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.

Answer 1

Clearly, $OA=OB=OC=10cm$. Let OB and AC intersect at P.

Since the diagonals of a rhombus bisect each other at right angles, we have $OP=5cm$ and $\angle OPC=90°.$

Now, $AC=2CP$ and $CP=\surd \left(OC\right)²-\left(OP\right)²$

$CP=\sqrt{{\left(10\right)}^2-{\left(5\right)}^2}$

$CP=\sqrt{75}$

$CP=5\sqrt{3}cm$

Therefore, $AC=(2\times 5\sqrt{3})cm$

$=10\sqrt{3}cm$

$=(10\times 1.732)cm$

$=17.31cm$

Now we calculate the area of a rhombus OABC,

Ar(Rhombus $OABC)=\frac{1}{2}\times OB\times AC$

$⟹(\frac{1}{2}\times 10\times 17.32){cm}^2$

$⟹86.6{cm}^2$ (Answer)