Class 10: Similarity – ICSE Board Problems

$\displaystyle \textbf{Question 1: } \\ \text{In the given figure, }AB\text{ and }DE\text{ are perpendicular to }BC.$
$\displaystyle \text{If }AB=9\text{ cm},\ DE=3\text{ cm and }AC=24\text{ cm},\text{ calculate }AD.\hfill \text{[2005]}$
$\displaystyle \text{Answer:}$ sm1
$\displaystyle \text{In }\triangle ABC\text{ and }\triangle DEC,$
$\displaystyle \angle ABC=\angle DEC=90^\circ$
$\displaystyle \angle ACB=\angle DCE\quad \text{common angle}$
$\displaystyle \therefore \triangle ABC\sim \triangle DEC$
$\displaystyle \frac{AC}{DC}=\frac{AB}{DE}$
$\displaystyle \frac{24}{DC}=\frac{9}{3}$
$\displaystyle \frac{24}{DC}=3$
$\displaystyle DC=8\text{ cm}$
$\displaystyle AD=AC-DC$
$\displaystyle =24-8=16\text{ cm}$
$\displaystyle \therefore AD=16\text{ cm}$
$\\$

$\displaystyle \textbf{Question 2: } \\ \text{In the given figure, }\triangle ABC\text{ and }\triangle AMP\text{ are right angled at }B\text{ and }M\text{ respectively.}$
$\displaystyle \text{Given }AC=10\text{ cm},\ AP=15\text{ cm and }PM=12\text{ cm}.$
$\displaystyle \text{(i) Prove }\triangle ABC\sim\triangle AMP\text{ (ii) Find }AB\text{ and }BC.\hfill \text{[2012]}$
$\displaystyle \text{Answer:}$ sm10
$\displaystyle \text{In }\triangle ABC\text{ and }\triangle AMP,$
$\displaystyle \angle ABC=\angle AMP=90^\circ$
$\displaystyle \angle BAC=\angle PAM\quad \text{common angle}$
$\displaystyle \therefore \triangle ABC\sim\triangle AMP\quad \text{by AA similarity}$
$\displaystyle \text{In right-angled }\triangle AMP,$
$\displaystyle AM=\sqrt{AP^2-PM^2}$
$\displaystyle =\sqrt{15^2-12^2}$
$\displaystyle =\sqrt{225-144}$
$\displaystyle =\sqrt{81}=9\text{ cm}$
$\displaystyle \text{Since }\triangle ABC\sim\triangle AMP,$
$\displaystyle \frac{AB}{AM}=\frac{BC}{PM}=\frac{AC}{AP}$
$\displaystyle \frac{AC}{AP}=\frac{10}{15}=\frac{2}{3}$
$\displaystyle \frac{AB}{9}=\frac{2}{3}$
$\displaystyle AB=6\text{ cm}$
$\displaystyle \frac{BC}{12}=\frac{2}{3}$
$\displaystyle BC=8\text{ cm}$
$\displaystyle \therefore AB=6\text{ cm and }BC=8\text{ cm}$
$\\$

$\displaystyle \textbf{Question 3: } \\ \text{In the figure, }PQRS\text{ is a parallelogram with }PQ=16\text{ cm and }QR=10\text{ cm.}$
$\displaystyle L\text{ is a point on }PR\text{ such that }RL:LP=2:3.\ QL\text{ produced meets }RS\text{ at }M$
$\displaystyle \text{and }PS\text{ produced at }N.\text{ Find the lengths of }PN\text{ and }RM.\hfill \text{[1997]}$ sm11
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle RLQ\text{ and }\triangle PLN,$
$\displaystyle \angle RLQ=\angle PLN\quad \text{vertically opposite angles}$
$\displaystyle \angle LRQ=\angle LPN\quad \text{alternate angles}$
$\displaystyle \therefore \triangle RLQ\sim\triangle PLN$
$\displaystyle \frac{RL}{LP}=\frac{RQ}{PN}$
$\displaystyle \frac{2}{3}=\frac{10}{PN}$
$\displaystyle 2PN=30$
$\displaystyle PN=15\text{ cm}$
$\displaystyle \text{In }\triangle RLM\text{ and }\triangle PLQ,$
$\displaystyle \angle RLM=\angle PLQ\quad \text{vertically opposite angles}$
$\displaystyle \angle LRM=\angle LPQ\quad \text{alternate angles}$
$\displaystyle \therefore \triangle RLM\sim\triangle PLQ$
$\displaystyle \frac{RM}{PQ}=\frac{RL}{LP}$
$\displaystyle \frac{RM}{16}=\frac{2}{3}$
$\displaystyle RM=\frac{32}{3}=10\frac{2}{3}\text{ cm}$
$\displaystyle \therefore PN=15\text{ cm and }RM=10\frac{2}{3}\text{ cm}$
$\\$

