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. i. ProveΔPQR∼ Δ
2. Find the lengths of QR and PS.
3. `(Area   of DeltaPQR)/(area   of Delta SPR)`

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 = 12 cm`

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

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

`=> SP = (8 xx 3)/6 = 4cm`

iii. `(area   of DeltaPQR)/(area   of Delta SPR) = (PQ^2)/(SP^2) = (8^2)/(4^2) = 64/16 = 4`