\displaystyle \textbf{Question 1: }\text{Triangles }ABC\text{ and }DEF\text{ are similar.}
\displaystyle \text{(i) If area }(\triangle ABC)=36\text{ cm}^2,\text{ area }(\triangle DEF)=64\text{ cm}^2\text{ and }DE=6.2\text{ cm, find }AB.
\displaystyle \text{(ii) If }AB=1.2\text{ cm and }DE=1.4\text{ cm, find the ratio of the areas of }\triangle ABC\text{ and }\triangle DEF.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\triangle ABC\sim\triangle DEF,
\displaystyle \frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle DEF)}=\frac{AB^2}{DE^2}

\displaystyle \text{(i) }\frac{36}{64}=\frac{AB^2}{(6.2)^2}
\displaystyle \frac{3^2}{4^2}=\frac{AB^2}{(6.2)^2}
\displaystyle \frac{3}{4}=\frac{AB}{6.2}
\displaystyle AB=\frac{3}{4}\times6.2=4.65\text{ cm}
\displaystyle \therefore AB=4.65\text{ cm.}

\displaystyle \text{(ii) }\frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle DEF)}=\frac{(1.2)^2}{(1.4)^2}
\displaystyle =\frac{144}{196}=\frac{36}{49}
\displaystyle \therefore \text{Area }(\triangle ABC):\text{Area }(\triangle DEF)=36:49.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In the adjoning figure, }\triangle ACB\sim\triangle APQ.\text{ If }BC=10\text{ cm},\ P
\displaystyle Q=5\text{ cm},  \ BA=6.5\text{ cm and }AP=2.8\text{ cm, find }CA\text{ and }AQ.
\displaystyle \text{ Also, find the area of }\triangle ACB: \ \text{area of }\triangle APQ. \displaystyle \text{Answer:}
\displaystyle \text{Given, }\triangle ACB\sim\triangle APQ
\displaystyle \therefore \frac{AC}{AP}=\frac{CB}{PQ}=\frac{AB}{AQ}
\displaystyle \frac{CB}{PQ}=\frac{10}{5}=2
\displaystyle \therefore \frac{AC}{2.8}=2
\displaystyle AC=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 \text{For similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding sides.}
\displaystyle \frac{\text{Area }(\triangle ACB)}{\text{Area }(\triangle APQ)}=\frac{CB^2}{PQ^2}
\displaystyle =\frac{10^2}{5^2}=\frac{100}{25}=4
\displaystyle \therefore \text{Area }(\triangle ACB):\text{Area }(\triangle APQ)=4:1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In the adjoning figure, }DE\parallel BC.
\displaystyle \text{(i) If }DE=4\text{ cm},\ BC=6\text{ cm and Area }(\triangle ADE)=16\text{ cm}^2,\text{ find the area of }\triangle ABC.
\displaystyle \text{(ii) If }DE=4\text{ cm},\ BC=8\text{ cm and Area }(\triangle ADE)=25\text{ cm}^2,\text{ find the area of }\triangle ABC.
\displaystyle \text{(iii) If }DE:BC=3:5.\text{ Calculate the ratio of the areas of }\triangle ADE\text{ and the trapezium }BCED.
\displaystyle \text{(iv) If }DE=6\text{ cm},\ BC=12\text{ cm, find the ratio of Area }(\triangle ADE)\text{ and Area (quad. }DECB). \displaystyle \text{Answer:}
\displaystyle \text{Since }DE\parallel BC,\ \triangle ADE\sim\triangle ABC.
\displaystyle \therefore \frac{\text{Area }(\triangle ADE)}{\text{Area }(\triangle ABC)}=\frac{DE^2}{BC^2}

\displaystyle \text{(i) }\frac{16}{\text{Area }(\triangle ABC)}=\frac{4^2}{6^2}=\frac{4}{9}
\displaystyle \text{Area }(\triangle ABC)=16\times\frac{9}{4}=36\text{ cm}^2
\displaystyle \therefore \text{Area }(\triangle ABC)=36\text{ cm}^2.