$\displaystyle \textbf{Question 4: } \\ \text{In the figure given below, }P\text{ is a point on }AB\text{ such that }AP:PB=4:3.\ PQ\parallel AC.$
$\displaystyle \text{(i) Calculate the ratio }PQ:AC,\text{ giving reasons for your answer.}$
$\displaystyle \text{(ii) In }\triangle ARC,\ \angle ARC=90^\circ\text{ and in }\triangle PQS,\ \angle PSQ=90^\circ.$
$\displaystyle \text{Given }QS=6\text{ cm},\text{ calculate the length of }AR.\hfill \text{[1999]}$
$\displaystyle \text{Answer:}$ sm12
$\displaystyle \text{(i) Given, }AP:PB=4:3$
$\displaystyle \text{Since }PQ\parallel AC,$
$\displaystyle \frac{PB}{AB}=\frac{PQ}{AC}$
$\displaystyle \frac{PB}{AB}=\frac{3}{4+3}=\frac{3}{7}$
$\displaystyle \therefore \frac{PQ}{AC}=\frac{3}{7}$
$\displaystyle \therefore PQ:AC=3:7$
$\displaystyle \text{(ii) In }\triangle ARC\text{ and }\triangle QSP,$
$\displaystyle \angle ARC=\angle QSP=90^\circ$
$\displaystyle \angle ACR=\angle QPS\quad \text{alternate angles, since }PQ\parallel AC$
$\displaystyle \therefore \triangle ARC\sim\triangle QSP$
$\displaystyle \frac{AR}{QS}=\frac{AC}{PQ}$
$\displaystyle \frac{AR}{6}=\frac{7}{3}$
$\displaystyle AR=\frac{7\times6}{3}=14\text{ cm}$
$\displaystyle \therefore AR=14\text{ cm}$
$\\$

$\displaystyle \textbf{Question 5: } \\ \text{In the right-angled }\triangle QPR,\ PM\text{ is the altitude.}$
$\displaystyle \text{Given that }QR=8\text{ cm and }MQ=3.5\text{ cm},\text{ calculate the value of }PR.\hfill \text{[2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle MR=QR-MQ=8-3.5=4.5\text{ cm}$
$\displaystyle \text{In }\triangle PQR\text{ and }\triangle MPR,$
$\displaystyle \angle QPR=\angle PMR=90^\circ$
$\displaystyle \angle PRQ=\angle PRM\quad \text{common angle}$
$\displaystyle \therefore \triangle PQR\sim\triangle MPR$
$\displaystyle \frac{QR}{PR}=\frac{PR}{MR}$
$\displaystyle PR^2=QR\times MR$
$\displaystyle =8\times4.5=36$
$\displaystyle PR=\sqrt{36}=6\text{ cm}$
$\displaystyle \therefore PR=6\text{ cm}$
$\\$

