Question

ABCD is a parallelogram. P is any point on CD. If ar($△$DPA) = 35 $c{m}^2$ and ar($△$APC) = 15 $c{m}^2$, then area($△$APB) is

• A. 50 $c{m}^2$
• B. 15 $c{m}^2$
• C. 70 $c{m}^2$
• D. 35 $c{m}^2$

Answer 1

The correct option is A 50 $c{m}^2$

Given: ABCD is a parallelogram.
area($△$DPA) = 35 $c{m}^2$
area($△$APC) = 15 $c{m}^2$
To find: area($△$APB)

$Area\left(\Delta ACD\right)=Area\left(\Delta DPA\right)+Area\left(\Delta APC\right)$
$=35c{m}^2+15c{m}^2$
$=50c{m}^2$

Now, ABCD is a parallelogram.
AC is the diagonal of parallelogram ABCD.
A diagonal of a parallelogram divides it into two congruent triangles.

$\Rightarrow Area\left(△ACD\right)=Area\left(△ACB\right)=50c{m}^2$

$△ACB$ and $△APB$ are on same base AB and between same parallels AB and DC.
Hence,
$Area\left(\Delta APB\right)=Area\left(\Delta ACB\right)$
$=50c{m}^2$