Recall that an isosceles right triangle is a \(\displaystyle 45-45-90\) triangle. That means that it looks like this:

This makes calculating the area very easy! Recall, the area of a triangle is defined as:

\(\displaystyle A = \frac{1}{2}bh\)

However, since \(\displaystyle b=h=x\) for our triangle, we know:

\(\displaystyle A=\frac{1}{2}x^2\)

Now, we know that \(\displaystyle A=80\:in^2\). Therefore, we can write:

\(\displaystyle 80=\frac{1}{2}x^2\)

Solving for \(\displaystyle x\), we get:

\(\displaystyle 160=x^2\)

\(\displaystyle x=\sqrt{160}= \sqrt{16}*\sqrt{10}=4\sqrt{10}\:in\)

This is the length of the height of the triangle for the side that is not the hypotenuse.

Based on the information given, you know that your triangle looks as follows:

This is a \(\displaystyle 45-45-90\) triangle. Recall your standard \(\displaystyle 45-45-90\) triangle:

You can set up the following ratio between these two figures:

\(\displaystyle \frac{x}{1}=\frac{18}{\sqrt{2}}\)

Now, the area of the triangle will merely be \(\displaystyle \frac{1}{2}x^2\) (since both the base and the height are \(\displaystyle x\)). For your data, this is:

\(\displaystyle A=\frac{1}{2}* \frac{18}{\sqrt{2}} * \frac{18}{\sqrt{2}}=\frac{18*18}{4}=9*9=81\:cm^2\)

To solve simply realize the hypotenuse of one of these triangles is of the form \(\displaystyle s\sqrt2\) where s is side length. Thus, our answer is \(\displaystyle 5\).

Recall that an isosceles right triangle is a \(\displaystyle 45-45-90\) triangle. That means that it looks like this:

This makes calculating the area very easy! Recall, the area of a triangle is defined as:

\(\displaystyle A = \frac{1}{2}bh\)

However, since \(\displaystyle b=h=x\) for our triangle, we know:

\(\displaystyle A=\frac{1}{2}x^2\)

Now, we know that \(\displaystyle A=100\:cm^2\). Therefore, we can write:

\(\displaystyle 100=\frac{1}{2}x^2\)

Solving for \(\displaystyle x\), we get:

\(\displaystyle 200=x^2\)

\(\displaystyle x=\sqrt{200}= \sqrt{2}*\sqrt{100}=10\sqrt{2}\)

However, be careful! Notice what the question asks: "What is its height that is correlative and perpendicular to this triangle's hypotenuse?" First, let's find the hypotenuse of the triangle. Recall your standard \(\displaystyle 45-45-90\) triangle:

Since one of your sides is \(\displaystyle 10\sqrt{2}\), your hypotenuse is \(\displaystyle 10\sqrt{2} * \sqrt{2}=20\:cm\).

Okay, what you are actually looking for is \(\displaystyle h\) in the following figure:

Therefore, since you know the area, you can say:

\(\displaystyle 100=\frac{1}{2}\ast h*20\)

Solving, you get: \(\displaystyle h=10\). 

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 \(\displaystyle 529\textup{ units}^{2}\).

Your right triangle is a \(\displaystyle 45-45-90\) triangle. It thus looks like this:

Now, you know that you also have a reference triangle for \(\displaystyle 45-45-90\) triangles. This is:

This means that you can set up a ratio to find \(\displaystyle x\). It would be:

\(\displaystyle \frac{x}{1}=\frac{20}{\sqrt{2}}\)

Your triangle thus could be drawn like this:

Now, notice that you can rationalize the denominator of \(\displaystyle \frac{20}{\sqrt{2}}\):

\(\displaystyle \frac{20}{\sqrt{2}} = \frac{20}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}=\frac{20\sqrt{2}}{2}=10\sqrt{2}\)

Thus, the perimeter of your figure is:

\(\displaystyle 10\sqrt{2} + 10\sqrt{2} + 20 = 20\sqrt{2}+20\:cm\)

Recall that an isosceles right triangle is also a \(\displaystyle 45-45-90\) triangle. Your reference figure for such a shape is:

 or 

Now, you know that the area of a triangle is:

\(\displaystyle A=\frac{1}{2}bh\)

For this triangle, though, the base and height are the same. So it is:

\(\displaystyle A=\frac{1}{2}x^2\)

Now, we have to be careful, given that our area contains \(\displaystyle x\). Let's use \(\displaystyle s\), for "side length":

\(\displaystyle 72x^2=\frac{1}{2}s^2\)

\(\displaystyle s^2=144x^2\)

Thus, \(\displaystyle s=12x\:in\). Now based on the reference figure above, you can easily see that your triangle is:

Therefore, your perimeter is:

\(\displaystyle 12x+12x+12x\sqrt{2} = 24x+12x\sqrt{2}\:in\)