Question

If the diagonals of a rhombus measure 32cm and 60cm,what is the perimeter of the rhombus

Answer 1

The information given can be represented in the diagram below

From the rhombus above, E is the perpendicular bisector of the diagonals AC and BD.

Let BD = 32cm, AC = 60cm (given)

Using Pythagoras theorem,

$AD^2=AE^2+DE^2$$\begin{gathered} AE=\frac{1}{2}\times AC \\ AE=\frac{1}{2}\times60\operatorname{cm} \\ AE=30\operatorname{cm} \end{gathered}$$\begin{gathered} DE=\frac{1}{2}\times BD \\ DE=\frac{1}{2}\times32\operatorname{cm} \\ DE=16\operatorname{cm} \end{gathered}$

Using Pythagoras theorem

$\begin{gathered} AD^2=AE^2+DE^2 \\ AD^2=30^2+16^2 \\ AD^2=900+256 \\ AD^2=1156 \\ AD=\sqrt[]{1156} \\ AD=34\operatorname{cm} \end{gathered}$

The perimeter of the rhombus would be

$\begin{gathered} P_{\rho\text{mbus}}=AB+BC+CD+AD \\ \text{Note,} \\ AB=BC=CD=AD=34\operatorname{cm} \\ P_{\rho\text{mbus}}=4AD \\ P_{\rho\text{mbus}}=4\times34 \\ P_{\rho\text{mbus}}=136\operatorname{cm} \end{gathered}$

Hence, the perimeter of the given rhombus is 136cm