The chord of a circle of radius `10cm` subtends a right angle at its centre. The length of the chord (in cm) is

Options

A. `(5)/(sqrt(2))`
B. `5sqrt(2)`
C. `10sqrt(2)`
D. `10sqrt(3)`

Explanation

To find the length of the chord that subtends a right angle at the center of a circle with a radius of 10 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circle and Chord**:
- We have a circle with a radius \( r = 10 \) cm.
- A chord subtends a right angle (90 degrees) at the center of the circle.

2. **Visualizing the Triangle**:
- When a chord subtends a right angle at the center, it forms an isosceles triangle with the two radii of the circle. Let's denote the endpoints of the chord as points A and B, and the center of the circle as point O.
- Therefore, triangle OAB is an isosceles triangle with OA = OB = radius = 10 cm and angle AOB = 90 degrees.

3. **Using the Pythagorean Theorem**:
- In triangle OAB, we can apply the Pythagorean theorem.
- The length of the chord AB can be found using the relationship in the right triangle formed by the radius and half of the chord.
- Let M be the midpoint of the chord AB. Then, OM is perpendicular to AB, and AM = MB = \( \frac{AB}{2} \).

4. **Applying the Pythagorean Theorem**:
- In triangle OAM:
\[ OA^2 = OM^2 + AM^2 \]
- Here, \( OA = 10 \) cm (radius), and \( OM \) is the height from the center to the midpoint of the chord. Since angle AOB is 90 degrees, OM can be calculated as:
\[ OM = OA \cdot \cos(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = 5\sqrt{2} \text{ cm} \]
- Now substituting back into the Pythagorean theorem:
\[ 10^2 = (5\sqrt{2})^2 + AM^2 \]
\[ 100 = 50 + AM^2 \]
\[ AM^2 = 100 - 50 = 50 \]
\[ AM = \sqrt{50} = 5\sqrt{2} \text{ cm} \]

5. **Finding the Length of the Chord**:
- Since \( AB = 2 \times AM \):
\[ AB = 2 \times 5\sqrt{2} = 10\sqrt{2} \text{ cm} \]

### Final Answer:
The length of the chord is \( 10\sqrt{2} \) cm.

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