If the perimeter of a right angled isosceles triangle is `sqrt(2) + 1`, then what is the length of the hypotenuse ?

Answer

To find the length of the hypotenuse of a right-angled isosceles triangle with a given perimeter of \( \sqrt{2} + 1 \), we can follow these steps:

### Step 1: Understand the properties of the triangle
In a right-angled isosceles triangle, the two legs are equal in length. Let's denote the length of each leg as \( A \). The hypotenuse can be calculated using the Pythagorean theorem.

### Step 2: Write the formula for the hypotenuse
According to the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Leg}^2 + \text{Leg}^2 \]
Since both legs are equal, we can write:
\[ \text{Hypotenuse}^2 = A^2 + A^2 = 2A^2 \]
Thus, the hypotenuse \( AC \) is:
\[ AC = \sqrt{2A^2} = A\sqrt{2} \]

### Step 3: Write the formula for the perimeter
The perimeter \( P \) of the triangle is the sum of all its sides:
\[ P = AB + BC + AC = A + A + A\sqrt{2} = 2A + A\sqrt{2} \]
We know from the problem statement that the perimeter is \( \sqrt{2} + 1 \):
\[ 2A + A\sqrt{2} = \sqrt{2} + 1 \]

### Step 4: Rearrange the equation
We can rearrange the equation to isolate \( A \):
\[ 2A + A\sqrt{2} = \sqrt{2} + 1 \]
This can be factored as:
\[ A(2 + \sqrt{2}) = \sqrt{2} + 1 \]

### Step 5: Solve for \( A \)
Now, we can solve for \( A \):
\[ A = \frac{\sqrt{2} + 1}{2 + \sqrt{2}} \]

### Step 6: Rationalize the denominator
To simplify \( A \), we can multiply the numerator and denominator by the conjugate of the denominator:
\[ A = \frac{(\sqrt{2} + 1)(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} \]
Calculating the denominator:
\[ (2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2 \]
Calculating the numerator:
\[ (\sqrt{2} + 1)(2 - \sqrt{2}) = 2\sqrt{2} - 2 + \sqrt{2} - 1 = 3 - \sqrt{2} \]
Thus, we have:
\[ A = \frac{3 - \sqrt{2}}{2} \]

### Step 7: Find the hypotenuse
Now we substitute \( A \) back into the hypotenuse formula:
\[ AC = A\sqrt{2} = \left(\frac{3 - \sqrt{2}}{2}\right)\sqrt{2} = \frac{(3 - \sqrt{2})\sqrt{2}}{2} \]
This simplifies to:
\[ AC = \frac{3\sqrt{2} - 2}{2} \]

### Step 8: Final answer
The length of the hypotenuse \( AC \) is:
\[ \frac{3\sqrt{2} - 2}{2} \]