Question

In the given figure, if parallelogram ABCD and rectangle ABEF are of equal area then :

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

Answer 1

Answer: (C)

Perimeter of ABCD > Perimeter of ABEF

Given that ABEF is a rectangle and ABCD is a parallelogram, with the same base AB and have equal areas.

Since $\parallel$gm & rectangle have equal areas with same base AB, thereore it will be between same set of $\parallel$al lines.

$AB=CD⟶\left(1\right)$ [opposite sides of the $\parallel$gm]

$AB=EF⟶\left(2\right)$ [opposite sides of rectangle]

from (1) & (2), we get

$CD=EF⟶\left(3\right)$

In $△ADF$,

$\angle AFD={90}^0$

$AD>AF$ [$\because$ Hypotenuse is greater than other sides.] $⟶\left(4\right)$

$BC>BE$ [Since BC=AD and BE=AF] $⟶\left(5\right)$

Perimeter of $\parallel$gm $ABCD$ $=AB+BC+CD+DA$

$=AB+BC+EF+DA$ [from equation (3)]

$>EF+FA+AB+BE$ [Using (5)]

which is the perimeter of rectangle ABEF.

Therefore perimeter of $\parallel$gm with the same base and with equal areas is greater than the perimeter of the rectangle.

$\therefore$ Perimeter $ABCD$ > Perimeter $ABEF$