In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find

(i) area of || gm ABCD

(ii) DP : PC.

(i) Since,

∆APB and || gm ABCD have same base AB and are between same parallel lines AB and DC. So,

area of ∆APB = $\dfrac{1}{2}$ area of ||gm ABCD

From figure,

⇒ $\dfrac{1}{2}$ area of ||gm ABCD = area of (∆DAP + ∆BCP)

⇒ $\dfrac{1}{2}$ area of ||gm ABCD = 25 + 15 = 40

⇒ area of ||gm ABCD = 2 × 40 = 80 cm².

Hence, area of ||gm ABCD = 80 cm².

(ii) From figure,

∆DAP and ∆BCP are on the same base CD and between the same parallel lines CD and AB.

$\Rightarrow \dfrac{\text{Area of ∆DAP}}{\text{Area of ∆BCP}} = \dfrac{DP}{PC} \\[1em] \Rightarrow \dfrac{25}{15} = \dfrac{DP}{PC} \\[1em] \Rightarrow \dfrac{DP}{PC} = \dfrac{5}{3}.$

Hence, DP : PC = 5 : 3.