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:

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

This is a right triangle where the rope is the hypotenuse. One leg is the radius of the circle, 5 feet. The other leg is half of the tree's height, 12 feet. We can now use the Pythagorean Theorem $\displaystyle a^2+b^2=c^2$ giving us $\displaystyle 5^2+12^2=169$. If $\displaystyle c^2=169$ then $\displaystyle c=\sqrt{169}=13$.

The ratio of sides to hypotenuse of an isosceles right triangle is always $\displaystyle x: x: x\sqrt2$. With this in mind, setting $\displaystyle x$ as our hypotenuse means we must have leg lengths equal to:

$\displaystyle \frac{x}{\sqrt2} = \frac{x\sqrt2}{2}$

Since the perimeter has two of these legs, we just need to multiply this by $\displaystyle 2$ and add the result to our hypothesis:

$\displaystyle P\Delta = x + 2\left(\frac{x\sqrt2}{2}\right) = x+ x\sqrt2$

So, our perimeter in terms of $\displaystyle x$ is: 

$\displaystyle x+ x\sqrt 2$

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.