 7.5 cm

(c) 8 cm

(d) 9 cm

Sol :

In the given figure,

PQ || CA

Let BC = x

$\therefore \frac{B Q}{Q C}=\frac{B P}{P A}$

$ \Rightarrow \frac{5}{x-5}=\frac{4}{2.4}$

$\Rightarrow 4 x-20=12$

$ \Rightarrow 4 x=12+20=32$

$\Rightarrow x=\frac{32}{4}=8$

∴BC=8 cm 

Ans (c)

Question 12

In the given figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is

(a) 4 cm

(b) 3.6 cm

(c) 2 cm

(d) 3 cm

Sol :

In the given figure, MN || QR

PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm

Let PM = x cm

$\therefore \frac{\mathrm{PM}}{\mathrm{MQ}}=\frac{\mathrm{PN}}{\mathrm{NR}} $

$\Rightarrow \frac{x}{5-x}=\frac{3.6}{2.4}$

$\Rightarrow \frac{x}{5-x}=\frac{36}{24}$

$ \Rightarrow 36(5-x)=24 x$

$\Rightarrow 180-36 x=24 x$

$\Rightarrow 180=24 x+36 x$

$ \Rightarrow 60 x=180$

$\Rightarrow x=\frac{180}{60}=3$

∴PM=3 cm

Ans (d)

Question 13

D and E are respectively the points on the sides AB and AC of a ∆ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then the length of DE is

(a) 2.5 cm

(b) 3 cm

(c) 5 cm

(d) 6 cm

Sol :

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

AD = 2 cm, BD = 3 cm,

BC = 7.5 cm

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

Let DE $=x \mathrm{~cm}$

In $\Delta \mathrm{ABC}, \mathrm{DE} \| \mathrm{BC}$

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

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

$\Rightarrow \frac{2}{2+3}=\frac{x}{7.5} $

$\Rightarrow \frac{2}{5}=\frac{x}{7.5}$

$\Rightarrow x=\frac{2 \times 7.5}{5}=\frac{15}{5}=3$

$\therefore \mathrm{DE}=3 \mathrm{~cm}$

Ans (b)

Question 14

It is given that ∆ABC ~ ∆PQR with $\frac{B O}{Q R}=\frac{1}{3}$ then $\begin{array}{lll}\text { area } & \text { of } & \Delta P Q R \\\hline \text { area } & \text { of } & \Delta A B C\end{array}$ equal to

(a) 9

(b) 3

(c) $\frac{1}{3}$

(d) $\frac{1}{9}$

Sol :

∆ABC ~ ∆PQR

$\frac{B C}{Q R}=\frac{1}{3}$

$\therefore \frac{\text { area of } \Delta \mathrm{PQR}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\mathrm{QR}^{2}}{\mathrm{BC}^{2}}$

$=\frac{(3)^{2}}{(1)^{2}}=\frac{9}{1}=9$

Ans (a)

Question 15

If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio

(a) 9 : 4

(b) 3 : 2

(c) 2 : 3

(d) 16 : 81

Sol :

Ratio in the areas of two similar triangles = 4 : 9

Ratio in their corresponding sides

= √4 : √9

= 2 : 3 (c)

Question 16

If ∆ABC ~ ∆PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of ∆ABC and ∆PQR is

(a) 8 : 6

(b) 3 : 4

(c) 9 : 16

(d) 16 : 9

Sol :

∆ABC ~ ∆PQR, BC = 8 cm, QR = 6 cm

$\begin{array}{ll}\text { area } & \text { of } \quad \Delta A B C \\\hline \text { area } & \text { of } \quad \Delta P Q R\end{array}=\frac{B C^{2}}{Q R^{2}}$

$=\frac{8^{2}}{6^{2}}=\frac{64}{36}=\frac{16}{9}$

Ans (d)

Question 17

If ∆ABC ~ ∆QRP $\begin{array}{lll}\text { area } & \text { of } & \Delta A B C \\\hline \text { area } & \text { of } & \Delta P Q R\end{array}$AB = 18 cm and BC = 15 cm, then the length of PR is equal to

(a) 10 cm

(b) 12 cm

(c) $\frac{20}{3}$

(d) 8 cm

Sol :

∆ABC ~ ∆QRP

$\begin{array}{ccc}\text { area } & \text { of } & \Delta A B C \\\hline \text { area } & \text { of } & \Delta P Q R\end{array}$

$A B=18 \mathrm{~cm}, B C=15 \mathrm{~cm}^{2}$

Length of PR

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{9}{4}=\frac{\mathrm{BC}^{2}}{\mathrm{PR}^{2}}$

$\Rightarrow \frac{9}{4}=\frac{15^{2}}{\mathrm{PR}^{2}}$

$ \Rightarrow \frac{3}{2}=\frac{15}{\mathrm{PR}}$

(Taking square root)

$\therefore P R=\frac{15 \times 2}{3}=10 \mathrm{~cm}$

Ans (a)

Question 18

If ∆ABC ~ ∆PQR, area of ∆ABC = 81 cm², area of ∆PQR = 144 cm² and QR = 6 cm, then length of BC is

(a) 4 cm

(b) 4.5 cm

(c) 9 cm

(d) 12 cm

Sol :

∆ABC ~ ∆PQR,

area of ∆ABC = 81 cm² and area of ∆PQR = 144 cm²,

QR = 6 cm, BC = ?

