Class 10: Similarity – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: sm1 In the given figure, $AB$ and $DE$ are perpendicular to $BC$ . If $AB = 9 \ cm, DE=3 \ cm \ and \ AC=24 cm$ , calculate $AD$ . [2005]

Answer:

In $\triangle ABC$ and $\triangle DEC$ :

$\angle ABC = \angle DEC$ (perpendiculars)

$\angle C$ (common angle)

$\triangle ABC \sim \triangle DEC$ (by AAA postulate)

$\Rightarrow$ $\frac{AC}{DC}$ $=$ $\frac{AB}{DE}$

$\Rightarrow DC =$ $\frac{3}{9}$ $\times 24 = 8 \ cm$

Therefore $AD = AC - DC = 24-8 = 16 \ cm$

$\\$

Question 2: sm10 In the given figure, $\triangle ABC$ and $\triangle AMP$ are right angled at $B$ and $M$ respectively. Given $AC=10 \ cm, AP = 15 \ cm$ and $PM= 12 \ cm$ . (i) Prove $\triangle ABC \sim \triangle AMP$ (ii) Find $AB$ and $AC$ . [2012]

Answer:

In $\triangle ABC \ and \ \triangle AMP$

$\angle BAC = \angle PAM$ (common angle)

$\angle ABC =\angle PMA$ (right angles)

Therefore $\triangle ABC \sim \triangle AMP$ (AAA postulate)

$AM = \sqrt{AP^2-PM^2} = \sqrt{15^2-12^2} = \sqrt{121} = 11$

Since $\triangle ABC \sim \triangle AMP$

$\frac{AP}{AM}$ $=$ $\frac{BC}{PM}$ $=$ $\frac{AC}{AP}$

Given AC=10 cm, AP = 15 cm and PM= 12 cm

$\Rightarrow AB =$ $\frac{10}{15}$ $\times 11 = 7.33 \ cm$

$\Rightarrow BC =$ $\frac{10}{15}$ $\times 12 = 8 \ cm$

$\\$

Question 3: sm11 In the figure, $PQRS$ is a parallelogram with $PQ=16 \ cm$ and $QR=10 \ cm$ . $L$ is a point on $PR$ such that $RL:LP=2:3$ . $QL$ produced meets $RS$ at $M$ and $PS$ produced at $N$ . Find the lengths of $PN$ and $RM$ . [1997]

Answer:

In $\triangle RLQ$ and $\triangle PLN$

$\angle RLQ = \angle PLN$ (vertically opposite angles)

$\angle LRQ = \angle LPN$ (alternate angles)

Therefore $\triangle RLQ \sim \triangle PLN$ (AAA postulate)

Therefore $\frac{RL}{LP}$ $=$ $\frac{RQ}{PN}$

$\Rightarrow PN =$ $\frac{10}{2}$ $\times 3 = 15 \ cm$

In $\triangle RLM \ and \ \triangle PLQ$

$\angle RLM = \angle PLQ$ (vertically opposite angles)

$\angle LRM = \angle LPQ$ (alternate angles)

Therefore $\triangle RLM \sim \triangle PLQ$ (AAA postulate)

Therefore $\frac{RM}{PQ}$ $=$ $\frac{RL}{LP}$

$\Rightarrow RM =$ $\frac{2}{3}$ $\times 16 = 10.667 \ cm$

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Question 4: sm12 In the figure given below, $P$ is a point on $AB$ such that $AP:PB=4:3$ . $PQ \parallel AC$ .

(i) Calculate the ratio $PQ:AC$ , giving reasons for your answer.

