We cannot compare the perimeter and the area as they have different dimensions. So we use only the magnitudes here.

Let the side of the square be A.

SO the perimeter is 4*A.

The area is A^2.

As the magnitude of the area is 45 more than that of the perimeter.

A^2 = 4A + 45

=> A^2 - 4A - 45 = 0

=> A^2 + 5A - 9A - 45 = 0

=> A ( A + 5) -9( A + 5) = 0

=> (A + 5)(A -9) = 0

A is either -5 or 9.

But as A is the side of the square it cannot be negative. So we consider only 9

**The side of the square is 9.**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now