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}=\frac{\mathrm{BC}^{2}}{(6)^{2}}=\frac{81}{144}$

(given)

$\Rightarrow \frac{\mathrm{BC}^{2}}{(6)^{2}}=\left(\frac{9}{12}\right)^{2} $

$\Rightarrow \frac{\mathrm{BC}}{6}=\frac{9}{12}$

$\therefore \mathrm{BC}=\frac{6 \times 9}{12}=\frac{9}{2}=4.5 \mathrm{~cm}$

Ans (b)

Question 19

In the given figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of ∆ABC to area of ∆BDE is

(a) 4 : 1

(b) 9 : 1

(c) 9 : 4

(d) 3 : 2

Sol :

In the given figure, DE || CA ,

D is a point on BC and BD : DC = 2 : 1

BD : BC = 2 : (2 + 1) = 2 : 3

In ∆ABC, DE || CA

∆ABC ~ ∆BDE

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{BDE}}=\frac{\mathrm{BC}^{2}}{\mathrm{BD}^{2}}=\frac{(3)^{2}}{(2)^{2}}=\frac{9}{4}$

∴Ratio=9 : 4

Ans (c)

Question 20

If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is

(a) 2 : 1

(b) 1 : 2

(c) 1 : 4

(d) 4 : 1

Sol :

∆ABC and ∆BDE are an equilateral triangles

Figure to be added

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

$\mathrm{D}$ is mid-point of $\mathrm{BC}$

$\therefore \mathrm{E}$ is mid-point of $\mathrm{AB}$

$\therefore \mathrm{DE} \| \mathrm{CA}=\frac{1}{2} \mathrm{CA}$

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{BDE}}=\frac{\mathrm{AC}^{2}}{\mathrm{ED}^{2}}=\frac{\mathrm{AC}^{2}}{\left(\frac{1}{2} \mathrm{AC}\right)^{2}}$

$=\frac{\mathrm{AC}^{2}}{\frac{1}{4} \mathrm{AC}^{2}}=\frac{4}{1}=4: 1$

$\therefore$ Ratio $=4: 1$

Ans (d)

Question 21

The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is

(a) 9 cm

(b) 7 cm

(c) 6 cm

(d) 4.5 cm

Sol :

Areas of two similar triangles are 81 cm² and 49 cm²

The altitude of the smaller triangle is 3.5 cm

Let the altitude of the bigger triangle is x, then

$\frac{\text { area of bigger triangle }}{\text { area of smaller triangle }}=\frac{81}{49}$

$=\frac{(\text { bigger altitude })^{2}}{(\text { smaller altitude })^{2}}$

$\Rightarrow \frac{81}{49}=\frac{x^{2}}{(3.5)^{2}}$

$ \Rightarrow \frac{9}{7}=\frac{x}{3.5}$

$\Rightarrow x=\frac{9 \times 3.5}{7}=4.5 \mathrm{~cm}$

$\therefore$ Altitude of bigger triangle $=4.5 \mathrm{~cm}$

Ans (d)

Question 22

Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm² and area of ∆PQR = 24 cm². If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is

(a) $\frac{49}{3} \mathrm{~cm}$

(b) $\frac{20}{3} \mathrm{~cm}$

(c) 15 cm

(d) 22.5 cm

Sol :

∆ABC ~ ∆PQR

area of ∆ABC = 54 cm²

and of ∆PQR = 24 cm²

AD and PM are their median respectively

PM = 10 cm

Let AD = x cm, then

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{AD}^{2}}{\mathrm{PM}^{2}}=\frac{x^{2}}{(10)^{2}}=\frac{54}{24}$

$\therefore \frac{x^{2}}{(10)^{2}}=\frac{54}{24}=\frac{9}{4}=\left(\frac{3}{2}\right)^{2}$

$\therefore \frac{x}{10}=\frac{3}{2}$

$ \Rightarrow x=\frac{3 \times 10}{2}=15 \mathrm{~cm}$

∴AD=15 cm

ML Aggarwal Solution Class 10 Chapter 13 Similarity Exercise 13.3

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20

 Exercise 13.3

Question 1

Given that ∆s ABC and PQR are similar.

Find:

(i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3.

(ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.

Sol :

(i) ∴ ∆ABC ~ ∆PQR

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

(By theorem 15.1)

But BC = QR =1 : 3

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{(1)^{2}}{(3)^{2}}=\frac{1}{9}$

Hence area of $\Delta \mathrm{ABC}:$ area of $\Delta \mathrm{PQR}$

=1: 9

(ii) $\because \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

(By theorem 15.1)

But area of ΔABC=area of ΔPQR

=25 : 36

$\therefore \frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}=\frac{25}{36} \Rightarrow\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{5}{6}\right)^{2}$

$\Rightarrow  \frac{B C}{Q R}=\frac{5}{6}$

$\Rightarrow \mathrm{BC}: \mathrm{QR}=5: 6$

Question 2

∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.

Sol :

Let EF = x

Given that

∆ABC ~ ∆DEF,

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

$\Rightarrow \frac{9}{16}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

$\Rightarrow \frac{(2 \cdot 1)^{2}}{x^{2}}=\frac{9}{16}$

$ \Rightarrow \frac{2 \cdot 1}{x}=\frac{3}{4}$

(Taking square root)

$\Rightarrow 3 x=4 \times 2 \cdot 1$

$ \Rightarrow x=\frac{4 \times 2 \cdot 1}{3}$

∴x=2.8

Hence EF=2.8 cm

Question 3

∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.

Sol :

∆ABC ~ ∆DEF

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

$\Rightarrow \frac{54}{\text { area of } \Delta \mathrm{DEF}}=\frac{(3)^{2}}{(4)^{2}}$

$\Rightarrow \frac{54}{\text { area of } \Delta \mathrm{DEF}}=\frac{9}{16}$

$\Rightarrow$ area of $\Delta \mathrm{DEF}=\frac{54 \times 16}{9}=6 \times 16$

=96 cm

Question 4

The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

Sol :

Let ABC ~ ∆DEF, AL and DM are their altitudes

then area of ∆ABC = 36 cm²

area of ∆DEF = 25 cm² and AL = 2.4 cm.

