A square and a rectangle have same perimeter. They differ in areas by 1 square cm. The length of the rectangle exceeds its breadth by

Explanation

Let the side of the square be a and the dimensions of the rectangle be L, B. Since perimeters are equal, 4a = 2(L+B) \implies 2a = L+B. The area of the square is always larger by the difference, so a^2 - LB = 1. Substituting B = 2a-L, we get a^2 - L(2a-L) = 1 \implies a^2 - 2aL + L^2 = 1 \implies (L-a)^2 = 1 \implies L = a+1. Thus B = 2a - (a+1) = a-1. Their difference L-B = (a+1) - (a-1) = 2\text{ cm}.

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