Question

In the figure, $OPQR$ is rhombus, three of whose vertices lie on the circle with centre $O$.
If the area of the rhombus is $32\sqrt{3}{cm}^2$, find the radius of the circle.

Answer 1

Let radius of circle be 'r'. Then $OP=OQ=OR=rcm$

OQ and RP are diagonals and these bisects each other at $90°$

Also, $RS=SP$

In right $△OSP,{OP}^2={OS}^2+{PS}^2\Rightarrow r^2={\left(\frac{r}{2}\right)}^2+{PS}^2$

$\Rightarrow {PS}^2=r^2-\frac{r^2}{4}=\frac{3r^2}{4}\Rightarrow PS=\frac{\sqrt{3}}{2}r\Rightarrow PR=2PS=2\times \frac{\sqrt{3}}{2}r=\sqrt{3}r$

Given, area of rhombus$=32\sqrt{3}cm$

$\Rightarrow \frac{1}{2}\times OQ\times RP=32\sqrt{3}$

$\Rightarrow \frac{1}{2}\times r\times \sqrt{3}r=32\sqrt{3}\Rightarrow r^2=64$

$\Rightarrow r=\sqrt{64}=8cm$

Therefore, radius of circle$=8cm$