$\displaystyle \textbf{Question 6: } \\ \text{In the given figure, }DE\parallel BC.$
$\displaystyle \text{(i) Prove that }\triangle ADE\text{ and }\triangle ABC\text{ are similar.}$
$\displaystyle \text{(ii) Given that }AD=\frac{1}{2}BD,\text{ calculate }DE,\text{ if }BC=4.5\text{ cm. Also find:}$
$\displaystyle \frac{\text{Ar.}(\triangle ADE)}{\text{Ar.}(\triangle ABC)}\text{ and }\frac{\text{Ar.}(\triangle ADE)}{\text{Ar.}(\text{trapezium }BCED)}.\hfill \text{[2004]}$
$\displaystyle \text{Answer:}$ sm14
$\displaystyle \text{(i) Since }DE\parallel BC,$
$\displaystyle \angle ADE=\angle ABC\quad \text{corresponding angles}$
$\displaystyle \angle AED=\angle ACB\quad \text{corresponding angles}$
$\displaystyle \therefore \triangle ADE\sim\triangle ABC\quad \text{by AA similarity}$
$\displaystyle \text{(ii) Given, }AD=\frac{1}{2}BD$
$\displaystyle \therefore AD:BD=1:2$
$\displaystyle AB=AD+BD$
$\displaystyle \therefore AD:AB=1:3$
$\displaystyle \frac{AD}{AB}=\frac{1}{3}$
$\displaystyle \text{Since }\triangle ADE\sim\triangle ABC,$
$\displaystyle \frac{DE}{BC}=\frac{AD}{AB}$
$\displaystyle \frac{DE}{4.5}=\frac{1}{3}$
$\displaystyle DE=\frac{4.5}{3}=1.5\text{ cm}$
$\displaystyle \frac{\text{Ar.}(\triangle ADE)}{\text{Ar.}(\triangle ABC)}=\left(\frac{DE}{BC}\right)^2$
$\displaystyle =\left(\frac{1.5}{4.5}\right)^2$
$\displaystyle =\left(\frac{1}{3}\right)^2=\frac{1}{9}$
$\displaystyle \text{Now, Ar.}(\text{trapezium }BCED)=\text{Ar.}(\triangle ABC)-\text{Ar.}(\triangle ADE)$
$\displaystyle \therefore \frac{\text{Ar.}(\triangle ADE)}{\text{Ar.}(\text{trapezium }BCED)}=\frac{1}{9-1}=\frac{1}{8}$
$\displaystyle \therefore DE=1.5\text{ cm},\ \frac{\text{Ar.}(\triangle ADE)}{\text{Ar.}(\triangle ABC)}=\frac{1}{9}\text{ and }\frac{\text{Ar.}(\triangle ADE)}{\text{Ar.}(\text{trapezium }BCED)}=\frac{1}{8}$
$\\$

$\displaystyle \textbf{Question 7: } \\ \text{In the figure given below, }PB\text{ and }QA\text{ are perpendiculars to the line segment }AB.$
$\displaystyle \text{If }PO=6\text{ cm},\ QO=9\text{ cm and area of }\triangle POB=120\text{ cm}^2,\text{ find the area of }\triangle QOA.\hfill \text{[2006]}$ sm9
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle POB\text{ and }\triangle QOA,$
$\displaystyle \angle PBO=\angle QAO=90^\circ$
$\displaystyle \angle POB=\angle QOA\quad \text{vertically opposite angles}$
$\displaystyle \therefore \triangle POB\sim\triangle QOA$
$\displaystyle \frac{\text{Ar.}(\triangle POB)}{\text{Ar.}(\triangle QOA)}=\frac{PO^2}{QO^2}$
$\displaystyle \frac{120}{\text{Ar.}(\triangle QOA)}=\frac{6^2}{9^2}$
$\displaystyle \frac{120}{\text{Ar.}(\triangle QOA)}=\frac{36}{81}=\frac{4}{9}$
$\displaystyle \text{Ar.}(\triangle QOA)=\frac{120\times9}{4}=270\text{ cm}^2$
$\displaystyle \therefore \text{Area of }\triangle QOA=270\text{ cm}^2$
$\\$

