\displaystyle \textbf{Question 1:}~\text{In the given figure, }\triangle ABC \sim \triangle PQR.~\text{If }AD\text{ and }PS\text{ are bisectors} \\ \text{of }\angle BAC\text{ and }\angle QPR\text{ respectively, then}~\text{[ICSE 2024]}
\displaystyle \text{(a) } \triangle ABC \sim \triangle PQS \qquad \text{(b) } \triangle ABD \sim \triangle PQS
\displaystyle \text{(c) } \triangle ABD \sim \triangle PSR \qquad \text{(d) } \triangle ABC \sim \triangle PSR

\displaystyle \text{Answer:}
\displaystyle \textbf{(b) In the given figure, }\triangle ABC \sim \triangle PQR.
\displaystyle \angle ABC=\angle PQR \quad \text{[corresponding angles]} \quad ...(i)
\displaystyle \angle BAC=\angle QPR \quad \text{[corresponding angles]}
\displaystyle \Rightarrow \frac{1}{2}\angle BAC=\frac{1}{2}\angle QPR \quad ...(ii)
\displaystyle \text{Now, }AD\text{ and }PS\text{ are angle bisectors of }\angle BAC\text{ and }\angle QPR\text{ respectively.}
\displaystyle \text{Thus, in }\triangle ABD\text{ and }\triangle PQS
\displaystyle \angle ABD=\angle PQS \quad \text{[from Eq. (i)]}
\displaystyle \angle BAD=\angle QPS \quad \text{[from Eq. (ii)]}
\displaystyle \angle BDA=\angle QSP
\displaystyle \therefore \triangle ABD \sim \triangle PQS \quad \text{[by AAA similarity]}

\displaystyle \textbf{Question 2:}~\text{In the given figure, }\angle BAP=\angle DCP=70^\circ,\, PC=6\text{ cm and } \\ CA=4\text{ cm, then }PD:DB\text{ is }~\text{[ICSE 2023]}
\displaystyle \text{(a) }5:3 \qquad \text{(b) }3:5 \qquad \text{(c) }3:2 \qquad \text{(d) }2:3

\displaystyle \text{Answer:}
\displaystyle \textbf{(c) In the given figure,}
\displaystyle \angle BAP=\angle DCP
\displaystyle \therefore AB \parallel CD \quad \text{[corresponding angles are equal]}
\displaystyle \Rightarrow \frac{PC}{CA}=\frac{PD}{DB} \quad \text{[by Basic Proportionality Theorem]}
\displaystyle \Rightarrow \frac{PD}{DB}=\frac{6}{4}=\frac{3}{2}
\displaystyle \text{Hence, }PD:DB=3:2.

\displaystyle \textbf{Question 3:}~\text{In the given figures the }\triangle ABC\text{ is similar to }\triangle DEF\text{ by the axiom } \\ \text{[ICSE 2023]}
\displaystyle \text{(a) SSS} \qquad \text{(b) SAS} \qquad \text{(c) AAA} \qquad \text{(d) RHS}

\displaystyle \text{Answer:}
\displaystyle \textbf{(b) In }\triangle ABC\text{ and }\triangle DEF,
\displaystyle \frac{AB}{DE}=\frac{4}{24}=\frac{1}{6}\quad \text{and}\quad \frac{BC}{EF}=\frac{3}{18}=\frac{1}{6}
\displaystyle \therefore \frac{AB}{DE}=\frac{BC}{EF}
\displaystyle \text{and }\angle B=\angle E \quad \text{[each }90^\circ\text{]}
\displaystyle \text{Hence, by SAS property, }\triangle ABC\text{ and }\triangle DEF\text{ are similar.}

\displaystyle \textbf{Question 4:}~ABCD\text{ is a trapezium with }AB\parallel DC.\,\text{Then }\triangle AOB\text{ is similar} \\ \text{to }~\text{[ICSE Semester I 2022]}
\displaystyle \text{(a) } \triangle ADB \qquad \text{(b) } \triangle ACB \qquad \text{(c) } \triangle COD \qquad \text{(d) } \triangle COB \displaystyle \text{Answer:}
\displaystyle \textbf{(c) Given that, }ABCD\text{ is a trapezium with }AB\parallel DC.
\displaystyle \text{So, in }\triangle AOB\text{ and }\triangle COD\text{ we have,}
\displaystyle \angle AOB=\angle COD \quad \text{[vertically opposite angles]}
\displaystyle \angle OAB=\angle OCD \quad \text{[since }AB\parallel DC,\text{ alternate angles]}
\displaystyle \therefore \triangle AOB \sim \triangle COD \quad \text{[by AA similarity]}
\displaystyle \text{Hence, }\triangle AOB\text{ is similar to }\triangle COD.

\displaystyle \textbf{Question 5:}~\text{If }\triangle ABC\sim\triangle QRP,\text{ then the corresponding proportional sides} \\ \text{are }~\text{[ICSE Semester I 2022]}
\displaystyle \text{(a) }\frac{AB}{QR}=\frac{BC}{RP} \qquad \text{(b) }\frac{AC}{QR}=\frac{BC}{RP}
\displaystyle \text{(c) }\frac{AB}{QR}=\frac{BC}{QP} \qquad \text{(d) }\frac{AB}{PQ}=\frac{BC}{RP}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a) Given, }\triangle ABC \sim \triangle QRP.
\displaystyle \therefore \angle A=\angle Q,\;\angle B=\angle R,\;\angle C=\angle P \quad \text{[corresponding angles]}
\displaystyle \Rightarrow \text{Sides of one triangle are proportional to the sides of the other triangle.}
\displaystyle \therefore \frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}
\displaystyle \text{Hence, }\frac{AB}{QR}=\frac{BC}{RP}.

