Question

In the given figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR.

Answer 1

Given: In ΔPQR, S and T are the points on the sides PQ and PR respectively such that PT = 2cm, TR = 4cm and ST is parallel to QR.

To find: Ratio of areas of ΔPST and ΔPQR

$$In∆PSTand∆PQR,$$
$$\mathrm{\angle }PST=\mathrm{\angle }Q\left(\text{Corresponding angles}\right)$$
$$\mathrm{\angle }P=\mathrm{\angle }P\left(\text{Common}\right)$$
$$\therefore ∆PST∆PQR\left(AA\text{Similarity}\right)$$

Now, we know that the areas of two similar triangles are in the ratio of the squares of the corresponding sides. Therefore,

${\displaystyle \frac{Area(\Delta PST)}{Area(\Delta PQR)}=\frac{P{T}^2}{P{R}^2}}$

${\displaystyle \frac{Area(\Delta PST)}{Area(\Delta PQR)}=\frac{P{T}^2}{{(PT+TR)}^2}}$

${\displaystyle \frac{Area(\Delta PST)}{Area(\Delta PQR)}=\frac{2^2}{{(2+4)}^2}}$

${\displaystyle \frac{Area(\Delta PST)}{Area(\Delta PQR)}=\frac{4}{36}=\frac{1}{9}}$