Question

The area of the parallelogram ABCD is $90c{m}^2$(see Figure). Find (i) $ar\left(ABEF\right)$(ii) $ar\left(ABD\right)$ (iii) $ar\left(BEF\right)$

Answer 1

According to the given details :

Area of parallelogram, $ABCD=90c{m}^2$

(i) We know that,

Parallelograms on the same base and between the same parallel lines are equal in areas.

The parallelograms $ABCD$ and $ABEF$ are on the same base $AB$ and between the same parallels $AB$and $CF$.

Therefore, $ar\left(ABEF\right)=ar\left(ABCD\right)=90c{m}^2$

(ii) We know that,

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

$\Delta ABD$and parallelogram $ABCD$ are on the same base$AB$ and between the same parallels $AB$ and $CD$.

Therefore, $ar\left(\Delta ABD\right)=\frac{1}{2}ar\left(ABCD\right)$

$=\frac{1}{2}\times 90c{m}^2=45c{m}^2$

(iii) We know that,

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of a triangle is equal to half of the area of the parallelogram.

Here,

$\Delta BEF$ and parallelogram$ABEF$ are on the same base $EF$ and between the same parallels$AB$ and $EF$.

Therefore, $ar\left(\Delta BEF\right)=\frac{1}{2}ar\left(ABEF\right)$

$=\frac{1}{2}\times 90c{m}^2=45c{m}^2$

Hence, $ar\left(ABEF\right)=90c{m}^2$, $ar\left(ABD\right)=45c{m}^2$and$ar\left(BEF\right)=45c{m}^2$