\displaystyle \textbf{Question 6:}~\text{In the given figure, }AB=24\text{ cm},\, AC=18\text{ cm},\, DE=12\text{ cm}, \\ DF=9\text{ cm and }\angle BAC=\angle EDF.\,\text{Find the required result. }~\text{[ICSE Semester I 2021]}
\displaystyle \text{Then, }\triangle ABC \sim \triangle DEF\text{ by the condition}~
\displaystyle \text{(a) AAA} \qquad \text{(b) SAS} \qquad \text{(c) SSS} \qquad \text{(d) AAS} \text{[ICSE Semester I 2021]}

\displaystyle \text{Answer:}
\displaystyle \textbf{(b) } \triangle ABC \sim \triangle DEF,\text{ then }\frac{AB}{AC}=\frac{DE}{DF}
\displaystyle \text{and }\angle BAC=\angle DEF \quad \text{[by SAS similarity criterion]}

\displaystyle \textbf{Question 7:}~\text{In the given figure, }PQ\parallel TR,\text{ then by using }\text{ properties of} \\ \text{similarity }~\text{[ICSE Semester I 2021]}
\displaystyle \text{(a) }\frac{PQ}{RT}=\frac{OP}{OT}=\frac{OQ}{OR}
\displaystyle \text{(b) }\frac{PQ}{RT}=\frac{OP}{OR}=\frac{OQ}{OT}
\displaystyle \text{(c) }\frac{PQ}{RT}=\frac{OP}{OT}=\frac{OQ}{OT}
\displaystyle \text{(d) }\frac{PQ}{RT}=\frac{OP}{OR}=\frac{OT}{OQ}

\displaystyle \text{Answer:}
\displaystyle \textbf{(b) Given, }PQ\parallel TR,
\displaystyle \Rightarrow \angle OPQ=\angle ORT
\displaystyle \text{and }\angle PQO=\angle OTR \quad \text{[corresponding angles]}
\displaystyle \therefore \triangle POQ \sim \triangle ROT \quad \text{[by AA similarity]}
\displaystyle \therefore \frac{PO}{RO}=\frac{OQ}{OT}=\frac{PQ}{RT}

\displaystyle \textbf{Question 8:}~\text{In }\triangle ABC\text{ and }\triangle EDC,\, AB\parallel ED.\, BD=\frac{1}{3}BC\text{ and } \\ AB=12.3\text{ cm.}~\text{[ICSE 2020]}
\displaystyle \text{(i) Prove that }\triangle ABC\sim\triangle EDC.
\displaystyle \text{(ii) Find }DE.
\displaystyle \text{(iii) Find }\frac{\mathrm{ar}(\triangle EDC)}{\mathrm{ar}(\triangle ABC)}.

\displaystyle \text{Answer:}
\displaystyle \textbf{According to the question,}
\displaystyle \text{(i) Since }AB\parallel DE
\displaystyle \angle 1=\angle 2\text{ and }\angle 3=\angle 4 \quad \text{[corresponding angles]}
\displaystyle \text{Now, in }\triangle ABC\text{ and }\triangle EDC
\displaystyle \angle 1=\angle 2
\displaystyle \angle 3=\angle 4 \quad \text{[proved above]}
\displaystyle \angle 5=\angle 5 \quad \text{[common]}
\displaystyle \therefore \text{By AAA criteria, }\triangle ABC \sim \triangle EDC.\quad \text{Hence proved.}
\displaystyle \text{(ii) Since, }\triangle ABC \sim \triangle EDC
\displaystyle \therefore \frac{AB}{ED}=\frac{BC}{DC}
\displaystyle \Rightarrow \frac{AB}{ED}=\frac{BC}{BC-DB}
\displaystyle \Rightarrow \frac{AB}{ED}=\frac{BC}{BC-\frac{1}{3}BC}\quad \text{[since }BD=\frac{1}{3}BC\text{]}
\displaystyle \Rightarrow \frac{AB}{ED}=\frac{BC}{\frac{2}{3}BC}
\displaystyle \Rightarrow \frac{AB}{ED}=\frac{3}{2}
\displaystyle \Rightarrow ED=\frac{2}{3}AB=\frac{2}{3}\times 12.3
\displaystyle =8.2\text{ cm}
\displaystyle \text{(iii) We know that if two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides.}
\displaystyle \therefore \frac{\mathrm{ar}(\triangle EDC)}{\mathrm{ar}(\triangle ABC)}=\left(\frac{ED}{AB}\right)^2
\displaystyle =\left(\frac{8.2}{12.3}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}

\displaystyle \textbf{Question 9:}~\text{In the given figure, }\angle PQR=\angle PST=90^\circ,\, PQ=5\text{ cm and } \\ PS=2\text{ cm. }~\text{[ICSE 2019]}
\displaystyle \text{Prove that }\triangle PQR\sim\triangle PST.

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }\angle PQR=\angle PST=90^\circ,\; PQ=5\text{ cm and }PS=2\text{ cm.}
\displaystyle \text{To prove }\triangle PQR\sim\triangle PST
\displaystyle \text{Proof: In }\triangle PQR\text{ and }\triangle PST,
\displaystyle \angle PQR=\angle PST=90^\circ \quad \text{[given]}
\displaystyle \angle QPR=\angle SPT \quad \text{[common angle]}
\displaystyle \therefore \triangle PQR\sim\triangle PST \quad \text{[by AA similarity criterion]}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 10:}~\text{In }\triangle PQR,\, MN\parallel QR\text{ and }\frac{PM}{MQ}=\frac{2}{3}.~\text{[ICSE 2018]}
\displaystyle \text{(i) Find }\frac{MN}{QR}.
\displaystyle \text{(ii) Prove that }\triangle OMN\sim\triangle ORQ.

