\displaystyle \textbf{Question 1. }\text{If in two triangles }ABC\text{ and }PQR,\ \frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}, \\ \text{then }\triangle PQR\sim\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }\triangle CBA.
\displaystyle \text{Since }\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ},\text{ the corresponding sides are proportional.}
\displaystyle AB\leftrightarrow QR,\quad BC\leftrightarrow PR,\quad CA\leftrightarrow PQ
\displaystyle \therefore A\leftrightarrow Q,\quad B\leftrightarrow R,\quad C\leftrightarrow P
\displaystyle \therefore \triangle PQR\sim\triangle CBA\qquad\text{[By SSS similarity criterion]}
\displaystyle \\

\displaystyle \textbf{Question 2. }D\text{ and }E\text{ are respectively the points on the sides }AB\text{ and }AC\text{ of }
\displaystyle \triangle ABC\text{ such that } \ AD=2\text{ cm},  \ BD=3\text{ cm},\ BC=7.5\text{ cm and }DE\parallel BC. \\ \text{Then }DE=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }3\text{ cm}.
\displaystyle AB=AD+DB=2+3=5\text{ cm}
\displaystyle \text{Since }DE\parallel BC,\ \triangle ADE\sim\triangle ABC.
\displaystyle \therefore \frac{DE}{BC}=\frac{AD}{AB}
\displaystyle \frac{DE}{7.5}=\frac{2}{5}
\displaystyle DE=\frac{2\times7.5}{5}=3\text{ cm}
\displaystyle \therefore DE=3\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 3. }\text{If in two triangles }DEF\text{ and }PQR,\ \angle D=\angle Q\text{ and }\angle R=\angle E,\text{ then } \\ \frac{DE}{QR}=\frac{DF}{PQ}=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }\frac{EF}{PR}.
\displaystyle \angle D=\angle Q,\quad \angle E=\angle R
\displaystyle \therefore \angle F=\angle P
\displaystyle \therefore \triangle DEF\sim\triangle QRP\qquad\text{[By AA similarity criterion]}
\displaystyle \therefore \frac{DE}{QR}=\frac{DF}{QP}=\frac{EF}{RP}
\displaystyle \therefore \frac{DE}{QR}=\frac{DF}{PQ}=\frac{EF}{PR}.
\displaystyle \\

\displaystyle \textbf{Question 4. }\text{If }\triangle ABC\sim\triangle EDF,\text{ then }\underline{\hspace{3cm}}.
\displaystyle \text{Answer: }\frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}.
\displaystyle \text{From }\triangle ABC\sim\triangle EDF,\text{ the corresponding vertices are }A\leftrightarrow E,\ B\leftrightarrow D,\ C\leftrightarrow F.
\displaystyle \therefore AB\leftrightarrow ED,\quad BC\leftrightarrow DF,\quad AC\leftrightarrow EF
\displaystyle \therefore \frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}.
\displaystyle \\

\displaystyle \textbf{Question 5. }\text{In }\triangle ABC\text{ and }\triangle DEF,\text{ if }\angle B=\angle E,\ \angle F=\angle C\text{ and }
\displaystyle AB=3DE.  \ \text{Then, the two triangles are }\underline{\hspace{1.5cm}}\text{ but not }\underline{\hspace{1.5cm}}\text{.}
\displaystyle \text{Answer: similar, congruent.}
\displaystyle \angle B=\angle E\text{ and }\angle C=\angle F
\displaystyle \therefore \triangle ABC\sim\triangle DEF\qquad\text{[By AA similarity criterion]}
\displaystyle \text{But }AB=3DE,\text{ so the corresponding sides are not equal.}
\displaystyle \therefore \text{The two triangles are similar but not congruent.}
\displaystyle \\

