\displaystyle \textbf{Note: Relation between the areas of two similar triangles:}

\displaystyle \text{If } \triangle ABC \sim \triangle DEF,\text{ then}

\displaystyle \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEF} =\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}

\displaystyle \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF} =\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle DEF}

\\

\displaystyle \textbf{Question 1: } \text{(i) The ratio between the corresponding sides of two similar triangles is } \\ 2:5.
\displaystyle \text{Find the ratio between the areas of these triangles.}
\displaystyle \text{(ii) The areas of two similar triangles are }98\text{ cm}^2\text{ and }128\text{ cm}^2.
\displaystyle \text{Find the ratio between the lengths of their corresponding sides.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Ratio of corresponding sides}=2:5
\displaystyle \text{Ratio of areas}=\frac{2^2}{5^2}=\frac{4}{25}
\displaystyle \text{(ii) Ratio of areas}=\frac{98}{128}=\frac{49}{64}
\displaystyle \text{Ratio of corresponding sides}=\sqrt{\frac{49}{64}}=\frac{7}{8}
\\

\displaystyle \textbf{Question 2: } \text{A line }PQ\text{ is drawn parallel to the base }BC\text{ of }\triangle ABC
\displaystyle \text{which meets sides }AB\text{ and }AC\text{ at points }P\text{ and }Q\text{ respectively.}
\displaystyle \text{If }AP=\frac{1}{2}PB,\text{ find the value of}
\displaystyle \text{(i) }\frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle APQ}\qquad \text{(ii) }\frac{\text{Area of }\triangle APQ}{\text{Area of trapezium }PBCQ}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AP=\frac{1}{2}PB
\displaystyle \Rightarrow PB=2AP
\displaystyle \therefore AB=AP+PB=AP+2AP=3AP
\displaystyle \text{Since }PQ\parallel BC,
\displaystyle \angle APQ=\angle ABC\text{ and }\angle AQP=\angle ACB
\displaystyle \therefore \triangle APQ\sim\triangle ABC
\displaystyle \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle APQ} =\frac{AB^2}{AP^2}
\displaystyle =\frac{(3AP)^2}{AP^2}=9
\displaystyle \therefore \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle APQ}=9
\displaystyle \text{Now, area of trapezium }PBCQ=\text{Area of }\triangle ABC-\text{Area of }\triangle APQ
\displaystyle \frac{\text{Area of }\triangle APQ}{\text{Area of trapezium }PBCQ} =\frac{1}{9-1}=\frac{1}{8}
\\

\displaystyle \textbf{Question 3: } \text{The perimeters of two similar triangles are }30\text{ cm and }24\text{ cm.}
\displaystyle \text{If one side of the first triangle is }12\text{ cm, determine the corresponding side of the second triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the two triangles are similar,}
\displaystyle \frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{\text{Perimeter of }\triangle ABC}{\text{Perimeter of }\triangle DEF}
\displaystyle \frac{12}{DE}=\frac{30}{24}
\displaystyle DE=\frac{12\times24}{30}=9.6\text{ cm}
\\

\displaystyle \textbf{Question 4: } \text{In the given figure, }AX:XB=3:5.\text{ Find:}
\displaystyle \text{(i) the length of }BC,\text{ if the length of }XY\text{ is }18\text{ cm.}
\displaystyle \text{(ii) the ratio between the areas of trapezium }XBCY\text{ and }\triangle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AX:XB=3:5
\displaystyle \therefore \frac{AX}{AB}=\frac{3}{3+5}=\frac{3}{8}
\displaystyle \text{Since }XY\parallel BC,
\displaystyle \angle AXY=\angle ABC\text{ and }\angle AYX=\angle ACB
\displaystyle \therefore \triangle AXY\sim\triangle ABC
\displaystyle \frac{AX}{AB}=\frac{XY}{BC}
\displaystyle \frac{3}{8}=\frac{18}{BC}
\displaystyle BC=\frac{8\times18}{3}=48\text{ cm}
\displaystyle \frac{\text{Area of }\triangle AXY}{\text{Area of }\triangle ABC}=\frac{AX^2}{AB^2}
\displaystyle =\left(\frac{3}{8}\right)^2=\frac{9}{64}
\displaystyle \frac{\text{Area of trapezium }XBCY}{\text{Area of }\triangle ABC}=1-\frac{9}{64}
\displaystyle =\frac{55}{64}
\displaystyle \therefore \text{Area of trapezium }XBCY:\text{Area of }\triangle ABC=55:64
\\

