Question

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

• A. $25$ $\sqrt{3}$ sq. cm.
• B. $50$ $\sqrt{3}$ sq. cm.
• C. $75$ $\sqrt{3}$ sq. cm.
• D. $20$ $\sqrt{3}$ sq. cm.

Answer 1

The correct option is B $50$ $\sqrt{3}$ sq. cm.

Since OABC is a rhombus
$OA=AB=BC=OC=10cm.$
Now, $OD\perp BC\Rightarrow CD=\frac{1}{2}BC=\frac{1}{2}\left(10\right)=5$ cm
By pyhthagoras theorem,
$O{C}^2=O{D}^2+D{C}^2$
$\Rightarrow O{D}^2=O{C}^2-D{C}^2=(10{)}^2-(5{)}^2=100-25=75$
$\Rightarrow OD=\sqrt{75}=5\sqrt{3}$
Therefore, area $\left(\Delta ABC\right)=\frac{1}{2}BC\times OD=\frac{1}{2}\left(10\right)\times 5\sqrt{3}=25\sqrt{3}$ sq.cm
So, area of rhombus $=2\left(\text{Area of}\Delta OABC\right)$
$=2\left(25\sqrt{3}\right)=50\sqrt{3}$ sq. cm