Question 1: 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]

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

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Question 2:  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]

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

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

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

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Question 3: 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]

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: 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]

(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)

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

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

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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]

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

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

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

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

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

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Question 6: 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]

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

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

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

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

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

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

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

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

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Question 7: 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]

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)

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

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

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Question 8: 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]

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

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

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

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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]

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

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

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Question 10: 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]

(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

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

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Question 11: 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]

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

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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]

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

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Question 13: 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]

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

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

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Question 14: 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]

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

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Question 15: 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]

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

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Question 16: In the given figure $AB$ and $DE$ are perpendiculars to $BC$.

(i) Prove that $\triangle ABC \sim \triangle DEC$

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

(iii) Find the ratio of the $area \ of \ \triangle ABC : area \ of \ \triangle DEC$  [2013]

(i) From  $\triangle ABC \ and \ \triangle DEC$

$\angle ABC = \angle DEC = 90^o$ (given)

$\angle ACB = \angle DCE$ (common angle)

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

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

In $\triangle ABC \ and \ \triangle DEC$,

$\frac{AB}{ DE} = \frac{AC}{ CD}$

$AB = 6 \ cm, DE = 4 \ cm$ and $AC = 15 \ cm$

Therefore $CD = \frac{15}{6} \times 4 = 10 \ cm$

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

$\frac{Area \ of \ \triangle ABC}{Area \ of \ \triangle DEC} = \frac{AB^2}{ DE^2} = \frac{6^2}{4^2}= \frac{9}{4}$

Therefore the $Area \ of \ \triangle ABC : Area \ of \ \triangle DEC = 9:4$

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