OABC is a rhombus whose three vertices. A, B and C lie on a circle with centre O.
(i) If the radiusof the circle is 10 cm, find the area of the rhombus.
(ii) If the area of the rhombus is `32sqrt(3)cm^(2)` find the radius of the circle.

Answer

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (i): Finding the Area of the Rhombus

1. **Identify the properties of the rhombus and the circle:**
- The rhombus OABC has vertices A, B, and C on the circle with center O.
- The radius of the circle is given as 10 cm.

2. **Determine the lengths of the diagonals:**
- The diagonals of the rhombus intersect at right angles and bisect each other.
- Let the diagonals be OB and AC.
- Since O is the center of the circle, OA, OB, and OC are all equal to the radius, which is 10 cm.

3. **Calculate the length of diagonal OB:**
- OB is a radius of the circle, so OB = 10 cm.

4. **Calculate the length of diagonal AC:**
- Let P be the intersection point of the diagonals.
- Since the diagonals bisect each other, OP = 10/2 = 5 cm.
- Using the Pythagorean theorem in triangle OAP:
\[ OA^2 = OP^2 + AP^2 \]
\[ 10^2 = 5^2 + AP^2 \]
\[ 100 = 25 + AP^2 \]
\[ AP^2 = 100 - 25 = 75 \]
\[ AP = \sqrt{75} = 5\sqrt{3} \text{ cm} \]
- Since AP = PC (because diagonals bisect each other), we have:
\[ AC = AP + PC = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3} \text{ cm} \]

5. **Calculate the area of the rhombus:**
- The area \( A \) of a rhombus can be calculated using the formula:
\[ A = \frac{1}{2} \times d_1 \times d_2 \]
where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.
- Here, \( d_1 = AC = 10\sqrt{3} \) and \( d_2 = OB = 10 \):
\[ A = \frac{1}{2} \times (10\sqrt{3}) \times 10 = 50\sqrt{3} \text{ cm}^2 \]

### Part (ii): Finding the Radius of the Circle Given the Area of the Rhombus

1. **Set up the equation with the given area:**
- The area of the rhombus is given as \( 32\sqrt{3} \) cm².
- Using the area formula:
\[ 32\sqrt{3} = \frac{1}{2} \times d_1 \times d_2 \]

2. **Express the diagonals in terms of the radius \( r \):**
- From the previous calculations, we know:
- \( d_2 = OB = r \)
- For diagonal \( AC \), we can express it in terms of \( r \):
- Using the triangle OAP:
\[ AP = \frac{\sqrt{3}}{2} r \]
- Thus, \( AC = 2 \times AP = \sqrt{3} r \).

3. **Substitute the diagonals into the area formula:**
- Substitute \( d_1 \) and \( d_2 \):
\[ 32\sqrt{3} = \frac{1}{2} \times (\sqrt{3} r) \times r \]
\[ 32\sqrt{3} = \frac{1}{2} \times \sqrt{3} r^2 \]

4. **Solve for \( r^2 \):**
- Multiply both sides by 2:
\[ 64\sqrt{3} = \sqrt{3} r^2 \]
- Divide both sides by \( \sqrt{3} \):
\[ 64 = r^2 \]

5. **Find the radius \( r \):**
- Taking the square root:
\[ r = \sqrt{64} = 8 \text{ cm} \]

### Final Answers:
- (i) The area of the rhombus is \( 50\sqrt{3} \) cm².
- (ii) The radius of the circle is \( 8 \) cm.