Let DM = x

Now ∆ABC ~ ∆DEF

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AL}^{2}}{\mathrm{DM}^{2}}$

$\Rightarrow \quad \frac{36}{25}=\frac{(2 \cdot 4)^{2}}{x^{2}}$

$\Rightarrow \quad 36 x^{2}=(2 \cdot 4)^{2} \times 25$

$\Rightarrow \quad x^{2}=\frac{(2 \cdot 4)^{2} \times 25}{36}=\frac{576 \times 25}{100 \times 36}=\frac{16}{4}$

$=4=(2)^{2}$

$\therefore x=2 \mathrm{~cm}$

Hence altitude of the other triangle

=2 cm

Question 5

(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA. (2006)

(b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.

(i) Prove that ∆OAB ~ ∆OCD.

(ii) Find CD and OB.

(iii) Find the ratio of areas of ∆OAB and ∆OCD.

Sol :

In ∆AOQ and ∆BOP, we have

∠ OAQ = ∠ OBP [Each = 90°]

∠ AOQ= ∠BOP

[Vertically opposite angles]

∆AOQ ~ ∆BOP [A.A. similarity]

$\Rightarrow \frac{\text { Area of } \Delta \mathrm{AOQ}}{\text { Area of } \Delta \mathrm{BOP}}=\frac{\mathrm{OQ}^{2}}{\mathrm{PO}^{2}}$

$[\because$ Area's of similar triangles are proportional to the squares of their corresponding sides]

$\frac{\text { Area of } \Delta \mathrm{AOQ}}{120}=\frac{(9)^{2}}{(6)^{2}}$

$\Rightarrow \frac{\text { Area of } \Delta \mathrm{AOQ}}{120}=\frac{81}{36}$

$\Rightarrow$ Area of $\Delta \mathrm{AOQ}=\frac{81 \times 120}{36}=9 \times 30 \mathrm{~cm}^{2}$

$=270 \mathrm{~cm}^{2}$

(b) Given : In the figure AB || CD

AO=10 cm, OC=5 cm

 AB=6.5 cm and OD=2.8 cm

To prove : (i) ΔOAB∼ΔOCD

(ii) Find CD and OB

(iii) Find the ratio of areas of ΔOAB and ΔOCD

Proof : In the ΔOAB and ΔOCD

(i) ∴∠AOB=∠COD

(vertically opposite angles)

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

$\angle \mathrm{OBA}=\angle \mathrm{ODC}$ (Alternate angles)

∴ΔOAB〜ΔOCD (AAA axiom)

(ii) ∴$\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{CD}}$

$\Rightarrow \frac{10}{5}=\frac{\mathrm{OB}}{2.8}=\frac{6.5}{\mathrm{CD}}$

(a) $\frac{\mathrm{OB}}{2.8}=\frac{10}{5}$

$\Rightarrow \mathrm{OB}=\frac{10}{5} \times 2.8=5.6 \mathrm{~cm}$

$\therefore \mathrm{OB}=5.6 \mathrm{~cm}$

(b) $\frac{6.5}{\mathrm{CD}}=\frac{10}{5}$

$\Rightarrow C D=\frac{6.5 \times 5}{10}$

$\mathrm{CD}=\frac{32.5}{10}=3.25 \mathrm{~cm}$

(iii) $\because \Delta \mathrm{OAB} \sim \Delta \mathrm{OCD}$ (Proved)

$\therefore \frac{\operatorname{ar}(\Delta \mathrm{OAB})}{\operatorname{ar}(\Delta \mathrm{OCD})}$

$=\frac{A B^{2}}{C D^{2}}=\frac{(6.5)^{2}}{(3.25)^{2}}$

$=\frac{6.5 \times 6.5}{3.25 \times 3.25}$

$=\frac{2 \times 2}{1}=\frac{4}{1}$

$\operatorname{ar}(\Delta \mathrm{OAB}): \operatorname{ar}(\Delta \mathrm{OCD})=4: 1$

Question 6

(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

(b) In the figure (iii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

Sol :

(a) In the figure,

DE || BC

∠D = ∠B and ∠E = ∠C

(Corresponding angles)

Now in $\Delta \mathrm{ADE}$ and $\Delta \mathrm{ABC}$

$\angle \mathrm{D}=\angle \mathrm{B}, \angle \mathrm{E}=\angle \mathrm{C}$ (proved)

$\angle \mathrm{A}=\angle \mathrm{A}$

$\therefore \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$ (AAA postulate)

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}$

$=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$

$\Rightarrow \frac{28}{\text { area of } \Delta \mathrm{ABC}}$

$=\frac{(6)^{2}}{(9)^{2}}=\frac{36}{81}$

$\Rightarrow$ area of $\Delta \mathrm{ABC}=\frac{28 \times 81}{36}=63$

$\therefore$ Area of $\Delta \mathrm{ABC}=63 \mathrm{~cm}^{2}$

(b) In the figure , DE || BC

$\therefore \angle \mathrm{D}=\angle \mathrm{B}$ and $\angle \mathrm{E}=\angle \mathrm{C}$

(corresponding angles)

Now in ΔADE and ΔABC

∠D=∠B, ∠E=∠C (proved)

∴∠A=∠A (common)

∴ΔADE〜ΔABC (AAA postulate)

But $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{1}{2}$

$\Rightarrow \frac{D B}{A D}=\frac{2}{1}$

Adding 1 both sides

$\frac{\mathrm{DB}}{\mathrm{AD}}+1=\frac{2}{1}+1 $

$\Rightarrow \frac{\mathrm{AD}+\mathrm{DB}}{\mathrm{AD}}=\frac{2+1}{1}$

$\Rightarrow \frac{A B}{A D}=\frac{3}{1} $

$\Rightarrow \frac{A D}{A B}=\frac{1}{3}$

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

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\mathrm{AD}^{2}}{\mathrm{AB}^{2}}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$

$\Rightarrow$ area of $\Delta \mathrm{ABC}=9$ area of $\Delta \mathrm{ADE}$

$\Rightarrow$ area of trapezium DBCE

$\Rightarrow$ area of $\Delta \mathrm{ABC}-$ area of $\Delta \mathrm{ADE}$

$=9$ area of $\Delta \mathrm{ADE}-$ area of $\Delta \mathrm{ADE}$

$=8$ area of $\Delta \mathrm{ADE}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of trapezium DBCE }}=\frac{1}{8}$

$\therefore$ area of $\Delta \mathrm{ADE}$ : area of trepezium DBCE

=1 : 8

Question 7

In the given figure, DE || BC.