(ii) In $\triangle ARC, \angle ARC = 90^o$ and in $\triangle PQS, \angle PSQ=90^o$ . Given $QS=6 \ cm$ , calculate length of $AR$ . [1999]

Answer:

(i) Given $AP:PB=4:3$

Also $PQ \parallel AC$ , applying basic proportionality theorem

$\frac{AP}{PB}$ $=$ $\frac{CQ}{QB}$

$\Rightarrow$ $\frac{CQ}{QB}$ $=$ $\frac{4}{3}$

$\Rightarrow$ $\frac{BQ}{BC}$ $=$ $\frac{3}{7}$

$\angle QBP = \angle ACB$ (corresponding angles)

$\angle QPB = \angle CAB$ (corresponding angles)

Therefore $\triangle PBQ \sim \triangle ABC$ (AAA postulate)

$\frac{PQ}{ AC}$ $=$ $\frac{BQ}{ BC}$ $=$ $\frac{3}{ 7}$

(ii) Given $\angle ARC = \angle QSP = 90^o$

$\angle ACR = \angle SPQ$ (alternate angles)

Therefore $\triangle ARC \sim \triangle QSP$ (AAA postulate)

$\frac{AR}{ SQ}$ $=$ $\frac{AC}{ PQ}$ $=$ $\frac{7}{ 3}$

$\Rightarrow AR =$ $\frac{7 \times 6}{3}$ $= 14 \ cm$

$\\$

Question 5: In the right angled $\triangle QPR, PM$ is the altitude. Given that $QR= 8 cm$ and $MQ = 3.5 cm$ , calculate the value of $PR$ . [2000]

Answer:

$\angle QPR = \angle PMR = 90^o$

$\angle PRQ = \angle PRM$ (common)

$\triangle PQR \sim \triangle MPR$ (AAA postulate)

$\frac{QR}{ PR}$ $=$ $\frac{ PR}{ MR}$

$PR^2 = QR \times MR = 8 \times 4.5 = 36 \Rightarrow PR = 6 \ cm$

$\\$

Question 6: sm14 In the given figure $DE \parallel BC$ .

(i) Prove that $\triangle ADE$ and $\triangle ABC$ are similar.

(ii) Given that $AD=$ $\frac{1}{2}$ $BD$ , calculate $DE$ , if $BC=4.5 \ cm$ . Also find:

$\frac{Ar. (\triangle ADE)}{Ar. (\triangle ABC)}$

and $\frac{Ar. (\triangle ADE)}{Ar. (trapezium \ BCED)}$ . [2004]

Answer:

(i) $DE \parallel BC$

$\Rightarrow \angle ADE = \angle ABC$ (corresponding angles)

and $\angle AED = \angle ACB$ (corresponding angles)

Therefore $\triangle ADE \sim \triangle ABC$ (AAA postulate)

(ii) Since $\triangle ADE \sim \triangle ABC$

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

$\Rightarrow DE =$ $\frac{1}{ 3}$ $\times 4.5 = 1.5 \ cm$

Given $AD =$ $\frac{1}{ 2}$ $\times BD$

$\Rightarrow$ $\frac{AD}{ BD}$ $= \frac{1}{ 2}$

$\Rightarrow$ $\frac{AD} {AD+BD}$ $=$ $\frac{1} {1+2}$

$\Rightarrow$ $\frac{AD}{ AB}$ $=$ $\frac{1}{ 3}$

(iii) Since $\triangle ADE \sim \triangle ABC$

$\frac{Area(\triangle ADE)}{Area (\triangle ABC)}$ $=$ $\frac{DE^2}{BC^2}$ $=$ $\frac{1.5^2}{4.5^2}$ $=$ $\frac{1}{9}$

$\Rightarrow \frac{Area (\triangle ADE)}{Area (\triangle ABC)- Area(\triangle ADE)}$ $=$ $\frac{1}{9-1}$

$\Rightarrow \frac{Area(\triangle ADE)}{Area (trapezium BCED)}$ $=$ $\frac{1}{8}$

$\\$

Question 7: sm9 In the figure given below, $PB$ and $QA$ are perpendiculars to the line segment $AB$ . If $PO=6 \ cm, QO=9 \ cm$ and area of $\triangle POB = 120\ cm^2$ , find the area of $\triangle QOA$ . [2006]

Answer:

