Question

In the given figure, $M$ and $N$ are the mid-points of the sides $DC$ and $AB$ respectively of the parallelogram $ABCD$. If the area of parallelogram $ABCD$ is $48c{m}^2$; then the area of the parallelogram $BNMC$ is equal to the area of the triangle $BEC$.

• A. True
• B. False

Answer 1

The correct option is A True

Parallelogram $ABCD$ and $△BEC$ have same base $BC$ and they are between same parallels $BC$ and $AD$.

$\therefore$ Area of $△BEC=\frac{1}{2}\text{area of parallelogram}ABCD$
$=\frac{1}{2}\times 48c{m}^2=24c{m}^2$

Since $N$ and $M$ are the midpoints of $AB$ and $DC$ respectively,

$\Rightarrow AN=NB,DM=MC$

$NB=\frac{1}{2}AB,MC=\frac{1}{2}DC$

But $AB=DC$ ( opposite side of parallelogram $ABCD$)

$\therefore NB=MC$ , also $NB\parallel MC$

$\therefore NBCM$ is a parallelogram

$\therefore A\left(NBCM\right)=\frac{1}{2}A\left(ABCD\right)=A\left(△BEC\right)$

$\therefore △BEC$ is equal in area of parallelogram $BNMC$