ABCD is a square with each side 12 cm. P is a point on BC such that area of Δ ABP : area of trapezium APCD = 1 : 5. Find the length of CP.

Square ABCD is shown in the figure below:

Given:

$\dfrac{\text{Area of Δ ABP}}{\text{Area of trapezium APCD}} = \dfrac{1}{5}$

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

= $\dfrac{1}{2}$ x 12 x (12 - CP)

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

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

$⇒ \dfrac{\dfrac{1}{2} \times 12 \times (12 - CP)}{\dfrac{1}{2} \times (12 + CP) \times 12} = \dfrac{1}{5}\\[1em] ⇒ \dfrac{\cancel{\dfrac{1}{2}} \times 12 \times (12 - CP)}{\cancel{\dfrac{1}{2}} \times (12 + CP) \times 12} = \dfrac{1}{5}\\[1em] ⇒ \dfrac{\cancel{12} \times (12 - CP)}{ (12 + CP) \times \cancel{12}} = \dfrac{1}{5}\\[1em] ⇒ \dfrac{(12 - CP)}{ (12 + CP)} = \dfrac{1}{5}\\[1em] ⇒ 5 \times (12 - CP) = 1 \times (12 + CP)\\[1em] ⇒ 60 - 5CP = 12 + CP\\[1em] ⇒ 60 - 12 = 5CP + CP\\[1em] ⇒ 48 = 6CP\\[1em] ⇒ CP = \dfrac{48}{6}\\[1em] ⇒ CP = 8 \text { cm}$

Hence, the length of CP is 8 cm.