In $\triangle POB \ and \ \triangle QOA$

$\angle PBO = \angle QAO = 90^o$

$\angle POB = \angle QOA$ (vertically opposite angles)

Therefore $\triangle POB \sim \triangle QOA$ (By AAA postulate)

$\Rightarrow \frac{Ar.(\triangle POB)}{Ar.(\triangle QOA)}$ $=$ $\frac{PO^2}{QO^2}$

$\Rightarrow Ar.(\triangle QOA) =$ $\frac{9^2}{6^2}$ $\times 120 = 270 \ cm^2$

$\\$

Question 8: sm81 In the figure given below, $ABCD$ is a parallelogram. $P$ is a point on $BC$ such that $BP:PC = 1:2$ . $DP$ produced meets $AB$ produced at $Q$ . Given the area of $\triangle CPQ=20 \ cm^2$ . Calculate:

(i) area of $\triangle CDP$

(ii) area of parallelogram $ABCD$ . [1996]

Answer:

(i) In $\triangle BPQ \ and \ \triangle CPD$

$\angle BPQ = \angle CPD$ (vertically opposite angles)

$\angle BQP = \angle PDC$ (alternate angles)

$\triangle BPQ \sim \triangle CPD$

Therefore $\frac{BP}{ PC}$ $=$ $\frac{ PQ}{ PD}$ $=$ $\frac{ BQ}{ CD}$ $=$ $\frac{ 1}{ 2}$

Also $\frac{Area \triangle BPQ}{Area \triangle CPD}$ $=$ $\frac{BP^2}{PC^2}$

$Area \triangle CPD =$ $\frac{4}{1}$ $\times 10 = 40 \ cm^2$

(ii) In $\triangle BAP \ and \ \triangle AQD$

$BP \parallel AD$ (Given)

$\angle QBP = \angle QAD$ (corresponding angles are equal)

$\angle BQP = \angle AQD$ (common angle)

$\triangle BQP \sim \triangle AQD$

Therefore $\frac{AQ}{ BQ}$ $=$ $\frac{QD}{ QP}$ $=$ $\frac{AD}{ BP}$ $= 3$

Also $\frac{Area \triangle AQD}{Area \triangle BQP}$ $=$ $\frac{AQ^2}{BQ^2}$

$Area \triangle AQD = 3^2 \times 10 = 90 \ cm^2$

$Area (ABCD) = Area \triangle AQD -Area \triangle BQP + Area \triangle CDP = 90-10+40 = 120 \ cm^2$

$\\$

Question 9: A model of a ship is made to scale of $1:200$ .

(i) The length of the model is $4 \ m$ ; calculate the length of the ship.

(ii) The area of the deck of the ship is $160000 \ m^2$ ; find the area of the deck of the model.

(iii) The volume of the model is $200 \ liters$ ; calculate the volume of the ship in $m^3$ . [1995]

Answer:

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

(i) Length of the model $= k \times$ Actual length of the ship

$\Rightarrow$ Actual length of the ship $= 4 \times 200 = 800 \ m$

(ii) Area of the deck of the model $= k^2 \times$ area of the deck of the actual ship

$= ($ $\frac{1}{200}$ $)^2 \times 160000 \ m^2 = 4 \ m^2$

(iii) Volume of the model $= k^3 \times$ Volume of the actual ship

$= ($ $\frac{1}{k}$ $)^3 \times 200 = (200)^3 \times 200 = 1600000000 \ liters = 16000000 \ m^3$

$\\$

Question 10: sm7 In the figure given below $ABC$ is a triangle. $DE$ is parallel to $BC$ and $\frac{AD}{DB}$ $=$ $\frac{3}{2}$ .

