The diagonals of a rhombus differ by 2 units and its perimeter exceeds the sum of the diagonals by 6 units. What is the area of the rhombus?

Explanation

Let diagonals be d_1 and d_2. Given d_1 - d_2 = 2. The side a = \frac{1}{2}\sqrt{d_1^2+d_2^2}. Perimeter = 2\sqrt{d_1^2+d_2^2}. We are given 2\sqrt{d_1^2+d_2^2} - (d_1+d_2) = 6. By substituting (d_1-d_2)^2 = d_1^2 + d_2^2 - 2d_1d_2 = 4 into this relation, solving gives d_1 = 8 and d_2 = 6. The area = \frac{1}{2} d_1 d_2 = \frac{1}{2}(8)(6) = 24.

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