\displaystyle \textbf{Question 5: } ABC\text{ is a triangle. }PQ\text{ is a line segment intersecting }AB\text{ in }P
\displaystyle \text{and }AC\text{ in }Q\text{ such that }PQ\parallel BC\text{ and divides }\triangle ABC
\displaystyle \text{into two parts equal in area. Find the value of ratio }BP:AB.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }PQ\parallel BC
\displaystyle \text{Since }PQ\text{ divides }\triangle ABC\text{ into two parts equal in area,}
\displaystyle \text{Area of }\triangle APQ=\frac{1}{2}\text{ Area of }\triangle ABC
\displaystyle \text{In }\triangle APQ\text{ and }\triangle ABC,
\displaystyle \angle APQ=\angle ABC\text{ and }\angle AQP=\angle ACB
\displaystyle \therefore \triangle APQ\sim\triangle ABC
\displaystyle \frac{\text{Area of }\triangle APQ}{\text{Area of }\triangle ABC}=\frac{AP^2}{AB^2}
\displaystyle \frac{1}{2}=\frac{AP^2}{AB^2}
\displaystyle \frac{AP}{AB}=\frac{1}{\sqrt{2}}
\displaystyle \frac{BP}{AB}=\frac{AB-AP}{AB}=1-\frac{AP}{AB}
\displaystyle =1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}
\displaystyle \therefore BP:AB=(\sqrt{2}-1):\sqrt{2}
\\

\displaystyle \textbf{Question 6: } \text{In the given }\triangle PQR,\ LM\parallel QR\text{ and }PM:MR=3:4.
\displaystyle \text{Calculate the value of the ratio:}
\displaystyle \text{(i) }\frac{PL}{PQ}\text{ and then }\frac{LM}{QR}
\displaystyle \text{(ii) }\frac{\text{Area of }\triangle LMN}{\text{Area of }\triangle MNR}
\displaystyle \text{(iii) }\frac{\text{Area of }\triangle LQM}{\text{Area of }\triangle LQN}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }PM:MR=3:4
\displaystyle \therefore PM:PR=3:7
\displaystyle \text{Since }LM\parallel QR,\ \triangle PLM\sim\triangle PQR
\displaystyle \therefore \frac{PL}{PQ}=\frac{LM}{QR}=\frac{PM}{PR}=\frac{3}{7}
\displaystyle \text{Thus, }\frac{PL}{PQ}=\frac{3}{7}\text{ and }\frac{LM}{QR}=\frac{3}{7}
\displaystyle \text{Now, }\triangle LMN\sim\triangle QRN
\displaystyle \therefore \frac{LN}{NR}=\frac{LM}{QR}=\frac{3}{7}
\displaystyle \text{Since }\triangle LMN\text{ and }\triangle MNR\text{ have the same altitude from }M,
\displaystyle \frac{\text{Area of }\triangle LMN}{\text{Area of }\triangle MNR}=\frac{LN}{NR}=\frac{3}{7}
\displaystyle \text{Also, }QN:NM=7:3
\displaystyle \therefore QM:QN=(7+3):7=10:7
\displaystyle \text{Since }\triangle LQM\text{ and }\triangle LQN\text{ have the same altitude from }L,
\displaystyle \frac{\text{Area of }\triangle LQM}{\text{Area of }\triangle LQN}=\frac{QM}{QN}=\frac{10}{7}
\\

\displaystyle \textbf{Question 7: } \text{The given diagram shows two isosceles triangles which are similar.}
\displaystyle PQ\text{ and }BC\text{ are not parallel. }PC=4,\ AQ=3,\ QB=12,\ BC=15\text{ and } \\ AP=PQ.
\displaystyle \text{Calculate:}
\displaystyle \text{(i) the length of }AP
\displaystyle \text{(ii) the ratio of the areas of }\triangle APQ\text{ and }\triangle ABC
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AQ=3,\ QB=12
\displaystyle \therefore AB=AQ+QB=3+12=15
\displaystyle \text{Also, }BC=15
\displaystyle \therefore AB=BC=15
\displaystyle \text{Let }AP=PQ=x
\displaystyle \text{Then }AC=AP+PC=x+4
\displaystyle \text{Since }\triangle APQ\sim\triangle ABC,
\displaystyle \frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}
\displaystyle \frac{x}{15}=\frac{3}{x+4}
\displaystyle x(x+4)=45
\displaystyle x^2+4x-45=0
\displaystyle (x+9)(x-5)=0
\displaystyle x=5
\displaystyle \therefore AP=5\text{ cm}
\displaystyle \frac{\text{Area of }\triangle APQ}{\text{Area of }\triangle ABC} =\frac{AQ^2}{AC^2}
\displaystyle =\frac{3^2}{(5+4)^2}=\frac{9}{81}=\frac{1}{9}
\displaystyle \therefore \text{Area of }\triangle APQ:\text{Area of }\triangle ABC=1:9
\\

