# In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then :

a. Perimeter of ABCD = Perimeter of ABEM

b. Perimeter of ABCD < Perimeter of ABEM

c. Perimeter of ABCD > Perimeter of ABEM

d. Perimeter of ABCD = 1/2 (Perimeter of ABEM)

**Solution:**

In rectangle ABEM,

AB = EM (sides of a rectangle)

In parallelogram ABCD,

CD = AB

By addition we get

AB + CD = EM + AB

As the perpendicular distance between two parallel sides of a parallelogram is less than the length of the other parallel sides.

BE < BC and AM < AD

In a right angled triangle, hypotenuse is greater than the other side

By adding the above inequalities

BE + AM < BC + AD

Or BC + AD > BE + AM

Let us add AB + CD on both sides

AB + CD + BC + AD > AB + CD + BE + AM

So we get

AB + BC + CD + AD > AB + BE + EM + AM (CD = AB = EM)

Perimeter of parallelogram ABCD > Perimeter of rectangle ABEM

Therefore, the perimeter of the parallelogram ABCD > perimeter of the rectangle ABEM.

**✦ Try This: **The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 10 cm and 5 cm is :

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.1 Problem 5**

## In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then : a. Perimeter of ABCD = Perimeter of ABEM, b. Perimeter of ABCD < Perimeter of ABEM, c. Perimeter of ABCD > Perimeter of ABEM, d. Perimeter of ABCD = 1/2 (Perimeter of ABEM)

**Summary:**

In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then perimeter of parallelogram ABCD > perimeter of rectangle ABEM

**☛ Related Questions:**

- The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a p . . . .
- Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is a. . . . .
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