Question

Find the perimeter of a rhombus, the lengths of whose diagonals are 16 cm and 30 cm.

Figure

Answer 1

We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, in right triangle AOB, we have:
AO = 8 cm
BO = 15 cm

By Pythagoras theorem in $∆$AOB:

$$\begin{gathered} A{B}^2=A{O}^2+B{O}^2 \\ \Rightarrow A{B}^2=8^2+{15}^2 \\ \Rightarrow A{B}^2=64+225 \\ \Rightarrow A{B}^2=289 \\ \Rightarrow A{B}^2={17}^2 \\ \Rightarrow AB=17\mathrm{cm} \end{gathered}$$

Now, as we know that all sides of a rhombus are equal.
∴ Perimeter of the rhombus = 4(side)
= 4(17)
= 68 cm