Each side of a square ABCD is 12 cm. A point P lies on side DC such that area of △ ADP : area of trapezium ABCP = 2 : 3. Find DP.

Given: AB = BC = CD = DA = 12 cm

Area of △ ADP : Area of trapezium ABCP = 2 : 3

Area of triangle = $\dfrac{1}{2}$ x base x height

Area of △ ADP = $\dfrac{1}{2}$ x DP x AD

= $\dfrac{1}{2}$ x DP x 12

Area od trapezium = $\dfrac{1}{2}$ x (sum of parallel sides) x height

Area of trapezium ABCP = $\dfrac{1}{2}$ x (AB + PC) x BC

= $\dfrac{1}{2}$ x (12 + PC) x 12

Now, substituting the values:

$\Rightarrow \dfrac{\text{area of △ ADP}}{\text{area of trapezium ABCP}} = \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{\dfrac{1}{2} \times DP \times 12}{\dfrac{1}{2} \times (12 + PC) \times 12} = \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{\cancel{\dfrac{1}{2}} \times DP \times \cancel{12}}{\cancel{\dfrac{1}{2}} \times (12 + PC) \times \cancel{12}} = \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{DP}{12 + PC} = \dfrac{2}{3}$

From figure, DC = DP + PC

⇒ 12 = DP + PC

⇒ PC = 12 - DP

Substituting this value,

$\Rightarrow \dfrac{DP}{12 + (12 - DP)} = \dfrac{2}{3}\\[1em] \Rightarrow \dfrac{DP}{24 - DP} = \dfrac{2}{3}\\[1em] \Rightarrow 3DP = 2(24 - DP)\\[1em] \Rightarrow 3DP = 48 - 2DP\\[1em] \Rightarrow 3DP + 2DP = 48\\[1em] \Rightarrow 5DP = 48\\[1em] \Rightarrow DP = \dfrac{48}{5}\\[1em] \Rightarrow DP = 9.6$

Hence, the length of DP = 9.6 cm.