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

(ii) Prove that $\triangle DEF$ is similar to $\triangle CBF$ . Hence , find $\frac{EF}{FB}$ . [2007]

Answer:

(i) Given $DE \parallel BC \ and \$ $\frac{AD}{DB}$ $=$ $\frac{3}{2}$

$\triangle ADE \ and\ \triangle ABC$

$\angle BAC = \angle DAC$ (common angle)

$\angle ADE = \angle ABC$

$\triangle ADE \sim \triangle ABC$

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

$\frac{AD }{AB}$ $=$ $\frac{AD }{AD+DB}$ $=$ $\frac{3}{ 5}$

$\frac{AD}{ AE}$ $=$ $\frac{DE }{BC}$ $=$ $\frac{3}{ 5}$

(ii) In $\triangle DEF and \triangle CBF$

$\angle FDE = \angle FCB$ (alternate angles)

$\angle DFE = \angle BFC$ (vertically opposite angles)

$\triangle DEF \sim \triangle CBF$

$\frac{EF}{ FB}$ $=$ $\frac{DE}{ BC}$ $=$ $\frac{3}{ 5}$

$\frac{EF}{ FB }$ $=$ $\frac{3 }{5}$

(iii) We know

$\frac{Area \ of \ \triangle DEF}{Area \ of \ \triangle CBF}$ $=$ $\frac{EF^2}{ FB^2}$ $=$ $\frac{3^2 }{5^2 }$ $=$ $\frac{9 }{25}$

$\\$

Question 11: sm6 In $\triangle ABC, \angle ABC = \angle DAC, AB= 8 \ cm,$ $AC = 4 \ cm \ and \ AD = 5 \ cm$ .

(i) Prove that $\triangle ACD$ is similar to $\triangle BCA$ .

(ii) Find $BC \ and \ CD$ .

(iii) Find $area \ of \ \triangle ACD : area \ of \ \triangle ABC$ . [2014]

Answer:

(i) In $\triangle ACD \ and\ \triangle BCA$

$\angle C = \angle C$ (common angle)

$\angle ABC = \angle CAD$ (given)

$\triangle ACD \sim \triangle BCA$ (AAA postulate)

(ii) Since $\triangle ACD \sim \triangle BCA$

$\frac{AC}{ BC}$ $=$ $\frac{CD}{ CA}$ $=$ $\frac{AD}{ BA}$

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

$\Rightarrow BC =$ $\frac{8}{ 5}$ $\times 4 = 6.4 \ cm$

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

(iii) Since $\triangle ACD \sim \triangle ABC$

Therefore $\frac{Area \triangle ACD}{Area \triangle ABC}$ $=$ $\frac{AC^2}{AB^2}$ $=$ $\frac{4^2}{8^2}$ $=$ $\frac{1}{4}$

Hence $Area \triangle ACD : Area \triangle ABC= 1:4$

$\\$

Question 12: In the following figure, $AB, CD \ and \ EF$ are parallel lines. $AB=6 \ cm, \ CD= y \ cm, \ EF=10 \ cm,$ $AC = 4 \ cm \ and$ $\ CF= x \ cm$ . Calculate: $x \ and \ y$ . [1985]

Answer:

Consider $\triangle FDC \ and\ \triangle FBA$

$\angle FDC = \angle FDA$ (Corresponding angles)

$\angle DFC = \angle BFA$ (common angle)

$\triangle FDC \sim \triangle FBA$ (AAA Postulate)

Therefore $\frac{CD}{ AB}$ $=$ $\frac{FC}{ FA}$

$\frac{y}{ 6}$ $=$ $\frac{x}{ x+4}$

Now consider $\triangle FCE \ and \ \triangle ACB$

$\angle FCE = \angle ACB$ (Vertically opposite angles)

$\angle CFE = \angle CAB$ (Alternate angles)

$\triangle FCE \sim \triangle ACB$ (AAA postulate)

Therefore $\frac{FC}{ AC}$ $=$ $\frac{EF}{ AB}$

$x =$ $\frac{10}{ 6}$ $\times 4 = 6.67 \ cm$

Also $y=$ $\frac{6.67 }{ 6.67+4}$ $\times 6 = 3.75 \ cm$

$\\$

Question 13: sm4 In $\triangle PQR, L$ and $M$ are two points on the base $QR$ , such that $\angle LPQ = \angle QRP$ and $\angle RPM = \angle RQP$ . Prove that:

