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The corresponding sides of $\triangle ABC$ and $\triangle PQR$ are in the ratio $3 : 5$. AD$\perp$BC and PS$\perp$QR as shown in the following figures :
(i) Prove that $\triangle ADC \sim \triangle PSR$
(ii) If $AD = 4$ cm, find the length of PS.
(iii) Using (ii) find ar ($\triangle ABC$) : ar ($\triangle PQR$)
(i) Prove that $\triangle ADC \sim \triangle PSR$
(ii) If $AD = 4$ cm, find the length of PS.
(iii) Using (ii) find ar ($\triangle ABC$) : ar ($\triangle PQR$)
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As, $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{3}{5}$
$\Rightarrow \triangle ABC \sim \triangle PQR$
$\angle C = \angle R$
(i) In $\triangle ADC$ and $\triangle PSR$,
$\angle ADC = \angle PSR$
and $\angle C = \angle R$
$\therefore \triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{AC}{PR} = \frac{3}{5}$
$\Rightarrow \frac{4}{PS} = \frac{3}{5}$
$\Rightarrow PS = \frac{20}{3}$ cm
(iii) $\frac{\text{ar (}\triangle ABC)}{\text{ar (}\triangle PQR)} = \frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PS}$
$= \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
$\therefore$ ar ($\triangle ABC$): ar ($\triangle PQR$) = $9 : 25$
$\Rightarrow \triangle ABC \sim \triangle PQR$
$\angle C = \angle R$
(i) In $\triangle ADC$ and $\triangle PSR$,
$\angle ADC = \angle PSR$
and $\angle C = \angle R$
$\therefore \triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{AC}{PR} = \frac{3}{5}$
$\Rightarrow \frac{4}{PS} = \frac{3}{5}$
$\Rightarrow PS = \frac{20}{3}$ cm
(iii) $\frac{\text{ar (}\triangle ABC)}{\text{ar (}\triangle PQR)} = \frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PS}$
$= \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
$\therefore$ ar ($\triangle ABC$): ar ($\triangle PQR$) = $9 : 25$