Question

If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

Answer 1

Let the other diagonal be d cm
Area of a rhombus = $\frac{1}{2}\left(\mathrm{Product}\mathrm{of}\mathrm{diagonals}\right)$
$$\begin{gathered} \Rightarrow 240=\frac{1}{2}\times \left(30\times d\right) \\ \Rightarrow d=\frac{240\times 2}{30}=16 \end{gathered}$$
AC = 30 cm
DB = 16 cm
Diagonals of a rhombus bisect at right angles.
In $△$AOB,
$$\begin{gathered} {\mathrm{AO}}^2+{\mathrm{OB}}^2={\mathrm{AB}}^2 \\ \Rightarrow {15}^2+8^2={\mathrm{AB}}^2 \\ \Rightarrow {\mathrm{AB}}^2=225+64=289 \\ \Rightarrow \mathrm{AB}=\sqrt{289}=17\mathrm{cm} \end{gathered}$$
Thus, the side of the rhombus = 17 cm
Perimeter = $4\times 17=68$ cm