Question

OABC is a rhombus whose three vertices A, B and C Iie on a circle with centre O. If the radius of the circle is $10$ cm, find the area of the rhombus.

Answer 1

OABC is rhombus whose three vertices A,B,C lies on circle with center O and radius 10cm.
Let diagonal of rhombus intersect at P.
Radius of circle $\left(r\right)=10cm$
$\therefore OA=OB=OC=10cm$
Diagonals of rhombus bisect each other at ${90}^o.$ Therefore $OP=PB=\frac{OB}{2}=\frac{10}{2}=5cm$
and $PC=PA.$ In right angle $\Delta OCP,{OC}^2={OP}^2+{PC}^2{\left(10\right)}^2={\left(5\right)}^2+{PC}^{2}PC=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}cm\therefore AC=2PC=2\times 5\sqrt{3}cm$
Area of rhombus$=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 10\times 10\sqrt{3}=50\sqrt{3}{cm}^2$