The area of a rhombus is \(720\ \text{cm}^2\) and the sum of its diagonals is \(98\ \text{cm}\). What is the perimeter of the rhombus?
\(164\ \text{cm}\)
Let the diagonals be \(d_1, d_2\). Area \(= \frac{1}{2}d_1 d_2 = 720 \Rightarrow d_1 d_2 = 1440\), and \(d_1 + d_2 = 98\). Side of the rhombus \(= \frac{1}{2}\sqrt{d_1^2+d_2^2}\). Now \(d_1^2+d_2^2 = (d_1+d_2)^2 - 2d_1 d_2 = 98^2 - 2(1440) = 9604 - 2880 = 6724\). So \(\sqrt{6724} = 82\), giving side \(= 41\ \text{cm}\). Perimeter \(= 4 \times 41 = 164\ \text{cm}\).
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