In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm.

(i) Prove that ΔPQR ∼ ΔPST.

(ii) Find area of ΔPQR : Area of quadrilateral SRQT.

(i) Considering ΔPQR and ΔPST.

∠P = ∠P [Common angles]

∠PQR = ∠PST [Both are equal to 90°]

∴ ΔPQR ∼ ΔPST (By A.A. axiom)

Hence, proved that ΔPQR ∼ ΔPST.

(ii) We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPST}} = \dfrac{{PQ}^2}{{PS}^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPST}} = \dfrac{5^2}{2^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPQR} - \text{Area of SRQT}} = \dfrac{25}{4} \\[1em] \Rightarrow 4\text{ Area of ΔPQR} = 25 (\text{ Area of ΔPQR} - \text{Area of SRQT}) \\[1em] \Rightarrow 4\text{ Area of ΔPQR} = 25 \text{ Area of ΔPQR} - 25\text{Area of SRQT} \\[1em] \Rightarrow 25\text{Area of SRQT} = 25 \text{ Area of ΔPQR} - 4\text{ Area of ΔPQR}\\[1em] \Rightarrow 25\text{Area of SRQT} = 21 \text{ Area of ΔPQR} \\[1em] \Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of SRQT}} = \dfrac{25}{21}.$

Hence, area of ΔPQR : Area of quadrilateral SRQT = 25 : 21.