Question

In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q.

If area of ΔCPQ = 20 cm², find:

1. area of ΔBPQ.
2. area of ΔCDP.

Answer 1

Given:

ABCD is a parallelogram. P is a point on BC such that, BP : PC = 1 : 2 and DP produced meets AB produced at Q.

Also,

ar⁡(△CQP) = 20 cm²

Since BP : PC = 1 : 2, point P divides BC in the ratio 1 : 2.

Hence, the perpendicular distance of P from AB is `1/3` of the distance between the parallel lines AB and DC.

Let, AB = DC = b and height of parallelogram = h

So, ar(ABCD) = bh

Now, △CQP has a base on the same line as AB, and from the figure, its area is `1/6` of the area of the parallelogram ABCD. Therefore,

ar(△CQP) = `1/6 bh`

But given, ar(△CQP) = 20

So, `1/6 bh = 20`

bh = 120

Hence, ar(ABCD) = 120 cm²

(i) Area of △BPQ:

In △BPQ,

• base BQ = `1/2 AB` = `b/2`
• height of P above AB = `h/3`

Therefore,

ar(△BPQ) = `1/2 xx b/2 xx h/3 = (bh)/12`

Substituting bh = 120,

ar(△BPQ) = `120/12`

∴ ar(△BPQ) = 10 cm²

(ii) Area of △CDP:

In △CDP,

• base CD = b
• perpendicular distance of P from CD = `(2h)/3`

Therefore,

ar⁡(△CDP) = `1/2 xx b xx (2h)/3 = (bh)/3`

Substituting bh = 120,

ar(△CDP) = `120/3`

∴ ar(△CDP) = 40 cm²