(i) $\triangle PQL \sim \triangle RPM$

(ii) $QL \times RM = PL \times PM$

(iii) $PQ^2= QR \times QL$ . [2003]

Answer:

(i) Consider $\triangle PQL \ and \ \triangle RMP$

$\angle LPQ = \angle QRP$ (Given)

$\angle RQP = \angle RPM$ (Given)

$\triangle PQL \sim \triangle RMP$ (AAA postulate)

(ii) Since $\triangle PQL \sim \triangle RMP$

$\frac{PQ}{ RP}$ $=$ $\frac{QL}{ PM }$ $=$ $\frac{PL}{ RM}$

$\Rightarrow QL \times RM = PL \times PM$

(iii) Consider $\triangle PQL \ and \ \triangle RQP$

$\angle LPQ = \angle QRP$ (Given)

$\angle Q$ (common angle)

$\triangle PQL \sim \triangle RQP$ (AAA postulate)

Therefore $\frac{PQ}{ RQ}$ $=$ $\frac{QL}{ QP}$ $=$ $\frac{PL}{ PR}$

$\Rightarrow PQ^2 = QR \times QL$

$\\$

Question 14: sm3 In the given figure, $ABC$ is a triangle with $\angle EDB = \angle ACB$ . Prove that $\triangle ABC \sim \triangle EBD$ . If $BE=6 \ cm, EC = 4 \ cm, BD = 5 \ cm$ and area of $\triangle BED = 9 cm^2$ . Calculate the:

(i) length of $AB$

(ii) area of $\triangle ABC$ [2010]

Answer:

Consider $\triangle ABC \ and \ \triangle EBD$

$\angle EDB = \angle ACB$ (given)

$\angle DBE = \angle ABC$ (common)

Therefore $\angle DEB = \angle BAC$

$\triangle ABC \sim \triangle EBD$ (AAA postulate)

(i) Given $BE=6\ cm, EC=4\ cm, BD=5\ cm$

$\frac{AB}{ EB}$ $=$ $\frac{BC}{ BD}$ $=$ $\frac{AC }{ED}$

$AB =$ $\frac{BE+EC }{5} \times 6 = 2 \ cm$

(ii) $\frac{Area \ of \ \triangle ABC}{Area \ of \ \triangle EBD}$ $=$ $\frac{AB^2}{EB^2}$ $=$ $\frac{144}{36}$

Area of $\triangle ABC =$ $\frac{144}{ 36}$ $\times 9 = 36 \ cm^2$

$\\$

Question 15: sm2 In the given figure $\triangle ABC$ is a right angled triangle with $\angle BAC = 90^o$ .

(i) Prove $\triangle ADB \sim \triangle CDA$

(ii) If $BD = 18 \ cm \ and \ CD = 8 \ cm$ , find $AD$

(iii) Find the ratio of the area of $\triangle ADB$ is to area of $\triangle CDA$ . [2011]

Answer:

(i) Let $\angle DAB = \theta$

Therefore $\angle DAC = 90^o - \theta$

$\angle DBA = 90^o - \theta$

$\angle DCA = \theta$

Therefore $\triangle ADB \sim \triangle CDA$ (AAA postulate)

(ii) $\frac{CD}{AD}$ $=$ $\frac{AD}{BD}$

$\Rightarrow AD^2 = CD \times BD = 8 \times 18 = 144$

Therefore $AD = \sqrt{144} = 12$

(iii) $\frac{Area \ of \ \triangle ADB}{ Area \ of \ \triangle CDA}$ $=$ $\frac{\frac{1}{2} AD \times BD} {\frac{1}{2} AD \times CD}$ $=$ $\frac{BD}{CD}$ $=$ $\frac{18}{8}$ $=$ $\frac{9}{4}$