\displaystyle \textbf{Question 1: }\text{In the adjoining figure, }\triangle ACB\sim\triangle APQ.\text{ If }BC=8\text{ cm},\ PQ=4\text{ cm},
\displaystyle BA=6.5\text{ cm and }AP=2.8\text{ cm, find }CA\text{ and }AQ. \displaystyle \text{Answer:}
\displaystyle \text{Since }\triangle ACB\sim\triangle APQ,
\displaystyle \frac{AC}{AP}=\frac{CB}{PQ}=\frac{AB}{AQ}
\displaystyle \frac{CB}{PQ}=\frac{8}{4}=2
\displaystyle \therefore \frac{AC}{AP}=2
\displaystyle AC=2\times AP=2\times2.8=5.6\text{ cm}
\displaystyle \text{Also, }\frac{AB}{AQ}=2
\displaystyle \frac{6.5}{AQ}=2
\displaystyle AQ=\frac{6.5}{2}=3.25\text{ cm}
\displaystyle \therefore CA=5.6\text{ cm and }AQ=3.25\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the adjoinng figure, }AB\parallel QR.\text{ Find the length of }PB.
\displaystyle \text{Answer:} \displaystyle \text{Since }AB\parallel QR,
\displaystyle \angle PAB=\angle PQR\text{ and }\angle PBA=\angle PRQ
\displaystyle \therefore \triangle PAB\sim\triangle PQR\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{PB}{PR}=\frac{AB}{QR}
\displaystyle \frac{PB}{6}=\frac{3}{9}=\frac{1}{3}
\displaystyle PB=\frac{6}{3}=2\text{ cm}
\displaystyle \therefore PB=2\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the adjoining figure, }XY\parallel BC.\text{ Find the length of }XY.

\displaystyle \text{Answer:}
\displaystyle \text{Given, }AX=1\text{ cm},\ XB=3\text{ cm and }BC=6\text{ cm}
\displaystyle AB=AX+XB=1+3=4\text{ cm}
\displaystyle \text{Since }XY\parallel BC,
\displaystyle \angle AXY=\angle ABC\text{ and }\angle AYX=\angle ACB
\displaystyle \therefore \triangle AXY\sim\triangle ABC\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{XY}{BC}=\frac{AX}{AB}
\displaystyle \frac{XY}{6}=\frac{1}{4}
\displaystyle XY=\frac{6}{4}=1.5\text{ cm}
\displaystyle \therefore XY=1.5\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the adjoining figure, }DE\parallel BC\text{ such that }AE=\frac{1}{4}AC.\text{ If }AB=6\text{ cm, find }AD.

\displaystyle \text{Answer:}
\displaystyle \text{Since }DE\parallel BC,
\displaystyle \angle ADE=\angle ABC\text{ and }\angle AED=\angle ACB
\displaystyle \therefore \triangle ADE\sim\triangle ABC\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{AD}{AB}=\frac{AE}{AC}
\displaystyle \frac{AD}{6}=\frac{1}{4}
\displaystyle AD=\frac{6}{4}=1.5\text{ cm}
\displaystyle \therefore AD=1.5\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In the adjoining figure, if }AB\perp BC,\ DC\perp BC\text{ and }DE\perp AC,\text{ prove that}
\displaystyle \triangle CED\sim\triangle ABC. \displaystyle \text{Answer:}
\displaystyle AB\perp BC\text{ and }DC\perp BC
\displaystyle \therefore AB\parallel DC
\displaystyle \text{Also, }DE\perp AC
\displaystyle \therefore \angle CED=90^\circ
\displaystyle \text{Since }AB\perp BC,\ \angle ABC=90^\circ
\displaystyle \therefore \angle CED=\angle ABC
\displaystyle \text{Also, }AB\parallel DC\text{ and }AC\text{ is a transversal.}
\displaystyle \therefore \angle ECD=\angle BAC
\displaystyle \therefore \triangle CED\sim\triangle ABC\qquad\text{[By AA criterion]}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Diagonals }AC\text{ and }BD\text{ of a trapezium }ABCD\text{ with }AB\parallel DC
\displaystyle \text{intersect each other at the point }O.\text{ Using similarity criterion for two triangles, show that}
\displaystyle \frac{OA}{OC}=\frac{OB}{OD}.
\displaystyle \text{Answer:} \displaystyle \text{Consider }\triangle AOB\text{ and }\triangle COD.
\displaystyle \angle AOB=\angle COD\qquad\text{[Vertically opposite angles]}
\displaystyle \angle ABO=\angle CDO\qquad\text{[Alternate interior angles, since }AB\parallel DC\text{]}
\displaystyle \therefore \triangle AOB\sim\triangle COD\qquad\text{[By AA criterion]}
\displaystyle \text{Therefore, corresponding sides of similar triangles are proportional.}
\displaystyle \frac{OA}{OC}=\frac{OB}{OD}=\frac{AB}{CD}
\displaystyle \therefore \frac{OA}{OC}=\frac{OB}{OD}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In a right angled triangle with sides }a\text{ and }b\text{ and hypotenuse }c,\text{ the altitude}
\displaystyle \text{drawn on the hypotenuse is }x.\text{ Prove that }ab=cx.
\displaystyle \text{Answer:} \displaystyle \text{Let }\triangle ABC\text{ be right angled at }A,\text{ where }AB=a,\ AC=b\text{ and }BC=c.
\displaystyle \text{Let }AD\perp BC\text{ such that }AD=x.
\displaystyle \text{Consider }\triangle ABD\text{ and }\triangle ACB.
\displaystyle \angle ADB=\angle CAB=90^\circ
\displaystyle \angle ABD=\angle ABC
\displaystyle \therefore \triangle ABD\sim\triangle CBA\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{AD}{AC}=\frac{AB}{BC}
\displaystyle \frac{x}{b}=\frac{a}{c}
\displaystyle cx=ab
\displaystyle \therefore ab=cx.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the adjoining figure, }\angle ABC=90^\circ\text{ and }BD\perp AC.
\displaystyle \text{(i) If }BD=8\text{ cm and }AD=4\text{ cm, find }CD.
\displaystyle \text{(ii) If }AD=4\text{ cm and }CD=5\text{ cm, find }BD\text{ and }AB.
\displaystyle \text{(iii) If }AB=5.7\text{ cm},\ BD=3.8\text{ cm and }CD=5.4\text{ cm, find }AC. \displaystyle \text{Answer:}
\displaystyle \text{Since }BD\perp AC,\text{ the triangles formed are similar, and hence}
\displaystyle BD^2=AD\times CD\text{ and }AB^2=AD\times AC.

