Question

The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 2 : 3,

$Ar\left(\Delta CPD\right):Ar\left(\Delta APB\right)=?$

• A. 4 : 9
• B. 2 : 3
• C. 3 : 2
• D. 9 : 4

Answer 1

The correct option is D 9 : 4

Given: AP : CP = 2 : 3

$\Rightarrow$ $\frac{CP}{AP}=\frac{3}{2}$

In $\Delta APB$ and $\Delta DPC,$

$\angle APB=\angle DPC$ (vertically opposite angles)

$\angle PAB=\angle PCD$ (alternate angles)

$\therefore \Delta APB\sim \Delta CPD$ (by AA similarity criterion)

$\therefore \frac{\text{area}\Delta CPD}{\text{area}\Delta APB}=\frac{C{P}^2}{A{P}^2}=\frac{3^2}{2^2}=\frac{9}{4}$

$\Rightarrow$ area $\Delta CPD$ : area $\Delta APB=9:4$