\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}
\displaystyle \text{Since }MN\parallel QR
\displaystyle \angle 1=\angle 2 \text{ and } \angle 3=\angle 4 \quad \text{[corresponding angles]}
\displaystyle \therefore \triangle PMN\sim\triangle PQR \quad \text{[by AA similarity criterion]}
\displaystyle \therefore \frac{MN}{QR}=\frac{PM}{PQ} \quad ...(i)
\displaystyle \text{We have } \frac{PM}{MQ}=\frac{2}{3}
\displaystyle \Rightarrow \frac{MQ}{PM}=\frac{3}{2}
\displaystyle \Rightarrow \frac{MQ}{PM}+1=\frac{3}{2}+1
\displaystyle \Rightarrow \frac{MQ+PM}{PM}=\frac{5}{2}
\displaystyle \Rightarrow \frac{PQ}{PM}=\frac{5}{2}
\displaystyle \Rightarrow \frac{PM}{PQ}=\frac{2}{5}
\displaystyle \text{Hence, from Eq. (i), }\frac{MN}{QR}=\frac{2}{5}.
\displaystyle \text{(ii) In }\triangle OMN\text{ and }\triangle ORQ,
\displaystyle \angle OMN=\angle ORQ \quad \text{[alternate interior angles]}
\displaystyle \angle ONM=\angle OQR \quad \text{[alternate interior angles]}
\displaystyle \therefore \triangle OMN\sim\triangle ORQ \quad \text{[by AA similarity criterion]}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 11:}~PQR\text{ is a triangle. }S\text{ is a point on the side }QR\text{ of }\triangle PQR \\ \text{such that }\angle PSR=\angle QPR.
\displaystyle \text{Given, }QP=8\text{ cm},\, PR=6\text{ cm and }SR=3\text{ cm.}~\text{[ICSE 2017]}
\displaystyle \text{(i) Prove that }\triangle PQR\sim\triangle SPR.
\displaystyle \text{(ii) Find the length of }QR\text{ and }PS.

\displaystyle \text{Answer:}
\displaystyle \textbf{(i) In }\triangle PQR\text{ and }\triangle SPR\text{ we have,}
\displaystyle \angle PSR=\angle QPR \quad \text{[given]}
\displaystyle \angle SRP=\angle PRQ \quad \text{[common in both triangles]}
\displaystyle \therefore \text{By AA axiom of similarity,}
\displaystyle \triangle PQR\sim\triangle SPR
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) }\triangle PQR\sim\triangle SPR
\displaystyle \therefore \frac{QR}{PR}=\frac{RP}{RS}
\displaystyle \Rightarrow \frac{QR}{6}=\frac{6}{3}
\displaystyle \Rightarrow QR=\frac{6\times 6}{3}=12\text{ cm}
\displaystyle \text{Also, }\frac{PQ}{PS}=\frac{RP}{RS}=\frac{8}{6}
\displaystyle \Rightarrow PS=\frac{8\times 3}{6}
\displaystyle \Rightarrow PS=4\text{ cm}

\displaystyle \textbf{Question 12:}~\text{In the figure given below, }ABCD\text{ is a parallelogram.}\,E\text{ is a point on }AB.
\displaystyle CE\text{ intersects the diagonal }BD\text{ at }G\text{ and }F\text{ is a point on }CE.\text{ If }AE:EB=1:2,\text{ then find }~\text{[ICSE 2017]}
\displaystyle \text{(i) }EF:AD
\displaystyle \text{(ii) }\mathrm{ar}(\triangle BEF):\mathrm{ar}(\triangle ABD)

\displaystyle \text{Answer:}
\displaystyle \textbf{Given, }ABCD\text{ is a parallelogram. }E\text{ is a point on }AB.
\displaystyle CE\text{ intersects the diagonal }BD\text{ at }G\text{ and }EF\parallel BC.
\displaystyle AE:EB=1:2
\displaystyle \text{(i) } \frac{AB}{BE}=\frac{AD}{AD}=\frac{EF}{AD}=\frac{BE}{AB} \quad ...(A)
\displaystyle \text{Given } \frac{AE}{EB}=\frac{1}{2}
\displaystyle \text{On adding 1 on both sides,}
\displaystyle \frac{AE}{EB}+1=\frac{1}{2}+1
\displaystyle \Rightarrow \frac{AE+EB}{EB}=\frac{3}{2}
\displaystyle \Rightarrow \frac{AB}{EB}=\frac{3}{2} \quad [AE+EB=AB]
\displaystyle \Rightarrow \frac{EB}{AB}=\frac{2}{3} \quad ...(B)
\displaystyle \therefore EF:AD=2:3
\displaystyle \text{(ii) }\frac{\mathrm{ar}(\triangle BEF)}{\mathrm{ar}(\triangle ABD)}=\left(\frac{EF}{AD}\right)^2
\displaystyle =\left(\frac{2}{3}\right)^2=\frac{4}{9}
\displaystyle \text{Hence, }\mathrm{ar}(\triangle BEF):\mathrm{ar}(\triangle ABD)=4:9

\displaystyle \textbf{Question 13:}~\text{In the given figure, }AB\text{ and }DE\text{ are perpendicular to }BC.~\text{[ICSE 2013]}
\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, then calculate }CD. \displaystyle \text{Answer:}
\displaystyle \textbf{Given }AB\perp BC\text{ and }DE\perp BC
\displaystyle \text{(i) To prove }\triangle ABC\sim\triangle DEC
\displaystyle \text{Proof: In }\triangle ABC\text{ and }\triangle DEC,
\displaystyle \angle ABC=\angle DEC=90^\circ \quad [AB\perp BC\text{ and }DE\perp BC]
\displaystyle \text{and } \angle ACB=\angle DCE \quad \text{[common angles]}
\displaystyle \therefore \triangle ABC\sim\triangle DEC \quad \text{[by AA axiom of similarity]}
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Given }AB=6\text{ cm},\; DE=4\text{ cm and }AC=15\text{ cm}
\displaystyle \text{Since }\triangle ABC\sim\triangle DEC
\displaystyle \therefore \frac{AB}{DE}=\frac{AC}{DC}
\displaystyle \Rightarrow \frac{6}{4}=\frac{15}{DC}
\displaystyle \Rightarrow DC=\frac{4\times 15}{6}
\displaystyle \Rightarrow DC=10\text{ cm}

\displaystyle \textbf{Question 14:}~\text{In the given figure, }\triangle ABC\text{ and }\triangle AMP\text{ are right angled at }B\text{ and }M \\ \text{respectively.}  \text{Given, }AC=10\text{ cm},\, AP=15\text{ cm and }PM=12\text{ cm. }~\text{[ICSE 2012]}
\displaystyle \text{(i) Prove that }\triangle ABC\sim\triangle AMP.
\displaystyle \text{(ii) Find }AB\text{ and }BC.

