\displaystyle \textbf{Question 1: } \text{In }\triangle ABC,\ \angle ABC=\angle DAC,\ AB=8\text{ cm},
\displaystyle AC=4\text{ cm and }AD=5\text{ cm}.
\displaystyle \text{(i) Prove that }\triangle ACD\text{ is similar to }\triangle BCA.
\displaystyle \text{(ii) Find }BC\text{ and }CD.
\displaystyle \text{(iii) Find Area of }\triangle ACD:\text{Area of }\triangle ABC.\hfill\text{[ICSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\triangle ACD\text{ and }\triangle BCA,
\displaystyle \angle ACD=\angle BCA\text{ (common angle)}
\displaystyle \angle CAD=\angle ABC\text{ (given)}
\displaystyle \therefore \triangle ACD\sim\triangle BCA
\displaystyle \text{(ii) Since }\triangle ACD\sim\triangle BCA,
\displaystyle \frac{AC}{BC}=\frac{CD}{CA}=\frac{AD}{BA}
\displaystyle \frac{4}{BC}=\frac{CD}{4}=\frac{5}{8}
\displaystyle BC=\frac{8\times4}{5}=6.4\text{ cm}
\displaystyle CD=\frac{5\times4}{8}=2.5\text{ cm}
\displaystyle \text{(iii) Since }\triangle ACD\sim\triangle BCA,
\displaystyle \frac{\text{Area of }\triangle ACD}{\text{Area of }\triangle BCA}=\frac{AD^2}{BA^2}
\displaystyle =\frac{5^2}{8^2}=\frac{25}{64}
\displaystyle \therefore \text{Area of }\triangle ACD:\text{Area of }\triangle ABC=25:64
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\displaystyle \textbf{Question 2: } \text{In the given triangle, }P,\ Q\text{ and }R\text{ are mid-points of sides}
\displaystyle AB,\ BC\text{ and }AC\text{ respectively. Prove that }\triangle PQR\sim\triangle ABC.
\displaystyle \text{Answer:}
\displaystyle \text{Since }P,\ Q\text{ and }R\text{ are mid-points of }AB,\ BC\text{ and }AC\text{ respectively,}
\displaystyle PQ\parallel AC,\ PR\parallel BC\text{ and }QR\parallel AB
\displaystyle \text{In }\triangle PQR\text{ and }\triangle ABC,
\displaystyle \angle PRQ=\angle B\text{ }(PR\parallel BC\text{ and }RQ\parallel BA)
\displaystyle \angle PQR=\angle C\text{ }(PQ\parallel AC\text{ and }QR\parallel AB)
\displaystyle \therefore \triangle PQR\sim\triangle ABC\text{ (AA postulate)}
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\displaystyle \textbf{Question 3: } \text{In the following figure }AD\text{ and }CE\text{ are medians of }\triangle ABC.
\displaystyle DF\parallel CE.\text{ Prove that:}
\displaystyle \text{(i) }EF=FB
\displaystyle \text{(ii) }AG:GD=2:1
\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AD\text{ and }CE\text{ are medians, }D\text{ and }E\text{ are mid-points of }BC\text{ and }AB.
