Now, this is really your standard $\displaystyle 45-45-90$ triangle. Since it is a right triangle, you know that you have at least one $\displaystyle 90$-degree angle. The other two angles must each be $\displaystyle 45$ degrees, because the triangle is isosceles.

Based on the description of your triangle, you can draw the following figure:

This is derived from your reference triangle for the $\displaystyle 45-45-90$ triangle:

For our triangle, we could call one of the legs $\displaystyle x$. We know, then:

$\displaystyle \frac{x}{26}=\frac{1}{\sqrt{2}}$

Thus, $\displaystyle x=\frac{26}{\sqrt{2}}$.

The area of your triangle is:

$\displaystyle A = \frac{1}{2}bh$

For your data, this is:

$\displaystyle \frac{1}{2}*\frac{26}{\sqrt{2}}*\frac{26}{\sqrt{2}}=\frac{26^2}{4}=169\:cm^2$

Now, this is really your standard $\displaystyle 45-45-90$ triangle. Since it is a right triangle, you know that you have at least one $\displaystyle 90$-degree angle. The other two angles must each be $\displaystyle 45$ degrees because the triangle is isosceles.

Based on the description of your triangle, you can draw the following figure:

This is derived from your reference triangle for the $\displaystyle 45-45-90$ triangle:

For our triangle, we could call one of the legs $\displaystyle x$. We know, then:

$\displaystyle \frac{x}{1}=\frac{9\sqrt{2}}{\sqrt{2}}$

Thus, $\displaystyle x=9$.

The area of your triangle is:

$\displaystyle A = \frac{1}{2}bh$

For your data, this is:

$\displaystyle \frac{1}{2}*9*9=40.5\:cm^2$

Now, this is really your standard $\displaystyle 45-45-90$ triangle. Since it is a right triangle, you know that you have at least one $\displaystyle 90$-degree angle. The other two angles must each be $\displaystyle 45$ degrees because the triangle is isosceles.

Based on the description of your triangle, you can draw the following figure:

This is derived from your reference triangle for the $\displaystyle 45-45-90$ triangle:

For our triangle, we could call one of the legs $\displaystyle x$. We know, then:

$\displaystyle \frac{x}{1}=\frac{7\sqrt{2}}{\sqrt{2}}$

Thus, $\displaystyle x=7$.

The area of your triangle is:

$\displaystyle A = \frac{1}{2}bh$

For your data, this is:

$\displaystyle \frac{1}{2}*7*7=24.5\:cm^2$

Right isosceles triangles (also called "45-45-90 right triangles") are special shapes. In a plane, they are exactly half of a square, and their sides can therefore be expressed as a ratio equal to the sides of a square and the square's diagonal:

$\displaystyle x: x: x\sqrt{2}$, where $\displaystyle x\sqrt{2}$ is the hypotenuse.

In this case, $\displaystyle 46$ maps to $\displaystyle x\sqrt{2}$, so to find the length of a side (so we can use the triangle area formula), just divide the hypotenuse by $\displaystyle \sqrt2$:

$\displaystyle \frac{46}{\sqrt2} = \frac{46}{\sqrt2} \cdot (\frac{\sqrt2}{\sqrt2}) = \frac{46\sqrt{2}}{2} = 23\sqrt{2}$

So, each side of the triangle is $\displaystyle 23\sqrt2$ long. Now, just follow your formula for area of a triangle:

$\displaystyle A\Delta = \frac{lh}{2} = \frac{(23\sqrt{2})(23\sqrt{2})}{2} = \frac{1058}{2} = 529$

Thus, the triangle has an area of $\displaystyle 529\textup{ units}^{2}$.