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }AC=10\text{ cm},\; AP=15\text{ cm},\; PM=12\text{ cm and }\angle ABC=\angle AMP=90^\circ.
\displaystyle \text{(i) To prove }\triangle ABC\sim\triangle AMP.
\displaystyle \text{Proof: In }\triangle ABC\text{ and }\triangle AMP,
\displaystyle \angle ABC=\angle AMP=90^\circ \quad \text{[given]}
\displaystyle \angle BAC=\angle PAM \quad \text{[common angles]}
\displaystyle \therefore \triangle ABC\sim\triangle AMP\quad \text{[by AA axiom of similarity]}
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Since }\triangle ABC\sim\triangle AMP,
\displaystyle \therefore \frac{AC}{AP}=\frac{BC}{MP}=\frac{AB}{AM}\quad \text{[if two triangles are similar, their corresponding sides are proportional]}
\displaystyle \text{Now, taking }\frac{AC}{AP}=\frac{BC}{MP}
\displaystyle \therefore \frac{10}{15}=\frac{BC}{12}
\displaystyle \Rightarrow BC=\frac{10\times 12}{15}=8\text{ cm}
\displaystyle \text{In }\triangle ABC,\text{ right angled at }B,\text{ using Pythagoras theorem,}
\displaystyle AB=\sqrt{AC^{2}-BC^{2}}
\displaystyle =\sqrt{(10)^{2}-(8)^{2}}=\sqrt{100-64}=\sqrt{36}=6\text{ cm}

\displaystyle \textbf{Question 15:}~\text{In the given figure, }ABC\text{ and }CEF\text{ are two triangles, where } \\ BA\parallel CE\text{ and }AF:AC=5:8.~\text{[ICSE 2009]}
\displaystyle \text{(i) Prove that }\triangle ADF\sim\triangle CEF.
\displaystyle \text{(ii) Find }AD,\text{ if }CE=6\text{ cm.}

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }BA\parallel CE\text{ and }AF:AC=5:8.
\displaystyle \text{(i) To prove }\triangle ADF\sim\triangle CEF.
\displaystyle \text{Proof: In }\triangle ADF\text{ and }\triangle CEF,
\displaystyle \angle AFD=\angle CFE \quad \text{[vertically opposite angles]}
\displaystyle \angle DAF=\angle FCE\quad \text{[alternate interior angles, since }BA\parallel CE\text{]}
\displaystyle \therefore \triangle ADF\sim\triangle CEF\quad \text{[by AA axiom of similarity]}
\displaystyle \text{(ii) Given, }CE=6\text{ cm}
\displaystyle \text{Since }\triangle ADF\sim\triangle CEF,
\displaystyle \therefore \frac{AD}{CE}=\frac{AF}{CF}
\displaystyle \text{But }\frac{AF}{AC}=\frac{5}{8}\Rightarrow \frac{AC}{AF}=\frac{8}{5}
\displaystyle \therefore \frac{AC-AF}{AF}=\frac{8-5}{5}=\frac{3}{5}
\displaystyle \therefore \frac{CF}{AF}=\frac{3}{5}\Rightarrow \frac{AF}{CF}=\frac{5}{3}
\displaystyle \therefore \frac{AD}{6}=\frac{5}{3}
\displaystyle \Rightarrow AD=\frac{5\times 6}{3}=10\text{ cm}

\displaystyle \textbf{Question 16:}~\text{In the given figure, }DE\parallel BC.~\text{[ICSE 2004]}
\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.}

\displaystyle \text{Answer:}
\displaystyle \textbf{16. Given }DE\parallel BC.
\displaystyle \text{(i) To prove }\triangle ADE\sim\triangle ABC.
\displaystyle \text{Proof: In }\triangle ADE\text{ and }\triangle ABC,
\displaystyle \angle DAE=\angle BAC \quad \text{[common angles]}
\displaystyle \angle ADE=\angle ABC\quad \text{[corresponding angles as }DE\parallel BC\text{]}
\displaystyle \therefore \triangle ADE\sim\triangle ABC\quad \text{[by AA axiom of similarity]}
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Since }\triangle ADE\sim\triangle ABC,
\displaystyle \therefore \frac{AD}{AB}=\frac{DE}{BC}
\displaystyle \text{Also }AB=AD+BD
\displaystyle \therefore \frac{AB}{AD}=\frac{BC}{DE}
\displaystyle \Rightarrow \frac{AD+BD}{AD}=\frac{BC}{DE}
\displaystyle \Rightarrow 1+\frac{BD}{AD}=\frac{BC}{DE}
\displaystyle \text{Given }AD=\frac{1}{2}BD
\displaystyle \Rightarrow \frac{BD}{AD}=2
\displaystyle \therefore 1+2=\frac{BC}{DE}
\displaystyle \Rightarrow \frac{BC}{DE}=3
\displaystyle \text{Given }BC=4.5\text{ cm}
\displaystyle \Rightarrow DE=\frac{4.5}{3}=1.5\text{ cm}

\displaystyle \textbf{Question 17:}~\text{In the given figure, }\triangle ADB\text{ and }\triangle ACB\text{ are two right angled} \\ \text{triangles with }\angle ADB=\angle BCA=90^\circ.
\displaystyle \text{Given, }AB=10\text{ cm},\, AD=6\text{ cm},\, BC=2.4\text{ cm and }DP=4.5\text{ cm.}~\text{[ICSE 2024]}
\displaystyle \text{(i) Prove that }\triangle APD\sim\triangle BPC.
\displaystyle \text{(ii) Find the length of }BD\text{ and }PB.
\displaystyle \text{(iii) Hence, find the length of }PA.
\displaystyle \text{(iv) Find }\mathrm{ar}(\triangle APD):\mathrm{ar}(\triangle BPC).

