Question

In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.

Answer 1

We have,

PQ || BC

And `"AP"/"PB"=1/2`

In ΔAPQ and ΔABC

∠A = ∠A [Common]

∠APQ = ∠B [Corresponding angles]

Then, ΔAPQ ~ΔABC [By AA similarity]

By area of similar triangle theorem

`("area"(triangleAPQ))/("area"(triangleABC))="AP"^2/"AB"^2`

`rArr("area"(triangleAPQ))/(("area"(triangleAPQ)+"area"("trap.BPQC")))=1^2/3^2`   `["AP"/"PB"=1/2]`

⇒ 9ar (APQ) = ar(ΔAPQ) + ar(trap. BPQC)

⇒ 9ar (APQ) − ar(ΔAPQ) = ar(trap. BPQC)

⇒ 8ar(APQ) = ar(trap. BPQC)

`rArr("area"(triangleAPQ))/("area"("trap.BPQC"))=1/8`