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Assume square A with side a; square B with side b. Perimeter of A is 4a; area of A is a2.

Perimeter of B is 4b; area of B is b2. Given the ratio of the perimeters equals the ratio of the areas, then 4a/4b = a2/b2; a/b = a2/b2 By cross-multiplication we get: ab2 = a2b Dividing both sides by ab we get: b = a This tells us that squares whose ratio of their perimeters equals the ratio of their areas have equal-length sides. (Side a of Square A = side b of Square B.) This appears to show, if not prove, that there are not two different-size squares meeting the condition.

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Q: What are two different size squares that the ratio of their perimeters is the same as the ratio of their areas?
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