In the figure,$ABCD$ is a parallelogram,$AE \perp DC$ and $CF \perp AD$. If $AB = 16 \, cm, AE = 8 \, cm$ and $CF = 10 \, cm$,find $AD$. (in $, cm$)

$P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP$. Show that $\operatorname{ar}(RQC) = \frac{3}{8} \operatorname{ar}(ABC)$.

If $E, F, G$ and $H$ are respectively the mid-points of the sides of a parallelogram $ABCD$,show that $\text{ar}(EFGH) = \frac{1}{2} \text{ar}(ABCD)$.

In the figure,$ABCD$ is a quadrilateral and $BE \parallel AC$. $BE$ meets $DC$ produced at $E$. Show that the area of $\Delta ADE$ is equal to the area of the quadrilateral $ABCD$.

In the figure,$D$ and $E$ are two points on $BC$ such that $BD = DE = EC$. Show that $ar(ABD) = ar(ADE) = ar(AEC)$.
Can you now answer the question that you have left in the 'Introduction' of this chapter,whether the field of Budhia has been actually divided into three parts of equal area?

Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that $ar(APB) \times ar(CPD) = ar(APD) \times ar(BPC)$.