\displaystyle \textbf{Question 6. }\text{If }\triangle ABC\sim\triangle QRP,\ \frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{9}{4},\ AB=18\text{ cm and } \\ BC=15\text{ cm, then }PR=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }10\text{ cm}.
\displaystyle \text{Since }\triangle ABC\sim\triangle QRP,\ BC\leftrightarrow RP.
\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\left(\frac{BC}{PR}\right)^2
\displaystyle \frac{9}{4}=\left(\frac{15}{PR}\right)^2
\displaystyle \frac{3}{2}=\frac{15}{PR}
\displaystyle PR=\frac{15\times2}{3}=10\text{ cm}
\displaystyle \therefore PR=10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 7. }\text{In an equilateral triangle }ABC,\ D\text{ is the mid-point of }AB\text{ and }E\text{ is the mid-point of }
\displaystyle AC.\text{ Then} \ ar(\triangle ABC):ar(\triangle ADE)=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }4:1.
\displaystyle \text{Since }D\text{ and }E\text{ are the mid-points of }AB\text{ and }AC,\ AD=\frac{AB}{2},\ AE=\frac{AC}{2}.
\displaystyle \therefore \triangle ADE\sim\triangle ABC
\displaystyle \frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\left(\frac{AD}{AB}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}
\displaystyle \therefore ar(\triangle ABC):ar(\triangle ADE)=4:1.
\displaystyle \\

\displaystyle \textbf{Question 8. }\text{If in }\triangle ABC\text{ and }\triangle DEF,\ \frac{AB}{DE}=\frac{BC}{FD},\text{ then they will be similar, when }\underline{\hspace{2cm}}.
\displaystyle \text{Answer: }\angle B=\angle D.
\displaystyle \text{Here, }AB\text{ and }BC\text{ include }\angle B,\text{ while }DE\text{ and }DF\text{ include }\angle D.
\displaystyle \text{Given }\frac{AB}{DE}=\frac{BC}{FD}\text{ and if }\angle B=\angle D,
\displaystyle \therefore \triangle ABC\sim\triangle EDF\qquad\text{[By SAS similarity criterion]}
\displaystyle \\

\displaystyle \textbf{Question 9. }\text{It is given that }\triangle ABC\sim\triangle PQR,\text{ with }\frac{BC}{QR}=\frac{1}{3},\text{ then } \\ \frac{ar(\triangle PRQ)}{ar(\triangle BCA)}=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }9.
\displaystyle \text{For similar triangles, the ratio of their areas is equal to the square of the ratio of}
\displaystyle \text{their corresponding sides.}
\displaystyle \frac{ar(\triangle PRQ)}{ar(\triangle BCA)}=\left(\frac{QR}{BC}\right)^2
\displaystyle =\left(\frac{3}{1}\right)^2=9
\displaystyle \therefore \frac{ar(\triangle PRQ)}{ar(\triangle BCA)}=9.
\displaystyle \\

\displaystyle \textbf{Question 10. }\text{It is given that }\triangle ABC\sim\triangle DFE,\ \angle A=30^\circ,\ \angle C=50^\circ,\ 
\displaystyle AB=5\text{ cm},  \ AC=8\text{ cm and }DF=7.5\text{ cm. Then }DE=\underline{\hspace{1.5cm}}\text{ and }\angle F=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }DE=12\text{ cm and }\angle F=100^\circ.
\displaystyle \text{Since }\triangle ABC\sim\triangle DFE,\ AB\leftrightarrow DF,\ AC\leftrightarrow DE\text{ and }\angle B\leftrightarrow\angle F.
\displaystyle \frac{AB}{DF}=\frac{AC}{DE}
\displaystyle \frac{5}{7.5}=\frac{8}{DE}
\displaystyle DE=\frac{8\times7.5}{5}=12\text{ cm}
\displaystyle \angle B=180^\circ-(30^\circ+50^\circ)=100^\circ
\displaystyle \therefore \angle F=\angle B=100^\circ.
\displaystyle \\

\displaystyle \textbf{Question 11. }\text{In }\triangle ABC,\text{ if }AB=24\text{ cm},\ BC=10\text{ cm and }AC=26\text{ cm, then } \\ \angle B=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }90^\circ.
\displaystyle AB^2+BC^2=24^2+10^2
\displaystyle =576+100=676=26^2=AC^2
\displaystyle \therefore \angle B=90^\circ\qquad\text{[By the converse of Pythagoras Theorem]}
\displaystyle \\

