In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of △DPC.

In △ABP and △DPC,

∠APB = ∠DPC (∵ vertically opposite angles are equal.)

∠ABP = ∠DCP (∵ angles in same segment are equal.)

△APB ~ △DPC. (By AA axiom)

We know that ratio of the area of similar triangles is the ratio of their corresponding sides.

$\therefore \dfrac{\text{Area of △APB}}{\text{Area of △DPC}} = \dfrac{AB^2}{CD^2} \\[1em] \Rightarrow \dfrac{24}{\text{Area of △DPC}} = \dfrac{64}{25} \\[1em] \Rightarrow \text{Area of △DPC} = \dfrac{24 \times 25}{64} \\[1em] \Rightarrow \text{Area of △DPC} = \dfrac{600}{64} \\[1em] \Rightarrow \text{Area of △DPC} = \dfrac{75}{8} \\[1em] \Rightarrow \text{Area of △DPC} = 9\dfrac{3}{8}. \\[1em]$

Hence, the area of △DPC = $9\dfrac{3}{8}$ cm².