\displaystyle \therefore BD=DC\text{ and }AE=EB
\displaystyle \text{In }\triangle BDF\text{ and }\triangle BCE,
\displaystyle DF\parallel CE
\displaystyle \angle BDF=\angle BCE\text{ and }\angle BFD=\angle BEC
\displaystyle \therefore \triangle BDF\sim\triangle BCE
\displaystyle \frac{BF}{BE}=\frac{BD}{BC}
\displaystyle \frac{BF}{BE}=\frac{1}{2}
\displaystyle \therefore BE=2BF
\displaystyle \text{But }BE=BF+FE
\displaystyle \therefore BF+FE=2BF
\displaystyle \therefore EF=FB
\displaystyle \text{(ii) Since }AE=EB\text{ and }EF=FB,
\displaystyle BE=EF+FB=2EF
\displaystyle \therefore AE=2EF
\displaystyle AF=AE+EF=2EF+EF=3EF
\displaystyle \therefore \frac{AF}{AE}=\frac{3EF}{2EF}=\frac{3}{2}
\displaystyle \text{In }\triangle AFD\text{ and }\triangle AEG,
\displaystyle FD\parallel EG
\displaystyle \therefore \triangle AFD\sim\triangle AEG
\displaystyle \frac{AD}{AG}=\frac{AF}{AE}=\frac{3}{2}
\displaystyle \frac{AG+GD}{AG}=\frac{3}{2}
\displaystyle 2AG+2GD=3AG
\displaystyle AG=2GD
\displaystyle \therefore AG:GD=2:1
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\displaystyle \textbf{Question 4: } \text{In the given figure, }\triangle ABC\sim\triangle PQR.
\displaystyle AM\text{ and }PN\text{ are altitudes, while }AX\text{ and }PY\text{ are medians.}
\displaystyle \text{Prove that }\frac{AM}{PN}=\frac{AX}{PY}.
\displaystyle \text{Answer:}
\displaystyle \text{The given solution is not completely correct. The corrected proof is as follows.}
\displaystyle \text{Since }\triangle ABC\sim\triangle PQR,
\displaystyle \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}
\displaystyle \text{Consider }\triangle ABM\text{ and }\triangle PQN.
\displaystyle \angle ABM=\angle PQN
\displaystyle \text{Since these are corresponding angles of the similar triangles.}
\displaystyle \angle AMB=\angle PNQ=90^\circ
\displaystyle \therefore \triangle ABM\sim\triangle PQN\qquad\text{(AA similarity)}
\displaystyle \therefore \frac{AM}{PN}=\frac{AB}{PQ}\qquad\ldots\text{(i)}
\displaystyle \text{Since }AX\text{ and }PY\text{ are medians, }X\text{ and }Y\text{ are midpoints.}
\displaystyle BX=\frac{BC}{2}\quad\text{and}\quad QY=\frac{QR}{2}
\displaystyle \therefore \frac{BX}{QY}=\frac{BC/2}{QR/2}=\frac{BC}{QR}
\displaystyle \text{But }\frac{BC}{QR}=\frac{AB}{PQ}
\displaystyle \therefore \frac{BX}{QY}=\frac{AB}{PQ}\qquad\ldots\text{(ii)}
\displaystyle \text{Now, in }\triangle ABX\text{ and }\triangle PQY,
\displaystyle \frac{AB}{PQ}=\frac{BX}{QY}
\displaystyle \angle ABX=\angle PQY
\displaystyle \text{Since }\angle ABX=\angle ABC\text{ and }\angle PQY=\angle PQR.
\displaystyle \therefore \triangle ABX\sim\triangle PQY\qquad\text{(SAS similarity)}
\displaystyle \therefore \frac{AX}{PY}=\frac{AB}{PQ}\qquad\ldots\text{(iii)}
\displaystyle \text{From (i) and (iii),}
\displaystyle \frac{AM}{PN}=\frac{AB}{PQ}=\frac{AX}{PY}
\displaystyle \therefore \frac{AM}{PN}=\frac{AX}{PY}
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\displaystyle \textbf{Question 5: } \text{Two similar triangles are equal in area. Prove that the triangles are congruent.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two triangles be }\triangle ABC\text{ and }\triangle PQR.
\displaystyle \text{Since }\triangle ABC\sim\triangle PQR,
\displaystyle \frac{\text{Ar. }\triangle ABC}{\text{Ar. }\triangle PQR}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}
\displaystyle \text{But the areas of the two triangles are equal.}
\displaystyle \therefore \frac{\text{Ar. }\triangle ABC}{\text{Ar. }\triangle PQR}=1
\displaystyle \therefore \frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}=1
\displaystyle \therefore AB=PQ,\quad BC=QR,\quad AC=PR
\displaystyle \therefore \triangle ABC\cong\triangle PQR\qquad\text{(SSS Congruence Criterion)}
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\displaystyle \textbf{Question 6: } \text{The ratio between the altitudes of two similar triangles is }3:5.
