In a parallelogram ABCD, point P lies in DC such that DP : PC = 3 : 2. If area of Δ DPB = 30 sq.cm, find the area of the parallelogram ABCD.

Given,

DP : PC = 3 : 2

We know that,

Ratio of the area of triangles with same vertex and bases along the same line is equal to the ratio of their respective bases.

$\Rightarrow \dfrac{\text{Area of Δ DPB}}{\text{Area of Δ PCB}} = \dfrac{DP}{PC} \\[1em] \Rightarrow \dfrac{\text{Area of Δ DPB}}{\text{Area of Δ PCB}} = \dfrac{3}{2} \\[1em] \Rightarrow \text{Area of Δ PCB} = \dfrac{2}{3} \times \text{Area of Δ DPB} \\[1em] \Rightarrow \text{Area of Δ PCB} = \dfrac{2}{3} \times 30 \\[1em] \Rightarrow \text{Area of Δ PCB} = 20 \text{ cm}^2.$

From figure,

Area of Δ CDB = Area of Δ PCB + Area of Δ DPB = 20 + 30 = 50 cm².

Since, diagonal divides a || gm into two triangles of equal area.

∴ Area of || gm ABCD = 2 Area of Δ CDB = 2 × 50 = 100 cm².

Hence, area of \\ gm ABCD = 100 cm².