If $\vartriangle ABC\sim\vartriangle QRP$, $\frac{ar( \vartriangle ABC)}{( \vartriangle QRP)}=\frac{9}{4}$, and $BC=15\ cm$, then find $PR$.

Given: $\vartriangle ABC\sim\vartriangle QRP$, $\frac{ar( \vartriangle ABC)}{( \vartriangle QRP)}=\frac{9}{4}$, and $BC=15\ cm$.

To do: To find $PR$.

Solution: 

$\because \vartriangle ABC\sim\vartriangle QRP$

$\frac{ar( \vartriangle ABC)}{ar( \vartriangle QRP)}=( \frac{BC}{PR})^{2}$

$\Rightarrow \frac{9}{4}=( \frac{15}{PR})^{2}$

$\Rightarrow \frac{225}{PR^{2}}=\frac{9}{4}$

$\Rightarrow PR^{2}=\frac{225\times4}{9}$ 

$\Rightarrow PR^{2}=100$

$\Rightarrow PR=\sqrt{100}$

$\Rightarrow PR=10\ cm$.

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Updated on: 10-Oct-2022

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