The area of a rhombus is 216 cm² and its one of the diagonals is 24 cm. Then its perimeter is –

Area of a rhombus $= \frac{1}{2} d_1 d_2 = 216\text{ cm}^2$. Since $d_1 = 24\text{ cm}$, we get $\frac{1}{2} \times 24 \times d_2 = 216 \implies d_2 = 18\text{ cm}$. The diagonals bisect each other at right angles with half-lengths $12\text{ cm}$ and $9\text{ cm}$, so each side is $a = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = 15\text{ cm}$. Therefore, perimeter $= 4a = 4 \times 15 = 60\text{ cm}$.