\displaystyle \text{(i) }BD^2=AD\times CD
\displaystyle 8^2=4\times CD
\displaystyle CD=\frac{64}{4}=16\text{ cm}
\displaystyle \therefore CD=16\text{ cm.}

\displaystyle \text{(ii) }BD^2=AD\times CD
\displaystyle BD^2=4\times5=20
\displaystyle \therefore BD=2\sqrt{5}\text{ cm}
\displaystyle AC=AD+CD=4+5=9\text{ cm}
\displaystyle AB^2=AD\times AC=4\times9=36
\displaystyle \therefore AB=6\text{ cm}
\displaystyle \therefore BD=2\sqrt{5}\text{ cm and }AB=6\text{ cm.}

\displaystyle \text{(iii) The data given in this part are not mutually consistent.}
\displaystyle \text{From }BD^2=AD\times CD,
\displaystyle AD=\frac{BD^2}{CD}=\frac{(3.8)^2}{5.4}=\frac{14.44}{5.4}\approx2.674\text{ cm}
\displaystyle AC=AD+CD\approx2.674+5.4=8.074\text{ cm}
\displaystyle \text{But then }AB^2=AD\times AC\approx2.674\times8.074,
\displaystyle AB\approx4.65\text{ cm},\text{ whereas the question gives }AB=5.7\text{ cm.}
\displaystyle \therefore \text{the numerical data in part (iii) contain an apparent error and do not determine a}
\displaystyle \text{consistent right triangle as stated.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }\angle A=\angle CED,\text{ prove that }\triangle CAB\sim\triangle CED.
\displaystyle \text{Also, find the value of }x. \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle CAB\text{ and }\triangle CED,
\displaystyle \angle CAB=\angle CED\qquad\text{[Given]}
\displaystyle \angle ACB=\angle ECD\qquad\text{[Common angle at }C\text{]}
\displaystyle \therefore \triangle CAB\sim\triangle CED\qquad\text{[By AA criterion]}
\displaystyle \text{Therefore, corresponding sides of similar triangles are proportional.}
\displaystyle \frac{CA}{CE}=\frac{AB}{ED}
\displaystyle CA=AD+DC=7+8=15\text{ cm}
\displaystyle CE=10\text{ cm},\ AB=9\text{ cm and }ED=x
\displaystyle \therefore \frac{15}{10}=\frac{9}{x}
\displaystyle 15x=90
\displaystyle x=6
\displaystyle \therefore x=6\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the adjoining figure, }\triangle ABC\text{ is right angled at }C\text{ and }DE\perp AB.
\displaystyle \text{Prove that }\triangle ABC\sim\triangle ADE\text{ and hence find the lengths of }AE\text{ and }DE. \displaystyle \text{Answer:}
\displaystyle \text{Given, }AD=3\text{ cm},\ DC=2\text{ cm and }BC=12\text{ cm}
\displaystyle AC=AD+DC=3+2=5\text{ cm}
\displaystyle \text{Since }\triangle ABC\text{ is right angled at }C,
\displaystyle AB=\sqrt{AC^2+BC^2}=\sqrt{5^2+12^2}=\sqrt{169}=13\text{ cm}
\displaystyle \text{In }\triangle ABC\text{ and }\triangle ADE,
\displaystyle \angle ACB=\angle AED=90^\circ
\displaystyle \angle BAC=\angle DAE\qquad\text{[Common angle at }A\text{]}
\displaystyle \therefore \triangle ABC\sim\triangle ADE\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}
\displaystyle \frac{13}{3}=\frac{5}{AE}
\displaystyle AE=\frac{15}{13}\text{ cm}
\displaystyle \text{Also, }\frac{13}{3}=\frac{12}{DE}
\displaystyle DE=\frac{36}{13}\text{ cm}
\displaystyle \therefore AE=\frac{15}{13}\text{ cm and }DE=\frac{36}{13}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The perimeters of two similar triangles are }25\text{ cm and }15\text{ cm respectively.}
\displaystyle \text{If one side of the first triangle is }9\text{ cm, what is the corresponding side of the other} \\ \text{triangle?}\hfill\text{[CBSE 2002 C]}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of corresponding sides is equal to the ratio of their perimeters.}
\displaystyle \therefore \frac{\text{Corresponding side of second triangle}}{9}=\frac{15}{25}
\displaystyle \text{Corresponding side of second triangle}=9\times\frac{15}{25}
\displaystyle =\frac{27}{5}=5.4\text{ cm}
\displaystyle \therefore \text{The corresponding side of the other triangle is }5.4\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A girl of height }90\text{ cm is walking away from the base of a lamp-post }
\displaystyle \text{at a speed of }1.2\text{ m/sec. If the lamp is }3.6\text{ m above the ground, find the length of} \\ \text{her shadow after }4\text{ seconds.}
\displaystyle \text{Answer:}
\displaystyle \text{Height of the girl}=90\text{ cm}=0.9\text{ m}
\displaystyle \text{Height of the lamp-post}=3.6\text{ m}
\displaystyle \text{Distance travelled by the girl in }4\text{ seconds}=1.2\times4=4.8\text{ m}
