The area of Square A is 121 square inches. The length of each side of Square B is 2 inches longer than the length of each side of Square A. How much greater is the area of Square B than the area of Square A? Show your work.​



Answer :

The area of the square B is a 169 and increase in the area w.r.t to the square A is 48.

According to the statement

We have to find that the Area of the square B.

So, For this purpose, we know that the

The Area is defined as the total space taken up by a flat (2-D) surface or shape of an object.

From the given information:

The area of Square A is 121 square inches. The length of each side of Square B is 2 inches longer than the length of each side of Square A.

Then

We know that the area of the square is a^2

Then

Given area of the square = 121.

Area = A^2

121 = A^2

A = (121)^1/2

A = 11.

The side of the square a is 11.

And

The side of the square B = 11+2

The side of the square B = 13.

Area of the square B = Side * Side

Area of the square B = 13* 13

Area of the square B = 169.

Increase in the area = 169 -121

Increase in the area = 48.

So, The area of the square B is a 169 and increase in the area w.r.t to the square A is 48.

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