Question

In the figure, if parallelogram ABCD and rectangle ABEM are of equal area, then

• A. Perimeter of ABCD $=$ Perimeter of ABEM
• B. Perimeter of ABCD $<$ Perimeter of ABEM
• C. Perimeter of ABCD $>$ Perimeter of ABEM
• D. Perimeter of ABCD $=\frac{1}{2}$ (Perimeter of ABEM)

Answer 1

The correct option is C Perimeter of ABCD $>$ Perimeter of ABEM

Parallelogram ABCD and rectangle ABEM are of equal areas having same base AB.

To find out-

The relation between the perimeters of the two given figures.

Solution-

Since the Parallelogram ABCD and rectangle ABEM are of equal areas having same base, they are within the same parallels $MC$ & $AB.$

Now, in ${\Delta }^{s}AMD$ & $BEC$ we have

$AM=BE$ ...( opposite sides of rectangle)

$AD=BC$ ...(opposite sides of parallelogram)

$\angle AMD=\angle BEC$ ...( both are right angles since they are corners of a rectangle)

$\therefore \Delta AMD$ & $\Delta BEC$ are congruent.

$⟹MD=EC$ ....(i)

Again $\Delta AMD$ is a right one.

$(\angle M={90}^o$ being corner of a rectangle.)

So, AD is the hypotenuse. i.e $AD>AM$........(ii)

Now, $ME=MD+DE$ & $DC=EC+DE=MD+DE$ ....(From i)

$⟹ME=DC$ ........(iii)

Now, perimeter $ABEM=2(AM+ME)$ .........(iv)

And perimeter $ABCD=2(DC+AD)=2(ME+AD)$ .......(v) ...[from iii]

Comparing (iv) & (v) AM is common to both but $AD>AM$.

$\therefore$ (v) >(iv) i.e perimeter $ABCD>$ perimeter $ABEM.$