In a circle of radius 10 cm, a chord is drawn 6 cm from the centre. If a chord half the length of the original chord were drawn, its distance in centimeters from the centre would be

Options

A. `sqrt(84)`
B. 9
C. 8
D. `3 pi`

Answer

To solve the problem step by step, we will follow these instructions:

### Given:
- Radius of the circle (r) = 10 cm
- Distance of the original chord from the center (d) = 6 cm

### Step 1: Find the length of the original chord
Using the Pythagorean theorem in triangle OAP (where O is the center, A is one endpoint of the chord, and P is the foot of the perpendicular from O to the chord):

- OA = radius = 10 cm
- OP = distance from the center to the chord = 6 cm
- AP = half the length of the chord

Using the Pythagorean theorem:
\[ OA^2 = OP^2 + AP^2 \]
\[ 10^2 = 6^2 + AP^2 \]
\[ 100 = 36 + AP^2 \]
\[ AP^2 = 100 - 36 \]
\[ AP^2 = 64 \]
\[ AP = \sqrt{64} = 8 \text{ cm} \]

Since AP is half the length of the chord AB, the total length of the chord AB is:
\[ AB = 2 \times AP = 2 \times 8 = 16 \text{ cm} \]

### Step 2: Find the distance from the center for the new chord
The new chord is half the length of the original chord:
\[ CD = \frac{1}{2} AB = \frac{1}{2} \times 16 = 8 \text{ cm} \]

Let N be the midpoint of the new chord CD. Thus, CN = DN = 4 cm.

### Step 3: Use the Pythagorean theorem again
Now we will find the distance ON from the center O to the new chord CD using triangle OCN:
- OC = radius = 10 cm
- CN = half the length of the new chord = 4 cm
- ON = distance from the center to the new chord

Using the Pythagorean theorem:
\[ OC^2 = ON^2 + CN^2 \]
\[ 10^2 = ON^2 + 4^2 \]
\[ 100 = ON^2 + 16 \]
\[ ON^2 = 100 - 16 \]
\[ ON^2 = 84 \]
\[ ON = \sqrt{84} \]

### Final Answer:
The distance from the center to the new chord is:
\[ ON = \sqrt{84} \text{ cm} \]