If the length of a chord of a circle , which makes a angle `45^@` with the tangent drawn at one end point of the chord , is 6 cm , then the radius of the circle is :

Options

A. `6 sqrt2` cm
B. 5 cm
C. `3 sqrt2` cm
D. 6 cm

Answer

To solve the problem, we need to find the radius of the circle given that the length of a chord is 6 cm and it makes an angle of 45 degrees with the tangent at one endpoint of the chord.

### Step-by-Step Solution:

1. **Draw the Circle and Chord**:
- Let’s denote the center of the circle as O, the endpoints of the chord as C and D, and the point where the tangent touches the circle as T. The chord CD makes an angle of 45 degrees with the tangent at point T.

2. **Identify the Right Triangle**:
- Since the angle between the chord and the tangent is 45 degrees, we can draw a radius OT to point T, which is perpendicular to the tangent at point T. This creates a right triangle OTD, where OT is the radius (R) and OD is the hypotenuse.

3. **Use the Properties of the Triangle**:
- In triangle OTD, angle ODT is 45 degrees. Therefore, triangle OTD is an isosceles right triangle (45-45-90 triangle). This means that the lengths of the legs (OT and OD) are equal.

4. **Relate the Chord Length to the Radius**:
- The length of the chord CD can be expressed in terms of the radius R. In a 45-45-90 triangle, the relationship between the legs and the hypotenuse is given by:
\[ CD = R \sqrt{2} \]
- We know that the length of the chord CD is 6 cm. Therefore, we can set up the equation:
\[ R \sqrt{2} = 6 \]

5. **Solve for the Radius**:
- To find R, we can rearrange the equation:
\[ R = \frac{6}{\sqrt{2}} \]
- Simplifying this gives:
\[ R = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2} \text{ cm} \]

6. **Conclusion**:
- The radius of the circle is \( 3\sqrt{2} \) cm.

### Final Answer:
The radius of the circle is \( 3\sqrt{2} \) cm.