\displaystyle \text{(ii) }\frac{25}{\text{Area }(\triangle ABC)}=\frac{4^2}{8^2}=\frac{1}{4}
\displaystyle \text{Area }(\triangle ABC)=25\times4=100\text{ cm}^2
\displaystyle \therefore \text{Area }(\triangle ABC)=100\text{ cm}^2.

\displaystyle \text{(iii) }\frac{\text{Area }(\triangle ADE)}{\text{Area }(\triangle ABC)}=\frac{DE^2}{BC^2}=\frac{3^2}{5^2}=\frac{9}{25}
\displaystyle \therefore \text{Area }(\triangle ADE):\text{Area }(\triangle ABC)=9:25
\displaystyle \text{Area of trapezium }BCED=25-9=16\text{ parts}
\displaystyle \therefore \text{Area }(\triangle ADE):\text{Area of trapezium }BCED=9:16.

\displaystyle \text{(iv) }\frac{\text{Area }(\triangle ADE)}{\text{Area }(\triangle ABC)}=\frac{6^2}{12^2}=\frac{1}{4}
\displaystyle \therefore \text{Area }(\triangle ADE):\text{Area }(\triangle ABC)=1:4
\displaystyle \text{Area of quadrilateral }DECB=4-1=3\text{ parts}
\displaystyle \therefore \text{Area }(\triangle ADE):\text{Area (quad. }DECB)=1:3.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The areas of two similar triangles are }81\text{ cm}^2\text{ and }49\text{ cm}^2\text{ respectively.}
\displaystyle \text{Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding heights.}
\displaystyle \therefore \frac{81}{49}=\left(\frac{h_1}{h_2}\right)^2
\displaystyle \frac{h_1}{h_2}=\sqrt{\frac{81}{49}}=\frac{9}{7}
\displaystyle \therefore \text{Ratio of their corresponding heights}=9:7.
\displaystyle \text{Also, the ratio of corresponding medians of two similar triangles is equal to the ratio}
\displaystyle \text{of their corresponding sides.}
\displaystyle \therefore \text{Ratio of their corresponding medians}=\sqrt{\frac{81}{49}}=9:7.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The areas of two similar triangles are }169\text{ cm}^2\text{ and }121\text{ cm}^2\text{ respectively.}
\displaystyle \text{If the longest side of the larger triangle is }26\text{ cm, find the longest side of the smaller triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding sides.}
\displaystyle \therefore \frac{169}{121}=\left(\frac{26}{x}\right)^2
\displaystyle \frac{13^2}{11^2}=\frac{26^2}{x^2}
\displaystyle \frac{13}{11}=\frac{26}{x}
\displaystyle 13x=26\times11
\displaystyle x=22\text{ cm}
\displaystyle \therefore \text{The longest side of the smaller triangle is }22\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The areas of two similar triangles are }25\text{ cm}^2\text{ and }36\text{ cm}^2\text{ respectively.}
\displaystyle \text{If the altitude of the first triangle is }2.4\text{ cm, find the corresponding altitude of the other.}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding altitudes.}
\displaystyle \therefore \frac{25}{36}=\left(\frac{2.4}{h}\right)^2
\displaystyle \frac{5}{6}=\frac{2.4}{h}
\displaystyle 5h=2.4\times6
\displaystyle h=\frac{14.4}{5}=2.88\text{ cm}
\displaystyle \therefore \text{The corresponding altitude of the other triangle is }2.88\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The corresponding altitudes of two similar triangles are }6\text{ cm and }
\displaystyle 9\text{ cm respectively.} \ \text{Find the ratio of their areas.}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding altitudes.}
\displaystyle \therefore \frac{\text{Area of first triangle}}{\text{Area of second triangle}}=\left(\frac{6}{9}\right)^2
\displaystyle =\left(\frac{2}{3}\right)^2=\frac{4}{9}
\displaystyle \therefore \text{The ratio of their areas}=4:9.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The areas of two similar triangles are }100\text{ cm}^2\text{ and }49\text{ cm}^2\text{ respectively.}
\displaystyle \text{If the altitude of the bigger triangle is }5\text{ cm, find the corresponding altitude of the other.}
\displaystyle \hfill\text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding altitudes.}
\displaystyle \therefore \frac{100}{49}=\left(\frac{5}{h}\right)^2
\displaystyle \frac{10}{7}=\frac{5}{h}
\displaystyle 10h=35
\displaystyle h=3.5\text{ cm}
\displaystyle \therefore \text{The corresponding altitude of the other triangle is }3.5\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The areas of two similar triangles are }121\text{ cm}^2\text{ and }64\text{ cm}^2\text{ respectively.}
\displaystyle \text{If the median of the first triangle is }12.1\text{ cm, find the corresponding median of the other.}
\displaystyle \hfill\text{[CBSE 2001]}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding medians.}
