It's helpful to remember upon coming across a 45/45/90 triangle that it's a special right triangle. This means that its sides can easily be calculated by using a derived side ratio:

$\displaystyle s:s:s\sqrt2$

Here, $\displaystyle s$ represents the length of one of the legs of the 45/45/90 triangle, and $\displaystyle s\sqrt2$ represents the length of the triangle's hypotenuse. Two sides are denoted as congruent lengths ($\displaystyle s$) because this special triangle is actually an isosceles triangle. This goes back to the fact that two of its angles are congruent. 

Therefore, using the side rules mentioned above, if $\displaystyle 1\sqrt2=s\sqrt2$, this problem can be resolved by solving for the value of $\displaystyle s$:

$\displaystyle 1\sqrt{2}=s\sqrt{2}$

$\displaystyle \frac{1\sqrt{2}}{\sqrt{2}}=s$

$\displaystyle 1=s$

Therefore, the length of one of the legs is 1.

If the hypotenuse of a 45-45-90 triangle is provided, its side length can only be one length, since the sides of all 45-45-90 triangles exist in a defined ratio of $\displaystyle 1:1:\sqrt2$, where $\displaystyle 1$ represents the length of one of the triangle's legs and $\displaystyle \sqrt2$ represents the length of the triangle's hypotenuse. Using this method, you can set up a proportion and solve for the length of one of the triangle's sides:

$\displaystyle \frac{1}{\sqrt2}=\frac{x}{5\:cm}$

Cross-multiply and solve for $\displaystyle x$.

$\displaystyle \sqrt2\cdot x=5\:cm$

$\displaystyle x=\frac{5}{\sqrt2}\:cm$

Rationalize the denominator.

$\displaystyle \frac{5}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{5\sqrt2}{2}\:cm$

You can also solve this problem using the Pythagorean Theorem.

$\displaystyle a^2+b^2=c^2$

In a 45-45-90 triangle, the side legs will be equal, so $\displaystyle a=b$. Substitute $\displaystyle a$ for $\displaystyle b$ and rewrite the formula.

$\displaystyle a^2 + a^2= c^2$

Substitute the provided length of the hypothenuse and solve for $\displaystyle a$.

$\displaystyle 2a^2= 5^2$

$\displaystyle 2a^2 = 25$

$\displaystyle a^2 = \frac{25}{2}$

$\displaystyle a=\sqrt{\frac{25}{2}}\:cm$

While the answer looks a little different from the result of our first method of solving this problem, the two represent the same value, just written in different ways.

$\displaystyle \sqrt{}\frac{25}{2}=\frac{\sqrt{5\cdot 5}}{\sqrt2}=\frac{5}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{5\sqrt2}{2}$

1. Remember that this is a special right triangle where the ratio of the sides is:

$\displaystyle x:x:x\sqrt{2}$

In this case that makes it:

$\displaystyle 5:5:5\sqrt{2}$

2. Find the perimeter by adding the side lengths together:

$\displaystyle 5+5+5\sqrt{2}=10+5\sqrt{2}=17.07$

Remember that this is a special right triangle where the ratio of the sides is:

$\displaystyle x:x:x\sqrt{2}$

In this case that makes it:

$\displaystyle 3:3:3\sqrt{2}$

Where  $\displaystyle 3\sqrt{2}$  is the length of the hypotenuse.