Note:  Relation between the areas of two similar triangles:  If $\triangle ABC \sim \triangle DEF$ then

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

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DE}=\frac{Perimeter \ \triangle ABC}{Perimeter \ \triangle DEF}$

Question 1: (i) The ratio between the corresponding sides of two similar triangles is $2 :5$. Find the ratios between the areas of these triangles.

(ii) Areas of two similar triangles is $98 \ cm^2$ and $128 \ cm^2$. Find the ratios between the length of their corresponding sides.

(i) Required ratio of their areas $= \frac{2^2}{5^2} = \frac{4}{25}$

(ii)  $\frac{Ar. \ of \ \triangle 1}{Ar. \ of \ \triangle 2} = \frac{side1 ^2}{side2^2}$

Therefore Required ratio $= \sqrt{\frac{98}{128}} = \frac{7}{8}$

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Question 2: A line $PQ$ is drawn parallel to the base $BC \ of \triangle ABC$ which meets sides $AB \ and \ AC$ at points $P \ and \ Q$ respectively. If $AP = \frac{1}{2} PB$; find the value of

(i) $\frac{Area \ of \ \triangle ABC}{Area \ of \ \triangle APQ}$

(ii) $\frac{Area \ of \ \triangle APQ}{Area \ of \ trapezium PBCQ}$

Given $AP = \frac{1}{2} PB$

Consider $\triangle APQ \ and\ \triangle ABC$

$\angle APQ = \angle ABC$ (alternate angles)

$\angle AQP =\angle ACB$ (alternate angles)

Therefore $\triangle APQ \sim \triangle ABC$   (AAA postulate)

Hence $\frac{Ar. \triangle ABC}{Ar. \triangle APQ} = \frac{AB^2}{AP^2} = \frac{(AP+PB)^2}{AP^2} = (\frac{1+\frac{PB}{AP}}{1})^2 = \frac{16}{1}$

Also $\frac{Ar. \ \triangle APQ}{Ar. \ trapezium PBCQ} = \frac{Ar. \ \triangle APQ}{Ar. \ \triangle ABC - Ar. \ \triangle APQ} = \frac{1}{16-1} = \frac{1}{15}$

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Question 3: The perimeter of two similar triangles are $30 \ cm$ and $24 \ cm$. If one side of the first triangle is $12 \ cm$, determine the corresponding side of the second triangle.

Since the two given triangles are similar, we have

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DE}=\frac{Perimeter \ \triangle ABC}{Perimeter \ \triangle DEF}$

$\frac{12}{DE}=\frac{30}{24}$

$\Rightarrow DE = \frac{12 \times 24}{30} = 9.6 \ cm$

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Question 4: In the given figure, $AX:XB=3:5$. Find:

(i) the length of $BC$, if the length of $XY$ is $18 \ cm$.

(ii) the ratio between the areas of trapezium $XBCY$ and $\triangle ABC$.

Given $AX:XB=3:5$

$\Rightarrow \frac{AX}{AB}=\frac{3}{8}$

Consider $\triangle ABC \ and\ \triangle AXY$

$\angle ABC = \angle AXY$ (alternate angles)

$\angle BAC =\angle XAY$ (common angle)

Therefore $\triangle ABC \sim \triangle AXY$

Therefore $\frac{AX}{AB}=\frac{XY}{BC} \Rightarrow BC = \frac{8 \times 18}{3} = 48 \ cm$

$\frac{Ar. \ \triangle AXY}{Ar. \ \triangle ABC}= \frac{AX^2}{AB^2} = \frac{9}{64}$

Also $\frac{Ar. \ \triangle ABC - Ar. \ \triangle AXY}{Ar. \ \triangle ABC} = 1- \frac{9}{64} = \frac{55}{64}$

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Question 5: $ABC$ is a triangle. $PQ$ is a line segment intersecting $AB \ in \ P$ and $AC \ in \ Q$ such that $PQ \parallel BC$ and divides $\triangle ABC$ into two parts equal in area. Find the value of ratio $BP:AB$.

