Question

In figure OPQR is a rhombus, three of whose vertices lie on the circle with Centre O. If the area of rhombus is \[32\sqrt{3}c{{m}^{2}}\], find the radius of the circle.

(a) 6 cm
(b) 2 cm
(c) 4cm
(d) 8 cm

Answer 1

Hint: As given in the question, OPQR is a rhombus, then RO = OP = OR = PQ because the sides of rhombus are equal. The length of side OQ is equal to ‘R’ that is the radius of the circle. We are taking two equilateral triangles$\Delta RQO\text{ and }\Delta QOP$ and adding their respective areas. By using this we can easily find the value of ‘R’ which is our radius.

Complete step-by-step answer:

In The circle given in the figure, we let the length of the radius of the circle be ‘R’. Inside the circle, one rhombus is given. In rhombus OPQR, the length of all sides are equal and the length of both diagonals are also equal.
$RO=OP=OR=PQ$(side of Rhombus)
$OQ=R$(Radius of the circle)

Now, we are taking two triangles, $\Delta RQO\text{ and }\Delta QOP$ are equilateral triangles. So, the area of rhombus is given by:
Area of Rhombus = Area of equilateral triangle RQO + Area of equilateral triangle QOP.

As we know that for an equilateral triangle, the area is $\dfrac{\sqrt{3}}{4}{{a}^{2}}$, where a is the dimension of the equilateral side.
$\begin{align}
  & \therefore \dfrac{\sqrt{3}}{4}{{R}^{2}}+\dfrac{\sqrt{3}}{4}{{R}^{2}}=32\sqrt{3} \\
 & \dfrac{\sqrt{3}}{2}{{R}^{2}}=32\sqrt{3} \\
 & \Rightarrow R=\pm 8cm \\
\end{align}$
Since, the dimension of a fixed quantity can never be negative so neglecting the negative value of radius.
Hence, the dimension of radius ‘R’ is 8cm.
Therefore, option (d) is correct.

Note: The key step in solving this problem is the knowledge of geometrical figures particularly rhombus and circle. By using some properties and visual observation we can easily identify the measurement of radius of circle with conclusive evidence. These observation skills will enhance the confidence of students.