# The area of a rectangle is 100 square inches. The perimeter of the rectangle is 40 inches.? A second rectangle has the same area but a different perimeter. Is the second rectangle a square?

Sep 6, 2016

No. The second rectangle is not a square.

#### Explanation:

The reason why the second rectangle isn't a square is because the first rectangle is the square. For instance, if the first rectangle (a.k.a. the square) has a perimeter of $100$ square inches and a perimeter of $40$ inches then one side must have a value of $10$.

With this being said, let's justify the statement above. If the first rectangle is indeed a square* then all of it's sides must be equal.
Moreover, this would actually make sense for the reason that if one of its sides is $10$ then all of its other sides must be $10$ as well. Thus, this would give this square a perimeter of $40$ inches.

Also, this would mean that the area must be $100$ ($10 \cdot 10$). In continuation, if the second square has the same area, but a different perimeter then it can't be a square because its features wouldn't match the ones of a square.

To clarify, what this means is that there wouldn't possible be a way of getting a square with an area of $100$ and still have a different perimeter form the first square (it would be like trying to get another combination of four numbers that have the same value, but that when you multiply two of them together they give you $100$).

In conclusion, that is why the second rectangle isn't (and can't be) a square.

*A square can be a rectangle, but a rectangle can't be a square so, the first rectangle was originally a square.