\displaystyle \textbf{Question 8: } \text{In the figure given below, }ABCD\text{ is a parallelogram. }P\text{ is a point on}
\displaystyle BC\text{ such that }BP:PC=1:2.\ DP\text{ produced meets }AB\text{ produced at }Q.
\displaystyle \text{Given the area of }\triangle CPQ=20\text{ cm}^2.\text{ Calculate:}
\displaystyle \text{(i) area of }\triangle CDP
\displaystyle \text{(ii) area of parallelogram }ABCD\hfill\text{[ICSE 1996]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }BP:PC=1:2
\displaystyle \text{Since }\triangle BPQ\text{ and }\triangle CPQ\text{ have the same altitude from }Q,
\displaystyle \frac{\text{Area of }\triangle BPQ}{\text{Area of }\triangle CPQ}=\frac{BP}{PC}=\frac{1}{2}
\displaystyle \therefore \text{Area of }\triangle BPQ=\frac{1}{2}\times20=10\text{ cm}^2
\displaystyle \text{In }\triangle BPQ\text{ and }\triangle CPD,
\displaystyle \angle BPQ=\angle CPD\text{ and }\angle BQP=\angle CDP
\displaystyle \therefore \triangle BPQ\sim\triangle CPD
\displaystyle \frac{\text{Area of }\triangle BPQ}{\text{Area of }\triangle CPD}=\frac{BP^2}{PC^2}=\frac{1^2}{2^2}=\frac{1}{4}
\displaystyle \therefore \text{Area of }\triangle CPD=4\times10=40\text{ cm}^2
\displaystyle \text{In }\triangle BQP\text{ and }\triangle AQD,
\displaystyle BP\parallel AD
\displaystyle \angle QBP=\angle QAD\text{ and }\angle BQP=\angle AQD
\displaystyle \therefore \triangle BQP\sim\triangle AQD
\displaystyle \frac{AD}{BP}=\frac{BC}{BP}=\frac{BP+PC}{BP}=\frac{1+2}{1}=3
\displaystyle \frac{\text{Area of }\triangle AQD}{\text{Area of }\triangle BQP}=\frac{AD^2}{BP^2}=3^2=9
\displaystyle \therefore \text{Area of }\triangle AQD=9\times10=90\text{ cm}^2
\displaystyle \text{Area of parallelogram }ABCD=\text{Area of }\triangle AQD-\text{Area of }\triangle BQP+\text{Area of }\triangle CDP
\displaystyle =90-10+40=120\text{ cm}^2
\displaystyle \therefore \text{Area of }\triangle CDP=40\text{ cm}^2
\displaystyle \therefore \text{Area of parallelogram }ABCD=120\text{ cm}^2
\\

\displaystyle \textbf{Question 9: } \text{In the given figure, }BC\parallel DE.\text{ Area of }\triangle ABC=25\text{ cm}^2,
\displaystyle \text{area of trapezium }BCED=24\text{ cm}^2\text{ and }DE=14\text{ cm. Calculate the length of }BC.
\displaystyle \text{Also find the area of }\triangle BCD.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }BC\parallel DE
\displaystyle \text{Area of }\triangle ADE=\text{Area of }\triangle ABC+\text{Area of trapezium }BCED
\displaystyle =25+24=49\text{ cm}^2
\displaystyle \text{In }\triangle ABC\text{ and }\triangle ADE,
\displaystyle \angle ABC=\angle ADE\text{ and }\angle ACB=\angle AED
\displaystyle \therefore \triangle ABC\sim\triangle ADE
\displaystyle \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle ADE}=\frac{BC^2}{DE^2}
\displaystyle \frac{25}{49}=\frac{BC^2}{14^2}
\displaystyle BC^2=\frac{25}{49}\times196=100
\displaystyle \therefore BC=10\text{ cm}
\displaystyle \text{Now, area of trapezium }BCED=\frac{1}{2}(BC+DE)\times\text{height}
\displaystyle 24=\frac{1}{2}(10+14)\times\text{height}
\displaystyle 24=12\times\text{height}
\displaystyle \therefore \text{height}=2\text{ cm}
\displaystyle \text{Area of }\triangle BCD=\frac{1}{2}\times BC\times\text{height}
\displaystyle =\frac{1}{2}\times10\times2=10\text{ cm}^2
\\

