Question

The diagonals of a rhombus measure \(16\) \(cm\) and \(30\) \(cm\). Find its perimeter.

Answer 1

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 = \(4 × Side \;of\; the\; rhombus\) 
=\( 4 × 17 = 68\) \(cm\)