Question

In the given figure, $ST\parallel RQ,PS=3$ cm and $SR=4$cm. Find the ratio of the areas of $△PST$ to the area of $△PRQ$

Answer 1

In figure ST || QR

and PS= 3 cm & RS= 4cm

In $\Delta PST$ & $PQR$

$QPR=\angle TPS$ [Common]

$\angle PRQ=\angle PST$ [corresponding]

$\Delta PQR\sim \Delta PST$ [AA similarity]

Now

In similar triangles ratios of the area of triangles is equal to the ratio of the square of corresponding sides.

$\frac{Ar.of\Delta PST}{Ar.of\Delta PRQ}$ $(\frac{3}{3+4}{)}^2$ $=\frac{9}{49}$

Hence $[\frac{Ar.of\Delta PST}{Ar.of\Delta PRQ}=\frac{9}{49}]$