If x be the area of a right angled `triangle`ABC in which `angleABC = 90^(@)` and BC = b, then the length of the altitude BN on the hypotenuse AC is__________

Options

A. `(2b)/sqrt(b^(4)+4x^(2))`
B. `(2bx)/sqrt(b^(4)-4x^(2))`
C. `(4bx)/sqrt(b^(2)+4x^(2))`
D. `(2bx)/sqrt(b^(4)+4x^(2))`

Answer

To find the length of the altitude \( BN \) from vertex \( B \) to the hypotenuse \( AC \) in the right-angled triangle \( ABC \) where \( \angle ABC = 90^\circ \) and \( BC = b \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Area of Triangle**:
The area \( X \) of triangle \( ABC \) can be expressed using the formula for the area of a triangle:
\[ X = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( BC \) as the base and \( AB \) as the height. Thus, we have:
\[ X = \frac{1}{2} \times b \times AB \]
Rearranging gives:
\[ AB = \frac{2X}{b} \]

2. **Apply Pythagorean Theorem**:
Since \( ABC \) is a right triangle, we can apply the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values we have:
\[ AC^2 = \left(\frac{2X}{b}\right)^2 + b^2 \]
Simplifying this gives:
\[ AC^2 = \frac{4X^2}{b^2} + b^2 \]

3. **Find the Length of Hypotenuse \( AC \)**:
To find \( AC \), we take the square root of both sides:
\[ AC = \sqrt{\frac{4X^2}{b^2} + b^2} \]
This can be rewritten as:
\[ AC = \sqrt{\frac{4X^2 + b^4}{b^2}} = \frac{\sqrt{4X^2 + b^4}}{b} \]

4. **Calculate the Length of Altitude \( BN \)**:
The area of triangle \( ABC \) can also be expressed in terms of the hypotenuse \( AC \) and the altitude \( BN \):
\[ X = \frac{1}{2} \times AC \times BN \]
Rearranging gives:
\[ BN = \frac{2X}{AC} \]
Substituting the expression for \( AC \):
\[ BN = \frac{2X}{\frac{\sqrt{4X^2 + b^4}}{b}} = \frac{2Xb}{\sqrt{4X^2 + b^4}} \]

5. **Final Expression for \( BN \)**:
Thus, the length of the altitude \( BN \) is given by:
\[ BN = \frac{2Xb}{\sqrt{4X^2 + b^4}} \]