\displaystyle \textbf{Question 12. }\text{The lengths of the diagonals of a rhombus are }16\text{ cm and }12\text{ cm. }
\displaystyle \text{Then the length of the side of the rhombus is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }10\text{ cm}.
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore \text{Half of the diagonals are }8\text{ cm and }6\text{ cm.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle \text{Side}^2=8^2+6^2=64+36=100
\displaystyle \therefore \text{Side}=\sqrt{100}=10\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 13. }\text{In an isosceles triangle }ABC,\text{ if }AC=BC\text{ and }AB^2=2AC^2,\text{ then } \\ \angle C=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }90^\circ.
\displaystyle AC=BC\Rightarrow AC^2=BC^2
\displaystyle AB^2=2AC^2=AC^2+BC^2
\displaystyle \therefore AB^2=AC^2+BC^2
\displaystyle \therefore \angle C=90^\circ\qquad\text{[By the converse of Pythagoras Theorem]}
\displaystyle \\

\displaystyle \textbf{Question 14. }\text{In the adjoinng figure, }\angle BAC=90^\circ\text{ and }AD\perp BC.\text{ Then, } \\ BD\cdot CD=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }AD^2.
\displaystyle \text{Since }AD\perp BC,\ \angle ADB=\angle ADC=90^\circ.
\displaystyle \angle BAD=\angle ACD
\displaystyle \therefore \triangle ABD\sim\triangle CAD\qquad\text{[By AA similarity criterion]}
\displaystyle \therefore \frac{BD}{AD}=\frac{AD}{CD}
\displaystyle \therefore BD\cdot CD=AD^2.
\displaystyle \\

\displaystyle \textbf{Question 15. }\text{In the adjoinng figure, }\angle ABC=90^\circ,\ BC=10\text{ cm},\ CD=6\text{ cm, then } \\ AD=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }\frac{32}{3}\text{ cm}.
\displaystyle \text{Since }BD\perp AC,\text{ by the right triangle altitude theorem,}
\displaystyle BC^2=AC\cdot CD
\displaystyle 10^2=AC\times6
\displaystyle AC=\frac{100}{6}=\frac{50}{3}\text{ cm}
\displaystyle AD=AC-CD
\displaystyle =\frac{50}{3}-6=\frac{50-18}{3}=\frac{32}{3}\text{ cm}
\displaystyle \therefore AD=\frac{32}{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 16. }\text{In the adjoinng figure, if }DE\parallel BC,\text{ then }\frac{ar(\triangle ADE)}{ar(DECB)}=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }\frac{1}{3}.
\displaystyle \text{Since }DE\parallel BC,\ \triangle ADE\sim\triangle ABC.
\displaystyle \therefore \frac{ar(\triangle ADE)}{ar(\triangle ABC)}=\left(\frac{DE}{BC}\right)^2
\displaystyle =\left(\frac{6}{12}\right)^2=\frac{1}{4}
\displaystyle \therefore ar(\triangle ADE)=\frac{1}{4}ar(\triangle ABC)
\displaystyle ar(DECB)=ar(\triangle ABC)-ar(\triangle ADE)
\displaystyle =\left(1-\frac{1}{4}\right)ar(\triangle ABC)=\frac{3}{4}ar(\triangle ABC)
\displaystyle \therefore \frac{ar(\triangle ADE)}{ar(DECB)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 17. }\text{If }\triangle ABC\sim\triangle DEF,\ AB=4\text{ cm},\ DE=6\text{ cm},\ 
\displaystyle EF=9\text{ cm and }FD=12\text{ cm, then the perimeter} \ \text{of }\triangle ABC\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }18\text{ cm}.
\displaystyle \text{Since }\triangle ABC\sim\triangle DEF,\ \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}.
\displaystyle \frac{AB}{DE}=\frac{4}{6}=\frac{2}{3}
\displaystyle BC=\frac{2}{3}\times9=6\text{ cm}
\displaystyle AC=\frac{2}{3}\times12=8\text{ cm}
\displaystyle \text{Perimeter of }\triangle ABC=AB+BC+CA
\displaystyle =4+6+8=18\text{ cm}
\displaystyle \therefore \text{Perimeter of }\triangle ABC=18\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 18. }\text{The altitude of an equilateral triangle of side }8\text{ cm is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }4\sqrt{3}\text{ cm}.
\displaystyle \text{The altitude of an equilateral triangle bisects the opposite side.}
\displaystyle \therefore \text{Half of the side}=\frac{8}{2}=4\text{ cm}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle h^2+4^2=8^2
\displaystyle h^2=64-16=48
\displaystyle h=\sqrt{48}=4\sqrt{3}\text{ cm}
\displaystyle \therefore \text{Altitude}=4\sqrt{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 19. }\text{Corresponding sides of two similar triangles are in the ratio }
\displaystyle 2:3.\text{ If the area of the smaller} \ \text{triangle is }48\text{ cm}^2,\text{ then the area of the} \\ \text{larger triangle is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }108\text{ cm}^2.
\displaystyle \frac{\text{Area of smaller triangle}}{\text{Area of larger triangle}}=\left(\frac{2}{3}\right)^2=\frac{4}{9}
\displaystyle \frac{48}{\text{Area of larger triangle}}=\frac{4}{9}
\displaystyle \text{Area of larger triangle}=\frac{48\times9}{4}=108\text{ cm}^2
\displaystyle \therefore \text{Area of the larger triangle}=108\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 20. }\text{Areas of two similar triangles are }36\text{ cm}^2\text{ and }100\text{ cm}^2.
\displaystyle \text{If the length of a side of the larger triangle} \ \text{is }20\text{ cm, then the length of the} \\ \text{corresponding side of the smaller triangle is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }12\text{ cm}.
\displaystyle \frac{\text{Area of smaller triangle}}{\text{Area of larger triangle}}=\left(\frac{\text{Corresponding side of smaller}}{\text{Corresponding side of larger}}\right)^2
\displaystyle \frac{36}{100}=\left(\frac{x}{20}\right)^2
\displaystyle \frac{6}{10}=\frac{x}{20}
\displaystyle x=\frac{6}{10}\times20=12\text{ cm}
\displaystyle \therefore \text{Length of the corresponding side of the smaller triangle}=12\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 21. }\text{Diagonals of a trapezium }PQRS\text{ intersect each other at the point }O.\text{ If }
\displaystyle PQ\parallel RS\text{ and }PQ=3RS,  \ \text{then }\frac{ar(\triangle POQ)}{ar(\triangle ROS)}=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }9.
\displaystyle \text{Since }PQ\parallel RS,\ \triangle POQ\sim\triangle ROS.
\displaystyle \therefore \frac{ar(\triangle POQ)}{ar(\triangle ROS)}=\left(\frac{PQ}{RS}\right)^2
\displaystyle =\left(\frac{3RS}{RS}\right)^2=3^2=9
\displaystyle \therefore \frac{ar(\triangle POQ)}{ar(\triangle ROS)}=9.
\displaystyle \\