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }\angle ADB=\angle BCA=90^\circ,
\displaystyle AB=10\text{ cm},\; AD=6\text{ cm},\; BC=2.4\text{ cm and }DP=4.5\text{ cm.}
\displaystyle \text{(i) In }\triangle ADP\text{ and }\triangle PCB,
\displaystyle \angle ADP=\angle BCP=90^\circ
\displaystyle \angle APD=\angle BPC \quad \text{[vertically opposite angles]}
\displaystyle \therefore \triangle APD\sim\triangle BPC\quad \text{[by AA similarity criterion]}
\displaystyle \text{(ii) Since }\triangle APD\sim\triangle BPC,
\displaystyle \therefore \frac{AD}{BC}=\frac{DP}{CP}
\displaystyle \Rightarrow \frac{6}{2.4}=\frac{4.5}{CP}
\displaystyle \Rightarrow CP=1.8\text{ cm}
\displaystyle \text{Now, in }\triangle BCP,\text{ using Pythagoras theorem,}
\displaystyle BC^{2}+CP^{2}=PB^{2}
\displaystyle \Rightarrow (2.4)^{2}+(1.8)^{2}=PB^{2}
\displaystyle \Rightarrow 5.76+3.24=PB^{2}=9
\displaystyle \Rightarrow PB=3\text{ cm}
\displaystyle \therefore BD=DP+PB=4.5+3=7.5\text{ cm}
\displaystyle \text{(iii) Since }\triangle ADP\sim\triangle BCP,
\displaystyle \therefore \frac{AD}{CB}=\frac{AP}{BP}
\displaystyle \Rightarrow \frac{6}{2.4}=\frac{AP}{3}
\displaystyle \Rightarrow AP=\frac{18}{2.4}=7.5\text{ cm}
\displaystyle \therefore PA=7.5\text{ cm}
\displaystyle \text{(iv) Since }\triangle APD\sim\triangle BPC,
\displaystyle \therefore \frac{\mathrm{ar}(\triangle APD)}{\mathrm{ar}(\triangle BPC)}=\left(\frac{AD}{BC}\right)^{2}
\displaystyle =\left(\frac{6}{2.4}\right)^{2}=\left(\frac{5}{2}\right)^{2}=\frac{25}{4}
\displaystyle \therefore \mathrm{ar}(\triangle APD):\mathrm{ar}(\triangle BPC)=25:4.

\displaystyle \textbf{Question 18:}~\text{In the given figure, }AC\parallel DE\parallel BF.  \text{If }AC=24\text{ cm}, \\ EG=8\text{ cm},\, GB=16\text{ cm and }BF=30\text{ cm.}~\text{[ICSE 2023]}
\displaystyle \text{(i) Prove }\triangle GED\sim\triangle GBF.
\displaystyle \text{(ii) Find }DE.
\displaystyle \text{(iii) }DB:AB.

\displaystyle \text{Answer:}
\displaystyle \textbf{Given, }AC\parallel DE\parallel BF,\; AC=24\text{ cm}, \\ EG=8\text{ cm},\; GB=16\text{ cm and }BF=30\text{ cm.}
\displaystyle \text{(i) From the figure, we have }\angle EGD=\angle BGF\quad \text{[vertically opposite angles]}
\displaystyle \angle DEG=\angle FBG \quad \text{[alternate interior angles]}
\displaystyle \therefore \triangle GED\sim\triangle GBF\quad \text{[by AA similarity criterion]}
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Since }\triangle GED\sim\triangle GBF,
\displaystyle \therefore \frac{GE}{GB}=\frac{DE}{FB}
\displaystyle \Rightarrow \frac{8}{16}=\frac{DE}{30}
\displaystyle \Rightarrow DE=\frac{8\times 30}{16}
\displaystyle \Rightarrow DE=15\text{ cm}
\displaystyle \text{(iii) We have, }ED\parallel AC
\displaystyle \therefore \triangle BED\sim\triangle BCA\quad \text{[by AA similarity criterion]}
\displaystyle \Rightarrow \frac{DB}{DE}=\frac{AB}{AC}
\displaystyle \Rightarrow \frac{DB}{AB}=\frac{DE}{AC}=\frac{15}{24}
\displaystyle \therefore DB:AB=5:8

\displaystyle \textbf{Question 19:}~\text{In the given figure, }ABC\text{ is a triangle and }BCFD\text{ is a parallelogram.}
\displaystyle \text{Given, }AD:DB=4:5\text{ and }EF=15\text{ cm. Find }~\text{[ICSE 2023]}
\displaystyle \text{(i) }AE:EC \qquad \text{(ii) }DE \qquad \text{(iii) }BC.

