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

(i) Prove that △PQR ~ △PST.

(ii) Find area of △PQR : area of quadrilateral SRQT.

(i) Considering △PQR and △PST.

∠P = ∠P (Common angles)

∠PQR = ∠PST (Both are equal to 90°)

Hence, by AA axiom △PQR ~ △PST.

(ii) $\dfrac{\text{Area of △PQR}}{\text{Area of △PST}} = \dfrac{PQ^2}{PS^2} = \dfrac{5^2}{2^2} = \dfrac{25}{4}$

or,

$\Rightarrow \dfrac{\text{Area of △PQR}}{\text{Area of △PQR - Area of SRQT}} = \dfrac{25}{4}$

⇒ 4 Area of △PQR = 25 Area of △PQR - 25 Area of SRQT
⇒ 25 Area of SRQT = 25 Area of △PQR - 4 Area of △PQR
⇒ 25 Area of SRQT = 21 Area of △PQR

$\Rightarrow \dfrac{\text{Area of △PQR}}{\text{Area of SRQT}} = \dfrac{25}{21}$

Hence, area of △PQR : area of quadrilateral SRQT is 25 : 21.