\displaystyle \textbf{Question 10: } \text{The given figure shows a trapezium in which }AB\parallel DC
\displaystyle \text{and diagonals }AC\text{ and }BD\text{ intersect at point }P.\text{ If }AP:CP=3:5.\text{ Find:}
\displaystyle \text{(i) }\triangle APB:\triangle CPB\qquad \text{(ii) }\triangle DPC:\triangle APB
\displaystyle \text{(iii) }\triangle ADP:\triangle APB\qquad \text{(iv) }\triangle APB:\triangle ADB
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AP:CP=3:5
\displaystyle \text{Since }AB\parallel DC,\text{ diagonals divide each other proportionally.}
\displaystyle \therefore AP:CP=BP:DP=3:5
\displaystyle \text{(i) }\triangle APB\text{ and }\triangle CPB\text{ have the same altitude from }B.
\displaystyle \therefore \frac{\text{Area of }\triangle APB}{\text{Area of }\triangle CPB}=\frac{AP}{CP}=\frac{3}{5}
\displaystyle \therefore \triangle APB:\triangle CPB=3:5
\displaystyle \text{(ii) Since }\triangle DPC\sim\triangle APB,
\displaystyle \frac{\text{Area of }\triangle DPC}{\text{Area of }\triangle APB}=\frac{PC^2}{AP^2}=\frac{5^2}{3^2}=\frac{25}{9}
\displaystyle \therefore \triangle DPC:\triangle APB=25:9
\displaystyle \text{(iii) }\triangle ADP\text{ and }\triangle APB\text{ have the same altitude from }A.
\displaystyle \therefore \frac{\text{Area of }\triangle ADP}{\text{Area of }\triangle APB}=\frac{DP}{PB}=\frac{5}{3}
\displaystyle \therefore \triangle ADP:\triangle APB=5:3
\displaystyle \text{(iv) }\triangle APB\text{ and }\triangle ADB\text{ have the same altitude from }A.
\displaystyle \therefore \frac{\text{Area of }\triangle APB}{\text{Area of }\triangle ADB}=\frac{PB}{DB}=\frac{3}{3+5}=\frac{3}{8}
\displaystyle \therefore \triangle APB:\triangle ADB=3:8
\\

\displaystyle \textbf{Question 11: } \text{On a map drawn to a scale of }1:2500000\text{ a triangular plot }PQR\text{ has}
\displaystyle PQ=3\text{ cm},\ QR=4\text{ cm}\text{ and }\angle PQR=90^\circ.\text{ Calculate:}
\displaystyle \text{(i) the actual lengths of }QR\text{ and }PR\text{ in kilometres}
\displaystyle \text{(ii) the actual area of the plot in }km^2.
\displaystyle \text{Answer:}
\displaystyle \text{Scale }=1:2500000
\displaystyle \text{Actual length of }PQ=3\times2500000=7500000\text{ cm}=75\text{ km}
\displaystyle \text{Actual length of }QR=4\times2500000=10000000\text{ cm}=100\text{ km}
\displaystyle \text{Since }\angle PQR=90^\circ,
\displaystyle PR=\sqrt{75^2+100^2}=\sqrt{15625}=125\text{ km}
\displaystyle \text{Area of the plot}=\frac{1}{2}\times75\times100
\displaystyle =3750\text{ km}^2
\\

\displaystyle \textbf{Question 12: } \text{A model of a ship is made to a 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{[ICSE 1995]}
\displaystyle \text{Answer:}
\displaystyle \text{Scale factor}=\frac{1}{200}
\displaystyle \text{(i) Length of ship}=4\times200=800\text{ m}
\displaystyle \text{(ii) Area of deck of model}=\left(\frac{1}{200}\right)^2\times160000
\displaystyle =\frac{160000}{40000}=4\text{ m}^2
\displaystyle \text{(iii) Volume of ship}=200^3\times200\text{ litres}
\displaystyle =8,000,000\times200=1,600,000,000\text{ litres}
\displaystyle =\frac{1,600,000,000}{1000}=1,600,000\text{ m}^3
\\

\displaystyle \textbf{Question 13: } \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{[ICSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }DE\parallel BC\text{ and }\frac{AD}{DB}=\frac{3}{2}
\displaystyle \therefore \frac{AD}{AB}=\frac{AD}{AD+DB}=\frac{3}{3+2}=\frac{3}{5}
\displaystyle \text{In }\triangle ADE\text{ and }\triangle ABC,
\displaystyle \angle ADE=\angle ABC\text{ and }\angle AED=\angle ACB
\displaystyle \therefore \triangle ADE\sim\triangle ABC
\displaystyle \therefore \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}
\displaystyle \therefore \frac{DE}{BC}=\frac{3}{5}
\displaystyle \text{In }\triangle DEF\text{ and }\triangle CBF,
\displaystyle \angle FDE=\angle FCB\text{ and }\angle DFE=\angle CFB
\displaystyle \therefore \triangle DEF\sim\triangle CBF
\displaystyle \therefore \frac{EF}{FB}=\frac{DE}{BC}=\frac{3}{5}
\\

