Question 1: In $\triangle ABC, \angle ABC = \angle DAC, AB= 8 \ cm,$ $AC = 4 \ cm \ and \ AD = 5 \ cm$.

(i) Prove that $\triangle ACD$ is similar to $\triangle BCA$.

(ii) Find $BC \ and \ CD$.

(iii) Find $area \ of \ \triangle ACD : area \ of \ \triangle ABC$   [2014]

(i) In $\triangle ACD \ and\ \triangle BCA$

$\angle C = \angle C$ (common angle)

$\angle ABC = \angle CAD$  (given)

$\triangle ACD \sim \triangle BCA$ (AAA postulate)

(ii) Since $\triangle ACD \sim \triangle BCA$

$\frac{AC}{ BC} = \frac{CD}{ CA} = \frac{AD}{ BA}$

$\frac{4 }{ BC}= \frac{CD}{ 4} = \frac{5}{ 8}$

$\Rightarrow BC = \frac{8}{ 5} \times 4 = 6.4 \ cm$

$\Rightarrow CD = \frac{5}{ 8} \times 4 = 2.5 \ cm$

(iii) Since $\triangle ACD \sim \triangle ABC$

Therefore $\frac{Area \triangle ACD}{Area \triangle ABC} = \frac{AC^2}{AB^2} = \frac{4^2}{8^2} = \frac{1}{4}$

Hence $Area \triangle ACD : Area \triangle ABC= 1:4$

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Question 2: In the given triangle $P, Q \ and \ R$ are mid points of sides $AB, BC \ and \ AC$ respectively. Prove that $\triangle PQR \sim \triangle ABC$.

In $\triangle ABC, PR \parallel BC$

Consider $\triangle ABC \ and\ \triangle PAR$

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

$\angle ARP =\angle ACB$ alternate)

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

$\frac{PR}{BC}=\frac{AP}{AB}$

Since $P$ is the mid point of $AB$

$AB = 2 AP$

$\frac{PR}{BC}=\frac{1}{2}$

Similarly,  we can prove $\frac{PQ}{AC}=\frac{1}{2} and \frac{RQ}{AB}=\frac{1}{2}$

Therefore $\frac{PR}{BC}=\frac{PQ}{AC}=\frac{RQ}{AB}$

Therefore $\triangle ABC \sim \triangle PQR$ (SSS postulate)

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Question 3: In the following figure $AD \ and \ CE$ are medians of $\triangle ABC. DF \parallel CE$. Prove that :

(i) $EF=FB$

(ii) $AG : GD = 2:1$

(i) Consider $\triangle BDF \ and\ \triangle BCE$

$DF \parallel CE$

$\angle BDF = \angle BCE$ (alternate angles)

$\angle BFD = \angle BEC$ (alternate angles)

Therefore $\triangle BDF \sim \triangle BCE$ (AAA postulate)

$\frac{BD}{BC}=\frac{BF}{BE}$

$\frac{BD}{2BD}=\frac{BF}{BE}$

$\Rightarrow 2 BF = BE$

$\Rightarrow BF = FE$

(ii)  Consider $\triangle AFD \ and \ \triangle AEG$

$FD \parallel EG$

$\angle DFA = \angle GEA$ (alternate angles)

$\angle FDA =\angle EGA$ (alternate angles)

Therefore $\triangle AFD \sim \triangle AEG$ (AAA postulate)

$\frac{AG}{GD}= \frac{AE}{FE}$ By basic proportionality theorem

Given $AE = EB$

$AE = 2 FE \Rightarrow \frac{AE}{FE} = 2$

$\frac{AG}{GD}= \frac{2}{1}$

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Question 4: In the given figure $\triangle ABC \sim \triangle PQR$. $AM \ and \ PN$ are altitudes  where as $AX \ and \ PY$ are medians. Prove:

$\frac{AM}{PN} = \frac{AX}{PY}$

(Since $\triangle ABC \sim \triangle PQR$

$\frac{AB}{PQ}= \frac{AC}{PR} = \frac{BC}{QR}$

Given $AX \ and \ PY$ are medians

Therefore $2BX = BC$ and $2 YR = QR$

$\frac{AB}{PQ}= \frac{BC}{QR} = \frac{2BX}{2QY} = \frac{BX}{QY}$  … … … … (i)

Consider $\triangle ABM \ and\ \triangle PQN$

$\angle ABN = \angle PQN$  Since $\triangle ABC \sim \triangle PQR$

$\angle AMB = \angle PNQ = 90^o$ (alternate angles)

Therefore $\triangle ABM \sim \triangle PQN$ (AAA postulate)

$\frac{AB}{PQ}= \frac{AM}{PN}$ … … … … (ii)

From (i) and (ii) we get

$\frac{AB}{PQ}= \frac{AM}{PN}$

and $\angle ABX = \angle PQY$

Therefore $\triangle ABX \sim \triangle PQY$

$\Rightarrow \frac{AB}{PQ}= \frac{AX}{PY}$

Hence $\frac{AM}{PN} = \frac{AX}{PY}$

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Question 5: Two similar triangles are equal in area. Prove that the triangles are congruent.

