Question

In triangle ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

Find:

1. area ΔAPO : area ΔABC.
2. area ΔAPO : area ΔCQO.

Answer 1

In ΔABC,

AP : PB = 2 : 3

PQ || BC and CQ || BA

∵ PQ || BC

∴ `(AP)/(PB) = (AO)/(OC) = 2/3`

i. In ΔAPQ ∼ ΔABC

∴ `(Area  ΔAPO)/(Area  ΔABC) = (AP^2)/(AB^2)`

= `(AP^2)/(AP + PB)^2`

= `(2)^2/(2 + 3)^2`

= `4/25`

∴ area ΔAPO : area ΔABC = 4 : 25

ii. In ΔAPO and ΔCQO

∠APO = ∠OQC  ...(Alternate angles)

∠AOP = ∠COQ  ...(Vertically opposite angles)

∴ ΔAPO ∼ ΔCQO  ...(AA axiom)

∴ `(Area  ΔAPO)/(Area  ΔCQO) = (AP^2)/(CQ^2)`

= `(AP^2)/(PB^2)`  {∵ PBCQ is a || gm}

= `(2)^2/(3)^2`

= `4/9`

∴ area ΔAPO : area ΔCQO = 4 : 9