\displaystyle \textbf{Question 22. }ABCD\text{ is a trapezium in which }AB\parallel DC\text{ and }P\text{ and }Q\text{ are the points on }
\displaystyle AD\text{ and }BC \ \text{respectively such that }PQ\parallel DC.\text{ If }PD=18\text{ cm},\ BQ=35\text{ cm and } \\ QC=15\text{ cm, then }AD=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }60\text{ cm}.
\displaystyle \text{Since }AB\parallel PQ\parallel DC,\text{ the non-parallel sides are divided proportionally.}
\displaystyle \therefore \frac{AP}{PD}=\frac{BQ}{QC}
\displaystyle \frac{AP}{18}=\frac{35}{15}=\frac{7}{3}
\displaystyle AP=18\times\frac{7}{3}=42\text{ cm}
\displaystyle AD=AP+PD=42+18=60\text{ cm}
\displaystyle \therefore AD=60\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 23. }\text{In the adjoinng figure, }PQR\text{ is a right triangle right angled at }Q\text{ and }
\displaystyle QS\perp PR.\text{ If }PQ=6\text{ cm and }PS=4\text{ cm}, \ \text{then perimeter of }\triangle QSR\text{ is }\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }5+5\sqrt{5}\text{ cm}.
\displaystyle \text{Since }QS\perp PR,\text{ using the right triangle altitude property,}
\displaystyle PQ^2=PS\cdot PR
\displaystyle 6^2=4\cdot PR\Rightarrow PR=9\text{ cm}
\displaystyle SR=PR-PS=9-4=5\text{ cm}
\displaystyle QS^2=PS\cdot SR=4\times5=20
\displaystyle \therefore QS=2\sqrt{5}\text{ cm}
\displaystyle QR^2=PR\cdot SR=9\times5=45
\displaystyle \therefore QR=3\sqrt{5}\text{ cm}
\displaystyle \text{Perimeter of }\triangle QSR=QS+SR+QR
\displaystyle =2\sqrt{5}+5+3\sqrt{5}=5+5\sqrt{5}\text{ cm}
\displaystyle \therefore \text{Perimeter of }\triangle QSR=5+5\sqrt{5}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 24. }\text{In the adjoinng figure, }ABC\text{ is a triangle right angled at }B\text{ and }
\displaystyle BD\perp AC.\text{ If }AD=4\text{ cm and }CD=5\text{ cm, then} \ BD=\underline{\hspace{1.5cm}}\text{ and } \\ AB=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }BD=2\sqrt{5}\text{ cm and }AB=6\text{ cm}.
\displaystyle AC=AD+DC=4+5=9\text{ cm}
\displaystyle \text{Since }BD\perp AC,\text{ using the right triangle altitude properties,}
\displaystyle BD^2=AD\cdot DC=4\times5=20
\displaystyle \therefore BD=2\sqrt{5}\text{ cm}
\displaystyle \text{Also, }AB^2=AD\cdot AC=4\times9=36
\displaystyle \therefore AB=6\text{ cm}
\displaystyle \therefore BD=2\sqrt{5}\text{ cm and }AB=6\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 25. }\text{In the adjoinng figure, if }\angle ACB=\angle CDA,\ AC=8\text{ cm and } \\ AD=3\text{ cm, then }BD=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }\frac{55}{3}\text{ cm}.
\displaystyle \text{Since }A,D,B\text{ are collinear, }\angle CAD=\angle CAB.
\displaystyle \text{Also, }\angle CDA=\angle ACB\qquad\text{[Given]}
\displaystyle \therefore \triangle ACD\sim\triangle ABC\qquad\text{[By AA similarity criterion]}
\displaystyle \therefore \frac{AD}{AC}=\frac{AC}{AB}
\displaystyle \frac{3}{8}=\frac{8}{AB}
\displaystyle AB=\frac{64}{3}\text{ cm}
\displaystyle BD=AB-AD=\frac{64}{3}-3=\frac{55}{3}\text{ cm}
\displaystyle \therefore BD=\frac{55}{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 26. }\text{A }15\text{ metres high tower casts a shadow }24\text{ metres long at a }
\displaystyle \text{certain time and at the same time, a telephone pole casts a shadow }16\text{ metres} \\ \text{long. The height of the telephone pole is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }10\text{ metres}.
\displaystyle \text{At the same time, the triangles formed by the tower and the telephone pole with their shadows are similar.}
\displaystyle \therefore \frac{15}{24}=\frac{h}{16}
\displaystyle h=\frac{15\times16}{24}=10\text{ metres}
\displaystyle \therefore \text{Height of the telephone pole}=10\text{ metres}.
\displaystyle \\

