Question

In fig. $S$ and $T$ are the points on the sides $PQ$ and $PR$ respectively of $\Delta PQR$, such that $PT=4$ cm, $TR=4$ cm & $ST\parallel QR$. Find the ratio of areas of $\Delta PST$ & $\Delta PQR$.

Answer 1

Given:

$ST\parallel QR$

PT= 4 cm

TR = 4cm

In $△PST$ and $△PQR$,

$\angle SPT=\angle QPR$(Common)

$\angle PST=\angle PQR$ (Corresponding angles)

$△PST\sim △PQR$(By AA similarity criterion)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{area△PST}{area△PQR}=\frac{P{T}^2}{P{R}^2}$

$\frac{area△PST}{area△PQR}=\frac{4^2}{(PT+TR{)}^2}$

$\frac{area△PST}{area△PQR}=\frac{16}{(4+4{)}^2}=\frac{16}{8^2}=\frac{16}{64}=\frac{1}{4}$

Thus, the ratio of the areas of $△PST$ and $△PQR$ is

$1:4$.