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

Answer 1

The correct option is B $8m$

Let the $r$ be the radius of the circle.
$\therefore OA=OB=OC=r$ ...(I)
$\square OABC$ is a rombus.
$\therefore OA=AB=BC=OC$ ...(II)
From (i) and (ii) we get,
$OA=OB=AB=BC=OC=r$
Thus,
$△OAB$ and $△OBC$ are equilateral triangle having each side $r$ meter.
Also, we know that diagonal of a parallelogram divides it into two triangles of equal areas.
$\therefore$ Area of $△OBC=$ Area of $△OAB$ ...(iii)
Area of a rhombus $OABC=32\sqrt{3}{m}^2$
Area of $△OAB+$ Area of $△OBC=32\sqrt{3}{m}^2$
$2\times$ Area of $△OAB=32\sqrt{3}{m}^2$
$2\times \frac{\sqrt{3}}{4}{r}^2=32\sqrt{3}{m}^2$
$=>r^2=64$
$=>r=8m$