\displaystyle \therefore \frac{121}{64}=\left(\frac{12.1}{m}\right)^2
\displaystyle \frac{11}{8}=\frac{12.1}{m}
\displaystyle 11m=12.1\times8
\displaystyle m=\frac{96.8}{11}=8.8\text{ cm}
\displaystyle \therefore \text{The corresponding median of the other triangle is }8.8\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The areas of two similar triangles }ABC\text{ and }PQR\text{ are in the ratio }
\displaystyle 9:16. \ \text{If }BC=4.5\text{ cm, find the length of }QR. \hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{For two similar triangles, the ratio of their areas is equal to the square of the ratio}
\displaystyle \text{of their corresponding sides.}
\displaystyle \therefore \frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle PQR)}=\frac{BC^2}{QR^2}
\displaystyle \frac{9}{16}=\left(\frac{4.5}{QR}\right)^2
\displaystyle \frac{3}{4}=\frac{4.5}{QR}
\displaystyle 3QR=4.5\times4
\displaystyle QR=6\text{ cm}
\displaystyle \therefore QR=6\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 11: }ABC\text{ is a triangle and }PQ\text{ is a straight line meeting }AB\text{ in }P\text{ and }
\displaystyle AC\text{ in }Q. \ \text{If }AP=1\text{ cm},\ PB=3\text{ cm},\ AQ=1.5\text{ cm and }QC=4.5\text{ cm, prove that area of }
\displaystyle \triangle APQ \ \text{is one-sixteenth of the area of }\triangle ABC.\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:} \displaystyle AB=AP+PB=1+3=4\text{ cm}
\displaystyle AC=AQ+QC=1.5+4.5=6\text{ cm}
\displaystyle \frac{AP}{AB}=\frac{1}{4}
\displaystyle \frac{AQ}{AC}=\frac{1.5}{6}=\frac{1}{4}
\displaystyle \therefore \frac{AP}{AB}=\frac{AQ}{AC}
\displaystyle \therefore PQ\parallel BC\qquad\text{[By the converse of Basic Proportionality Theorem]}
\displaystyle \therefore \triangle APQ\sim\triangle ABC
\displaystyle \frac{\text{Area }(\triangle APQ)}{\text{Area }(\triangle ABC)}=\frac{AP^2}{AB^2}
\displaystyle =\left(\frac{1}{4}\right)^2=\frac{1}{16}
\displaystyle \therefore \text{Area }(\triangle APQ)=\frac{1}{16}\text{Area }(\triangle ABC).
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }D\text{ is a point on the side }AB\text{ of }\triangle ABC\text{ such that }
\displaystyle AD:DB=3:2 \ \text{and }E\text{ is a point on }BC\text{ such that }DE\parallel AC,
\displaystyle \text{ find the ratio of areas of }\triangle ABC \ \text{and }\triangle BDE.\hfill\text{[CBSE 2006 C]}
\displaystyle \text{Answer:} \displaystyle AD:DB=3:2
\displaystyle \therefore AB:DB=(3+2):2=5:2
\displaystyle \text{Since }DE\parallel AC,
\displaystyle \triangle BDE\sim\triangle BAC\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{\text{Area }(\triangle BDE)}{\text{Area }(\triangle BAC)}=\frac{BD^2}{BA^2}
\displaystyle =\left(\frac{2}{5}\right)^2=\frac{4}{25}
\displaystyle \therefore \text{Area }(\triangle ABC):\text{Area }(\triangle BDE)=25:4.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\triangle ABC\text{ and }\triangle BDE\text{ are equilateral triangles, where }
\displaystyle D\text{ is the mid-point}  \ \text{of }BC,\text{ find the ratio of areas of }\triangle ABC\text{ and }\triangle BDE.\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Since }D\text{ is the mid-point of }BC,
\displaystyle BD=\frac{1}{2}BC
\displaystyle \text{Also, }\triangle ABC\text{ and }\triangle BDE\text{ are equilateral, hence they are similar.}
\displaystyle \therefore \frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle BDE)}=\frac{BC^2}{BD^2}
\displaystyle =\left(\frac{BC}{\frac{1}{2}BC}\right)^2=2^2=4
\displaystyle \therefore \text{Area }(\triangle ABC):\text{Area }(\triangle BDE)=4:1.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Diagonals of a trapezium }PQRS\text{ intersect each other at the point }O,\ 
\displaystyle PQ\parallel RS \ \text{and }PQ=3RS.\text{ Find the ratio of the areas of triangles }POQ\text{ and }ROS.
\displaystyle \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:} \displaystyle \text{Since }PQ\parallel RS,
\displaystyle \angle PQO=\angle RSO\text{ and }\angle QPO=\angle SRO
\displaystyle \therefore \triangle POQ\sim\triangle ROS\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{\text{Area }(\triangle POQ)}{\text{Area }(\triangle ROS)}=\frac{PQ^2}{RS^2}
\displaystyle \text{Given, }PQ=3RS
\displaystyle \therefore \frac{\text{Area }(\triangle POQ)}{\text{Area }(\triangle ROS)}=\left(\frac{3RS}{RS}\right)^2=9
\displaystyle \therefore \text{Area }(\triangle POQ):\text{Area }(\triangle ROS)=9:1.
\displaystyle \\

