Answer:
![A(square) = x*x= x^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/87th4umd9xok8vk835ta11agkl69e5yc8o.png)
![A(rectangle) = (a-2) (a+2) = a^2 +2a- 2a-4 = a^2 -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/zjnqq9t8ud03gpyabje507uu4jqbhvv53s.png)
![A(rectangle) = A(square) -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/46f1mth0ig9ae91irszpkx3pm8ls5n4fpa.png)
Explanation:
For this case we assume that the square have sides of length x.
So then the area for this square is given by
![A(square) = x*x= x^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/87th4umd9xok8vk835ta11agkl69e5yc8o.png)
We have a rectangle and on this case we have this information :" the rectangle has side lengths 2 less and 2 more than a square"
So then the sides of the rectangle needs to have lengths a-2 and a+2, and if we find the area for the rectangle we got:
![A(rectangle) = (a-2) (a+2) = a^2 +2a- 2a-4 = a^2 -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/zjnqq9t8ud03gpyabje507uu4jqbhvv53s.png)
And as we can see th area for the rectangl is 4 units less than the area for the rectangle:
![A(rectangle) = A(square) -4](https://img.qammunity.org/2021/formulas/mathematics/high-school/46f1mth0ig9ae91irszpkx3pm8ls5n4fpa.png)