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 $= \sqrt{\frac{98}{128}}$ $=$ $\frac{7}{8}$

$\\$

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

$\\$

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

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

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

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

$\\$

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

$\\$

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

$\\$

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

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

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

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

$\\$

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

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

$\\$

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

$\\$

Question 11: On a map drawn to a scale of $1:2500000$ a triangular plot of $PQR$ of land has the following measurements: $PQ = 3 \ cm , QR=4 \ cm \ and \ \angle PQR=90^o$. Calculate:

(i) the actual length of $QR \ and \ PR$ in kilometers

(ii)  the actual area of the plot in $km^2$

Scale factor $k =\frac{1}{2500000}$

(i) Length of side $PQ$  in the map = $k \times$ the actual length of the side $PQ$ in the land

$\Rightarrow 3 \ cm = \frac{1}{2500000} \times actual \ length \ of \ PQ$

$\Rightarrow Actual \ length \ of \ PQ = 3 \times 2500000 = 7500000 \ m = 7.5 \ km$

Length of side $QR$  in the map $= k \times$ the actual length of the side $QR$ in the land

$\Rightarrow 4 \ cm = \frac{1}{2500000} \times$ actual length of $QR$

$\Rightarrow \ Actual \ length \ of \ QR = 4 \times 2500000 = 10000000 \ m = 10 \ km$

Hence $PR = \sqrt{7.5^2+10^2} = \sqrt{156.5} = 12.5 \ km$

(ii) Area of the plot $= \frac{1}{2} \times {base} \times {height} = \frac{1}{2} \times 7.5 \times 10 = 37.5 \ km^2$

$\\$

Question 12: A model of a ship is made to scale of $1:200$.

(i) The length of the model is $4 \ m$; calculate the length of the ship.

(ii) The area of the deck of the ship is $160000 \ m^2$; find the area of the deck of the model.

(iii) The volume of the model is $200 \ liters$; calculate the volume of the ship in $m^3$  [1995]

Scale factor $= \frac{1}{k}$

(i) Length of the model $= k \times$ Actual length of the ship

$\Rightarrow$ Actual length of the ship $= 4 \times 200 = 800 \ m$

(ii) Area of the deck of the model $= k^2 \times$ area of the deck of the actual ship

$= (\frac{1}{200})^2 \times 160000 \ m^2 = 4 \ m^2$

(iii) Volume of the model $= k^3 \times$ Volume of the actual ship

$= (\frac{1}{k})^3 \times 200 = (200)^3 \times 200 = 1600000000 \ liters = 16000000 \ m^3$

$\\$

Question 13: In the figure given below $ABC$ is a triangle. $DE$ is parallel to $BC$ and $\frac{AD}{DB}=\frac{3}{2}$.

(i) Determine the ratios $\frac{AD}{AB} \ and \ \frac{DE}{BC}$

(ii) Prove that $\triangle DEF$is similar to $\triangle CBF$. Hence , find $\frac{EF}{FB}$  [2007]

(i)   Given $DE \parallel BC \ and \$ $\frac{AD}{DB}=\frac{3}{2}$

$\triangle ADE \ and\ \triangle ABC$

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

$\angle ADE = \angle ABC$

$\triangle ADE \sim \triangle ABC$

$\frac{AD}{ AB} = \frac{AE }{AC} = \frac{DE}{ BC}$

$\frac{AD }{AB} = \frac{AD }{AD+DB} = \frac{3}{ 5}$

$\frac{AD}{ AE} = \frac{DE }{BC} = \frac{3}{ 5}$

(ii) In $\triangle DEF and \triangle CBF$

$\angle FDE = \angle FCB$ (alternate angles)

$\angle DFE = \angle BFC$ (vertically opposite angles)

$\triangle DEF \sim \triangle CBF$

$\frac{EF}{ FB} = \frac{DE}{ BC} = \frac{3}{ 5}$

$\frac{EF}{ FB }= \frac{3 }{5}$

(iii) We know

$\frac{Area \ of \ \triangle DEF}{Area \ of \ \triangle CBF} = \frac{EF^2}{ FB^2} = \frac{3^2 }{5^2 }= \frac{9 }{25}$

$\\$

Question 14: In the given figure $\angle B = \angle E$, $\angle ACD = \angle BCE$ . $AB = 10.4 \ cm \ and \ DE=7.8 \ cm$. Find the ratio between the area of the $\triangle ABC \ and \ \triangle DEC$.

Consider $\triangle ABC and \triangle CDE$

$\angle CBA = \angle CED (given)$

$\angle ECD = \angle BCA$ (Since $\angle BCD$ is common and $\angle ACD = \angle BCE$ is given)

Therefore $\triangle ABC \sim \triangle CDE$

We know that for similar triangles

$\frac{Ar. \ \triangle ABC}{Ar. \ \triangle DEC} = \frac{AB^2}{DE^2} = \frac{10.4^2}{7.8^2} = \frac{16}{9}$ $= 1.778$

$\\$

Question 15: $\triangle ABC$ is an isosceles triangle in which $AB=AC=13 \ cm$ and $BC=10 \ cm.$ $AD \perp BC$. If $CE = 8 \ cm \ and \ EF \perp AB$, find:

(i) $\frac{Ar. \ \triangle ADC}{Ar. \ \triangle FEB}$

(ii)  $\frac{Ar. \ \triangle FEB}{Ar. \ \triangle ABC}$

(i) Consider $\triangle ADC and \triangle FEB$

$\angle BFE = \angle ADC = 90^o (given)$

$\angle ACD = \angle FBE$ (Since $AB=AC \Rightarrow$ angles opposite equal sides are also equal)

Therefore $\triangle ADC \sim \triangle FEB$

We know that for similar triangles

$\frac{Ar. \ \triangle ADC}{Ar. \ \triangle FEB} = \frac{AC^2}{BE^2} = \frac{13^2}{18^2} = \frac{169}{324}$

(ii)  Now Consider $\triangle ABD and \triangle ACD$

$\angle ABC = \angle ACB (given)$

$\angle ADB = \angle ADC = 90^o$ (given)

Therefore $\triangle ABD \cong \triangle ACD$

Therefore $\frac{Ar. \ \triangle FEB}{Ar. \ \triangle ABC} = \frac{Ar. \ \triangle FEB}{2 \times Ar. \ \triangle ADC} = = \frac{324}{2 \times 169} = \frac{162}{169}$

$\\$

Question 16: An airplane is $30 \ m$ long and its model is $15 \ cm$ long. If the total outer surface area of the model is $150 \ cm^2$ , find the cost of painting the outer surface of the airplane at the rate of $Rs. 120 /m^2$ . Given that $50 \ m^2$ of the surface of the airplane sis left for windows.

$15 \ cm$ of the model represent $30 \ m$ of actual airplane

Therefore $1 \ cm$ wold represent $2 \ m$ of actual airplane

Hence $1 \ cm^2$ would represent $4 \ m^2$ of the surface are of the actual airplane

Given that the surface area of the model $= 150 \ cm^2$ .

Therefore the actual surface are of the actual airplane $= 150 \times 4 = 600 \ m^2$

Area to be painted $= 600-50 = 550 \ m^2$

Cost of painting $= 120 Rs. /m^2$

Therefore the total cost of painting $= 120 \times 550 = 66000 \ Rs.$