\displaystyle \text{Answer:}
\displaystyle \textbf{Given, }AD:DB=4:5\text{ and }EF=15\text{ cm.}
\displaystyle \text{(i) In }\triangle ABC,
\displaystyle \frac{AD}{DB}=\frac{AE}{EC} \quad \text{[by basic proportionality theorem]}
\displaystyle \Rightarrow \frac{4}{5}=\frac{AE}{EC}
\displaystyle \therefore AE:EC=4:5
\displaystyle \text{(ii) In }\triangle AED\text{ and }\triangle CEF,
\displaystyle \angle AED=\angle CEF \quad \text{[vertically opposite angles]}
\displaystyle \angle ADE=\angle EFC \quad \text{[alternate angles]}
\displaystyle \therefore \triangle AED\sim\triangle CEF \quad \text{[by AA rule]}
\displaystyle \text{So, }\frac{DE}{EF}=\frac{AE}{CE}=\frac{AD}{CF}
\displaystyle \Rightarrow \frac{DE}{15}=\frac{4}{5}\quad \text{[since }EF=15\text{ cm and }AE:EC=4:5\text{]}
\displaystyle \Rightarrow DE=12\text{ cm}
\displaystyle \text{(iii) }DF=BC\text{ and }DF=DE+EF
\displaystyle \therefore BC=12+15=27\text{ cm}

\displaystyle \textbf{Question 20:}~\text{In the given }\triangle PQR,\; AB\parallel QR,\; QP\parallel CB\text{ and }AR\text{ intersects }CB\text{ at }O.
\displaystyle \text{Using the given figure answer the following questions }~\text{[ICSE Semester I 2022]}
\displaystyle \text{(i) The triangle similar to }\triangle ARQ\text{ is }
\displaystyle \text{(a) }\triangle ORC \qquad \text{(b) }\triangle ARP \qquad \text{(c) }\triangle OBR \qquad \text{(d) }\triangle QRP.
\displaystyle \text{(ii) }\triangle PQR\sim\triangle BCR\text{ by axiom }
\displaystyle \text{(a) SAS} \qquad \text{(b) AAA} \qquad \text{(c) SSS} \qquad \text{(d) AAS.}
\displaystyle \text{(iii) If }QC=6\text{ cm},\, CR=4\text{ cm},\, BR=3\text{ cm. The length of }RP\text{ is }
\displaystyle \text{(a) }4.5\text{ cm} \qquad \text{(b) }8\text{ cm} \qquad \text{(c) }7.5\text{ cm} \qquad \text{(d) }5\text{ cm.}
\displaystyle \text{(iv) The ratio }PQ:BC\text{ is }
\displaystyle \text{(a) }2:3 \qquad \text{(b) }3:2 \qquad \text{(c) }5:2 \qquad \text{(d) }2:5.

\displaystyle \text{Answer:}
\displaystyle \textbf{(i) (a) In }\triangle ARQ\text{ and }\triangle ORC,\text{ we have }
\displaystyle AQ\parallel OC \quad \text{[since, given }QP\parallel CB\text{]}
\displaystyle \text{So, }\angle QAR=\angle COR \quad \text{[corresponding angles]}
\displaystyle \angle ARQ=\angle ORC \quad \text{[common angles]}
\displaystyle \text{and }\angle AQR=\angle OCR \quad \text{[corresponding angles]}
\displaystyle \therefore \triangle ARQ\sim\triangle ORC\quad \text{[by AAA similarity criterion]}
\displaystyle \text{(ii) (b) In }\triangle PQR\text{ and }\triangle BCR,\text{ we have,}
\displaystyle QP\parallel CB \quad \text{[given]}
\displaystyle \text{So, }\angle QPR=\angle CBR \quad \text{[corresponding angles]}
\displaystyle \angle PRQ=\angle BRC \quad \text{[common angles]}
\displaystyle \text{and }\angle PQR=\angle BCR \quad \text{[corresponding angles]}
\displaystyle \therefore \triangle PQR\sim\triangle BCR\quad \text{[by AAA similarity criterion]}
\displaystyle \text{(iii) (c) From part (ii), we get }
\displaystyle \triangle PQR\sim\triangle BCR
\displaystyle \text{So, }\frac{PQ}{BC}=\frac{QR}{CR}=\frac{PR}{BR}\quad \text{[corresponding sides of similar triangles]} \quad ...(i)
\displaystyle \Rightarrow \frac{QC+CR}{CR}=\frac{PB+BR}{BR}
\displaystyle \quad \text{[since }QR=QC+CR\text{ and }PR=PB+BR\text{]}
\displaystyle \text{Let }PB=x.
\displaystyle \text{On putting all these values in Eq. (i), we get }
\displaystyle \frac{6+4}{4}=\frac{(x+3)}{3}
\displaystyle \Rightarrow x=4.5\Rightarrow PB=4.5\text{ cm}
\displaystyle \therefore RP=BR+PB
\displaystyle \Rightarrow RP=3+4.5=7.5\text{ cm}
\displaystyle \text{(iv) (d) From part (ii), we have }\triangle PQR\sim\triangle BCR
\displaystyle \therefore \frac{PQ}{BC}=\frac{QR}{CR}=\frac{PR}{BR}\quad \text{[corresponding sides of similar triangles]}
\displaystyle \text{So, }\frac{PQ}{BC}=\frac{PR}{BR}=\frac{7.5}{3}
\displaystyle \Rightarrow \frac{PQ}{BC}=\frac{7.5}{3}=\frac{5}{2}
\displaystyle \therefore PQ:BC=5:2

