The given figure shows a parallelogram ABCD with area 324 sq.cm. P is a point in AB such that AP : PB = 1 : 2. Find the area of △ APD.

Join BD.

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.

$\therefore \dfrac{\text{Area of △ APD}}{\text{Area of △ BPD}} = \dfrac{AP}{BP} \\[1em] \Rightarrow \dfrac{\text{Area of △ APD}}{\text{Area of △ BPD}} = \dfrac{1}{2} \\[1em] \Rightarrow \text{Area of △ BPD} = 2\text{Area of △ APD}.$

We know that,

The area of triangle is half that of a parallelogram on the same base and between the same parallels.

△ ABD and || gm ABCD lie on same base AB and between same parallel lines AB and DC.

∴ Area of △ ABD = $\dfrac{1}{2}$ Area of || gm ABCD = $\dfrac{1}{2} \times 324$ = 162 cm².

From figure,

⇒ Area of △ ABD = Area of △ APD + Area of △ BPD

⇒ 162 = Area of △ APD + 2 Area of △ APD

⇒ 3 Area of △ APD = 162

⇒ Area of △ APD = $\dfrac{162}{3}$ = 54 cm².

Hence, area of △ APD = 54 cm².