Question

ABCD is a parallelogram in which BC is Produced to P such that CP=BC , as shown in the adjoining figure AP intersects CD at M. If ar(DMB) =$7c{m}^2$ find the area of parallelogram ABCD.

Answer 1

Consider $△ADM$ and $△PCM$

From the figure we know that $\angle ADM$ and $\angle PCM$ are alternate angles

$\angle ADM=\angle PCM$

We know that $AD=BC=CP$

It can be written as

$AD=CP$

$\angle AMD$ and $\angle PMC$ are vertically opposite angles

$\angle AMD=\angle PMC$

By ASA congruence criterion

$△ADM\simeq △PCM$

So we get

Area of $△ADM$ = Area of $△PCM$

$DM=CM$

We know that $BM$ is the median of $△BDC$

So we get

Area of $△DMB$ = Area of $△CMB$

We get

Area of $triangleBDC=2$ ( Area of $△DMB$)

By substituting the value

Area of $△BDC=2\times 7$

By multiplication

Area of $△BDC=14c{m}^2$

We know that

Area of parallelogram $ABCD=2\times 14$

By multiplication

Area of parallelogram $ABCD=28c{m}^2$

Therefore, area of parallelogram $ABCD$ is $28c{m}^2$.