(i) Prove that ∆ADE and ∆ABC are similar.

(ii) Given that $A D=\frac{1}{2} B D$, calculate DE if BC=4.5 cm

(iii) If area of ∆ABC = 18cm², find the area of trapezium DBCE

Sol :

(i) Given : In ∆ABC, DE || BC.

To prove : ∆ADE ~ ∆ABC

Proof: In ∆ADE and ∆ABC,

∠A = ∠A (common)

∠ADE = ∠ABC (corresponding angles)

∴ ∆ADE ~ ∆ABC. (AA axiom)

(ii) ∴ ∆ADE ~ ∆ABC

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

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

$\Rightarrow \frac{\frac{1}{2} B D}{\frac{1}{2} B D+B D}=\frac{D E}{4.5}$

(∵BC=4.5)

$\Rightarrow  \frac{\frac{1}{2} B D}{\frac{3}{2} B D}=\frac{D E}{4 \cdot 5}$

$\Rightarrow  \frac{1}{2} \times \frac{2}{3}=\frac{\mathrm{DE}}{4 \cdot 5}$

$\Rightarrow \frac{1}{3}=\frac{\mathrm{DE}}{4 \cdot 5}$

$\therefore \mathrm{DE}=\frac{4 \cdot 5}{3}=1 \cdot 5 \mathrm{~cm}$ 

(iii) Area of $\Delta \mathrm{ABC}=18 \mathrm{~cm}^{2}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

(Area of similar triangles are proportional to the square of their corresponding sides)

$\frac{\text { area of } \Delta \mathrm{ADE}}{18}=\left(\frac{\mathrm{DE}}{\mathrm{BC}}\right)^{2}$

$\Rightarrow \quad \frac{\text { area of } \Delta \mathrm{ADE}}{18}=\left(\frac{\mathrm{AD}}{\mathrm{AB}}\right)^{2}$

$\Rightarrow \quad \frac{\text { area of } \Delta \mathrm{ADE}}{18}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$

[proved in (ii)]

$\Rightarrow$ area of $\Delta \mathrm{ADE}=18 \times \frac{1}{9}=2$

$\therefore$ area of tripezium DBCE

=area of ΔABC-area of ΔADE

=18-2=16$\mathrm{cm}^{2}$

Question 8

In the given figure, AB and DE are perpendicular to BC.

(i) Prove that ∆ABC ~ ∆DEC

(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.

(iii) Find the ratio of the area of ∆ABC : area of ∆DEC

Sol :

(i) To prove : ∆ABC ~ ∆DEC

In ∆ABC and ∆DEC

$\angle \mathrm{ABC}=\angle \mathrm{DEC}=90^{\circ}$ 

$\angle \mathrm{C}=\angle \mathrm{C}$ (common)

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{DEC}$ (by AA axiom)

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

(Corresponding sides of similar triangles are proportional)

$\frac{15}{\mathrm{CD}}=\frac{6}{4}$

$\therefore \mathrm{CD}=\frac{15 \times 4}{6}$

(iii) $\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEC}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}$

$=\frac{6^{2}}{4^{2}}=\frac{36}{16}=\frac{9}{4}=9: 4$

Question 9

In the adjoining figure, ABC is a triangle.

$\mathrm{DE}$ is parallel to $\mathrm{BC}$ and $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$

(i) Determine the ratios $\frac{\mathrm{AD}}{\mathrm{AB}}, \frac{\mathrm{DE}}{\mathrm{BC}}$

(ii) Prove that ∆DEF is similar to ∆CBF.

Hence, find $\frac{E F}{F B}$

(iii) What is the ratio of the areas of ∆DEF and ∆CBF ? (2007)

Sol :

(i) $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$ (given)

$\therefore \frac{\mathrm{DB}}{\mathrm{AD}}=\frac{2}{3}$

or $\frac{\mathrm{DB}}{\mathrm{AD}}+1=\frac{2}{3}+1$

or $\frac{\mathrm{DB}+\mathrm{AD}}{\mathrm{AD}}=\frac{2+3}{3}$

or $\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{5}{3} $

$\Rightarrow \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{3}{5}$

In ΔADE and ΔABC

∠ADE=∠B (Corresponding ∠s)

∠AED=∠C (Corresponding ∠s)

∴By AA similarity

ΔADE〜ΔABC

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

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

(ii) In ΔDEF and ΔCBF

∠1=∠2 (Alternate ∠s)

∠3=∠4 (Alternate ∠s)

∠5=∠6 (Vertically oppo. ∠s)

∴ΔDEF〜ΔCBF

$\frac{E F}{F B}=\frac{D E}{B C}=\frac{3}{5}$

(iii) As the ratio of the area of two similar triangles is equal to the ratio of the square of any two corresponding sides.

$\therefore \frac{\text { Area of } \Delta \mathrm{DFE}}{\text { Area of } \Delta \mathrm{BFC}}=\frac{\mathrm{DE}^{2}}{\mathrm{BC}^{2}}$

$=\left(\frac{\mathrm{DE}}{\mathrm{BC}}\right)^{2}=\left(\frac{3}{5}\right)^{2}=\frac{9}{25}$

Question 10

In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:

(i) area ∆APO : area ∆ABC.

(ii) area ∆APO : area ∆CQO. (2008)

Sol :

In the figure,

PQ || BC and PO is produced to Q such that CQ || BA

and AP : PB = 2 : 3.