\displaystyle \textbf{Question 27. }\text{In the adjoinng figure, if }\angle A=\angle C,\ AB=6\text{ cm},\ BP=15\text{ cm},\
\displaystyle AP=12\text{ cm and }CP=4\text{ cm, then }PD=\underline{\hspace{1.5cm}}  \ \text{and }CD=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }PD=5\text{ cm and }CD=2\text{ cm}.
\displaystyle \angle APB=\angle CPD\qquad\text{[Vertically opposite angles]}
\displaystyle \angle A=\angle C\qquad\text{[Given]}
\displaystyle \therefore \triangle APB\sim\triangle CPD\qquad\text{[By AA similarity criterion]}
\displaystyle \therefore \frac{AP}{CP}=\frac{BP}{PD}=\frac{AB}{CD}
\displaystyle \frac{12}{4}=\frac{15}{PD}=\frac{6}{CD}
\displaystyle 3=\frac{15}{PD}\Rightarrow PD=5\text{ cm}
\displaystyle 3=\frac{6}{CD}\Rightarrow CD=2\text{ cm}
\displaystyle \therefore PD=5\text{ cm and }CD=2\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 28. }\text{A flag pole }18\text{ m high casts a shadow }9.6\text{ m long. The distance }
\displaystyle \text{of the top of the pole from the far end of the shadow is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }20.4\text{ m}.
\displaystyle \text{The pole and its shadow form a right triangle.}
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle d^2=18^2+(9.6)^2
\displaystyle =324+92.16=416.16
\displaystyle d=\sqrt{416.16}=20.4\text{ m}
\displaystyle \therefore \text{The required distance is }20.4\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 29. }\text{If it is given that }\triangle ABC\sim\triangle EDF\text{ such that }AB=5\text{ cm},\ 
\displaystyle AC=7\text{ cm}, DF=15\text{ cm and }DE=12\text{ cm. Then, }BC=\underline{\hspace{1.5cm}}\text{ and }EF=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }BC=\frac{25}{4}\text{ cm and }EF=\frac{84}{5}\text{ cm}.
\displaystyle \text{Since }\triangle ABC\sim\triangle EDF,\ AB\leftrightarrow ED,\ BC\leftrightarrow DF,\ AC\leftrightarrow EF.
\displaystyle \therefore \frac{AB}{DE}=\frac{BC}{DF}=\frac{AC}{EF}
\displaystyle \frac{5}{12}=\frac{BC}{15}
\displaystyle BC=\frac{5\times15}{12}=\frac{25}{4}\text{ cm}
\displaystyle \frac{5}{12}=\frac{7}{EF}
\displaystyle EF=\frac{7\times12}{5}=\frac{84}{5}\text{ cm}
\displaystyle \therefore BC=\frac{25}{4}\text{ cm and }EF=\frac{84}{5}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 30. }\text{In two triangles }ABC\text{ and }DEF,\ \angle A=\angle D\text{ and the sum of the angles }A\text{ and}
\displaystyle B\text{ is equal to the sum} \ \text{of the angles }D\text{ and }E.\text{ If }BC=6\text{ cm and }EF=8\text{ cm, then } \\ ar(\triangle ABC):ar(\triangle DEF)=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }9:16.
\displaystyle \angle A+\angle B=\angle D+\angle E\text{ and }\angle A=\angle D
\displaystyle \therefore \angle B=\angle E
\displaystyle \therefore \triangle ABC\sim\triangle DEF\qquad\text{[By AA similarity criterion]}
\displaystyle \therefore \frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\left(\frac{BC}{EF}\right)^2
\displaystyle =\left(\frac{6}{8}\right)^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}
\displaystyle \therefore ar(\triangle ABC):ar(\triangle DEF)=9:16.
\displaystyle \\

