Question
$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}, \mathrm{A}(\triangle \mathrm{ABC})=16, \mathrm{~A}(\triangle \mathrm{PQP})=25$, then find the value of ratio $\frac{\mathrm{AB}}{\mathrm{PQ}}$.

Answer

$: \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$
$\therefore \frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\text{ .......... theorem of areas of similar triangles}$
$\therefore \frac{16}{25}=\frac{\mathrm{AB}^2}{\mathrm{PQ}^2} \quad \therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{5}\text{.......... taking square roots} $
theorem of areas of similar triangles taking square roots

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free