Question

In the given figure, $ABCD$ is a parallelogram, then $ar(△AFB)$ is

Answer 1

From the given figure $AB$ is the common base to the parallelogram $\text{ABCD}$ and $∆\text{AFB}$, and they are between the parallel line segment $\text{AB}$ and $\text{CD}$.

As we know that the area of a triangle is half of the area of a parallelogram on the same base and between the same parallels.

$\therefore ar(△AFB)$$=\frac{1}{2}ar\left(\parallel gmABCD\right)$

$ar(△AFB)=\frac{1}{2}\times 4\times 4=8c{m}^2$ (Using formula, area of parallelogram$=base\times height$)

Thus the area of $△AFB$ is $8c{m}^2$.