What is the radius of a sphere, with centre at the origin, that passes through point (2,3,4)?

Options

A. 3
B. 3.31
C. 3.32
D. 5.3

Explanation

To find the radius of a sphere centered at the origin (0, 0, 0) that passes through the point (2, 3, 4), we can use the distance formula in three-dimensional space.

### Step-by-Step Solution:

1. **Identify the center and the point on the sphere**:
- The center of the sphere is at the origin, which is the point (0, 0, 0).
- The point through which the sphere passes is (2, 3, 4).

2. **Use the distance formula**:
- The radius \( r \) of the sphere can be calculated using the distance formula between the center of the sphere and the point on the sphere:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
- Here, \( (x_1, y_1, z_1) = (0, 0, 0) \) and \( (x_2, y_2, z_2) = (2, 3, 4) \).

3. **Substitute the coordinates into the formula**:
\[ r = \sqrt{(2 - 0)^2 + (3 - 0)^2 + (4 - 0)^2} \]

4. **Calculate each term**:
- Calculate \( (2 - 0)^2 = 2^2 = 4 \)
- Calculate \( (3 - 0)^2 = 3^2 = 9 \)
- Calculate \( (4 - 0)^2 = 4^2 = 16 \)

5. **Add the squared terms**:
\[ r = \sqrt{4 + 9 + 16} \]
\[ r = \sqrt{29} \]

6. **Approximate the value of the radius**:
- The approximate value of \( \sqrt{29} \) is about 5.385.
- Rounding this to one decimal place gives approximately 5.4.

### Final Answer:
The radius of the sphere is \( \sqrt{29} \) or approximately 5.4.