The area of a rhombus is 336 square cm. If the length of one of its diagonals is 48 cm, then what is the perimeter of the rhombus?

Explanation

Area of a rhombus = \frac{1}{2} d_1 d_2 \implies 336 = \frac{1}{2}(48) d_2 \implies d_2 = \frac{336}{24} = 14\text{ cm}. Diagonals bisect each other at 90^{\circ}, so side length a = \sqrt{(\frac{48}{2})^2 + (\frac{14}{2})^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = 25\text{ cm}. The perimeter is 4a = 4 \times 25 = 100\text{ cm}.

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