NCERT Solutions for Chapter: Areas of Parallelograms and Triangles, Exercise 2: Exercise

$\mathrm{ABCD}$ is a parallelogram and $\mathrm{X}$ is the midpoint of $\mathrm{AB}$.  If $ar\left(\mathrm{AXCD}\right)=24 {\mathrm{cm}}^2$, then $\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=24{\mathrm{cm}}^2$.

$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \mathrm{cm}$. $A$ is any point on $PQ$. If $PS=5 \mathrm{cm},$ then $\text{ar}\left(PAS\right)=30 {\mathrm{cm}}^2$.

$PQRS$ is a parallelogram whose area is $180 {\mathrm{cm}}^2$and $A$ is any point on the diagonal $QS$. Then, area of $\Delta ASR=90 {\mathrm{cm}}^2$.

$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the midpoint of $BC$. Then, $ar\left(\Delta BDE\right)=\frac{1}{4}ar\left(\Delta ABC\right)$.

In the given figure, $ABCD$ and $EFGD$ are two parallelograms and $G$is the midpoint of $CD$, then $ar(△DPC)=\frac{1}{2}ar\left(EFGD\right)$.