\displaystyle \text{Write the ratios between their (i) medians (ii) perimeters (iii) areas.}
\displaystyle \text{Answer:}
\displaystyle \text{In similar triangles, the ratio of altitudes equals the ratio of corresponding sides.}
\displaystyle \therefore \text{Ratio of corresponding sides}=3:5
\displaystyle \text{(i) Ratio of their medians}=3:5
\displaystyle \text{(ii) Ratio of their perimeters}=3:5
\displaystyle \text{(iii) Ratio of their areas}=3^2:5^2=9:25
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\displaystyle \textbf{Question 7: } \text{The ratio between the altitudes of two similar triangles is }16:25.
\displaystyle \text{Find the ratio between their (i) perimeters (ii) altitudes (iii) medians.}
\displaystyle \text{Answer:}
\displaystyle \text{The given solution is incorrect.}
\displaystyle \text{In similar triangles, the ratio of corresponding altitudes equals the ratio of corresponding sides.}
\displaystyle \therefore \text{Ratio of corresponding sides}=16:25
\displaystyle \text{(i) Ratio of their perimeters}=16:25
\displaystyle \text{(ii) Ratio of their altitudes}=16:25
\displaystyle \text{(iii) Ratio of their medians}=16:25
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\displaystyle \textbf{Question 8: } \text{The following figure shows a }\triangle PQR\text{ in which }XY\parallel QR.
\displaystyle \text{If }PX:XQ=1:3\text{ and }QR=9\text{ cm, find the length of }XY.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }PX:XQ=1:3\text{ and }QR=9\text{ cm}
\displaystyle \text{Since }XY\parallel QR,
\displaystyle \angle PXY=\angle PQR\qquad\text{(corresponding angles)}
\displaystyle \angle PYX=\angle PRQ\qquad\text{(corresponding angles)}
\displaystyle \therefore \triangle PXY\sim\triangle PQR\qquad\text{(AA similarity)}
\displaystyle \therefore \frac{PX}{PQ}=\frac{XY}{QR}
\displaystyle \text{Now, }PQ=PX+XQ
\displaystyle \therefore \frac{PX}{PX+XQ}=\frac{XY}{QR}
\displaystyle \frac{PX/XQ}{PX/XQ+1}=\frac{XY}{QR}
\displaystyle \frac{1}{1+3}=\frac{XY}{9}
\displaystyle \frac{1}{4}=\frac{XY}{9}
\displaystyle XY=\frac{9}{4}\text{ cm}=2.25\text{ cm}
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\displaystyle \textbf{Question 9: } \text{In the following figure, }AB,\ CD\text{ and }EF\text{ are parallel lines.}
\displaystyle AB=6\text{ cm},\ CD=y\text{ cm},\ EF=10\text{ cm},\ AC=4\text{ cm and }CF=x\text{ cm}.
\displaystyle \text{Calculate }x\text{ and }y.\hfill\text{[ICSE 1985]}
\displaystyle \text{Answer:}
\displaystyle \text{The solution is correct, with minor corrections as shown below.}
\displaystyle \text{Consider }\triangle FDC\text{ and }\triangle FBA.
\displaystyle \angle FDC=\angle FBA\qquad\text{(corresponding angles, since }CD\parallel AB\text{)}
\displaystyle \angle DFC=\angle BFA\qquad\text{(common angle)}
\displaystyle \therefore \triangle FDC\sim\triangle FBA\qquad\text{(AA similarity)}
\displaystyle \therefore \frac{CD}{AB}=\frac{FC}{FA}
\displaystyle \frac{y}{6}=\frac{x}{x+4}\qquad\ldots\text{(i)}
\displaystyle \text{Now, consider }\triangle FCE\text{ and }\triangle ACB.
