OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.

(i) If the radius of the circle is 10 cm, find the area of the rhombus.

(ii) If the area of the rhombus is $32\sqrt{3}$ cm², find the radius of the circle.

(i) Given, radius = 10 cm

In rhombus OABC,

OC = 10 cm

As, diagonal of rhombus bisect each other.

OE = $\dfrac{1}{2}$ x OB = $\dfrac{1}{2}$ x 10 = 5 cm.

Now, in right ∆OCE,

⇒ OC² = OE² + EC²

⇒ 10² = 5² + EC²

⇒ EC² = 100 - 25 = 75

⇒ EC = $\sqrt{75} = 5\sqrt{3}$.

Hence, AC = 2 x EC = 2 x $5\sqrt{3} = 10\sqrt{3}$ cm.

We know that,

Area of rhombus = $\dfrac{1}{2} \times d$1 \times d2

= $\dfrac{1}{2} \times OB \times AC$

= $\dfrac{1}{2} \times 10 \times 10\sqrt{3}$

= $50\sqrt{3}$ = 86.60 cm².

Hence, area of rhombus = 86.60 cm².

(ii) Given, area of rhombus = $32\sqrt{3}$ cm²

We know that,

Diagonal of rhombus divides it into two equilateral triangles.

∴ Area of rhombus OABC = 2 x area of ∆OAB

⇒ Area of rhombus OABC = 2 x $\dfrac{\sqrt{3}}{4}r^2$

$\Rightarrow 32\sqrt{3} = 2 \times \dfrac{\sqrt{3}}{4}r^2 \\[1em] \Rightarrow r^2 = \dfrac{32\sqrt{3} \times 4}{2\sqrt{3}} \\[1em] \Rightarrow r^2 = 64 \\[1em] \Rightarrow r = \sqrt{64} = 8 \text{ cm}.$

Hence, radius of circle = 8 cm.