Question

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

Answer 1

From the figure, O is the centre of the circle and OPQR is a rhombus.
Let the diagonals OQ and PR intersect at S

Given area of rhombus OPQR $=32\sqrt{3}c{m}^2$

Let $OP=OQ=OR=rcmOS=SQ=\left(\frac{r}{2}\right)cmandRS=PS$

In right $\Delta OSP,O{P}^2=O{S}^2+P{S}^2\text{(By Pythagoras theorem)}\Rightarrow r^2=(\frac{r}{2}{)}^2+P{S}^2\Rightarrow P{S}^2{=}^{r}2–(\frac{r}{2}{)}^2=\frac{3r^2}{4}\therefore PS=\frac{\sqrt{3}r}{2}\Rightarrow PR=2PS=\sqrt{3}r$

Area of rhombus OPQR $=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times OQ\times PR\Rightarrow 32\sqrt{3}=\frac{1}{2}\times r\times \sqrt{3}r\Rightarrow 32=\frac{1}{2}\times r^2{r}^2=64r=8cm$

Radius of circle = 8 cm