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