AB is a diameter of the circumcircle of `DeltaAPB`, N is the foot of the perpendicular drawn from the point P on AB. If AP=8 cm and BP = 6cm , then the length of BN is

Options

A. 3.6 cm
B. 3 cm
C. 3.4 cm
D. 3.5 cm

Answer

To solve the problem step by step, we will use the properties of triangles and the Pythagorean theorem.

### Given:
- \( AP = 8 \) cm
- \( BP = 6 \) cm
- \( AB \) is the diameter of the circumcircle of triangle \( \Delta APB \)
- \( N \) is the foot of the perpendicular from point \( P \) to line \( AB \)

### To Find:
- Length of \( BN \)

### Step 1: Find the length of \( AB \)
Since \( A \) and \( B \) are endpoints of the diameter and \( P \) lies on the circle, we can use the Pythagorean theorem in triangle \( APB \).

Using the Pythagorean theorem:
\[ AB^2 = AP^2 + BP^2 \]
Substituting the given values:
\[ AB^2 = 8^2 + 6^2 = 64 + 36 = 100 \]
Thus,
\[ AB = \sqrt{100} = 10 \text{ cm} \]

### Step 2: Set up the relationship between segments
Let \( BN = x \). Then, since \( AB = 10 \) cm,
\[ AN = AB - BN = 10 - x \]

### Step 3: Use the Pythagorean theorem in triangles \( APN \) and \( PBN \)
In triangle \( APN \):
\[ AP^2 = AN^2 + PN^2 \]
Substituting the known values:
\[ 8^2 = (10 - x)^2 + PN^2 \]
This simplifies to:
\[ 64 = (10 - x)^2 + PN^2 \quad \text{(1)} \]

In triangle \( PBN \):
\[ PB^2 = BN^2 + PN^2 \]
Substituting the known values:
\[ 6^2 = x^2 + PN^2 \]
This simplifies to:
\[ 36 = x^2 + PN^2 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (2), we can express \( PN^2 \):
\[ PN^2 = 36 - x^2 \quad \text{(3)} \]

Substituting equation (3) into equation (1):
\[ 64 = (10 - x)^2 + (36 - x^2) \]
Expanding \( (10 - x)^2 \):
\[ 64 = 100 - 20x + x^2 + 36 - x^2 \]
This simplifies to:
\[ 64 = 136 - 20x \]
Rearranging gives:
\[ 20x = 136 - 64 \]
\[ 20x = 72 \]
\[ x = \frac{72}{20} = 3.6 \text{ cm} \]

### Conclusion
The length of \( BN \) is \( 3.6 \) cm.