Suppose AB is a diameter of a circle, whose centre is at O and C be any point on the circle. If CD is perpendicular to AB and CD = 12 cm, AD = 16 cm, then BD is equal to

Options

A. 10 cm
B. 12 cm
C. 8 cm
D. 9 cm

Answer

To solve the problem, we need to find the length of BD given the following information:

- CD is perpendicular to AB (the diameter of the circle).
- CD = 12 cm.
- AD = 16 cm.

Let’s denote the lengths as follows:
- Let BD = x cm.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have a circle with diameter AB, where O is the center of the circle.
- Point C is on the circumference of the circle, and CD is perpendicular to AB at point D.

2. **Use the Pythagorean Theorem**:
- In triangle ACD, we can apply the Pythagorean theorem since CD is perpendicular to AB.
- According to the theorem:
\[ AC^2 = AD^2 + CD^2 \]

3. **Substitute the Known Values**:
- We know AD = 16 cm and CD = 12 cm.
- Substitute these values into the equation:
\[ AC^2 = 16^2 + 12^2 \]

4. **Calculate the Squares**:
- Calculate \(16^2\) and \(12^2\):
\[ 16^2 = 256 \]
\[ 12^2 = 144 \]

5. **Add the Squares**:
- Now add these two results:
\[ AC^2 = 256 + 144 = 400 \]

6. **Find AC**:
- Take the square root to find AC:
\[ AC = \sqrt{400} = 20 \text{ cm} \]

7. **Use the Length of AB**:
- Since AB is the diameter, and we know that AD + BD = AB, we can express AB as:
\[ AB = AD + BD = 16 + x \]

8. **Relate AC and AB**:
- Since AC is the radius, and AB is the diameter:
\[ AB = 2 \times AC = 2 \times 20 = 40 \text{ cm} \]

9. **Set Up the Equation**:
- Now we can set up the equation:
\[ 16 + x = 40 \]

10. **Solve for x (BD)**:
- Rearranging gives:
\[ x = 40 - 16 = 24 \text{ cm} \]

11. **Find BD**:
- Since BD is the remaining part of the diameter:
\[ BD = 24 \text{ cm} \]

### Final Answer:
BD = 24 cm.