The area of the parallelogram $ABCD$ is $90 \, cm^{2}$ (see figure). Find:
$(i) \; ar(ABEF)$
$(ii) \; ar(ABD)$
$(iii) \; ar(BEF)$

(N/A) $(i)$ Since parallelograms on the same base and between the same parallels are equal in area,we have:
$ar(ABEF) = ar(ABCD)$
Therefore,$ar(ABEF) = 90 \, cm^{2}$.
$(ii)$ $ar(ABD) = \frac{1}{2} \times ar(ABCD)$
[Since a diagonal of a parallelogram divides it into two triangles of equal area]
$ar(ABD) = \frac{1}{2} \times 90 \, cm^{2} = 45 \, cm^{2}$.
$(iii)$ $ar(BEF) = \frac{1}{2} \times ar(ABEF)$
[Since a triangle and a parallelogram on the same base and between the same parallels have the area of the triangle equal to half the area of the parallelogram]
$ar(BEF) = \frac{1}{2} \times 90 \, cm^{2} = 45 \, cm^{2}$.