\displaystyle \text{Let the length of her shadow}=x\text{ m}.
\displaystyle \text{Distance from the lamp-post to the tip of the shadow}=(4.8+x)\text{ m}.
\displaystyle \text{The triangles formed by the lamp-post and the girl with their shadows are similar.}
\displaystyle \therefore \frac{3.6}{4.8+x}=\frac{0.9}{x}
\displaystyle 3.6x=0.9(4.8+x)
\displaystyle 3.6x=4.32+0.9x
\displaystyle 2.7x=4.32
\displaystyle x=1.6
\displaystyle \therefore \text{The length of her shadow after }4\text{ seconds is }1.6\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A vertical stick of length }6\text{ m casts a shadow }4\text{ m long on the ground}
\displaystyle \text{and at the same time a tower casts a shadow }28\text{ m long. Find the height of the tower.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower}=h\text{ m}.
\displaystyle \text{Since the shadows are cast at the same time, the triangles formed by the stick and tower are similar.}
\displaystyle \therefore \frac{h}{28}=\frac{6}{4}
\displaystyle h=\frac{6}{4}\times28=42\text{ m}
\displaystyle \therefore \text{The height of the tower is }42\text{ m.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the adjoining figure, }PA,\ QB\text{ and }RC\text{ are each perpendicular to }AC.
\displaystyle \text{Prove that }\frac{1}{x}+\frac{1}{z}=\frac{1}{y}.\hfill\text{[CBSE 2024, 2025]} \displaystyle \text{Answer:}
\displaystyle \text{Since }PA\perp AC,\ QB\perp AC\text{ and }RC\perp AC,
\displaystyle PA\parallel QB\parallel RC.
\displaystyle \text{In }\triangle PCA,\ QB\parallel PA.
\displaystyle \therefore \triangle QBC\sim\triangle PAC\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{QB}{PA}=\frac{BC}{AC}
\displaystyle \therefore \frac{y}{x}=\frac{BC}{AC}\qquad ...(i)
\displaystyle \text{In }\triangle ACR,\ QB\parallel CR.
\displaystyle \therefore \triangle ABQ\sim\triangle ACR\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{QB}{CR}=\frac{AB}{AC}
\displaystyle \therefore \frac{y}{z}=\frac{AB}{AC}\qquad ...(ii)
\displaystyle \text{Adding (i) and (ii),}
\displaystyle \frac{y}{x}+\frac{y}{z}=\frac{BC}{AC}+\frac{AB}{AC}
\displaystyle y\left(\frac{1}{x}+\frac{1}{z}\right)=\frac{AB+BC}{AC}
\displaystyle y\left(\frac{1}{x}+\frac{1}{z}\right)=\frac{AC}{AC}=1
\displaystyle \therefore \frac{1}{x}+\frac{1}{z}=\frac{1}{y}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the adjoining figure, we have }AB\parallel CD\parallel EF.\text{ If }AB=6\text{ cm},
\displaystyle CD=x\text{ cm},\ EF=10\text{ cm},\ BD=4\text{ cm and }DE=y\text{ cm, calculate }x\text{ and }y. \displaystyle \text{Answer:}
\displaystyle \text{Since }AB\parallel EF,\text{ in }\triangle ABD\text{ and }\triangle FED,
\displaystyle \angle ABD=\angle FED\text{ and }\angle BAD=\angle EFD
\displaystyle \therefore \triangle ABD\sim\triangle FED\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{AB}{EF}=\frac{BD}{DE}
\displaystyle \frac{6}{10}=\frac{4}{y}
\displaystyle 6y=40
\displaystyle y=\frac{20}{3}\text{ cm}
\displaystyle BE=BD+DE=4+\frac{20}{3}=\frac{32}{3}\text{ cm}
\displaystyle \text{Now, in }\triangle ABE,\ CD\parallel AB.
\displaystyle \therefore \triangle CDE\sim\triangle ABE\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{CD}{AB}=\frac{DE}{BE}
\displaystyle \frac{x}{6}=\frac{\frac{20}{3}}{\frac{32}{3}}=\frac{5}{8}
\displaystyle x=6\times\frac{5}{8}=\frac{15}{4}=3.75\text{ cm}
\displaystyle \therefore x=3.75\text{ cm and }y=\frac{20}{3}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The corresponding sides of }\triangle ABC\text{ and }\triangle PQR\text{ are in the ratio }3:5.
\displaystyle AD\perp BC\text{ and }PS\perp QR\text{ as shown in the following figures.}\hfill\text{[CBSE 2025]}
\displaystyle \text{(i) Prove that }\triangle ADC\sim\triangle PSR.
\displaystyle \text{(ii) If }AD=4\text{ cm, find the length of }PS.
\displaystyle \text{(iii) Using (ii) find }ar(\triangle ABC):ar(\triangle PQR). \displaystyle \text{Answer:}
\displaystyle \text{Since the corresponding sides of }\triangle ABC\text{ and }\triangle PQR\text{ are proportional,}
\displaystyle \triangle ABC\sim\triangle PQR\qquad\text{[By SSS criterion]}
\displaystyle \therefore \angle ACB=\angle PRQ

