The area of trapezium $\mathrm{ABCD}$ in the given figure is

In the given figure, $\mathrm{ABCD}$ is a parallelogram in which $\mathrm{AB}=\mathrm{CD}=5 \mathrm{cm}$ and $\mathrm{BD}\perp \mathrm{DC}$ such that $\mathrm{BD}=6.8 \mathrm{cm}$. Then, the area of $\mathrm{ABCD}=?$

In the given figure, $\mathrm{ABCD}$ is a parallelogram in which diagonals $\mathrm{AC}$ and $\mathrm{BD}$ intersect at $\mathrm{O}$. If $\mathrm{ar}(//\mathrm{gm} \mathrm{ABCD})$ is $52 {\mathrm{cm}}^2$ then the $\mathrm{ar}(∆\mathrm{AOB})=?$

In the given figure, $\mathrm{ABCD}$ is a $\mathrm{parallelogram}$ in which $\mathrm{DL}\perp \mathrm{AB}$. If $\mathrm{AB}=10 \mathrm{cm}$ and $\mathrm{DL}=4 \mathrm{cm}$ then the $\mathrm{ar}(\mathrm{parallelogram} \mathrm{ABCD})=?$

The area of parallelogram $ABCD$ is

In the given figure, $ABCD$ and $ABPQ$ are two parallelograms and $M$ is a point on $AQ$ and $BMP$ is a triangle. Then, $ar(△BMP)=\frac{1}{2}ar( ▱ABCD)$ is 

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to