NCERT Solutions for Chapter: Areas of Parallelograms and Triangles, Exercise 3: Exercise

If the area of the parallelogram $ABCD$ is $90 {\mathrm{cm}}^2$ and $ar\left(ABEF\right)=k {\mathrm{cm}}^2$, then find $k$.

The area of the parallelogram $ABCD$ is $90 {\mathrm{cm}}^2$ . If the $ar \left(BEF\right)$ is $k {\mathrm{cm}}^2$ then find the value of $k$.

In $\Delta \mathrm{ABC},D$ is the midpoint of $AB$and $P$ is any point on $\mathrm{BC}$. If $CQ\left|\right|PD$meets $AB$ in $Q$, then prove that $ar\left(\Delta BPQ\right)=\frac{1}{2}ar\left(\Delta ABC\right)$.

$ABCD$is a square. $E$ and $F$ are respectively the midpoints of $\mathrm{BC}$ and $CD$. If $R$ is the mid-point of $EF$ as shown in the figure, prove that $$$\mathrm{ar} \left(\Delta AER\right)=ar \left(AFR\right)$.

 $O$ is any point on the diagonal $PR$ of a parallelogram $PQRS$ in the given figure. Prove that $ar \left(PSO\right) = ar \left(PQO\right).$

$ABCD$ is a parallelogram in which $BC$ is produced to $\mathrm{E}$ such that $\mathrm{CE}=\mathrm{BC}$ as shown in the figure. $AE$intersects $\mathrm{CD}$ at $\mathrm{F}$. If $ar\left(DFB\right)=3 c{m}^2$, find the area of the parallelogram $ABCD$.

 In trapezium $ABCD, AB\left|\right|DC$ and  $L$ is the midpoint of $\mathrm{BC}$. Through $L$, a line $PQ\left|\right|AD$ has been drawn which meets $AB$  in $P$ and $DC$ produced in $Q$ as shown in the figure. Prove that $ar \left(\mathrm{ABCD}\right)=ar \left(\mathrm{APQD}\right)$.

 If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral. [Hint: Join $BD$and draw perpendicular from$A$ on $BD$.]