Given $Ar. \ \triangle APQ = \frac{1}{2} Ar. \ \triangle ABC$

Consider $\triangle APQ \ and\ \triangle ABC$

$\angle APB = \angle ABC$ (corresponding angles)

$\angle BAC =\angle PAQ$ (common angle)

Therefore $\triangle APQ \sim \triangle ABC$

Therefore $\frac{Ar. \ \triangle APQ}{Ar. \ \triangle ABC}= \frac{AP^2}{AB^2}$

$\frac{AP^2}{AB^2} = \frac{1}{2}$

$(1+\frac{PB}{AB})^2 = 2$

$\Rightarrow \frac{PB}{AB} = (\sqrt{2}-1)$

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Question 6: In the given $\triangle PQR, LM \parallel QR$ and $PM:MR=3:4$. Calculate the value of the ratio:

(i) $\frac{PL}{PQ}$ and then $\frac{LM}{QR}$

(ii) $\frac{Area \ of \triangle LMN}{Area \ of \triangle MNR}$

(iii) $\frac{Area \ of \triangle LQM}{Area \ of \triangle LQN}$

(i) Given $\triangle PQR, LM \parallel QR$ and $PM:MR=3:4$.

$\Rightarrow PM:PR=3:7$

Consider $\triangle PLM \ and\ \triangle PQR$

$\angle PLM = \angle PQR$ (alternate angles)

$\angle LPM =\angle QPR$ (common angle)

Therefore $\triangle PLM \sim \triangle PQR$

Therefore $\frac{LM}{QR}=\frac{PL}{PQ}=\frac{PM}{PR}$

$\Rightarrow \frac{PL}{PQ}=\frac{3}{7}$

(ii)  Since $\triangle LMN \ and \ \triangle MNR$ have common vertex $L$ and their bases $LN \ and \ LR$ are along the same straight line

$\frac{Ar. \ \triangle LMN}{Ar. \ \triangle MNR} =\frac{LN}{NR}$

Consider $\triangle LMN \ and\ \triangle QRN$

$\angle NLM = \angle NRQ$ (alternate angles)

$\angle LMN =\angle NQR$ (common angle)

Therefore $\triangle LMN \sim \triangle QRN$

Therefore $\frac{LM}{NR}=\frac{LM}{QR}=\frac{MN}{QN} =\frac{3}{7}$

$\frac{Area \ of \triangle LMN}{Area \ of \triangle MNR} = \frac{3}{7}$

(iii) Since $\triangle LQN \ and \ \triangle LQM$ have common vertex $L$ and their bases $MN \ and \ MQ$ are along the same straight line

Therefore $\frac{Area \ of \triangle LQM}{Area \ of \triangle LQN} = \frac{10}{7}$

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Question 7: The given diagram shows two isosceles triangles which are similar. $PQ \ and \ BC$ are not parallel. $PC=4, AQ=3, QB-=12, BC=15$ and $AP=PQ$ . Calculate:

(i) the length of $AP$

(ii) the ratio of the areas of $\triangle APQ \ and \ \triangle ABC$

Given $AB = AC = 15$

$AP = PQ = x$

Since $\triangle APQ \sim \triangle ABC$

$\frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}$

$\frac{AP}{15}=\frac{x}{15}=\frac{3}{x+4}$

Solving $x^2+4x-45 = 0 \Rightarrow x = 5 \ cm$

(ii) $\frac{Area \ of \triangle APQ}{Area \ of \triangle ABC} = \frac{AQ^2}{AC^2} = \frac{3^2}{9^2} = \frac{1}{9}$

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Question 8:  In the figure given below, $ABCD$ is a parallelogram. $P$ is a point on $BC$ such that $BP:PC = 1:2$. $DP$ produced meets $AB$ produced at $Q$. Given the area of $\triangle CPQ=20 \ cm^2$. Calculate:

(i) area of $\triangle CDP$

(ii) area of parallelogram $ABCD$ [1996]

(i)   In $\triangle BPQ \ and \ \triangle CPD$

$\angle BPQ = \angle CPD$ (vertically opposite angles)

$\angle BQP = \angle PDC$ (alternate angles)

