What if you were given an isosceles right triangle and the length of one of its sides? How could you figure out the lengths of its other sides? After completing this Concept, you'll be able to use the 45-45-90 Theorem to solve problems like this one.

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CK-12 Foundation: Chapter8454590RightTrianglesA

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James Sousa: Trigonometric Function Values of Special Angles

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James Sousa: Solving Special Right Triangles

### Guidance

There are two types of special right triangles, based on their angle measures. The first is an isosceles right triangle. Here, the legs are congruent and, by the Base Angles Theorem, the base angles will also be congruent. Therefore, the angle measures will be \begin{align*}90^\circ, 45^\circ,\end{align*} and \begin{align*}45^\circ\end{align*}. You will also hear an isosceles right triangle called a 45-45-90 triangle. Because the three angles are always the same, all isosceles right triangles are similar.

##### Investigation: Properties of an Isosceles Right Triangle

Tools Needed: Pencil, paper, compass, ruler, protractor

- Construct an isosceles right triangle with 2 in legs. Use the SAS construction that you learned in Chapter 4.
- Find the measure of the hypotenuse. What is it? Simplify the radical.
- Now, let’s say the legs are of length \begin{align*}x\end{align*} and the hypotenuse is \begin{align*}h\end{align*}. Use the Pythagorean Theorem to find the hypotenuse. What is it? How is this similar to your answer in #2?

\begin{align*}x^2 + x^2 & = h^2\\ 2x^2 & = h^2\\ x \sqrt{2} & = h\end{align*}

**45-45-90 Corollary:** If a triangle is an isosceles right triangle, then its sides are in the extended ratio \begin{align*}x : x : x \sqrt{2}\end{align*}.

Step 3 in the above investigation proves the 45-45-90 Triangle Theorem. So, anytime you have a right triangle with congruent legs or congruent angles, then the sides will always be in the ratio \begin{align*}x : x : x \sqrt{2}\end{align*}. The hypotenuse is always \begin{align*}x \sqrt{2}\end{align*} because that is the longest length. This is a specific case of the Pythagorean Theorem, so it will still work, if for some reason you forget this corollary.

#### Example A

Find the length of the missing sides.

Use the \begin{align*}x : x : x \sqrt{2}\end{align*} ratio.

\begin{align*}TV = 6\end{align*} because it is equal to \begin{align*}ST\end{align*}. So, \begin{align*}SV = 6 \sqrt{2}\end{align*} .

#### Example B

Find the length of \begin{align*}x\end{align*}.

Again, use the \begin{align*}x : x : x \sqrt{2}\end{align*} ratio. We are given the hypotenuse, so we need to solve for \begin{align*}x\end{align*} in the ratio.

\begin{align*}x \sqrt{2} &= 16\\ x &= \frac{16}{ \sqrt{2}} \cdot \frac{ \sqrt{2}}{\sqrt{2}} \\ x&== \frac{16 \sqrt{2}}{2} \\x&= 8 \sqrt{2}\end{align*}

Note that we ** rationalized the denominator**. Whenever there is a radical in the denominator of a fraction, multiply the top and bottom by that radical. This will cancel out the radical from the denominator and reduce the fraction.

#### Example C

A square has a diagonal with length 10, what are the lengths of the sides?

Draw a picture.

We know half of a square is a 45-45-90 triangle, so \begin{align*}10=s \sqrt{2}\end{align*}.

\begin{align*}s \sqrt{2} &= 10\\ s &= \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{10 \sqrt{2}}{2}=5 \sqrt{2}\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter8454590RightTrianglesB

### Vocabulary

A ** right triangle** is a triangle with a \begin{align*}90^\circ\end{align*} angle. A

**is a right triangle with angle measures of \begin{align*}45^\circ, 45^\circ\end{align*}, and \begin{align*}90^\circ\end{align*}.**

*45-45-90 triangle*### Guided Practice

1. Find the length of the missing sides.

2. Find the length of \begin{align*}x\end{align*}.

3. \begin{align*}x\end{align*} is the hypotenuse of a 45-45-90 triangle with leg lengths of \begin{align*}5\sqrt{3}\end{align*}.

**Answers:**

1. Use the \begin{align*}x : x : x \sqrt{2}\end{align*} ratio. \begin{align*}AB = 9 \sqrt{2}\end{align*} because it is equal to \begin{align*}AC\end{align*}. So, \begin{align*}BC = 9 \sqrt{2} \cdot \sqrt{2} = 9 \cdot 2 = 18\end{align*}.

2. Use the \begin{align*}x : x : x \sqrt{2}\end{align*} ratio. We need to solve for \begin{align*}x\end{align*} in the ratio.

\begin{align*}12 \sqrt{2} &= x \sqrt{2}\\ 12 &= x\end{align*}

3. \begin{align*}x=5\sqrt{3}\cdot \sqrt{2}=5\sqrt{6}\end{align*}

### Practice

- In an isosceles right triangle, if a leg is \begin{align*}x\end{align*}, then the hypotenuse is __________.
- In an isosceles right triangle, if the hypotenuse is \begin{align*}x\end{align*}, then each leg is __________.
- A square has sides of length 15. What is the length of the diagonal?
- A square’s diagonal is 22. What is the length of each side?
- A square has sides of length \begin{align*}6\sqrt{2}\end{align*}. What is the length of the diagonal?
- A square has sides of length \begin{align*}4 \sqrt{3}\end{align*}. What is the length of the diagonal?
- A baseball diamond is a square with 90 foot sides. What is the distance from home base to second base? (HINT: It’s the length of the diagonal).
- Four isosceles triangles are formed when both diagonals are drawn in a square. If the length of each side in the square is \begin{align*}s\end{align*}, what are the lengths of the legs of the isosceles triangles?

Find the lengths of the missing sides. Simplify all radicals.