Question

In the given figure ABCD is a parallelogram, P is the point on CD. What is the area of $\Delta APB$ if the area of $\Delta DPA$ is $25c{m}^2$ and $\Delta APC$ is $30c{m}^2$.

• A. $30c{m}^2$
• B. $110c{m}^2$
• C. $55c{m}^2$
• D. $25c{m}^2$

Answer 1

The correct option is C

$55c{m}^2$

Area of $\Delta ADC=$ Area of $\Delta DPA$ + Area of $\Delta APC$
$\Rightarrow$ Area of $\Delta ADC=30+25=55c{m}^2$
AC is the diagonal of the parallelogram ABCD and it divides the area of ABCD into two equal parts.
$\Rightarrow$ Area of $\Delta ADC=$ Area of $\Delta ACB$
So, Area of $\Delta ACB=55c{m}^2$
Triangles in parallelogram having same bases having same areas
$\Rightarrow$ Area of $\Delta APB$ = Area of $\Delta ACB=55c{m}^2$