If the sum of the lengths of the hypotenuse and another side of a right-angled triangle is given, the area of the triangle us maximum when the angle between these sides is

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To solve the problem, we need to find the angle between the hypotenuse and another side of a right-angled triangle when the area of the triangle is maximized, given that the sum of the lengths of the hypotenuse and the other side is constant.

### Step-by-Step Solution:

1. **Define the Variables:**
Let the length of the hypotenuse be \( x \) and the length of the other side (base) be \( y \). We are given that \( x + y = k \), where \( k \) is a constant.

2. **Express \( y \) in Terms of \( x \):**
From the equation \( x + y = k \), we can express \( y \) as:
\[ y = k - x \]

3. **Area of the Triangle:**
The area \( A \) of a right-angled triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \( y \) as the base and the height can be derived from the Pythagorean theorem. The height can be expressed as:
\[ \text{height} = \sqrt{x^2 - y^2} \]
Therefore, the area becomes:
\[ A = \frac{1}{2} \times y \times \sqrt{x^2 - y^2} \]

4. **Substituting \( y \):**
Substituting \( y = k - x \) into the area formula:
\[ A = \frac{1}{2} \times (k - x) \times \sqrt{x^2 - (k - x)^2} \]

5. **Simplifying the Area Expression:**
We simplify the expression under the square root:
\[ x^2 - (k - x)^2 = x^2 - (k^2 - 2kx + x^2) = 2kx - k^2 \]
Thus, the area becomes:
\[ A = \frac{1}{2} (k - x) \sqrt{2kx - k^2} \]

6. **Maximizing the Area:**
To maximize the area, we differentiate \( A \) with respect to \( x \) and set the derivative equal to zero:
\[ \frac{dA}{dx} = 0 \]

7. **Finding Critical Points:**
After differentiating and simplifying, we will find the value of \( x \) that maximizes the area. The calculations lead to:
\[ x = \frac{2k}{3} \]
Substituting back to find \( y \):
\[ y = k - x = k - \frac{2k}{3} = \frac{k}{3} \]

8. **Finding the Angle \( \theta \):**
To find the angle \( \theta \) between the hypotenuse and the side \( y \), we use the cosine ratio:
\[ \cos \theta = \frac{y}{x} = \frac{\frac{k}{3}}{\frac{2k}{3}} = \frac{1}{2} \]
Thus, we find:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer:
The angle between the hypotenuse and the other side when the area of the triangle is maximized is \( 60^\circ \).