Question

ABCD ia a parallelogram. P is any point on CD. If $ar\left(\Delta DPA\right)=15c{m}^{2}andar\left(\Delta APC\right)=20c{m}^2$, then find the ar $\left(\Delta APB\right).$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Given, P is any point on CD
$ar\left(\Delta DPA\right)=15c{m}^2$
$ar\left(\Delta APC\right)=20c{m}^2$

AC is a diagonal of parallelogram ABCD, then
$\Delta ADC=\Delta ABC$ ------(i)
We know, if parallelogram and triangle o the same base and between same parallels, the area of triangle is equal to half of the area of parallelogram.
$\Delta APBand\Delta ACB$ are on same base AB and between same parallels.
So, ar $\Delta APB=\Delta ACB$ ----- (ii)
We know , from fig
$ar\left(\Delta ACB\right)=ar\left(\Delta APD\right)+ar\left(\Delta APC\right)$
from eq (2) and (3),
$ar\left(\Delta APB\right)=15c{m}^2+20c{m}^2$
$ar\left(\Delta APB\right)=35c{m}^2$
Hence, correct option is (c).