PQR is a triangle, S is a point on the side QR of ΔPQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

(i) Prove ΔPQR ∼ ΔSPR.

(ii) Find the length of QR and PS.

(iii) Find $\dfrac{\text{area of ΔPQR}}{\text{area of ΔSPR}}$.

(i) In ΔPQR and ΔSPR,

⇒ ∠PSR = ∠QPR [Given ]

⇒ ∠PRQ = ∠PRS [Common angle]

∴ ΔPQR ∼ ΔSPR by AA similarity.

Hence, proved that ΔPQR ∼ ΔSPR.

(ii) Since, ΔPQR ∼ ΔSPR and corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{QR}{PR} = \dfrac{PR}{SR} \\[1em] \Rightarrow \dfrac{QR}{6} = \dfrac{6}{3} \\[1em] \Rightarrow \dfrac{QR}{6} = \dfrac{36}{3} = 12 \text{ cm.}$

Also,

$\Rightarrow \dfrac{PQ}{SP} = \dfrac{PR}{SR} \\[1em] \Rightarrow \dfrac{8}{SP} = \dfrac{6}{3} \\[1em] \Rightarrow SP = \dfrac{8 \times 6}{3} \\[1em] \Rightarrow SP = \dfrac{24}{3} = 4 \text{ cm.}$

Hence, QR = 12 cm and PS = 4 cm.

(iii) We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$\therefore \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPQR}} = \Big(\dfrac{PQ}{SP}\Big)^2 \\[1em] = \Big(\dfrac{8}{4}\Big)^2 \\[1em] = (2)^2 \\[1em] = 4$

Hence, $\dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPQR}} = \dfrac{4}{1}$