\displaystyle \textbf{Question 21:}~\text{In the given figure, }ABCD\text{ is a trapezium, where }DC\parallel AB.\, AB=16\text{ cm and }DC=8\text{ cm},\, OD=5\text{ cm},\, OB=(y+3)\text{ cm},\, OA=11\text{ cm and }OC=(x-1)\text{ cm. }~\text{[ICSE Semester I 2022]}
\displaystyle \text{Using the given information answer the following questions.}
\displaystyle \text{(i) From the given figure, name the pair of similar triangles.}
\displaystyle \text{(a) }\triangle OAB,\triangle OBC \qquad \text{(b) }\triangle COD,\triangle AOB \qquad \text{(c) }\triangle ADB,\triangle ACB \qquad \text{(d) }\triangle COD,\triangle COB
\displaystyle \text{(ii) The corresponding proportional sides with respect to the pair of similar} \\ \text{triangles obtained in (i).}
\displaystyle \text{(a) }\frac{CD}{AB}=\frac{OC}{OA}=\frac{OD}{OB}
\displaystyle \text{(b) }\frac{AD}{BC}=\frac{OC}{OA}=\frac{OD}{OB}
\displaystyle \text{(c) }\frac{AD}{BC}=\frac{BD}{AC}=\frac{AB}{DC}
\displaystyle \text{(d) }\frac{OD}{OB}=\frac{CD}{CB}=\frac{OC}{OA}
\displaystyle \text{(iii) The ratio of the sides of the pair of similar triangles is}
\displaystyle \text{(a) }1:3 \qquad \text{(b) }1:2 \qquad \text{(c) }2:3 \qquad \text{(d) }3:1
\displaystyle \text{(iv) Using the ratio of sides of the pair of similar triangles the values of }x\text{ and }y \\ \text{are respectively.}
\displaystyle \text{(a) }x=4.6,\;y=7 \qquad \text{(b) }x=7,\;y=7 \qquad \text{(c) }x=6.5,\;y=7 \qquad \text{(d) }x=6.5,\;y=2

\displaystyle \text{Answer:}
\displaystyle \textbf{(i) (b) } \triangle COD \sim \triangle AOB\quad [\because CD\parallel AB\text{ and by AA similarity}]
\displaystyle \text{(ii) (a) We have, } \triangle COD \sim \triangle AOB,\text{ then }\frac{CO}{AO}=\frac{OD}{OB}=\frac{CD}{AB}
\displaystyle \text{(iii) We have } CD=8\text{ cm and }AB=16\text{ cm}
\displaystyle \therefore \text{From part (ii), }\frac{CD}{AB}=\frac{8}{16}=\frac{1}{2}\text{ or }1:2
\displaystyle \text{(iv) (c) We have, }AB=16\text{ cm}, AO=11\text{ cm}, OB=(y+3)\text{ cm}, OD=5\text{ cm}, OC=(x-1)\text{ cm and }CD=8\text{ cm}
\displaystyle \text{From part (ii), }\frac{CO}{AO}=\frac{OD}{OB}=\frac{CD}{AB}\Rightarrow \frac{x-1}{11}=\frac{5}{y+3}=\frac{8}{16}
\displaystyle \text{By taking (I) and (III) terms, }\frac{x-1}{11}=\frac{8}{16}\Rightarrow x-1=\frac{11}{2}
\displaystyle \Rightarrow x=\frac{11}{2}+1=\frac{13}{2}\Rightarrow x=6.5
\displaystyle \text{By taking (II) and (III) terms, }\frac{5}{y+3}=\frac{8}{16}\Rightarrow \frac{5}{y+3}=\frac{1}{2}
\displaystyle \Rightarrow 10=y+3\Rightarrow y=7

\displaystyle \textbf{Question 22:}~ABC\text{ is a right angled triangle with }\angle ABC=90^\circ.\, D\text{ is any point} \\ \text{on }AB\text{ and }DE\text{ is perpendicular to }AC.~\text{[ICSE 2015]}
\displaystyle \text{(i) Prove that }\triangle ADE\sim\triangle ACB.
\displaystyle \text{(ii) If }AC=13\text{ cm},\, BC=5\text{ cm and }AE=4\text{ cm, then find }DE\text{ and }AD.

\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\triangle ADE\text{ and }\triangle ACB,\; \angle E=\angle B\quad [90^\circ\text{ each}]
\displaystyle \angle DAE=\angle BAC\quad [\text{common angle}]
\displaystyle \therefore \triangle ADE\sim\triangle ACB\quad [\text{by AA similarity}]
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) In }\triangle ABC,\; AC=13\text{ cm}, BC=5\text{ cm}
\displaystyle \text{Using Pythagoras theorem in }\triangle ABC,
\displaystyle AC^2=AB^2+BC^2
\displaystyle \Rightarrow AB=\sqrt{AC^2-BC^2}
\displaystyle =\sqrt{(13)^2-(5)^2}=\sqrt{169-25}=12\text{ cm}
\displaystyle \text{Since }\triangle ADE\sim\triangle ACB,
\displaystyle \frac{AD}{AC}=\frac{AE}{CB}
\displaystyle \Rightarrow \frac{AD}{13}=\frac{4}{5}\Rightarrow AD=\frac{52}{5}=10.4\text{ cm}
\displaystyle \frac{DE}{AB}=\frac{AE}{AC}
\displaystyle \Rightarrow \frac{DE}{12}=\frac{4}{13}\Rightarrow DE=\frac{48}{13}\approx3.69\text{ cm}

\displaystyle \textbf{Question 23:}~\text{In }\triangle ABC,\;\angle ABC=\angle DAC,\; AB=8\text{ cm},\; AC=4\text{ cm and } \\ AD=5\text{ cm.}~\text{[ICSE 2014]}
\displaystyle \text{(i) Prove that }\triangle ACD\sim\triangle BCA.
\displaystyle \text{(ii) Find }BC\text{ and }CD.

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }\angle ABC=\angle DAC,\; AB=8\text{ cm},\; AC=4\text{ cm}\text{ and }AD=5\text{ cm}
\displaystyle \text{(i) To prove }\triangle ACD\sim\triangle BCA
\displaystyle \text{Proof: In }\triangle ACD\text{ and }\triangle BCA,
\displaystyle \angle ACD=\angle BCA\quad [\text{common angle}]
\displaystyle \angle ABC=\angle DAC\quad [\text{given}]
\displaystyle \therefore \triangle ACD\sim\triangle BCA\quad [\text{by AA axiom of similarity}]
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Since }\triangle ACD\sim\triangle BCA,
\displaystyle \frac{AC}{BC}=\frac{CD}{CA}=\frac{AD}{BA}
\displaystyle \Rightarrow \frac{4}{BC}=\frac{5}{8}
\displaystyle \Rightarrow BC=\frac{32}{5}=6.4\text{ cm}
\displaystyle \frac{CD}{4}=\frac{5}{8}\Rightarrow CD=\frac{20}{8}=2.5\text{ cm}

