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.

Find -

(i) $\Delta APB:\Delta CPB$

(ii) $\Delta DPC:\Delta APB$

(iii) $\Delta ADP:\Delta APB$

• A. (i) 3 : 2, (ii) 4 : 9, (iii) 3 : 2
• B. (i) 2 : 3, (ii) 9 : 4, (iii) 3 : 2
• C. (i) 1 : 2, (ii) 3 : 2, (iii) 2 : 3
• D. (i) 9 : 4, (ii) 2 : 3, (iii) 5 : 3

Answer 1

The correct option is B

(i) 2 : 3, (ii) 9 : 4, (iii) 3 : 2

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

(i) Now in $\Delta APB$ and $\Delta CPB,$

$\because$ These triangles have same vertex and their bases are in the same straight line

$\therefore$ area $\Delta APB$: area $\Delta CPB=AP:PC=2:3$

Or $\Delta APB:\Delta CPB=2:3$

(ii) In $\Delta APB$ and $\Delta DPC.$

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

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

$\therefore \Delta APB\sim \Delta DPC$ (AA postulate)

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

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

Or $\Delta DPC:\Delta APB=9:4$

(iii) In $\Delta ADP$ and $\Delta APB,$

$\because$ There have the same vertex and their bases are in the same straight line.

$\therefore$ area $\Delta ADP$ : area $\Delta APB=DP:PB$

But PC : AP = 3 : 2

$\therefore \Delta ADP:\Delta APB=3:2$