Question

A rhombus $OABC$ is drawn inside a circle whose centre is at $O$ in such a way that the vertices $A,B$ and $C$ of the rhombus are on the circle. If the area of the rhombus, is $32\sqrt{3}{m}^2$, then the radius of the circle is:

• A. $64$ m
• B. $8$ m
• C. $32$ m
• D. $46$ m

Answer 1

The correct option is B $8$ m

Let $AC=2x$, $OB=2y$ and D be the point of the intersection of $AC$ and $OB$.
Now, consider triangle $ODC$,
$O{D}^2+D{A}^2=O{A}^2$
$OA=OB=2y$.....(radius)
$(2y{)}^2=y^2+x^2$
$x=\sqrt{3}y$....(1)
We know Area of rhombus is $\frac{d_1{d}_2}{2}$ where $d_1$ and $d_2$ are the diagonals of rhombus.
$32\sqrt{3}=\frac{AC\times OB}{2}$
$32\sqrt{3}=\frac{2x\times 2y}{2}$
$xy=16\sqrt{3}$....(2)
From (1) and (2), we get
$x=4\sqrt{3}$ and $y=4$
Radius $=2y=2\times 4=8$
Hence, option $B$ is the correct answer.