# Properties of Right Triangles

Right triangles are special.  They are also useful.  So much of geometry is taking a tiny bit of information and using it to generate all kinds of other information (e.g., the measure of one angle telling you the measure of all of the other angles in the figure!).  Right triangles are super helpful in this process for two reasons:

1. There are lots of rules about right triangles that make it easy to turn a small amount of information into a lot of information.
2. Right triangles are everywhere!  Even if it doesn't look like your figure has a right triangle: rectangles can be cut into two right triangles; many non-right triangles can be split in right triangles; all regular polygons can be split into right triangles; even in circles, radii are often used to create right triangles!

So, if you want to be able to solve a lot of geometry problems, it helps to really know how to use right triangles.  First, know the basic parts of the right triangle:

• Right triangle: A triangle with one right angle.
• Hypotenuse: The side of a right triangle that is opposite the right angle (the hypotenuse is always the longest side).  We refer to this side as $c$.
• Legs: The sides of a right triangle are adjacent to the right angle (the sides that make up the arms of the right angle). One is $a$ and one is $b$, it doesn't matter which one is which.
Once you know those parts, you can apply the two main rules about right triangles:
• Pythagorean Theorem: The rule that says that if you square each of the legs of a right triangle, their sum will equal the square of the hypotenuse.
• Special right triangles: Right triangles whose non-right angles either equal 45º and 45º or 30º and 60º.  If you have the length of even one sides of these special triangles, you can find the lengths of all of the sides.

As we mentioned before, every right triangle has a hypotenuse, which is the longest side, which is always directly across from the right angle (draw a line from the right angle, and you'll cross the hypotenuse). For the purposes of the Pythagorean Theorem, we call the hypotenuse $c$.

The other two sides are called legs.  They are referred to as $a$ and $b$ and it does not matter which one is $a$ and which one is $b$.

Pythagorean Theorem states that $a^2+b^2=c^2$

If you are given the lengths of any two sides of a right triangle, you can figure out the length of the third side using the Pythagorean Theorem.

Example: A right triangle has two legs with length 10.  What is the length of the third side?

Figure out what you know:

\eqalign{\text{Leg: } a&=10\\\text{Leg: } b&=10\\\text{Hypotenuse: }c&=\text{?}}

Plug what you know into the equation and solve:

\eqalign{a^2 + b^2&=c^2\\10^2+10^2&=c^2\\100+100&=c^2\\200&=c^2\\\sqrt{200}&=\sqrt{c^2}\\sqrt{200}&=c \text{ or } 10\sqrt{2}=c}

Sometimes, you won't be given the legs, but the formula works anyway!

Example: A right triangle a hypotenuse of 4 and a leg of 1.  What is the length of the other leg?

Figure out what you know:

\eqalign{\text{Leg: } a&=1\\\text{Leg: } b&=\text{?}\\\text{Hypotenuse: }c&=4}

Plug what you know into the equation and solve:

\eqalign{a^2 + b^2&=c^2\\1^2+b^2&=4^2\\1+b^2&=16\\-1\;\;&=-1\\^2b&=15\\\sqrt{b}&=\sqrt{15}\\b&=\sqrt{15}}

The Pythagorean Theorem works on ALL right triangles.  But there are some other rules that work on special right triangles!  When you're working with special right triangles you can figure out the length of any side of the triangle with just one other length.

There are two special right triangles:

 45-45-90 TrianglesIsosceles right triangles have two $45^{\circ}$ angles. The legs are equal ($s=s$) and the hypotenuse is $s\sqrt{2}$ (which is s times the square root of 2). 30-60-90 Triangles30-60-90 triangles have one 30º angle and one 60º angle. The short leg (which is across from the $30^{\circ}$ angle is $x$. The long leg (across from the $60^{\circ}$ angles is $x\sqrt{3}$. The hypotenuse is $2x$.

Using the special right triangle rules is really a matter of finding the rule that applies to your question, plugging in, and solving.

Example: You have an isosceles right triangle.  The hypotenuse is 15 units long.  How long are the other legs?

First, figure out what kind of triangle you are working with.

This is an isosceles triangle, so it's a 45-45-90.

Next, figure out what leg you have been given.

You have the hypotenuse.

Use the right formula:

Hypotenuse is $s\sqrt{2}$

You know that the hypotenuse is 15, so:

$15=s\sqrt{2}$

Solve for s (which is the length of the other sides -- what you're looking for!).

\eqalign{15&=s\sqrt{2}\\\div \sqrt{2}&=\;\;\div\sqrt{2}\\\dfrac{15}{\sqrt{2}}&=s \text{ or } 10.6=s}

The length of the legs is 10.6.

Sometimes you'll have to solve for one variable and then plug back in to find the answer you're looking for.

Example: You have a right triangle.  One angle has a measure of $60^{\circ}$The longer of the two legs is equal to 27.  What is the length of the hypotenuse?

First, figure out what kind of triangle you are working with.

One of the legs is longer than the other.  This is not an isosceles triangle.  Plus one angle is $60^{\circ}$, so the other must be $30^{\circ}$.  This is a 30-60-90 triangle.

Next, figure out what leg you have been given.

You have the longer leg.

Use the right formula:

Longer leg is is $x\sqrt{3}$

You know that the longer leg is 27, so:

$27=x\sqrt{3}$

Solve for x:

\eqalign{27&=x\sqrt{3}\\\div \sqrt{3}&=\;\;\div\sqrt{3}\\\dfrac{27}{\sqrt{3}}&=x\\\dfrac{3\sqrt{3}}{\sqrt{3}}&=x&&\text{Simplify radicals if you can!}\\3&=x}

You now have the length of the shorter leg.  But look at what the question wants: the hypotenuse.  Solve for the hypotenuse:

Hypotenuse$=2x=2(3)=6$

The hypotenuse is 6 units long.

Overall, keep an eye out for right triangles and special right triangles when you're solving geometry problems.  Pythagorean Theorem and the Special Right Triangle rules are often the keys that unlock an entire problem!

Note: SAT provides Pythagorean Theorem and the Special Right Triangle rules on the first page of every math section.  But, most students should memorize them anyway!