\displaystyle \angle FCE=\angle ACB\qquad\text{(vertically opposite angles)}
\displaystyle \angle CFE=\angle CAB\qquad\text{(alternate interior angles, since }EF\parallel AB\text{)}
\displaystyle \therefore \triangle FCE\sim\triangle ACB\qquad\text{(AA similarity)}
\displaystyle \therefore \frac{FC}{AC}=\frac{EF}{AB}
\displaystyle \frac{x}{4}=\frac{10}{6}
\displaystyle x=\frac{40}{6}=\frac{20}{3}\text{ cm}
\displaystyle \text{Substituting }x=\frac{20}{3}\text{ in (i),}
\displaystyle \frac{y}{6}=\frac{\frac{20}{3}}{\frac{20}{3}+4}
\displaystyle \frac{y}{6}=\frac{\frac{20}{3}}{\frac{32}{3}}=\frac{5}{8}
\displaystyle y=6\times\frac{5}{8}=\frac{15}{4}\text{ cm}
\displaystyle \therefore x=\frac{20}{3}\text{ cm and }y=\frac{15}{4}\text{ cm}
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\displaystyle \textbf{Question 10: } \text{On a map, drawn to a scale of }1:20000,\text{ a rectangular plot }ABCD
\displaystyle \text{has }AB=24\text{ cm and }BC=32\text{ cm. Calculate:}
\displaystyle \text{(i) the diagonal distance of the plot in km}
\displaystyle \text{(ii) the area of the plot in }\text{km}^2.
\displaystyle \text{Answer:}
\displaystyle \text{The given solution is correct except for the unit in part (ii).}
\displaystyle \text{Scale}=\frac{1}{20000}
\displaystyle \text{Actual length of }AB=24\times20000\text{ cm}=480000\text{ cm}=4.8\text{ km}
\displaystyle \text{Actual length of }BC=32\times20000\text{ cm}=640000\text{ cm}=6.4\text{ km}
\displaystyle \text{(i) Diagonal}=\sqrt{4.8^2+6.4^2}=\sqrt{23.04+40.96}=\sqrt{64}=8\text{ km}
\displaystyle \text{(ii) Area}=4.8\times6.4=30.72\text{ km}^2
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\displaystyle \textbf{Question 11: } \text{The dimensions of a model of a multi-storeyed building are}
\displaystyle 1\text{ m}\times60\text{ cm}\times1.20\text{ m}.
\displaystyle \text{If the scale factor is }1:50,\text{ find the actual dimensions of the building.}
\displaystyle \text{Also find:}
\displaystyle \text{(i) the floor area of a room, if the area of the corresponding model room is }50\text{ cm}^2.
\displaystyle \text{(ii) the space inside the model room, if the corresponding actual space is }90\text{ m}^3.