Let the two triangles be $ABC \ and \ PQR$

Since the two triangles are similar, We know

$\frac{Ar. \ \triangle ABC}{Ar. \ \triangle PQR} = \frac{AB^2}{PQ^2}=\frac{BC^2}{Qr^2}=\frac{AC^2}{PR^2}$

Since the are of the two triangles is equal

$\Rightarrow AB=PQ, BC=QR \ and \ AC = PR$

Therefore $\triangle ABC \cong \triangle PQR$

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Question 6: The ratio between the altitudes of two similar triangles is $3:5$. Write the ratios between their (i) medians (ii) perimeters (iii) areas.

The ratio of the altitude of two similar triangles  is the same as the ratio of their sides. Given ratio $= 3:5$

(i) Ratio between their median $= 3:5$

(ii) Ratio between their perimeter $= 3:5$

(iii) Ratio between their areas $= 3^2:5^2 = 9:25$

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Question 7: The ratio between the altitudes of two similar triangles is $16:25$. Find the ration between their: (i) perimeters (ii) altitudes (iii) medians

The ratio between the altitudes of two similar triangles is $16:25$.

This means that the ratio of the sides of the triangles = 4:5

(i)  Ratio between their perimeter $= 4:5$

(ii) Ratio between their altitude $= 4:5$

(iii) Ratio between their median $= 4:5$

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Question 8: The following figure shows a $\triangle PQR$ in which $XY \parallel QR$. If $PX:XQ=1:3$ and $QR =9 \ cm$, find the length of $XY$.

Given $PX:XQ=1:3$ and $QR =9 \ cm$

Consider $\triangle PXY \ and \ \triangle PQR$

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

$\angle PYX =\angle PRQ$ (alternate angles)

Therefore $\triangle PXY \sim \triangle PQR$ (AAA postulate)

$\frac{PX}{PQ}= \frac{XY}{QR}$

$\frac{PX}{PX+XQ}= \frac{XY}{QR}$

$\frac{PX/XQ}{PX/XQ+1}= \frac{XY}{QR}$

$\frac{1}{4}= \frac{XY}{9}$

$\Rightarrow XY = \frac{9}{P4}$

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Question 9: In the following figure, $AB, CD \ and \ EF$ are parallel lines. $AB=6 \ cm, \ CD= y \ cm, \ EF=10 \ cm,$ $AC = 4 \ cm \ and$ $\ CF= x \ cm$. Calculate: $x \ and \ y$  [1985]

Consider $\triangle FDC \ and\ \triangle FBA$

$\angle FDC = \angle FDA$ (Corresponding angles)

$\angle DFC = \angle BFA$ (common angle)

$\triangle FDC \sim \triangle FBA$ (AAA Postulate)

Therefore $\frac{CD}{ AB} = \frac{FC}{ FA}$

$\frac{y}{ 6} = \frac{x}{ x+4}$

Now consider $\triangle FCE \ and \ \triangle ACB$

$\angle FCE = \angle ACB$ (Vertically opposite angles)

$\angle CFE = \angle CAB$ (Alternate angles)

$\triangle FCE \sim \triangle ACB$ (AAA postulate)

Therefore $\frac{FC}{ AC} = \frac{EF}{ AB}$

$x = \frac{10}{ 6} \times 4 = 6.67 \ cm$

Also $y= \frac{6.67 }{ 6.67+4} \times 6 = 3.75 \ cm$

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Question 10: On a map, drawn to a scale of $1:20000$, a rectangular plot of land $ABCD$ has  $AB = 24 \ cm and BC = 32 \ cm$. Calculate:

(i) the diagonal distance  of the plot in km

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

$k = \frac{1}{20000}$

Length of AB on map $= k \times$ actual length of AB

Actual length of  $AB = 24 \times 20000 \ cm = 4.8 \ km$

Similarly Actual length of $BC = 32 \times 20000 \ cm = 6.4 \ km$

(i) Therefore the diagonal $= \sqrt{4.8^2+6.4^2} = 8 \ km$

(ii) Area of the plot $= 4.8 \times 6.4 = 30.72 \ km$

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