\displaystyle \textbf{Question 24:}~\text{In the following figure, }ABC\text{ is a right angled triangle }\text{with } \\ \angle BAC=90^\circ.
\displaystyle \text{(i) Prove that }\triangle ADB\sim\triangle CDA.
\displaystyle \text{(ii) If }BD=18\text{ cm},\, CD=8\text{ cm, then find }AD.~\text{[ICSE 2011]}

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }\angle BAC=90^\circ
\displaystyle \text{Let }\angle DAB=\theta
\displaystyle \text{(i) To prove }\triangle ADB\sim\triangle CDA
\displaystyle \text{Proof: In }\triangle ADB\text{ and }\triangle CDA,
\displaystyle \angle DAB=\angle DCA=\theta
\displaystyle \angle CAD=\angle ABD=90^\circ-\theta
\displaystyle \therefore \triangle ADB\sim\triangle CDA\quad [\text{by AA axiom of similarity}]
\displaystyle \text{Hence proved.}
\displaystyle \text{(ii) Since }\triangle ADB\sim\triangle CDA,
\displaystyle \therefore \frac{AD}{CD}=\frac{BD}{AD}\quad [\because \text{if two triangles are similar, their corresponding sides are proportional}]
\displaystyle \Rightarrow AD^{2}=BD\times CD
\displaystyle \Rightarrow AD^{2}=18\times 8=144\quad [\because BD=18\text{ cm and }CD=8\text{ cm}]
\displaystyle \Rightarrow AD=\sqrt{144}=12\text{ cm}
\displaystyle \quad [\text{taking positive square root}]

\displaystyle \textbf{Question 25:}~\text{In the given figure, }ABC\text{ is a triangle. }DE\parallel BC\text{ and } \\ \frac{AD}{BD}=\frac{3}{2}.~\text{[ICSE 2007]}
\displaystyle \text{(i) Determine the ratio }\frac{AD}{AB}\text{ and }\frac{DE}{BC}.
\displaystyle \text{(ii) Prove that }\triangle DEF\sim\triangle CBF.\;\text{Also find }\frac{EF}{FB}.

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }DE\parallel BC\text{ and }\frac{AD}{DB}=\frac{3}{2}
\displaystyle \text{(i) Now, }\frac{AD}{DB}=\frac{3}{2}
\displaystyle \Rightarrow \frac{DB}{AD}=\frac{2}{3}
\displaystyle \Rightarrow \frac{DB}{AD}+1=\frac{2}{3}+1\quad [\text{on adding 1 to both sides}]
\displaystyle \Rightarrow \frac{DB+AD}{AD}=\frac{5}{3}
\displaystyle \Rightarrow \frac{AB}{AD}=\frac{5}{3}\quad [\because DB+AD=AB]
\displaystyle \Rightarrow \frac{AD}{AB}=\frac{3}{5}
\displaystyle \text{In }\triangle ADE\text{ and }\triangle ABC,\text{ we get}
\displaystyle \angle DAE=\angle BAC \quad [\text{common angle}]
\displaystyle \angle ADE=\angle ABC\quad [\text{corresponding angles as }DE\parallel BC]
\displaystyle \therefore \triangle ADE\sim\triangle ABC\quad [\text{by AA axiom of similarity}]
\displaystyle \text{Then, }\frac{DE}{BC}=\frac{AD}{AB}
\displaystyle \quad [\because \text{if two triangles are similar, their corresponding sides are proportional}]
\displaystyle \Rightarrow \frac{DE}{BC}=\frac{3}{5} \quad ...(i)
\displaystyle \text{(ii) To prove }\triangle DEF\sim\triangle CBF
\displaystyle \text{Proof: In }\triangle DEF\text{ and }\triangle CBF,\text{ we have}
\displaystyle \angle FED=\angle BFC\quad [\text{alternate interior angles as }DE\parallel BC]
\displaystyle \angle DFE=\angle BCF\quad [\text{vertically opposite angles}]
\displaystyle \therefore \triangle DEF\sim\triangle CBF\quad [\text{by AA axiom of similarity}]
\displaystyle \text{Then, }\frac{DE}{BC}=\frac{FE}{FB}
\displaystyle \quad [\because \text{if two triangles are similar, their corresponding sides are proportional}]
\displaystyle \therefore \frac{FE}{FB}=\frac{3}{5}\quad [\text{from Eq. (i)}]

\displaystyle \textbf{Question 26:}~\text{In the given figure, }AB\text{ and }DE\text{ are perpendicular }\text{to } \\ BC.\; AB=9\text{ cm},\; DE=3\text{ cm and }AC=24\text{ cm, then calculate }AD.~\text{[ICSE 2005]}

\displaystyle \text{Answer:}
\displaystyle \textbf{Given }AB=9\text{ cm},\; DE=3\text{ cm and }AC=24\text{ cm}
\displaystyle \text{In }\triangle ABC\text{ and }\triangle DEC,
\displaystyle \angle B=\angle E=90^\circ
\displaystyle \quad [\because AB\perp BC\text{ and }DE\perp BC,\text{ given}]
\displaystyle \angle ACB=\angle DCE \quad [\text{common angle}]
\displaystyle \therefore \triangle ABC\sim\triangle DEC\quad [\text{by AA axiom of similarity}]
\displaystyle \Rightarrow \frac{AB}{DE}=\frac{AC}{DC}
\displaystyle \quad [\because \text{if two triangles are similar, their corresponding sides are proportional}]
\displaystyle \Rightarrow \frac{9}{3}=\frac{24}{DC}
\displaystyle \Rightarrow DC=\frac{24\times 3}{9}=8\text{ cm}
\displaystyle \text{Now, }AD=AC-DC
\displaystyle \Rightarrow AD=24-8=16\text{ cm}

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