If ABC-AORP, arRE.and BC -15 cm, …

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Sia ? 6 years, 6 months ago
{tex}\frac{\operatorname{arc}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{QRP})}{/tex} = {tex}\left(\frac{B C}{R P}\right)^{2}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{9}{4}{/tex} = {tex}\left(\frac{15}{P R}\right)^{2}{/tex}
{tex}\Rightarrow{/tex} PR = 10 cm
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