Question

In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 $c{m}^2$.
Calculate : (i) area of triangle CDP, (ii) area of parallelogram ABCD.

Answer 1

Draw CL ⊥ DQ

Area of △ CPQ = 1/2 × base × altitude

Area of △ CPQ = 1/2 × PQ × CL ............(1)

Area of △ DCP = 1/2 × base × altitude

Area of △ DCP = 1/2 × DP × CL ............(2)

Dividing (1) by (2), we get,
Area of △ CPQ/Area of △ DCP = PQ/DP ...............(3)

Since DC ∥ AQ and BC is a transversal
⇒ ∠ DCP = ∠ QBP
[Alternate interior angles]

In △ DCP and △ QBP,
∠ DCP = ∠QBP
[Proved above]

∠ DPC = ∠ QPB
[Vertically opposite angles]
⇒ △ DCP ~ △ QBP [AA similarity]
⇒ PC/BP = DP/PQ = CD/BQ [Corresponding sides of similar △'s are proportional]
⇒ PC/BP = DP/PQ
⇒ 2/1 = DP/PQ
⇒ PQ/DP = 1/2 ............(4)
Now, from (3),
Area of △ CPQ/Area of △ DCP = PQ/DP = 1/2
⇒ 20/Area (△ DCP) = 1/2
⇒ Area (△ DCP) = 40 cm^2

Since ∥gm ABCD & △ DCQ have same base DC & between same parallels DC & AQ, then,
area of ∥gm ABCD = 2 × ar (△ DCQ) = 2 × [ar (△ DCP) + ar (△ CPQ)] = 2 × [40 + 20]
= 2 × 60 = 120 cm^2