Question

In the given figure, OABC is a rhombus, three of whose vertices lie on a circle with center O.

If the area of the rhombus is $50\sqrt{3}c{m}^2$, then find the area of the circle. (Take $\pi =3.14$)

Answer 1

Join OB.

We know, OA = OC = OB

(Radius of same circle)

Also, OA = AB = BC = CO

(ABCD is a rhombus )

Hence the rhombus can be divided into two equilateral $△$'s.

$\begin{array}{cc}\text{Area of Rhombus}& =2\left(\frac{\sqrt{3}}{4}\right)r^2\\ 50\sqrt{3}& =2\left(\frac{\sqrt{3}}{4}\right)r^2\\ r^2& =50\sqrt{3}\times \frac{2}{\sqrt{3}}\\ r^2& =100\\ \Rightarrow r& =10cm\end{array}$

$\begin{array}{cc}\text{Area of circle}& =\pi r^2\\ & =\left(3.14\right)(10{)}^2\\ & =3.14\times 100\\ & =314c{m}^2\end{array}$