Question

in rhombus ABCD diagonals AC=10cm and BD=24cm. find the perimeter of the rhombus​

Answer 1

$side = \sqrt{ { (\frac{10}{2}) }^{2} + {(\frac{24}{2}) }^{2} } \\ = > side = \sqrt{ {5}^{2} + {12}^{2} } \\ = > side = \sqrt{25 + 144} \\ = > side = \sqrt{169} \\ = > side = 13 \: cm$

A/Q

$perimeter = 4 \times side \\ = > 4 \times 13 \\ = > 52 \: cm$