\displaystyle \text{(i) In }\triangle ADC\text{ and }\triangle PSR,
\displaystyle \angle ADC=\angle PSR=90^\circ
\displaystyle \angle ACD=\angle PRS\qquad[\because \angle ACB=\angle PRQ]
\displaystyle \therefore \triangle ADC\sim\triangle PSR\qquad\text{[By AA criterion]}

\displaystyle \text{(ii) Since }\triangle ADC\sim\triangle PSR,
\displaystyle \frac{AD}{PS}=\frac{AC}{PR}=\frac{3}{5}
\displaystyle \frac{4}{PS}=\frac{3}{5}
\displaystyle PS=\frac{4\times5}{3}=\frac{20}{3}\text{ cm}
\displaystyle \therefore PS=\frac{20}{3}\text{ cm.}

\displaystyle \text{(iii) }\frac{BC}{QR}=\frac{3}{5}\text{ and }\frac{AD}{PS}=\frac{3}{5}
\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PS}
\displaystyle =\frac{BC}{QR}\times\frac{AD}{PS}=\frac{3}{5}\times\frac{3}{5}=\frac{9}{25}
\displaystyle \therefore ar(\triangle ABC):ar(\triangle PQR)=9:25.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }\triangle CAB\text{ is a right triangle, right angled at }
\displaystyle A\text{ and }AD\perp BC.  \ \text{Prove that }\triangle ADB\sim\triangle CDA.\text{ Further, if } BC=10\text{ cm and }
\displaystyle CD=2\text{ cm, find the length of }AD.  \hfill\text{[CBSE 2025]} \displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ADB\text{ and }\triangle CDA,
\displaystyle \angle ADB=\angle CDA=90^\circ\qquad[\because AD\perp BC]
\displaystyle \angle ABD=\angle CAD
\displaystyle \text{Therefore, }\triangle ADB\sim\triangle CDA\qquad\text{[By AA criterion]}
\displaystyle \text{Hence, }\frac{BD}{AD}=\frac{AD}{CD}
\displaystyle \therefore AD^2=BD\times CD
\displaystyle \text{Now, }BD=BC-CD=10-2=8\text{ cm}
\displaystyle \therefore AD^2=8\times2=16
\displaystyle AD=\sqrt{16}=4\text{ cm}
\displaystyle \therefore AD=4\text{ cm.}
\displaystyle \\

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