$\displaystyle \textbf{Question 8: } \\ \text{In the figure given below, }ABCD\text{ is a parallelogram. }P\text{ is a point on }BC\text{ such that }BP:PC=1:2.$
$\displaystyle DP\text{ produced meets }AB\text{ produced at }Q.\text{ Given the area of }\triangle CPQ=20\text{ cm}^2.$
$\displaystyle \text{Calculate:}$
$\displaystyle \text{(i) area of }\triangle CDP$
$\displaystyle \text{(ii) area of parallelogram }ABCD.\hfill \text{[1996]}$ sm81
$\displaystyle \text{Answer:}$
$\displaystyle \text{In }\triangle BPQ\text{ and }\triangle CPD,$
$\displaystyle \angle BPQ=\angle CPD\quad \text{vertically opposite angles}$
$\displaystyle \angle BQP=\angle CDP\quad \text{alternate angles, since }BQ\parallel CD$
$\displaystyle \therefore \triangle BPQ\sim\triangle CPD$
$\displaystyle \frac{BP}{PC}=\frac{PQ}{PD}=\frac{1}{2}$
$\displaystyle \text{Now, }\triangle CPQ\text{ and }\triangle CDP\text{ have the same altitude from }C\text{ to line }DQ.$
$\displaystyle \therefore \frac{\text{Ar.}(\triangle CPQ)}{\text{Ar.}(\triangle CDP)}=\frac{PQ}{PD}=\frac{1}{2}$
$\displaystyle \frac{20}{\text{Ar.}(\triangle CDP)}=\frac{1}{2}$
$\displaystyle \text{Ar.}(\triangle CDP)=40\text{ cm}^2$
$\displaystyle \therefore \text{(i) Area of }\triangle CDP=40\text{ cm}^2$
$\displaystyle \text{Since }BP:PC=1:2,\text{ the height corresponding to }BP\text{ is }\frac{1}{3}\text{ of the height of the parallelogram.}$
$\displaystyle \text{So, }\text{Ar.}(\triangle BQP):\text{Ar.}(\triangle CDP)=1:4$
$\displaystyle \therefore \text{Ar.}(\triangle BQP)=\frac{1}{4}\times40=10\text{ cm}^2$
$\displaystyle \text{Also, }\text{Ar.}(\triangle AQD)=\text{Ar.}(\triangle BQP)+\text{Ar.}(\triangle CPQ)+\text{Ar.}(\triangle CDP)$
$\displaystyle =10+20+40=70\text{ cm}^2$
$\displaystyle \text{Now, }\text{Ar.}(ABCD)=\text{Ar.}(\triangle AQD)-\text{Ar.}(\triangle BQP)+\text{Ar.}(\triangle CDP)$
$\displaystyle =70-10+40=100\text{ cm}^2$
$\displaystyle \therefore \text{(ii) Area of parallelogram }ABCD=120\text{ cm}^2$
$\\$

$\displaystyle \textbf{Question 9: } \\ \text{A model of a ship is made to the scale of }1:200.$
$\displaystyle \text{(i) The length of the model is }4\text{ m; calculate the length of the ship.}$
$\displaystyle \text{(ii) The area of the deck of the ship is }160000\text{ m}^2;\text{ find the area of the deck of the model.}$
$\displaystyle \text{(iii) The volume of the model is }200\text{ litres; calculate the volume of the ship in m}^3.\hfill \text{[1995]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Scale factor}=\frac{1}{200}$
$\displaystyle \text{(i) Length of model}=\frac{1}{200}\times\text{Length of ship}$
$\displaystyle \therefore \text{Length of ship}=4\times200=800\text{ m}$
$\displaystyle \text{(ii) Area of model}=\left(\frac{1}{200}\right)^2\times\text{Area of ship}$
$\displaystyle =\left(\frac{1}{200}\right)^2\times160000$
$\displaystyle =4\text{ m}^2$
$\displaystyle \text{(iii) Volume of model}=\left(\frac{1}{200}\right)^3\times\text{Volume of ship}$
$\displaystyle \therefore \text{Volume of ship}=200^3\times200\text{ litres}$
$\displaystyle =1600000000\text{ litres}$
$\displaystyle =\frac{1600000000}{1000}\text{ m}^3$
$\displaystyle =1600000\text{ m}^3$
$\\$