\displaystyle \textbf{Question 31. }\text{In }\triangle ABC,\text{ sides }AB\text{ and }AC\text{ are extended to }D\text{ and }E\text{ respectively, such that }
\displaystyle AB=BD\text{ and} \ AC=CE.\text{ If }BC=6\text{ cm, then }DE=\underline{\hspace{1.5cm}}.

\displaystyle \text{Answer: }12\text{ cm}.
\displaystyle AB=BD\Rightarrow B\text{ is the mid-point of }AD.
\displaystyle AC=CE\Rightarrow C\text{ is the mid-point of }AE.
\displaystyle \text{In }\triangle ADE,\text{ the line joining the mid-points }B\text{ and }C\text{ is parallel to }DE
\displaystyle \text{and }BC=\frac{1}{2}DE.
\displaystyle 6=\frac{1}{2}DE
\displaystyle DE=12\text{ cm}
\displaystyle \therefore DE=12\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 32. }\text{In the adjoining figure, }AB\parallel PR\text{ and }ar(\triangle PAB):ar(\triangle PQR)=1:2.\text{ Then }PQ:AQ=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }(2+\sqrt{2}):1.
\displaystyle \text{Since }AB\parallel QR,\ \triangle PAB\sim\triangle PQR.
\displaystyle \therefore \frac{ar(\triangle PAB)}{ar(\triangle PQR)}=\left(\frac{PA}{PQ}\right)^2
\displaystyle \frac{1}{2}=\left(\frac{PA}{PQ}\right)^2
\displaystyle \therefore \frac{PA}{PQ}=\frac{1}{\sqrt{2}}
\displaystyle AQ=PQ-PA=PQ\left(1-\frac{1}{\sqrt{2}}\right)
\displaystyle \therefore \frac{PQ}{AQ}=\frac{1}{1-\frac{1}{\sqrt{2}}}=\frac{\sqrt{2}}{\sqrt{2}-1}=2+\sqrt{2}
\displaystyle \therefore PQ:AQ=(2+\sqrt{2}):1.
\displaystyle \\