\displaystyle \text{Answer:}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{Dimensions of the model}=100\text{ cm}\times60\text{ cm}\times120\text{ cm}
\displaystyle \text{Scale factor}=\frac{\text{model dimension}}{\text{actual dimension}}=\frac{1}{50}
\displaystyle \text{Actual length}=100\times50=5000\text{ cm}=50\text{ m}
\displaystyle \text{Actual breadth}=60\times50=3000\text{ cm}=30\text{ m}
\displaystyle \text{Actual height}=120\times50=6000\text{ cm}=60\text{ m}
\displaystyle \therefore \text{Actual dimensions}=50\text{ m}\times30\text{ m}\times60\text{ m}
\displaystyle \text{(i) Ratio of areas}=1:50^2
\displaystyle \text{Actual floor area}=50\times50^2=125000\text{ cm}^2
\displaystyle =\frac{125000}{10000}\text{ m}^2=12.5\text{ m}^2
\displaystyle \text{(ii) Ratio of volumes}=1:50^3
\displaystyle \text{Volume of the model room}=\frac{90}{50^3}\text{ m}^3
\displaystyle =\frac{90}{125000}\text{ m}^3=0.00072\text{ m}^3
\displaystyle =0.00072\times1000000\text{ cm}^3=720\text{ cm}^3
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\displaystyle \textbf{Question 12: } \text{In }\triangle PQR,\ L\text{ and }M\text{ are points on the base }QR,
\displaystyle \text{such that }\angle LPQ=\angle QRP\text{ and }\angle RPM=\angle RQP.\text{ Prove that:}
\displaystyle \text{(i) }\triangle PQL\sim\triangle RPM
\displaystyle \text{(ii) }QL\times RM=PL\times PM
\displaystyle \text{(iii) }PQ^2=QR\times QL\hfill\text{[ICSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle PQL\text{ and }\triangle RPM.
\displaystyle \angle LPQ=\angle PRM\qquad\text{(given, since }Q,L,M,R\text{ are collinear)}
\displaystyle \angle PQL=\angle RPM\qquad\text{(given, since }Q,L,M,R\text{ are collinear)}
\displaystyle \therefore \triangle PQL\sim\triangle RPM\qquad\text{(AA similarity)}
\displaystyle \text{(ii) Since }\triangle PQL\sim\triangle RPM,
\displaystyle \frac{PQ}{RP}=\frac{QL}{PM}=\frac{PL}{RM}
\displaystyle \frac{QL}{PM}=\frac{PL}{RM}
\displaystyle \therefore QL\times RM=PL\times PM
\displaystyle \text{(iii) Consider }\triangle PQL\text{ and }\triangle RQP.
\displaystyle \angle LPQ=\angle QRP\qquad\text{(given)}
\displaystyle \angle PQL=\angle RQP\qquad\text{(since }Q,L,R\text{ are collinear)}
\displaystyle \therefore \triangle PQL\sim\triangle RQP\qquad\text{(AA similarity)}
\displaystyle \therefore \frac{PQ}{RQ}=\frac{QL}{QP}
\displaystyle PQ\times QP=RQ\times QL
\displaystyle \therefore PQ^2=QR\times QL
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\displaystyle \textbf{Question 13: } \text{In }\triangle ABC,\ \angle ACB=90^\circ\text{ and }CD\perp AB.
\displaystyle \text{Prove that }\frac{BC^2}{AC^2}=\frac{BD}{AD}.
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle BCD\text{ and }\triangle ACD.
\displaystyle \angle BDC=\angle ADC=90^\circ
\displaystyle \angle BCD=\angle CAD
\displaystyle \therefore \triangle BCD\sim\triangle ACD\qquad\text{(AA similarity)}
\displaystyle \therefore \frac{BC}{AC}=\frac{BD}{CD}=\frac{CD}{AD}
\displaystyle \frac{BC}{AC}=\frac{BD}{CD}\qquad\ldots\text{(i)}
\displaystyle \frac{BC}{AC}=\frac{CD}{AD}\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying (i) and (ii),}
\displaystyle \frac{BC^2}{AC^2}=\frac{BD}{CD}\times\frac{CD}{AD}
\displaystyle \therefore \frac{BC^2}{AC^2}=\frac{BD}{AD}
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\displaystyle \textbf{Question 14: } \text{A }\triangle ABC\text{ with }AB=3\text{ cm},\ BC=6\text{ cm and }AC=4\text{ cm}
\displaystyle \text{is enlarged to }\triangle DEF\text{ such that its longest side is }9\text{ cm}.
\displaystyle \text{Find the scale factor and hence the lengths of the other sides of }\triangle DEF.
\displaystyle \text{Answer:}
\displaystyle \text{The longest side of }\triangle ABC\text{ is }BC=6\text{ cm}.