\displaystyle \textbf{Question 14: } \text{In the given figure }\angle B=\angle E,\ \angle ACD=\angle BCE.
\displaystyle AB=10.4\text{ cm and }DE=7.8\text{ cm. Find the ratio between the areas of }
\displaystyle \triangle ABC\text{ and }\triangle DEC.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle ABC\text{ and }\triangle CDE.
\displaystyle \angle CBA=\angle CED\text{ (given)}
\displaystyle \angle ECD=\angle BCA\text{ (since }\angle BCD\text{ is common and }\angle ACD=\angle BCE\text{)}
\displaystyle \therefore \triangle ABC\sim\triangle CDE
\displaystyle \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle DEC}=\frac{AB^2}{DE^2}
\displaystyle =\frac{10.4^2}{7.8^2}=\left(\frac{104}{78}\right)^2=\left(\frac{4}{3}\right)^2=\frac{16}{9}
\displaystyle \therefore \text{Area of }\triangle ABC:\text{Area of }\triangle DEC=16:9
\\

\displaystyle \textbf{Question 15: } \triangle ABC\text{ is an isosceles triangle in which }AB=AC=13\text{ cm}
\displaystyle \text{and }BC=10\text{ cm. }AD\perp BC.\text{ If }CE=8\text{ cm and }EF\perp AB,\text{ find:}
\displaystyle \text{(i) }\frac{\text{Area of }\triangle ADC}{\text{Area of }\triangle FEB}\qquad \text{(ii) }\frac{\text{Area of }\triangle FEB}{\text{Area of }\triangle ABC}
\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=AC,\text{ and }AD\perp BC,\text{ therefore }BD=DC=5\text{ cm}
\displaystyle \text{Also, }BE=BC+CE=10+8=18\text{ cm}
\displaystyle \text{In }\triangle ADC\text{ and }\triangle FEB,
\displaystyle \angle ADC=\angle BFE=90^\circ
\displaystyle \angle ACD=\angle FBE
\displaystyle \therefore \triangle ADC\sim\triangle FEB
\displaystyle \frac{\text{Area of }\triangle ADC}{\text{Area of }\triangle FEB}=\frac{AC^2}{BE^2}
\displaystyle =\frac{13^2}{18^2}=\frac{169}{324}
\displaystyle \therefore \text{Area of }\triangle ADC:\text{Area of }\triangle FEB=169:324
\displaystyle \text{Now, area of }\triangle ABC=2\times\text{Area of }\triangle ADC
\displaystyle \frac{\text{Area of }\triangle FEB}{\text{Area of }\triangle ABC} =\frac{\text{Area of }\triangle FEB}{2\times\text{Area of }\triangle ADC}
\displaystyle =\frac{324}{2\times169}=\frac{162}{169}
\displaystyle \therefore \text{Area of }\triangle FEB:\text{Area of }\triangle ABC=162:169
\\

\displaystyle \textbf{Question 16: } \text{An airplane is }30\text{ m long and its model is }15\text{ cm long.}
\displaystyle \text{If the total outer surface area of the model is }150\text{ cm}^2,\text{ find the cost of painting}
\displaystyle \text{the outer surface of the airplane at the rate of }\text{Rs. }120/\text{m}^2.
\displaystyle \text{Given that }50\text{ m}^2\text{ of the surface of the airplane is left for windows.}
\displaystyle \text{Answer:}
\displaystyle 15\text{ cm of the model represents }30\text{ m of the actual airplane}
\displaystyle \therefore 1\text{ cm of the model represents }2\text{ m of the actual airplane}
\displaystyle \therefore 1\text{ cm}^2\text{ of the model represents }4\text{ m}^2\text{ of the actual airplane}
\displaystyle \text{Surface area of the model}=150\text{ cm}^2
\displaystyle \therefore \text{Surface area of the actual airplane}=150\times4=600\text{ m}^2
\displaystyle \text{Area to be painted}=600-50=550\text{ m}^2
\displaystyle \text{Cost of painting}=\text{Rs. }120\text{ per m}^2
\displaystyle \therefore \text{Total cost}=120\times550=\text{Rs. }66000
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