The sides of an isosceles right triangle (aka a 45-45-90 special right triangle) are in the ratio \(1 : 1 : \sqrt{x}\)

So let's assign the following lengths to the right triangle


Since the perimeter is \(1+\sqrt{2}\), we can write: \(x + x + x\sqrt{2} = 1+\sqrt{2}\)

Simplify the left side: \(2x + x\sqrt{2} = 1+\sqrt{2}\)

Factor the left side: \((2 + \sqrt{2})x = 1+\sqrt{2}\)

Solve for: \(x =\frac{1+\sqrt{2}}{2 + \sqrt{2}}\)

Now that we know the value of x, we can find the area of the triangle.

Area of triangle \(= \frac{1}{2}(base)(height)\)

\(= \frac{1}{2}(\frac{1+\sqrt{2}}{2 + \sqrt{2}})(\frac{1+\sqrt{2}}{2 + \sqrt{2}})\)

\(= \frac{1}{2}(\frac{1+2\sqrt{2} + 2}{4 + 4\sqrt{2} + 2})\)

\(= \frac{1}{2}(\frac{3+2\sqrt{2}}{6 + 4\sqrt{2}})\)

\(= \frac{1}{2}(\frac{3+2\sqrt{2}}{2(3+2\sqrt{2})})\)

\(= \frac{1}{2}(\frac{1}{2})\)

\(=\frac{1}{4}\)

Answer: A

We know, sides of an Isosceles Right Triangle are: \(x\), \(x\), \(\sqrt{2}x\)

So, \(x + x + \sqrt{2}x = 1 + \sqrt{2}\)
\(2x + \sqrt{2}x = 1 + \sqrt{2}\)
\(\sqrt{2}x (1 + \sqrt{2}) = 1 + \sqrt{2}\)
\(\sqrt{2}x = 1\)
\(x = \frac{1}{\sqrt{2}}\)

Area = \(\frac{1}{2}x^2 = \frac{1}{2}(\frac{1}{\sqrt{2}})^2 = \frac{1}{4}\)

Hence, option A