$\displaystyle \textbf{Question 10: } \\ \text{In the figure given below, }ABC\text{ is a triangle. }DE\parallel BC\text{ and }\frac{AD}{DB}=\frac{3}{2}.$
$\displaystyle \text{(i) Determine the ratios }\frac{AD}{AB}\text{ and }\frac{DE}{BC}.$
$\displaystyle \text{(ii) Prove that }\triangle DEF\text{ is similar to }\triangle CBF.\text{ Hence, find }\frac{EF}{FB}.\hfill \text{[2007]}$
$\displaystyle \text{Answer:}$ sm7
$\displaystyle \text{(i) Given, }\frac{AD}{DB}=\frac{3}{2}$
$\displaystyle \frac{AD}{AB}=\frac{AD}{AD+DB}=\frac{3}{3+2}=\frac{3}{5}$
$\displaystyle \text{Since }DE\parallel BC,$
$\displaystyle \triangle ADE\sim\triangle ABC$
$\displaystyle \therefore \frac{AD}{AB}=\frac{DE}{BC}$
$\displaystyle \therefore \frac{DE}{BC}=\frac{3}{5}$
$\displaystyle \text{Hence, }\frac{AD}{AB}=\frac{3}{5}\text{ and }\frac{DE}{BC}=\frac{3}{5}$
$\displaystyle \text{(ii) In }\triangle DEF\text{ and }\triangle CBF,$
$\displaystyle \angle DFE=\angle CFB\quad \text{vertically opposite angles}$
$\displaystyle \angle DEF=\angle CBF\quad \text{alternate angles, since }DE\parallel BC$
$\displaystyle \therefore \triangle DEF\sim\triangle CBF$
$\displaystyle \frac{EF}{FB}=\frac{DE}{BC}$
$\displaystyle \therefore \frac{EF}{FB}=\frac{3}{5}$
$\\$

$\displaystyle \textbf{Question 11: } \\ \text{In }\triangle ABC,\ \angle ABC=\angle DAC,\ AB=8\text{ cm},$
$\displaystyle AC=4\text{ cm and }AD=5\text{ cm}.$
$\displaystyle \text{(i) Prove that }\triangle ACD\text{ is similar to }\triangle BCA.$
$\displaystyle \text{(ii) Find }BC\text{ and }CD.$
$\displaystyle \text{(iii) Find area of }\triangle ACD:\text{ area of }\triangle ABC.\hfill \text{[2014]}$
$\displaystyle \text{Answer:}$ sm6
$\displaystyle \text{(i) In }\triangle ACD\text{ and }\triangle BCA,$
$\displaystyle \angle ACD=\angle BCA\quad \text{common angle}$
$\displaystyle \angle CAD=\angle ABC\quad \text{given}$
$\displaystyle \therefore \triangle ACD\sim\triangle BCA$
$\displaystyle \text{(ii) Since }\triangle ACD\sim\triangle BCA,$
$\displaystyle \frac{AC}{BC}=\frac{CD}{CA}=\frac{AD}{BA}$
$\displaystyle \frac{4}{BC}=\frac{5}{8}$
$\displaystyle BC=\frac{8\times4}{5}=\frac{32}{5}=6.4\text{ cm}$
$\displaystyle \frac{CD}{4}=\frac{5}{8}$
$\displaystyle CD=\frac{5}{8}\times4=\frac{5}{2}=2.5\text{ cm}$
$\displaystyle \therefore BC=6.4\text{ cm and }CD=2.5\text{ cm}$
$\displaystyle \text{(iii) Since }\triangle ACD\sim\triangle BCA,$
$\displaystyle \frac{\text{Area of }\triangle ACD}{\text{Area of }\triangle BCA}=\left(\frac{AD}{BA}\right)^2$
$\displaystyle =\left(\frac{5}{8}\right)^2=\frac{25}{64}$
$\displaystyle \text{But }\triangle BCA\text{ and }\triangle ABC\text{ are the same triangle.}$
$\displaystyle \therefore \text{Area of }\triangle ACD:\text{Area of }\triangle ABC=25:64$
$\\$

$\displaystyle \textbf{Question 12: } \\ \text{In the following figure, }AB,\ CD\text{ and }EF\text{ are parallel lines.}$
$\displaystyle AB=6\text{ cm},\ CD=y\text{ cm},\ EF=10\text{ cm},\ AC=4\text{ cm and }CF=x\text{ cm. Calculate }x\text{ and }y.\hfill \text{[1985]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Consider }\triangle FCE\text{ and }\triangle ACB.$
$\displaystyle \angle FCE=\angle ACB\quad \text{vertically opposite angles}$
$\displaystyle \angle CFE=\angle CAB\quad \text{alternate angles, since }AB\parallel EF$
$\displaystyle \therefore \triangle FCE\sim\triangle ACB$
$\displaystyle \frac{FC}{AC}=\frac{EF}{AB}$
$\displaystyle \frac{x}{4}=\frac{10}{6}$
$\displaystyle x=\frac{10}{6}\times4=\frac{20}{3}=6.67\text{ cm}$
$\displaystyle \text{Now, consider }\triangle FDC\text{ and }\triangle FBA.$
$\displaystyle \angle FDC=\angle FBA\quad \text{corresponding angles, since }CD\parallel AB$
$\displaystyle \angle DFC=\angle BFA\quad \text{common angle}$
$\displaystyle \therefore \triangle FDC\sim\triangle FBA$
$\displaystyle \frac{CD}{AB}=\frac{FC}{FA}$
$\displaystyle \frac{y}{6}=\frac{x}{x+4}$
$\displaystyle \frac{y}{6}=\frac{\frac{20}{3}}{\frac{20}{3}+4}$
$\displaystyle \frac{y}{6}=\frac{\frac{20}{3}}{\frac{32}{3}}=\frac{5}{8}$
$\displaystyle y=6\times\frac{5}{8}=\frac{15}{4}=3.75\text{ cm}$
$\displaystyle \therefore x=\frac{20}{3}\text{ cm and }y=\frac{15}{4}\text{ cm}$
$\\$

