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

(i) Prove △PQR ~ △SPR.

(ii) Find the lengths of QR and PS.

(iii) $\dfrac{\text{area of △PQR}}{\text{area of △SPR}}$

(i) In △PQR and △SPR,

⇒ ∠PSR = ∠QPR [Given]

⇒ ∠PRQ = ∠PRS [Common angle]

∴ △PQR ~ △SPR [By AA]

Hence, proved that △PQR ~ △SPR.

(ii) Since, △PQR ~ △SPR and corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{QR}{PR} = \dfrac{PR}{SR} \\[1em] \Rightarrow \dfrac{QR}{6} = \dfrac{6}{3} \\[1em] \Rightarrow QR = \dfrac{36}{3} = 12 \text{ cm}. \\[1em] \text{ Also}, \\[1em] \Rightarrow \dfrac{PQ}{SP} = \dfrac{PR}{SR} \\[1em] \Rightarrow \dfrac{8}{SP} = \dfrac{6}{3} \\[1em] \Rightarrow SP = \dfrac{24}{6} = 4 \text{ cm}.$

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.

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

Hence, $\dfrac{\text{area of △PQR}}{\text{area of △SPR}}$ = 4 : 1.