\displaystyle \textbf{Question 33. }\text{In the adjoinng figure, }\angle ABC=90^\circ,\ AD=15\text{ cm and }DC=20\text{ cm. If }
\displaystyle BD\text{ is the bisector of }\angle ABC,\text{ then} \ \text{perimeter of }\triangle ABC\text{ is }\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }84\text{ cm}.
\displaystyle AC=AD+DC=15+20=35\text{ cm}
\displaystyle \text{Since }BD\text{ bisects }\angle ABC,\text{ by the Angle Bisector Theorem,}
\displaystyle \frac{AB}{BC}=\frac{AD}{DC}=\frac{15}{20}=\frac{3}{4}
\displaystyle \text{Let }AB=3x\text{ and }BC=4x.
\displaystyle \text{Since }\angle ABC=90^\circ,\text{ by Pythagoras Theorem,}
\displaystyle AC^2=AB^2+BC^2
\displaystyle 35^2=(3x)^2+(4x)^2
\displaystyle 1225=25x^2
\displaystyle x=7
\displaystyle \therefore AB=21\text{ cm and }BC=28\text{ cm}
\displaystyle \text{Perimeter of }\triangle ABC=AB+BC+AC=21+28+35=84\text{ cm}
\displaystyle \therefore \text{Perimeter of }\triangle ABC=84\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 34. }\text{In the adjoining figure, }\angle ABC=90^\circ, \\ AB:BD:DC=3:1:3.\text{ If }AC=20\text{ cm, then }AD=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }4\sqrt{10}\text{ cm}.
\displaystyle \text{Let }AB=3x,\ BD=x\text{ and }DC=3x.
\displaystyle \therefore BC=BD+DC=x+3x=4x
\displaystyle \text{Since }\angle ABC=90^\circ,\text{ by Pythagoras Theorem,}
\displaystyle AC^2=AB^2+BC^2
\displaystyle 20^2=(3x)^2+(4x)^2
\displaystyle 400=25x^2
\displaystyle x=4
\displaystyle \therefore AB=12\text{ cm and }BD=4\text{ cm}
\displaystyle \text{In right }\triangle ABD,\text{ by Pythagoras Theorem,}
\displaystyle AD^2=AB^2+BD^2
\displaystyle =12^2+4^2=144+16=160
\displaystyle AD=\sqrt{160}=4\sqrt{10}\text{ cm}
\displaystyle \therefore AD=4\sqrt{10}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 35. }\text{In the adjoining figure, }PQ\parallel BC\text{ and }PQ:BC=1:3. \\ \text{ If }ar(\triangle ABC)=144\text{ cm}^2,\text{ then }ar(\triangle APQ)=\underline{\hspace{1.5cm}}. \displaystyle \text{Answer: }16\text{ cm}^2.
\displaystyle \text{Since }PQ\parallel BC,\ \triangle APQ\sim\triangle ABC.
\displaystyle \therefore \frac{ar(\triangle APQ)}{ar(\triangle ABC)}=\left(\frac{PQ}{BC}\right)^2
\displaystyle =\left(\frac{1}{3}\right)^2=\frac{1}{9}
\displaystyle \therefore ar(\triangle APQ)=\frac{1}{9}\times144=16\text{ cm}^2
\displaystyle \therefore ar(\triangle APQ)=16\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 36. }\text{If }ABC\text{ is an equilateral triangle of side }2a,\text{ then the length} \\ \text{of one of its altitudes is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer: }a\sqrt{3}.
\displaystyle \text{The altitude of an equilateral triangle bisects the opposite side.}
\displaystyle \therefore \text{Half of the side}=\frac{2a}{2}=a
\displaystyle \text{By Pythagoras Theorem,}
\displaystyle h^2+a^2=(2a)^2
\displaystyle h^2=4a^2-a^2=3a^2
\displaystyle h=a\sqrt{3}
\displaystyle \therefore \text{Length of the altitude}=a\sqrt{3}.
\displaystyle \\

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