Question

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.

1. Prove ΔPQR ∼ ΔSPR.
2. Find the lengths of QR and PS.
3. Find `(Area  of ΔPQR)/(Area  of ΔSPR)`

Answer 1

(i)

In ∆PQR and ∆SPR,

∠PSR = ∠QPR     ....given

∠PRQ = ∠PRS     ....common angle

⇒ ∆PQR ∼ ∆SPR  (AA Test)

(ii)

Find the lengths of QR and PS.

Since ∆PQR ∼ ∆SPR     ....from (i)

`(PQ)/(SP) = (QR)/(PR) = (PR)/(SR)`     ....(a)

⇒ `(QR)/(PR) = (PR)/(SR)`     ....from (a)

`(QR)/6 = 6/3`

`QR = (6 xx 6)/3`

∴ QR = 12 cm

⇒ `(PQ)/(SP) = (PR)/(SR)`     ....from (a)

`8/(SP) = 6/3`

`SP = (8 xx 3)/6`

∴ SP = 4 cm

(iii)

`(Area  of ΔPQR)/(Area  of ΔSPR) = (PQ^2)/(SP^2)`

`(Area  of ΔPQR)/(Area  of ΔSPR) = (8^2)/(4^2)`

`(Area  of ΔPQR)/(Area  of ΔSPR) = 64/16`

∴ `(Area  of ΔPQR)/(Area  of ΔSPR) = 4/1`

Hence,

1. ∆PQR ∼ ∆SPR
2. QR = 12 cm and SP = 4 cm
3. (Area of ΔPQR) : (Area of ΔSPR) = 4 : 1