Figure to be added

(i) Now in ΔAPO and ΔABC

∠A=∠A (Common)

∠APO=∠ABC (Corresponding angles)

∴ΔAPO~ΔABC (AA axiom)

∵Areas of the similar triangle are proportional to the square of their corresponding sides

$\therefore \frac{\operatorname{area}(\Delta \mathrm{APO})}{\operatorname{area}(\Delta \mathrm{ABC})}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}=\frac{\mathrm{AP}^{2}}{(\mathrm{AP}+\mathrm{PB})^{2}}$

$=\frac{(2)^{2}}{(2+3)^{2}}=\frac{4}{(5)^{2}}=\frac{4}{25}$

area $(\triangle \mathrm{APO}):$ area $(\Delta \mathrm{ABC})=4: 25$

(ii) In $\Delta \mathrm{APO}$ and $\Delta \mathrm{CQO}$

$\angle \mathrm{AOP}=\angle \mathrm{COQ}$

(vertically opposite angles)

$\angle \mathrm{APQ}=\angle \mathrm{OQC}$ (Alternate angles)

∴ΔAPQ～ΔCQO

$\therefore \frac{\operatorname{area}(\Delta \mathrm{APO})}{\operatorname{area}(\Delta C Q O)}=\frac{A P^{2}}{C Q^{2}}=\frac{A P^{2}}{P B^{2}}$

$(\because P B=C Q)$

$=\frac{(2)^{2}}{(3)^{2}}=\frac{4}{9}$

area (ΔAPO) : area (ΔCQO)

=4 : 9

Question 11

(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.

(b) In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find

(i) AB

(ii) BC

(iii) area of ∆ADM : area of ∆ANB.

(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find

(i) EF : AD

(ii) area of ∆BEF : area of ∆ABD

(iii) area of ∆ABD : area of trap. AFED

(iv) area of ∆FEO : area of ∆OBC.

Sol :

(a) In trapezium ABCD, AB || DC.

∠OAB = ∠OCD [alternate angles]

∠OBA = ∠ODC

∆AOB ~ ∆COD

$\therefore \frac{\text { area of } \Delta \mathrm{AOB}}{\text { area of } \Delta \mathrm{COD}}=\frac{\mathrm{AB}^{2}}{\mathrm{CD}^{2}}$

$=\frac{(2 \mathrm{CD})^{2}}{\mathrm{CD}^{2}}$ $(\because A B=2 C D)$

$=\frac{4 \mathrm{CD}^{2}}{\mathrm{CD}^{2}}=\frac{4}{1}$

∴area of ΔAOB : area of ΔCOD= 4 : 1

(b) In ||gm ABCD, AM⊥DC and AN⊥CB 

Now area of ||gm ABCD= DC×AM or

BC×AN

∴DC×AM=BC×AN=area of ||gm

⇒DC×6=BC×10=45

(i) ∴$\mathrm{DC}=\frac{45}{6}=\frac{15}{2}=7 \cdot 5 \mathrm{~cm}$

∴AB=7.5 cm (∵AB=DC)

(ii) Now in ΔADM and ΔABN

∠D=∠B (opposite angles of a ||gm)

∠M=∠N (each 90°)

$\therefore \Delta \mathrm{ADM} \sim \Delta \mathrm{ABN}$

$\therefore \frac{\text { area of } \Delta \mathrm{ADM}}{\text { area of } \Delta \mathrm{ABN}}=\frac{\mathrm{AD}^{2}}{\mathrm{AB}^{2}}$

$=\frac{B C^{2}}{A B^{2}}=\frac{(4 \cdot 5)^{2}}{(7 \cdot 5)^{2}}$

$=\frac{20 \cdot 25}{56 \cdot 25}=\frac{2025}{5625}$

$=\frac{81}{225}=\frac{9}{25}$

$\therefore$ area of $\Delta$ ADM : area of $\Delta$ ABN

= 9 : 25

(c) In ||gm ABCD, E is a point on AB , CE intersects the diagonal BD at O

EF || BC and AE : EB=2 : 3

In $\Delta \mathrm{ABD}, \mathrm{EF} \| \mathrm{BC}$ or $\mathrm{AD}$

(i) $\therefore \frac{\mathrm{AB}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{EF}}$

$\Rightarrow \frac{\mathrm{EF}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{AB}}$

But $\frac{A E}{E B}=\frac{2}{3}$

$\Rightarrow \frac{A E}{E B}+1=\frac{2}{3}+1$

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

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{EB}}=\frac{5}{3}$

$ \Rightarrow \frac{\mathrm{BE}}{\mathrm{AB}}=\frac{3}{5}$

⇒EF : AD= 3 : 5

(ii) ∵ΔBEF~ΔABD

$\frac{\text { area of } \Delta \mathrm{BEF}}{\text { area of } \Delta \mathrm{ABD}}=\frac{(\mathrm{EF})^{2}}{(\mathrm{AD})^{2}}$

$=\frac{(3)^{2}}{(5)^{2}}=\frac{9}{25}$

$\therefore$ area of $\Delta \mathrm{BEF}:$ area of $\Delta \mathrm{ABD}=9: 25$

(iii) $\frac{\text { area of } \Delta \mathrm{ABD}}{\text { area of } \Delta \mathrm{BEF}}=\frac{25}{9}$ (from (ii))

25 area of $\Delta \mathrm{BEF}=9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ (area of $\Delta \mathrm{ABD}$ - area of trap AEFD)

$=9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ area of $\Delta \mathrm{ABD}-25$ area of AFED

$=9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ area of trap AEFD $=25$ area of $\Delta \mathrm{ABD}-9$ area of $\Delta \mathrm{ABD}$

$\Rightarrow 25$ area of $A E F D=16$ area of $\Delta A B D$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{ABD}}{\text { area of trap } \mathrm{AEFD}}=\frac{25}{16}$

