Is there a unique solution to this geometry problem?

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This problem actually has two distinct solutions in the integers, but an infinite number of solutions if you permit non-integer solutions. Here's how to solve it: Suppose the width is x and the length is y. If area equals perimeter, then

xy = 2x + 2y

Since you have one equation in two variables, you won't get a unique solution. For any value of x that you pick (except x = 2) the corresponding y value is

y = 2x/(x-2)

which you get from isolating y in the first expression. However, if you want just the integer solutions, you have to guess and check with different whole number values of x and y. The only two that work are

x = 4, y = 4

x = 6, y = 3

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Thank you.

Yes. The answer is both the length and width = 4.

area = length * width = 4 * 4 = 16.

perimeter = 2 * length + 2 * width = 2 * 4 + 2 * 4 = 16.

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Some other shapes whose area and perimeter are equal:

• Circle with diameter of 4 (radius of 2) has circumference and area equal to 4*pi.

• Right triangle with integer side lengths 5, 12, 13 has perimeter and area equal to 30.

• Right triangle with integer side lengths 6, 8, 10 has perimeter and area equal to 24

• Equilateral triangle with a side length of 4*sqrt(3) has a perimeter and area equal to 12*sqrt(3).

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