\displaystyle \textbf{Question 15: }AD\text{ is an altitude of an equilateral triangle }ABC.\text{ On }AD\text{ as base, another}
\displaystyle \text{equilateral triangle }ADE\text{ is constructed. Prove that Area }
\displaystyle (\triangle ADE):\text{Area }(\triangle ABC)=3:4. \ \hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let each side of equilateral }\triangle ABC=a.
\displaystyle \text{Since }AD\text{ is an altitude of an equilateral triangle,}
\displaystyle AD=\frac{\sqrt{3}}{2}a
\displaystyle \text{Since }\triangle ABC\text{ and }\triangle ADE\text{ are equilateral, they are similar.}
\displaystyle \therefore \frac{\text{Area }(\triangle ADE)}{\text{Area }(\triangle ABC)}=\frac{AD^2}{AB^2}
\displaystyle =\left(\frac{\frac{\sqrt{3}}{2}a}{a}\right)^2=\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}
\displaystyle \therefore \text{Area }(\triangle ADE):\text{Area }(\triangle ABC)=3:4.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the adjoning figure, }\triangle ABC\text{ and }\triangle DBC\text{ are on the same base }BC.
\displaystyle \text{If }AD\text{ and }BC\text{ intersect at }O,\text{ prove that }\frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle DBC)}=\frac{AO}{DO}.
\displaystyle \hfill\text{[CBSE 2000, 2005]} \displaystyle \text{Answer:}
\displaystyle \text{Draw }AL\perp BC\text{ and }DM\perp BC.
\displaystyle \text{In }\triangle AOL\text{ and }\triangle DOM,
\displaystyle \angle ALO=\angle DMO=90^\circ
\displaystyle \angle AOL=\angle DOM\qquad\text{[Vertically opposite angles]}
\displaystyle \therefore \triangle AOL\sim\triangle DOM\qquad\text{[By AA criterion]}
\displaystyle \therefore \frac{AL}{DM}=\frac{AO}{DO}
\displaystyle \text{Now, }\frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle DBC)}
\displaystyle =\frac{\frac{1}{2}\times BC\times AL}{\frac{1}{2}\times BC\times DM}
\displaystyle =\frac{AL}{DM}
\displaystyle =\frac{AO}{DO}
\displaystyle \therefore \frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle DBC)}=\frac{AO}{DO}.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.