$\Rightarrow$ area of $\Delta \mathrm{ABD}:$ area of trap $\mathrm{AEFD}$

= 25 : 16

(iv) In $\Delta$ FEO and $\Delta$ OBC

$\angle \mathrm{EOF}=\angle \mathrm{BOC}$

(Vertically opposite angles)

$\angle \mathrm{F}=\angle \mathrm{OBC}$ (Alternate angles)

$\therefore \Delta \mathrm{FEO} \sim \Delta \mathrm{OBC}$

$\therefore \frac{\text { area of FEO }}{\text { area of } \Delta \mathrm{OBC}}=\frac{\mathrm{EF}^{2}}{\mathrm{BC}^{2}}$

$=\frac{E F^{2}}{A D^{2}}=\frac{9}{25}$ [From (i)]

$\therefore$ area of $\Delta \mathrm{FEO}:$ area of $\Delta \mathrm{OBC}=9: 25$

Question 12

In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find

(i) area of ∆BPQ.

(ii) area ∆CDP.

(iii) area of || gm ABCD.

Sol :

In the figure, ABCD is a parallelogram.

P is a point on BC such that BP : PC = 1 : 2

and DP is produced to meet ABC produced at Q.

Area ∆CPQ = 20 cm²

Draw QN $\perp$ CB (Produced)

$\frac{\operatorname{area} \Delta \mathrm{BPQ}}{\text { area } \Delta \mathrm{CPQ}}=\frac{\frac{1}{2} \mathrm{BP} \times \mathrm{QN}}{\frac{1}{2} \mathrm{PC} \times \mathrm{QN}}$

$=\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{1}{2}$ (given)

(i) ∴Area $\Delta \mathrm{BPQ}=\frac{1}{2}$ area $\Delta \mathrm{CPQ}=\frac{1}{2} \times 20$ $\mathrm{cm}^{2}=10 \mathrm{~cm}^{2}$

(ii) Now in $\Delta$ CDP and $\Delta$ BQP,

$\angle \mathrm{CPD}=\angle \mathrm{QPB}$ (Vertically opposite angles)

$\angle \mathrm{PDC}=\angle \mathrm{POB} $ (Alternate angles)

$\therefore \Delta \mathrm{CDP} \sim \Delta \mathrm{BQP}$ (AA axiom)

$\therefore \frac{\text { area } \Delta \mathrm{CDP}}{\text { area } \Delta \mathrm{BQP}}=\frac{\mathrm{PC}^{2}}{\mathrm{BP}^{2}}$

⇒$\frac{\operatorname{areaCDP}}{\operatorname{areaBQP}}=\frac{(2)^{2}}{(1)^{2}}$

$=\frac{4}{1}$

$\therefore$ area $\Delta \mathrm{CDP}=4 \times$ area $\Delta \mathrm{BQP}$

$=4 \times 10=40 \mathrm{~cm}^{2}$

(iii) Now area of ||gm ABCD=2 Area ΔDCQ

{ΔDCQ and ||gm ABCD are one the same base and between the same parallels}

=2 area (ΔDCP+ΔCPQ)

=2(40+20)$\mathrm{cm}^{2}$

$=2 \times 60 \mathrm{~cm}^{2}=120 \mathrm{~cm}^{2}$

Question 13

(a) In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of ∆EDC : area of trapezium ABCD.

Sol :

(a) In ∆ABC, DE || BC

Now in ∆ABC and ∆ADE

∠A = ∠A (common)

∠D = ∠B and ∠E = ∠C

(Corresponding angles)

∆ADE ~ ∆ABC

$\therefore \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}$...(i)

But $\frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of trap } \mathrm{DBCE}}=\frac{4}{5}$

$\Rightarrow \frac{\text { area of trap } \mathrm{DBCE}}{\text { area of } \Delta \mathrm{ADE}}=\frac{5}{4}$

$\Rightarrow \frac{\text { area of trap } \mathrm{DBCE}}{\text { area of } \Delta \mathrm{ADE}}+1$

$=\frac{5}{4}+1$

(Adding 1 both sides)

$\Rightarrow \frac{\text { area of trap } \mathrm{DBCE}+\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ADE}}$

$=\frac{5+4}{4}=\frac{9}{4}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{ADE}}=\frac{9}{4}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{ADE}}{\text { area of } \Delta \mathrm{ABC}}=\frac{4}{9}$

Now from (i) $\frac{(\mathrm{DE})^{2}}{(\mathrm{BC})^{2}}=\frac{4}{9}$

$\Rightarrow \frac{(2)^{2}}{(3)^{2}}$

$\Rightarrow \frac{\mathrm{DE}}{\mathrm{BC}}=\frac{2}{3}$

$\Rightarrow \mathrm{DE}: \mathrm{BC}=2: 3$

(b) In the figure , DC||AB, AB=2 DC

AD=3 cm, BC=4 cm

In ΔEAB, DC||AB

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

$=\frac{A B}{D C}=\frac{2 D C}{D C}=\frac{2}{1}$

(i) ∴EA=2, DA=2×3=6

ED=EA-DA=6-3=3cm

(ii) $\frac{E B}{C B}=\frac{2}{1}$

$ \Rightarrow E B=2 C B=2 \times 4=8 \mathrm{~cm}$

$\therefore B E=8 \mathrm{~cm}$

(iii) $\therefore \operatorname{In} \Delta \mathrm{EAB}, \mathrm{DC} \| \mathrm{AB}$

$\therefore \Delta \mathrm{EDC} \sim \Delta \mathrm{EAB}$

$\therefore \frac{\text { area of } \Delta \mathrm{EDC}}{\text { area of } \Delta \mathrm{ABE}}=\frac{\mathrm{DC}^{2}}{\mathrm{AB}^{2}}$

