Question

In the given figure, a || gm ABCD and a 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 $ABCD=\frac{1}{2}(\mathrm{perimeter}\mathrm{of}ABEF)$

Answer 1

(c) perimeter of ABCD > perimeter of ABEF

Parallelogram ABCD and rectangle ABEF lie on the same base AB, i.e., one side is common in both the figures.
In ||gm ABCD, we have:
AD is the hypotenuse of right angled triangle ADF.
So, AD > AF
∴ Perimeter of ABCD > perimeter of ABEF