N is the foot of the perpendicular from a point P of a circle with radius 7 cm, on a diameter AB of the circle. If the length of the chord PB is 12 cm, the distance of the point N from the point B is

Options

A. `6 5/7 cm`
B. `12 2/7 cm`
C. `3 5/7 cm`
D. `10 2/7 cm`

Answer

To solve the problem, we need to find the distance of point N from point B, given the radius of the circle and the length of the chord PB.

### Step-by-Step Solution:

1. **Understand the Circle and its Components**:
- We have a circle with radius \( r = 7 \) cm.
- The diameter \( AB \) of the circle is \( 2r = 14 \) cm.
- Point \( P \) is outside the diameter \( AB \), and \( N \) is the foot of the perpendicular from point \( P \) to line \( AB \).

2. **Identify the Chord**:
- The length of the chord \( PB \) is given as \( 12 \) cm.

3. **Use the Right Triangle**:
- Since \( N \) is the foot of the perpendicular from \( P \) to \( AB \), we can form a right triangle \( PBN \) where:
- \( PB \) is the hypotenuse (12 cm).
- \( BN \) is one leg (which we need to find).
- \( PN \) is the other leg.

4. **Apply the Pythagorean Theorem**:
- According to the Pythagorean theorem:
\[ PB^2 = PN^2 + BN^2 \]
- Plugging in the values we have:
\[ 12^2 = PN^2 + BN^2 \]
- This simplifies to:
\[ 144 = PN^2 + BN^2 \]

5. **Find the Length of PN**:
- Since \( P \) is on the circle and \( N \) is the foot of the perpendicular, we can also use the radius:
- The radius \( PA \) is \( 7 \) cm.
- In triangle \( PAN \) (another right triangle):
\[ PA^2 = PN^2 + AN^2 \]
- We know \( PA = 7 \) cm, so:
\[ 7^2 = PN^2 + AN^2 \]
- This simplifies to:
\[ 49 = PN^2 + AN^2 \]

6. **Express AN in terms of BN**:
- Since \( AB = AN + BN \) and \( AB = 14 \) cm, we can express \( AN \) as:
\[ AN = 14 - BN \]

7. **Substitute AN in the Equation**:
- Substitute \( AN \) into the equation from step 5:
\[ 49 = PN^2 + (14 - BN)^2 \]
- Expanding this gives:
\[ 49 = PN^2 + (196 - 28BN + BN^2) \]
- Rearranging gives:
\[ PN^2 + BN^2 - 28BN + 196 - 49 = 0 \]
- This simplifies to:
\[ PN^2 + BN^2 - 28BN + 147 = 0 \]

8. **Substituting PN^2 from Step 4**:
- Substitute \( PN^2 = 144 - BN^2 \) into the equation:
\[ 144 - BN^2 + BN^2 - 28BN + 147 = 0 \]
- This simplifies to:
\[ 291 - 28BN = 0 \]
- Solving for \( BN \):
\[ 28BN = 291 \implies BN = \frac{291}{28} \approx 10.39 \text{ cm} \]

### Final Answer:
The distance of point N from point B is approximately \( 10.39 \) cm.