$=\frac{\mathrm{DC}^{2}}{(2 \mathrm{DC})^{2}}$

$=\frac{D C^{2}}{4 D C^{2}}=\frac{1}{4}$

$=\frac{1}{4}$

$\therefore$ area of $\mathrm{ABE}=4$ area of $\Delta \mathrm{EDC}$

$\Rightarrow$ area of $\Delta \mathrm{EDC}+$ area of trap $\mathrm{ABCD}=4$ area of $\Delta$ EDC

area of $\operatorname{trap} A B C D=4$ area of $\Delta E D C-$ area of $\Delta$ EDC

$\Rightarrow$ area of $\operatorname{trap} \mathrm{ABCD}=3$ area of $\Delta \mathrm{EDC}$

$\Rightarrow \frac{\text { area of } \Delta \mathrm{EDC}}{\text { area of trap } \mathrm{ABCD}}=\frac{1}{3}$

$\Rightarrow$ area of $\Delta E D C:$ area of trap. $A B C D=1: 3$

Question 14

(a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find

(i) BP

(ii) the ratio of areas of ∆APB and ∆DPC.

(b) In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.

(i) Prove that ∆ACD is similar to ∆BCA

(ii) Find BC and CD

(iii) Find the area of ∆ACD : area of ∆ABC.

Sol :

(a) In trapezium ABCD, DC || AB

AB=9 cm, DC=6 cm, BD=12 cm

(i) In $\Delta \mathrm{APB}$ and $\Delta \mathrm{CPD}$

$\angle \mathrm{APB}=\angle \mathrm{CPD}$ 

(Vertically opposite angles)

$\angle \mathrm{PAB}=\angle \mathrm{PCD}$ (alternate angles)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{CPD}$ (alternate angles)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{CPD}$ (AA postulate)

$\therefore \frac{\mathrm{BP}}{\mathrm{PD}}=\frac{\mathrm{AB}}{\mathrm{CD}}$

$ \Rightarrow \frac{\mathrm{BP}}{12-\mathrm{BP}}=\frac{9}{6}$

$\Rightarrow  6 \mathrm{BP}=108-9 \mathrm{BP}$

$\Rightarrow  6 \mathrm{BP}+9 \mathrm{BP}=108$

$\Rightarrow 15 \mathrm{BP}=108$

$ \Rightarrow \mathrm{BP}=\frac{108}{15}=7 \cdot 2 \mathrm{~cm}$

(ii) Again 

∵ΔAPB～ΔCPD

$\therefore \frac{\text { area of } \Delta \mathrm{APB}}{\text { area of } \Delta \mathrm{CPD}}=\frac{\mathrm{AB}^{2}}{\mathrm{CD}^{2}}$

$=\frac{(9)^{2}}{(6)^{2}}=\frac{81}{36}=\frac{9}{4}$

(b) In ΔACD and ΔBCA

∠C=∠C (common)

∠ABC=∠CAD (Given)

∴ΔACD≅ΔBCA (by AA axiom)

$\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{CD}}{\mathrm{CA}}=\frac{\mathrm{AD}}{\mathrm{AB}}$

$\Rightarrow \frac{4}{\mathrm{BC}}=\frac{\mathrm{CD}}{4}=\frac{5}{8}$

$\therefore \frac{4}{\mathrm{BC}}=\frac{5}{8}$

$\mathrm{BC}=\frac{4 \times 8}{5}=\frac{32}{5}=6.4 \mathrm{~cm}$

$\therefore \frac{C D}{4}=\frac{5}{8}$

$\Rightarrow \mathrm{CD}=\frac{5}{8} \times 4=\frac{5}{2}=2.5 \mathrm{~cm}$

(iii) $\because \Delta \mathrm{ACD} \cong \Delta \mathrm{BCA}$

$\therefore \frac{\operatorname{area}(\Delta \mathrm{ACB})}{\operatorname{area}(\Delta \mathrm{BCA})}=\frac{\mathrm{AC}^{2}}{\mathrm{AB}^{2}}$

$=\frac{(4)^{2}}{(8)^{2}}=\frac{16}{64}=\frac{1}{4}$

area $(\Delta \mathrm{ACD}):$ area $(\Delta \mathrm{ABC})=1: 4$

Question 15

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:

(i) ∆ADE ~ ∆ACB.

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

(iii) Find, area of ∆ADE : area of quadrilateral BCED.

Sol :

(i) Consider DADE and DACB

∠A = ∠A (Common)

m∠B = m∠E = 90°

Thus by angle-angle similarity, triangles,

∆ACB ~ ∆ADE

(ii) Consider ∆ADE and ∆ACB

Since they are similar triangles,

the sides are proportional Thus, we have,

$\frac{\mathrm{AE}}{\mathrm{AB}}=\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}$..(i)

Consider $\Delta \mathrm{ABC}$

By applying Pythagoras Theorem, we have, $\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$

$\Rightarrow \mathrm{AB}^{2}+5^{2}=13^{2} \Rightarrow \mathrm{AB}^{2}+25=169$

$\Rightarrow \mathrm{AB}^{2}=169-25=144$

$\Rightarrow \mathrm{AB}=12 \mathrm{~cm}$

From equation (1), we have

$\frac{4}{12}=\frac{\mathrm{AD}}{13}=\frac{\mathrm{DE}}{5}$

$\Rightarrow \frac{1}{3}=\frac{\mathrm{AD}}{13}$

$\Rightarrow \mathrm{AD}=\frac{13}{3} \mathrm{~cm}=4.33 \mathrm{~cm}$

Also, $\frac{4}{12}=\frac{\mathrm{DE}}{5}$

$\Rightarrow \mathrm{DE}=\frac{20}{12}=\frac{5}{3} \mathrm{~cm}=1.67 \mathrm{~cm}$

(iii) We need to find the area of ΔADE and quadrilateral BCED

Area of $\Delta \mathrm{ADE}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{DE}$