\displaystyle \text{Therefore, the corresponding longest side of }\triangle DEF\text{ is }EF=9\text{ cm}.
\displaystyle \text{Scale factor }k=\frac{EF}{BC}=\frac{9}{6}=\frac{3}{2}=1.5
\displaystyle \frac{DE}{AB}=1.5
\displaystyle DE=3\times1.5=4.5\text{ cm}
\displaystyle \frac{DF}{AC}=1.5
\displaystyle DF=4\times1.5=6\text{ cm}
\displaystyle \therefore \text{The other sides of }\triangle DEF\text{ are }4.5\text{ cm and }6\text{ cm}.
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\displaystyle \textbf{Question 15: } \text{Two isosceles triangles have equal vertex angles. Show that they are similar.}
\displaystyle \text{If the ratio of their areas is }16:25,\text{ find the ratio of their corresponding altitudes.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the isosceles triangles }\triangle ABC\text{ and }\triangle DEF,
\displaystyle \text{where }AB=AC,\ DE=DF\text{ and }\angle BAC=\angle EDF.
\displaystyle \frac{AB}{AC}=1\quad\text{and}\quad\frac{DE}{DF}=1
\displaystyle \therefore \frac{AB}{AC}=\frac{DE}{DF}
\displaystyle \angle BAC=\angle EDF\qquad\text{(given)}
\displaystyle \therefore \triangle ABC\sim\triangle DEF\qquad\text{(SAS similarity)}
\displaystyle \text{Let }AM\text{ and }DN\text{ be their corresponding altitudes.}
\displaystyle \text{For similar triangles, the ratio of their areas is the square of the ratio}
\displaystyle \text{of their corresponding altitudes.}
\displaystyle \frac{\text{Ar. }\triangle ABC}{\text{Ar. }\triangle DEF}=\frac{AM^2}{DN^2}
\displaystyle \frac{16}{25}=\left(\frac{AM}{DN}\right)^2
\displaystyle \frac{AM}{DN}=\sqrt{\frac{16}{25}}=\frac{4}{5}
\displaystyle \therefore \text{The ratio of their corresponding altitudes is }4:5.
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\displaystyle \textbf{Question 16: } \text{In }\triangle ABC,\ AP:PB=2:3.\text{ A line through }P\text{ parallel to }BC
\displaystyle \text{meets }AC\text{ at }O\text{ and is extended to }Q\text{ such that }CQ\parallel BA.\text{ Find:}
\displaystyle \text{(i) Ar. }\triangle APO:\text{Ar. }\triangle ABC
\displaystyle \text{(ii) Ar. }\triangle APO:\text{Ar. }\triangle CQO
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle APO\text{ and }\triangle ABC.
\displaystyle \angle APO=\angle ABC\qquad\text{(corresponding angles, since }PO\parallel BC\text{)}
\displaystyle \angle AOP=\angle ACB\qquad\text{(corresponding angles, since }PO\parallel BC\text{)}
\displaystyle \therefore \triangle APO\sim\triangle ABC\qquad\text{(AA similarity)}
\displaystyle AP:PB=2:3
\displaystyle \therefore AP:AB=2:(2+3)=2:5
\displaystyle \frac{\text{Ar. }\triangle APO}{\text{Ar. }\triangle ABC}=\left(\frac{AP}{AB}\right)^2
\displaystyle =\left(\frac{2}{5}\right)^2=\frac{4}{25}
\displaystyle \therefore \text{Ar. }\triangle APO:\text{Ar. }\triangle ABC=4:25
\displaystyle \text{(ii) Since }PQ\parallel BC\text{ and }PB\parallel CQ,
\displaystyle PBCQ\text{ is a parallelogram.}
\displaystyle \therefore CQ=PB
\displaystyle \text{Consider }\triangle APO\text{ and }\triangle CQO.
