A rhombus whose 3 points lie …

Since the sides of a rhombus are of equal length.Therefore,

OP = OR = OQ =r
Let OQ and PR intersect at S
We know the diagonals of a rhombus bisect each other at right angle.
Therefore, we have
{tex}O S = \frac { 1 } { 2 } r{/tex}
and {tex}\angle O S R = 90 ^ { \circ }{/tex}

By pythagoras theorm, we have,
{tex}S R = \sqrt { O R ^ { 2 } - O S ^ { 2 } }{/tex}
{tex}= \sqrt { r ^ { 2 } - \frac { r ^ { 2 } } { 4 } }{/tex}
{tex}= \frac { \sqrt { 3 } } { 2 }r{/tex}
PR = {tex}2 \times S \mathrm { R }{/tex}
{tex}= \sqrt { 3 } r{/tex}
Area of rhombus {tex}= \frac { 1 } { 2 } \times \mathrm { OQ } \times \mathrm { PR }{/tex}
{tex}= \frac { 1 } { 2 } \times r \times \sqrt { 3 } r{/tex}
{tex}= \frac { \sqrt { 3 } r ^ { 2 } } { 2 }{/tex}

But Area of rhombus={tex}32\sqrt{3}{/tex}.
Therefore,
{tex}\frac { \sqrt { 3 } r ^ { 2 } } { 2 } = 32 \sqrt { 3 }{/tex}
{tex}\Rightarrow r ^ { 2 } = \frac { 32 \sqrt { 3 } } { \sqrt { 3 } } \times 2{/tex} =64cm
r = 8 cm