$=\frac{1}{2} \times 4 \times \frac{5}{3}$

$=\frac{10}{3} \mathrm{~cm}^{2}$

Area of quadrilateral BCED=Area of ΔABC-Area of ΔADE

$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}-\frac{10}{3}$

Area of quad. $\mathrm{BCED}=$ Area of $\Delta \mathrm{ABC}-$ Area of $\Delta \mathrm{ADE}$

$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}-\frac{10}{3}$

$=\frac{1}{2} \times 5 \times 12-\frac{10}{3}$

$=30-\frac{10}{3}$

$=\frac{90-10}{3}=\frac{80}{3} \mathrm{~cm}^{2}$

Thus ratio of areas of $\Delta \mathrm{ADE}$ to quadrilateral

$\mathrm{BCED}=\frac{\frac{10}{3}}{\frac{80}{3}}=\frac{1}{8}$

Question 16

Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding height.

Sol :

In two isosceles ∆s ABC and DEF

$\angle \mathrm{A}=\angle \mathrm{D}$ (given)

$\therefore \angle \mathrm{B}+\angle \mathrm{C}=\angle \mathrm{E}+\angle \mathrm{F}$

But $\angle \mathrm{B}=\angle \mathrm{C}$ and $\angle \mathrm{E}=\angle \mathrm{F}$

(opposite angles of equal sides)

$\therefore \angle B=\angle E$ and $\angle C=\angle F$

$\therefore \Delta A B C \sim \Delta D E F$

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AL}^{2}}{\mathrm{DM}^{2}}$

(Cor. of theorem 13.1)

$\Rightarrow \frac{\mathrm{AL}^{2}}{\mathrm{DM}^{2}}=\frac{7}{16} $

$\Rightarrow \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\sqrt{7}}{4}$

Hence $\mathrm{AL}: \mathrm{DM}=\sqrt{7}: 4$

Question 17

On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements :

AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate

(i) the actual length of AB in km.

(ii) the area of the plot in sq. km:

Sol :

Scale factor k = 1 : 250000 = $\frac{1}{250000}$

Length on map,

AB = 3 cm, BC = 4 cm

$\therefore$ Length of $A B$ of Actual plot $=\frac{1}{k}$ (Length of AB on the map)

=250000 (3 cm)

$=\frac{250000 \times 3}{100 \times 1000} \mathrm{~km}=\frac{15}{2}=7 \cdot 5 \mathrm{~km}$

(ii) Area of plot on the map $=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$

$=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^{2}$

$\therefore$ Area of actual plot $=\frac{1}{k^{2}} \times 6 \mathrm{~cm}^{2}$

$=(250000)^{2} \times 6 \mathrm{~cm}^{2}$

$=\frac{250000 \times 250000 \times 6}{100000 \times 100000} \mathrm{~km}^{2}$

$=\frac{25}{4} \times 6=\frac{75}{2}=37 \cdot 5 \mathrm{~km}^{2}$

Question 18

On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm.

Calculate:

(i) the distance of a diagonal of the plot in km.

(ii) the area of the plot in sq. km.

Sol :

Scale factor $(\mathrm{k})=\frac{1}{25000}$

Measurements of plot ABCD on the map are

AB = 12 cm and BC = 16 cm.

Diagonal $A C=\sqrt{A B^{2}+B C^{2}}$

$=\sqrt{(12)^{2}+(16)^{2}}=\sqrt{144+256}$

$=\sqrt{400}=20 \mathrm{~cm}$

and area=AB×BC

$=12 \mathrm{~cm} \times 16 \mathrm{~cm}=192 \mathrm{~cm}^{2}$

(i) Now actual length of $\mathrm{AC}=\frac{1}{k}$

(Length of AC on map)

$=25000 \times 20 \mathrm{~cm}=\frac{25000 \times 20}{100 \times 1000} \mathrm{~km}$

=5 km

(ii) Area of plot $=\left(\frac{1}{k}\right)^{2}$ (Area of plot on map)

Question 19

The model of a building is constructed with the scale factor 1 : 30.

(i) If the height of the model is 80 cm, find the actual height of the building in metres.

(ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model. (2009)

Sol :

(i) $\frac{\text { Height of model }}{\text { Height of actual building }}=\frac{1}{30}$

$\frac{80}{\mathrm{H}}=\frac{1}{30}$

$\Rightarrow \mathrm{H}=2400 \mathrm{~cm}=24 \mathrm{~m}$

(ii) $\frac{\text { Volume of model }}{\text { Volume of tank }}=\left(\frac{1}{30}\right)^{3}$

$\frac{\mathrm{V}}{27}=\frac{1}{27000}$

$\Rightarrow \mathrm{V}=\frac{1}{1000} \mathrm{~m}^{3}=1000 \mathrm{~cm}^{3}$

Question 20

A model of a ship is made to a scale of 1 : 200.

(i) If the length of the model is 4 m, find the length of the ship.

(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.

(iii) If the volume of the model is 200 litres, find the volume of the ship in m³.

(100 litres = m³)

Sol :

Scale = 1 : 200

(i) Length of a model of ship = 4 m

$\therefore$ Length of the ship $=\frac{4 \times 200}{1}=800 \mathrm{~m}$

(ii) Area of deck of ship $=160000 \mathrm{~m}^{2}$

$\therefore$ Area of deck of the model

$=160000 \times\left(\frac{1}{200}\right)^{2} \mathrm{~m}^{2}$

$=160000 \times \frac{1}{40000}=4 \mathrm{~m}^{2}$

(iii) Volume of the model of the ship =200 l

Volume of ship $=200 \times\left(\frac{200}{1}\right)^{3} l$

$=200 \times 8000000 l$

$=\frac{200 \times 8000000}{1000} \mathrm{~m}^{3}=1600000 \mathrm{~m}^{3}$