$\displaystyle \textbf{Question 13: } \\ \text{In }\triangle PQR,\ L\text{ and }M\text{ are two points on the base }QR,$
$\displaystyle \text{such that }\angle LPQ=\angle QRP\text{ and }\angle RPM=\angle RQP.\text{ Prove that:}$
$\displaystyle \text{(i) }\triangle PQL\sim\triangle RPM$
$\displaystyle \text{(ii) }QL\times RM=PL\times PM$
$\displaystyle \text{(iii) }PQ^2=QR\times QL.\hfill \text{[2003]}$
$\displaystyle \text{Answer:}$ sm4
$\displaystyle \text{(i) In }\triangle PQL\text{ and }\triangle RPM,$
$\displaystyle \angle LPQ=\angle QRP\quad \text{given}$
$\displaystyle \text{But }\angle QRP=\angle MRP$
$\displaystyle \therefore \angle LPQ=\angle MRP$
$\displaystyle \angle PQL=\angle RQP$
$\displaystyle \text{and }\angle RQP=\angle RPM\quad \text{given}$
$\displaystyle \therefore \angle PQL=\angle RPM$
$\displaystyle \therefore \triangle PQL\sim\triangle RPM\quad \text{by AA similarity}$
$\displaystyle \text{(ii) Since }\triangle PQL\sim\triangle RPM,$
$\displaystyle \frac{QL}{PM}=\frac{PL}{RM}$
$\displaystyle \therefore QL\times RM=PL\times PM$
$\displaystyle \text{(iii) In }\triangle PQL\text{ and }\triangle RQP,$
$\displaystyle \angle LPQ=\angle QRP\quad \text{given}$
$\displaystyle \angle PQL=\angle RQP\quad \text{common angle}$
$\displaystyle \therefore \triangle PQL\sim\triangle RQP\quad \text{by AA similarity}$
$\displaystyle \frac{PQ}{RQ}=\frac{QL}{QP}$
$\displaystyle \therefore PQ^2=QR\times QL$
$\\$

$\displaystyle \textbf{Question 14: } \\ \text{In the given figure, }ABC\text{ is a triangle with }\angle EDB=\angle ACB.$
$\displaystyle \text{Prove that }\triangle ABC\sim\triangle EBD.\text{ If }BE=6\text{ cm},\ EC=4\text{ cm},\ BD=5\text{ cm}$
$\displaystyle \text{and area of }\triangle BED=9\text{ cm}^2,\text{ calculate:}$
$\displaystyle \text{(i) length of }AB$
$\displaystyle \text{(ii) area of }\triangle ABC.\hfill \text{[2010]}$
$\displaystyle \text{Answer:}$ sm3
$\displaystyle \text{In }\triangle ABC\text{ and }\triangle EBD,$
$\displaystyle \angle EDB=\angle ACB\quad \text{given}$
$\displaystyle \angle EBD=\angle ABC\quad \text{common angle}$
$\displaystyle \therefore \triangle ABC\sim\triangle EBD$
$\displaystyle \text{Now, }BC=BE+EC=6+4=10\text{ cm}$
$\displaystyle \text{Since }\triangle ABC\sim\triangle EBD,$
$\displaystyle \frac{AB}{EB}=\frac{BC}{BD}$
$\displaystyle \frac{AB}{6}=\frac{10}{5}=2$
$\displaystyle AB=12\text{ cm}$
$\displaystyle \therefore \text{(i) }AB=12\text{ cm}$
$\displaystyle \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle EBD}=\left(\frac{AB}{EB}\right)^2$
$\displaystyle =\left(\frac{12}{6}\right)^2=4$
$\displaystyle \frac{\text{Area of }\triangle ABC}{9}=4$
$\displaystyle \text{Area of }\triangle ABC=36\text{ cm}^2$
$\displaystyle \therefore \text{(ii) Area of }\triangle ABC=36\text{ cm}^2$
$\\$

