(c) In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB On adding, both equations, we get AB + CD = EM + AB …(i) We know that, the 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 [since, 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] Perimeter of parallelogram ABCD > perimeter of rectangle ABEM