Question

In a parallelogram ABCD, P is a point on DC such that DP : PC = 2 : 3. If area of ΔDPB = 40 cm², find the area of ◻ABCD.

Answer 1

Given:

Parallelogram ABCD with P on DC so that DP : PC = 2 : 3.

Area (ΔDPB) = 40 cm².

Step-wise calculation:

1. Triangles DPB and CPB have the same altitude from B to line DC.

So, their areas are proportional to their bases DP and PC.

Hence, Area (ΔDPB) : Area (ΔCPB)

= DP : PC

= 2 : 3

2. Let the common part be k.

Then Area (ΔDPB) = 2k

= 40

⇒ k = 20

So, Area (ΔCPB) = 3k = 60.

3. Area of ΔBCD with base DC and vertex B

= Area (ΔDPB) + Area (ΔCPB)

= 40 + 60

= 100 cm²

4. A diagonal (BD) divides a parallelogram into two equal-area triangles.

So, Area (parallelogram ABCD)

= 2 × Area (ΔBCD) 

= 2 × 100

= 200 cm²