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 length of QR and PS.
3. `"area of ΔPQR"/"area of ΔSPR"`

Answer 1

i. In ΔPQR and ΔSPR, we have

∠QPR = ∠PSR   ...(Given)

∠PRQ = ∠PRS  ...(Common)

So, by AA-axiom similarity, we have

ΔPQR ∼ ΔSPR  ...(Proved)

ii. Since ΔPQR ∼ ΔSPR  ...(Proved)

`=> (PQ)/(SP) = (QR)/(PR) = (PR)/(SR)`

Consider `(QR)/(PR) = (PR)/(SR)`  ...(From 1)

`=> (QR)/6 = 6/3`

`=> QR = (6 xx 6)/3 = 12  cm`

Also, `(PQ)/(SP) = (PR)/(SR)`

`=> 8/(SP) = 6/3`

`=>8/(SP) = 2`

`=> SP = 8/2 = 4  cm`

iii. `"area of ΔPQR"/"area of ΔSPR" = (PQ^2)/(SP^2)`

= `8^2/4^2`

= `64/16`

= `4/1`

`"area of ΔPQR"/"area of ΔSPR" = 4 : 1`