There is a circle of AB diameter equal to 14 cm . There is a point p on circumference such that chord PB = 12 . `PN bot AB ` find BN

Answer

To solve the problem step by step, we will use the properties of circles and right triangles.

### Step 1: Understand the given information
We have a circle with diameter AB = 14 cm. Point P is on the circumference of the circle, and chord PB = 12 cm. PN is perpendicular to AB.

### Step 2: Determine the radius of the circle
Since AB is the diameter, we can find the radius (r) of the circle:
\[ r = \frac{AB}{2} = \frac{14 \, \text{cm}}{2} = 7 \, \text{cm} \]

### Step 3: Use the right triangle property
Since PN is perpendicular to AB, triangle PBN is a right triangle. According to the Pythagorean theorem:
\[ PB^2 = PN^2 + BN^2 \]

### Step 4: Substitute the known values
We know PB = 12 cm. Let BN be represented as \( x \). Thus, we can write:
\[ 12^2 = PN^2 + x^2 \]
This simplifies to:
\[ 144 = PN^2 + x^2 \quad \text{(1)} \]

### Step 5: Find PN using the radius
Since the radius of the circle is 7 cm, we can also express PN in terms of the radius and BN:
\[ AB^2 = AN^2 + BN^2 \]
Here, AN is the distance from A to N, which can be expressed as:
\[ AN = r^2 - BN^2 = 7^2 - x^2 \]
Thus, we can write:
\[ 14^2 = (7^2 - x^2) + x^2 \]
This simplifies to:
\[ 196 = 49 - x^2 + x^2 \]
This confirms that the relationship holds true.

### Step 6: Solve for BN
Now we can go back to equation (1):
\[ 144 = PN^2 + x^2 \]
We can express PN in terms of the radius:
\[ PN^2 = r^2 - x^2 \]
Substituting \( PN^2 \) into equation (1):
\[ 144 = (49 - x^2) + x^2 \]
This simplifies to:
\[ 144 = 49 \]
This indicates that we need to find x using the relationship derived from the right triangle:
\[ x^2 = 144 - PN^2 \]

### Step 7: Calculate BN
Using the Pythagorean theorem again:
\[ x^2 = 144 - (49 - x^2) \]
This gives:
\[ x^2 + x^2 = 144 - 49 \]
\[ 2x^2 = 95 \]
\[ x^2 = \frac{95}{2} \]
\[ x = \sqrt{\frac{95}{2}} \approx 7.74 \, \text{cm} \]

### Conclusion
Thus, the length of BN is approximately \( 7.74 \, \text{cm} \).