\displaystyle \angle AOP=\angle COQ\qquad\text{(vertically opposite angles)}
\displaystyle \angle PAO=\angle OCQ\qquad\text{(alternate interior angles, since }AP\parallel CQ\text{)}
\displaystyle \therefore \triangle APO\sim\triangle CQO\qquad\text{(AA similarity)}
\displaystyle \frac{\text{Ar. }\triangle APO}{\text{Ar. }\triangle CQO}=\left(\frac{AP}{CQ}\right)^2
\displaystyle =\left(\frac{AP}{PB}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}
\displaystyle \therefore \text{Ar. }\triangle APO:\text{Ar. }\triangle CQO=4:9
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\displaystyle \textbf{Question 17: } \text{The following figure shows a }\triangle ABC\text{ in which }AD\perp BC
\displaystyle \text{and }BE\perp AC.\text{ Show that:}
\displaystyle \text{(i) }\triangle ADC\sim\triangle BEC\qquad\text{(ii) }CA\times CE=CB\times CD
\displaystyle \text{(iii) }\triangle ABC\sim\triangle DEC\qquad\text{(iv) }CD\times AB=CA\times DE
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle ADC\text{ and }\triangle BEC.
\displaystyle \angle ADC=\angle BEC=90^\circ\qquad\text{(given)}
\displaystyle \angle ACD=\angle BCE
\displaystyle \text{Since }C,D,B\text{ are collinear and }C,E,A\text{ are collinear.}
\displaystyle \therefore \triangle ADC\sim\triangle BEC\qquad\text{(AA similarity)}
\displaystyle \text{(ii) Since }\triangle ADC\sim\triangle BEC,
\displaystyle \frac{CA}{CB}=\frac{CD}{CE}
\displaystyle \therefore CA\times CE=CB\times CD
\displaystyle \text{(iii) Consider }\triangle ABC\text{ and }\triangle DEC.
\displaystyle \angle ACB=\angle DCE
\displaystyle \text{Since }C,D,B\text{ are collinear and }C,E,A\text{ are collinear.}
\displaystyle \text{From part (ii), }CA\times CE=CB\times CD
\displaystyle \therefore \frac{CA}{CD}=\frac{CB}{CE}
\displaystyle \therefore \triangle ABC\sim\triangle DEC\qquad\text{(SAS similarity)}
\displaystyle \text{(iv) Since }\triangle ABC\sim\triangle DEC,
\displaystyle \frac{AB}{DE}=\frac{CA}{CD}
\displaystyle \therefore CD\times AB=CA\times DE
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\displaystyle \textbf{Question 18: } \text{In the given figure, }D\text{ lies on }AB\text{ and }E\text{ lies on }BC,
\displaystyle \text{such that }\angle EDB=\angle ACB.\text{ Prove that }\triangle ABC\sim\triangle EBD.
\displaystyle \text{If }BE=6\text{ cm},\ EC=4\text{ cm},\ BD=5\text{ cm and}
\displaystyle \text{Ar. }\triangle BED=9\text{ cm}^2,\text{ calculate:}
\displaystyle \text{(i) the length of }AB\qquad\text{(ii) the area of }\triangle ABC.\hfill\text{[ICSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Consider }\triangle ABC\text{ and }\triangle EBD.
