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 $32\sqrt{3}$, find the radius of the circle.

Answer 1

In a rhombus, all sides are congruent to each other.

Thus, we have:
$OP=PQ=QR=RO$

Now, consider $∆QOP$.

$OQ\mathit{=}OP\mathit{}(\mathrm{Both}\mathrm{are}\mathrm{radii}.)$

Therefore, $∆QOP$ is equilateral.

Similarly, $∆QOR$ is also equilateral and $∆QOP\cong ∆QOR$.

$\mathrm{Ar}.\left(QROP\right)=\mathrm{Ar}.\left(∆QOP\right)+A\left(∆QOR\right)=2\mathrm{Ar}.\left[∆QOP\right]$

$$\begin{gathered} \mathrm{Ar}.\left(∆\mathrm{QOP}\right)=\frac{1}{2}\times 32\sqrt{3}=16\sqrt{3} \\ Or, \\ 16\sqrt{3}=\frac{\sqrt{3}}{4}{s}^2(\mathrm{where}s\mathrm{is}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{rhombus}) \\ Or, \\ {s}^2=16\times 4=64 \\ \Rightarrow s=8\mathrm{cm} \end{gathered}$$

∴ OQ = 8 cm

Hence, the radius of the circle is 8 cm.