(C) Perimeter of ABCD > Perimeter of ABEM
Explanation:
In rectangle ABEM,
AB = EM …(eq.1) [sides of rectangle]
In parallelogram ABCD,
CD = AB …(eq.2)
Adding, equations (1) and (2),
We get
AB + CD = EM + AB …(i)
We know that,
Perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.
BE < BC and AM < AD
[because, in a right angled triangle, the hypotenuse is greater than the other side]
On adding both above inequalities, we get
SE + AM <BC + AD or BC + AD> BE + AM
On adding AB + CD both sides, we get
AB + CD + BC + AD> AB + CD + BE + AM
⇒ AB+BC + CD + AD> AB+BE + EM+ AM [∴ CD = AB = EM]
Hence,
We get,
Perimeter of parallelogram ABCD > perimeter of rectangle ABEM
Hence, option (C) is the correct answer.