$\triangle BPQ \sim \triangle CPD$

Therefore $\frac{BP}{ PC}= \frac{ PQ}{ PD}= \frac{ BQ}{ CD}= \frac{ 1}{ 2}$

Also $\frac{Area \ \triangle BPQ}{Area \ \triangle CPD} = \frac{BP^2}{PC^2}$

$Area \ \triangle CPD = \frac{4}{1} \times 10 = 40 \ cm^2$

(ii) In $\triangle BAP \ and \ \triangle AQD$

$BP \parallel AD$ (Given)

$\angle QBP = \angle QAD$ (corresponding angles are equal)

$\angle BQP = \angle AQD$ (common angle)

$\triangle BQP \sim \triangle AQD$

Therefore $\frac{AQ}{ BQ} =\frac{QD}{ QP} =\frac{AD}{ BP} = 3$

Also $\frac{Area \ \triangle AQD}{Area \ \triangle BQP} = \frac{AQ^2}{BQ^2}$

$Area \ \triangle AQD = 3^2 \times 10 = 90 \ cm^2$

$Area (ABCD) = Area \ \triangle AQD -Area \ \triangle BQP + Area \ \triangle CDP$

$= 90-10+40 = 120 \ cm^2$

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Question 9: In the given figure, $BC \parallel DE$. Area of $\triangle ABC = 25 \ cm^2$, area of trapezium  $BCED = 24 \ cm^2$ and $DE = 14 \ cm$. Calculate the length of $BC$. Also find the area of $\triangle BCD$.

Given  $BC \parallel DE$

Consider $\triangle ABC \ and\ \triangle ADE$

$\angle ABC = \angle ADE$ (alternate angles)

$\angle ACB =\angle AED$ (common angle)

Therefore $\triangle ABC \sim \triangle ADE$

$\frac{Area \triangle ABC}{Area \triangle ADE} = \frac{BC^2}{DE^2}$

$\frac{Area \triangle ABC}{Area \triangle ABC - Area \ of \ trapezium BCDE} = \frac{BC^2}{DE^2}$

$\frac{25}{25+24} = \frac{BC^2}{14^2}$

$\Rightarrow BC = 10 \ cm$

Now Area of  trapezium $BCED = \frac{1}{2} (sum \ of \ parallel \ sides) \times height$

$\Rightarrow 24 = \frac{1}{2} (10+14) \times height$

$\Rightarrow height =\frac{24 \times 2}{24} = 2 \ cm$

Area of $\triangle BCD = \frac{1}{2} \times base \times height$

$= \frac{1}{2} \times 10 \times 2 = 10 \ cm^2$

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Question 10: The given figure shows a trapezium in which $AB \parallel DC$ and diagonals $AC \ and \ BD$ intersect at point $P$. If $AP:CP=3:5$. Find:

(i) $\triangle APB : \triangle CPB$

(ii) $\triangle DPC : \triangle APB$

(iii) $\triangle ADP : \triangle APB$

(iv) $\triangle APB : \triangle ADB$

(i) Since $\triangle APB \ and \ \triangle CPB$ have common vertex $B$ and their bases $AP \ and \ PC$ are along the same straight line

Therefore $\frac{Area \ of \triangle APB}{Area \ of \triangle CPB} = \frac{AP}{PC} = \frac{3}{5}$

(ii) Since $\triangle DPC \ \sim \ \triangle BAP$

Therefore $\frac{Area \ of \triangle DPC}{Area \ of \triangle BPA} = \frac{PC^2}{AP^2} = \frac{25}{9}$

(iii) Since $\triangle ADP \ and \ \triangle APB$ have common vertex $A$ and their bases $DP \ and \ PB$ are along the same straight line

Therefore $\frac{Area \ of \triangle ADP}{Area \ of \triangle APB} = \frac{DP}{PB} = \frac{5}{3}$

(iv) Since $\triangle APB \ and \ \triangle ADB$ have common vertex $A$ and their bases $BP \ and \ BD$ are along the same straight line

Therefore $\frac{Area \ of \triangle APB}{Area \ of \triangle ADB} = \frac{PB}{BD} = \frac{3}{8}$

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