If parallelogram ABCD and rectangle ABEM are of equal area, then:

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If parallelogram ABCD and rectangle ABEM are of equal area, then:

Options:

A . Perimeter of AMEB = Perimeter of ADCB

B . Perimeter of AMEB

<

Perimeter of ADCB

C . Perimeter of AMEB

>

Perimeter of ADCB

D . Perimeter of AMEB =

\frac{1}{2}

Perimeter of ADCB

Answer: Option B
:
B

In right angled $Δ A M D$
$A D^{2}$ = $A M^{2}$ + $M D^{2}$ (by Pythagoras theorem)
$\Rightarrow A D > A M$
Similarly, In right angled $Δ B E C$
$B C^{2}$ = $B E^{2}$ + $E C^{2}$
$\Rightarrow B C > B E$
AB = DC = ME (opposite sides of rectangle and parallelogram)
Perimeter of rectangle = AM + ME + EB + AB.
Perimeter of parallelogram = AB + BC + DC + AD.
Hence Perimeter of ADCB $>$ Perimeter of AMEB.

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