# Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

**Solution:**

Given: The parallelogram and the rectangle have the same base and equal areas, therefore, they will also lie between the same parallels.

Consider the parallelogram ABCD and rectangle ABEF as follows.

Here, it can be observed that parallelogram ABCD and rectangle ABEF are lying between the same parallels AB and CF.

It is given that the opposite sides of a parallelogram and a rectangle are of equal lengths.

Therefore, AB = EF (Opposite sides of a rectangle are equal.)

AB = CD (Opposite sides of a parallelogram are equal.)

∴ CD = EF

∴ AB + CD = AB + EF ...(1)

Now, in a right-angled triangle AFD, AD is a hypotenuse.

∴ AF < AD ...(2)

Similarly, in a right angled triangle EBC, EB is altitude and BC is a hypotenuse

∴ BE < BC ...(3)

Adding equation (2) and (3),

∴ AF + BE < AD + BC ...(4)

Now, from equations (1) and (4), we obtain

AB + EF + AF + BE < AD + BC + AB + CD

Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD.

**Video Solution:**

## Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

### Maths NCERT Solutions Class 9 - Chapter 9 Exercise 9.4 Question 1:

**Summary:**

If parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas, then the perimeter of the parallelogram is greater than that of the rectangle.