Let \(ABCD\) be a rhombus (all sides are of equal length) and its diagonals, \(AC\) and \(BD\), are intersecting each other at point \(O\).
Diagonals in a rhombus bisect each other at \(90 \degree\).
It can be observed that
\(AO= \frac{AC}{2}=\frac{16}{2}=8\;cm\)
\(BO = \frac{BD}{2}=\frac{30}{2}=15\;cm\)
By applying Pythagoras theorem in \(Δ\) \(AOB\),
\(OA^2 + OB^2= AB^2\)
\(8^ 2 + 15^2 = AB^2\)
\(64 + 225 = AB^2\)
\(289 = AB^2\)
\(AB = 17\)
Therefore, the length of the side of rhombus is \(17\) \(cm\).
Perimeter of rhombus = \(4 × Side \;of\; the\; rhombus\)
=\( 4 × 17 = 68\) \(cm\)