The length of the common chord of two circles of radii 15 cm and 20 cm whose centres are 25 cm apart is (in cm).

Options

A. 24
B. 25
C. 15
D. 20

Answer

To find the length of the common chord of two circles with radii 15 cm and 20 cm, and whose centers are 25 cm apart, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Radius of the first circle (r1) = 15 cm
- Radius of the second circle (r2) = 20 cm
- Distance between the centers of the circles (d) = 25 cm

2. **Draw the Diagram**:
- Draw two circles with centers O and O' such that the distance OO' = 25 cm.
- Draw the common chord AB where it intersects the line joining the centers at point C.

3. **Set Up Right Triangles**:
- The line segments OA and O'A are the radii of the circles, and OC and O'C are the segments from the centers to the point C where the chord intersects.
- Since the radius is perpendicular to the chord at the point of intersection, triangles OAC and O'A'C are right triangles.

4. **Use the Pythagorean Theorem**:
- For triangle OAC:
\[ OA^2 = OC^2 + AC^2 \]
\[ 15^2 = OC^2 + AC^2 \quad \text{(1)} \]
- For triangle O'A'C:
\[ O'A'^2 = O'C^2 + AC^2 \]
\[ 20^2 = O'C^2 + AC^2 \quad \text{(2)} \]

5. **Express OC and O'C**:
- Let OC = x. Then, O'C = d - OC = 25 - x.

6. **Substituting in the Equations**:
- From equation (1):
\[ 15^2 = x^2 + AC^2 \implies 225 = x^2 + AC^2 \quad \text{(3)} \]
- From equation (2):
\[ 20^2 = (25 - x)^2 + AC^2 \implies 400 = (25 - x)^2 + AC^2 \quad \text{(4)} \]

7. **Expanding Equation (4)**:
- Expand (25 - x)^2:
\[ 400 = 625 - 50x + x^2 + AC^2 \]
- Rearranging gives:
\[ 400 = 625 - 50x + x^2 + AC^2 \implies -225 = -50x + x^2 + AC^2 \]
- Thus:
\[ x^2 + AC^2 - 50x + 225 = 0 \quad \text{(5)} \]

8. **Substituting AC^2 from Equation (3) into Equation (5)**:
- From (3), we have:
\[ AC^2 = 225 - x^2 \]
- Substitute into (5):
\[ x^2 + (225 - x^2) - 50x + 225 = 0 \]
- Simplifying gives:
\[ 450 - 50x = 0 \implies 50x = 450 \implies x = 9 \]

9. **Finding AC**:
- Substitute x back into (3):
\[ 225 = 9^2 + AC^2 \implies 225 = 81 + AC^2 \implies AC^2 = 144 \implies AC = 12 \]

10. **Finding the Length of the Common Chord AB**:
- Since AB = 2 * AC:
\[ AB = 2 \times 12 = 24 \text{ cm} \]

### Final Answer:
The length of the common chord is **24 cm**.