\displaystyle \angle ACB=\angle EDB\qquad\text{(given)}
\displaystyle \angle ABC=\angle EBD
\displaystyle \text{Since }A,D,B\text{ are collinear and }B,E,C\text{ are collinear.}
\displaystyle \therefore \triangle ABC\sim\triangle EBD\qquad\text{(AA similarity)}
\displaystyle \text{(i) }BC=BE+EC=6+4=10\text{ cm}
\displaystyle \text{Since }\triangle ABC\sim\triangle EBD,
\displaystyle \frac{AB}{EB}=\frac{BC}{BD}
\displaystyle \frac{AB}{6}=\frac{10}{5}=2
\displaystyle AB=6\times2=12\text{ cm}
\displaystyle \text{(ii) }\frac{\text{Ar. }\triangle ABC}{\text{Ar. }\triangle EBD}=\left(\frac{AB}{EB}\right)^2
\displaystyle \frac{\text{Ar. }\triangle ABC}{9}=\left(\frac{12}{6}\right)^2=4
\displaystyle \text{Ar. }\triangle ABC=9\times4=36\text{ cm}^2
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\displaystyle \textbf{Question 19: } \text{In the given figure, }\triangle ABC\text{ is right-angled at }A,
\displaystyle \angle BAC=90^\circ,\text{ and }AD\perp BC.\text{ Prove that:}
\displaystyle \text{(i) }\triangle ADB\sim\triangle CDA
\displaystyle \text{(ii) If }BD=18\text{ cm and }CD=8\text{ cm, find }AD.
\displaystyle \text{(iii) Find }\text{Ar. }\triangle ADB:\text{Ar. }\triangle CDA.\hfill\text{[ICSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle ADB\text{ and }\triangle CDA.
\displaystyle \angle ADB=\angle CDA=90^\circ\qquad\text{(since }AD\perp BC\text{)}
\displaystyle \text{Let }\angle DAB=\theta.
\displaystyle \therefore \angle DAC=90^\circ-\theta
\displaystyle \text{In }\triangle ADB,\ \angle DBA=90^\circ-\theta
\displaystyle \therefore \angle DBA=\angle DAC
\displaystyle \therefore \triangle ADB\sim\triangle CDA\qquad\text{(AA similarity)}
\displaystyle \text{(ii) Since }\triangle ADB\sim\triangle CDA,
\displaystyle \frac{BD}{AD}=\frac{AD}{CD}
\displaystyle AD^2=BD\times CD
\displaystyle AD^2=18\times8=144
\displaystyle AD=\sqrt{144}=12\text{ cm}
\displaystyle \text{(iii) }\frac{\text{Ar. }\triangle ADB}{\text{Ar. }\triangle CDA}
\displaystyle =\frac{\frac{1}{2}\times BD\times AD}{\frac{1}{2}\times CD\times AD}
\displaystyle =\frac{BD}{CD}=\frac{18}{8}=\frac{9}{4}
\displaystyle \therefore \text{Ar. }\triangle ADB:\text{Ar. }\triangle CDA=9:4
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\displaystyle \textbf{Question 20: } \text{In the given figure, }AB\text{ and }DE\text{ are perpendicular to }BC.
\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},\text{ calculate }CD.
\displaystyle \text{(iii) Find }\text{Ar. }\triangle ABC:\text{Ar. }\triangle DEC.\hfill\text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Consider }\triangle ABC\text{ and }\triangle DEC.
\displaystyle \angle ABC=\angle DEC=90^\circ\qquad\text{(since }AB\perp BC\text{ and }DE\perp BC\text{)}
\displaystyle \angle ACB=\angle DCE
\displaystyle \text{Since }A,D,C\text{ are collinear and }B,E,C\text{ are collinear.}
\displaystyle \therefore \triangle ABC\sim\triangle DEC\qquad\text{(AA similarity)}
\displaystyle \text{(ii) Since }\triangle ABC\sim\triangle DEC,
\displaystyle \frac{AB}{DE}=\frac{AC}{DC}
\displaystyle \frac{6}{4}=\frac{15}{DC}
\displaystyle 6DC=4\times15
\displaystyle DC=\frac{60}{6}=10\text{ cm}
\displaystyle \text{(iii) }\frac{\text{Ar. }\triangle ABC}{\text{Ar. }\triangle DEC}=\left(\frac{AB}{DE}\right)^2
\displaystyle =\left(\frac{6}{4}\right)^2=\frac{36}{16}=\frac{9}{4}
\displaystyle \therefore \text{Ar. }\triangle ABC:\text{Ar. }\triangle DEC=9:4
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