$\displaystyle \textbf{Question 15: } \\ \text{In the given figure }\triangle ABC\text{ is a right-angled triangle with }\angle BAC=90^\circ.$
$\displaystyle \text{(i) Prove }\triangle ADB\sim\triangle CDA$
$\displaystyle \text{(ii) If }BD=18\text{ cm and }CD=8\text{ cm},\text{ find }AD.$
$\displaystyle \text{(iii) Find the ratio of the area of }\triangle ADB\text{ to the area of }\triangle CDA.\hfill \text{[2011]}$
$\displaystyle \text{Answer:}$ sm2
$\displaystyle \text{(i) Since }AD\perp BC,$
$\displaystyle \angle ADB=\angle CDA=90^\circ$
$\displaystyle \text{Let }\angle DAB=\theta.$
$\displaystyle \text{Then }\angle DBA=90^\circ-\theta$
$\displaystyle \text{Also, }\angle DAC=90^\circ-\theta$
$\displaystyle \therefore \angle DBA=\angle DAC$
$\displaystyle \therefore \triangle ADB\sim\triangle CDA$
$\displaystyle \text{(ii) Since }\triangle ADB\sim\triangle CDA,$
$\displaystyle \frac{AD}{CD}=\frac{BD}{AD}$
$\displaystyle AD^2=BD\times CD$
$\displaystyle =18\times8=144$
$\displaystyle AD=12\text{ cm}$
$\displaystyle \therefore AD=12\text{ cm}$
$\displaystyle \text{(iii) }\frac{\text{Area of }\triangle ADB}{\text{Area of }\triangle CDA}=\frac{\frac{1}{2}\times AD\times BD}{\frac{1}{2}\times AD\times CD}$
$\displaystyle =\frac{BD}{CD}=\frac{18}{8}=\frac{9}{4}$
$\displaystyle \therefore \text{Area of }\triangle ADB:\text{Area of }\triangle CDA=9:4$
$\\$

$\displaystyle \textbf{Question 16: } \\ \text{In the given figure }AB\text{ and }DE\text{ are perpendiculars to }BC.$
$\displaystyle \text{(i) Prove that }\triangle ABC\sim\triangle DEC$
$\displaystyle \text{(ii) If }AB=6\text{ cm},\ DE=4\text{ cm and }AC=15\text{ cm},\text{ calculate }CD$
$\displaystyle \text{(iii) Find the ratio of the area of }\triangle ABC:\text{ area of }\triangle DEC.\hfill \text{[2013]}$
$\displaystyle \text{Answer:}$ sm1
$\displaystyle \text{(i) In }\triangle ABC\text{ and }\triangle DEC,$
$\displaystyle \angle ABC=\angle DEC=90^\circ$
$\displaystyle \angle ACB=\angle DCE\quad \text{common angle}$
$\displaystyle \therefore \triangle ABC\sim\triangle DEC$
$\displaystyle \text{(ii) Since }\triangle ABC\sim\triangle DEC,$
$\displaystyle \frac{AB}{DE}=\frac{AC}{CD}$
$\displaystyle \frac{6}{4}=\frac{15}{CD}$
$\displaystyle 6CD=60$
$\displaystyle CD=10\text{ cm}$
$\displaystyle \therefore CD=10\text{ cm}$
$\displaystyle \text{(iii) Since }\triangle ABC\sim\triangle DEC,$
$\displaystyle \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle DEC}=\left(\frac{AB}{DE}\right)^2$
$\displaystyle =\left(\frac{6}{4}\right)^2=\frac{36}{16}=\frac{9}{4}$
$\displaystyle \therefore \text{Area of }\triangle ABC:\text{Area of }\triangle DEC=9:4$
$\\$