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45-45-90 Right Triangles

Leg times sqrt(2) equals hypotenuse.

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45-45-90 Triangles
License: CC BY-NC 3.0

Audrey, has decided to make her classroom look more like a Math class by putting wall paper squares on the doors of the storage cabinets. She has chosen paper that shows squares, rectangles and triangles.

License: CC BY-NC 3.0

If the length of one of the legs of the \begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangles in the wallpaper is 8 inches in length, then how can Audrey figure out the length of the diagonal of the wallpaper square?

In this concept, you will learn to identify isosceles right triangles.

Isosceles Triangle

A triangle that has two sides equal in length and two angles at the base of these sides equal in measure is called an isosceles triangle. If this concept is applied to a right triangle, then the two sides equal in length have to be the legs of the right triangle. Remember the longest side of the right triangle is called the hypotenuse and it is opposite the 90° angle. The two acute angles of a right triangle have a sum of 90° angle. If the legs are equal in length then the two acute angles of the triangle will each measure 45°. The following is an isosceles right triangle.

License: CC BY-NC 3.0

The angles of an isosceles right triangle will always have the measures of \begin{align*} 45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*}. This triangle is often referred to as a \begin{align*}45-45-90\end{align*} right triangle. The legs of an isosceles right triangle are equal in measure such that \begin{align*}a = b\end{align*}. The length of the hypotenuse can be found using the Pythagorean Theorem.

First, determine the values of \begin{align*}(a, b, c)\end{align*} for the Pythagorean Theorem.

\begin{align*}\begin{array}{rcl} a &=& a \\ b &=& a \\ c &=& ? \end{array}\end{align*}

Next, substitute the values for \begin{align*}(a, b, c)\end{align*} into the Pythagorean Theorem.

\begin{align*}\begin{array}{rcl} c^2 &=& a^2 + b ^2 \\ c^2 &=& (a) ^ 2 + (a)^2 \\ \end{array}\end{align*}

Next, perform the indicated squaring on the right side of the equation and simplify.

\begin{align*}\begin{array}{rcl} c^2 &=& (a) ^ 2 + (a)^2 \\ c^2 &=& a^2 + a ^2 \\ c^2 &=& 2a^2 \\ \end{array}\end{align*}

Then, solve for the variable ‘\begin{align*}c\end{align*}’ by taking the square root of both sides of the equation. Express the answer as a radical.

\begin{align*}\begin{array}{rcl} c^2 &=& 2a^2 \\ \sqrt{c^2} &=& \sqrt{2} \cdot \sqrt{a^2} \\ c&=& \sqrt{2} \cdot a \\ c&=& a \sqrt{2} \\ \end{array}\end{align*}

The answer is \begin{align*}a \sqrt{2}.\end{align*}

The length of the hypotenuse is \begin{align*}a \sqrt{2} \ cm\end{align*}. Since \begin{align*}a = b \end{align*} the length of the hypotenuse could be expressed as \begin{align*}b \sqrt{2} \ cm\end{align*}.

The coefficient ‘\begin{align*}a\end{align*}’ in front of \begin{align*}\sqrt{2} \end{align*} is the length of the legs of the given triangle. The length of the hypotenuse of an isosceles right triangle will always equal the product of the leg length and \begin{align*}\sqrt{2}\end{align*}.

If a \begin{align*} 45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} right triangle had a leg length of 5 inches then the length of the hypotenuse would be \begin{align*}5 \times \sqrt{2} = 5 \sqrt{2}\ in.\end{align*}

Examples

Example 1

Earlier, you were given a problem about Audrey and the wallpaper squares. She wants to figure out the length of the diagonal of the wallpaper square.

First, write down what you know.

\begin{align*}\begin{array}{rcl} a &=& 8 \ in = \ 8 \\ a &=& b \quad \ = \ 8 \\ c &=& ? \\ \end{array}\end{align*}

Next, write down an equation to show the relationship between the length of the hypotenuse and the length of a leg.

\begin{align*}c = a\sqrt{2}\end{align*}

Next, substitute what you know into the equation.

\begin{align*}\begin{array}{rcl} c &=& a\sqrt{2} \\ c &=& 8\sqrt{2} \\ \end{array}\end{align*}

Remember there are four \begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangles along the diagonal of the wallpaper square.

Then, multiply the value of ‘\begin{align*}c\end{align*}’ by the number of triangles.

\begin{align*}\begin{array}{rcl} c &=& 8\sqrt{2} \\ c &=& 4(8\sqrt{2}) \\ c &=& 32 \sqrt{2} \\ \end{array} \end{align*}The answer is \begin{align*}32 \sqrt{2}\end{align*}.

The length of the diagonal of the entire wallpaper square is \begin{align*}32 \sqrt{2}\end{align*} inches.

Example 2

The length of the leg of the hypotenuse of a \begin{align*} 45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangle is 9 centimeters. What is the length of the hypotenuse?

First, write down what you know.

For the given \begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangle \begin{align*}a = b = 9\end{align*}.

The length of the hypotenuse ‘\begin{align*}c\end{align*}’ is \begin{align*}a\sqrt{2}\end{align*}.

Then, substitute ‘9’ for ‘\begin{align*}a\end{align*}’ to find the length of the hypotenuse.

\begin{align*}\begin{array}{rcl} c &=& a\sqrt{2} \\ c &=& 9\sqrt{2} \\ \end{array}\end{align*}

The answer is \begin{align*}9 \sqrt{2}.\end{align*}

The length of the hypotenuse is \begin{align*} 9 \sqrt{2} \ cm\end{align*}.

Example 3

For the given isosceles, right triangle, find the length of a leg.

License: CC BY-NC 3.0

First, write down what you know.

\begin{align*}\begin{array}{rcl} AC &=& a \ = \ ? \ m \\ a &=& b \ \ = \ ? \ m \\ c &=& 10 \ m \ = \ 10 \\ \end{array}\end{align*}

Next, write down an equation to show the relationship between the length of the hypotenuse and the length of a leg.

\begin{align*}c= AC \sqrt{2}\end{align*}Next, substitute what you know into the equation.

\begin{align*}\begin{array}{rcl} c&=& AC \sqrt{2} \\ 10&=& a \sqrt{2} \\ \end{array}\end{align*}

Next, isolate the variable ‘\begin{align*}a\end{align*}’ by dividing both sides of the equation by \begin{align*}\sqrt{2}\end{align*} and simplify the equation.

\begin{align*}\begin{array}{rcl} 10 &=& a\sqrt{2} \\ \frac{10}{\sqrt{2}} &=& \frac{a{\cancel{\sqrt{2}}}} { \cancel{\sqrt{2}}} \\ \frac{10}{\sqrt{2}} &=& a \\ \end{array}\end{align*}

Next, multiply the left side of the denominator by \begin{align*}\frac{\sqrt{2}}{\sqrt{2}}\end{align*} to clear the radical from the denominator.

\begin{align*}\begin{array}{rcl} \frac{10}{\sqrt{2}} &=& a \\ \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} &=& a \\ \frac{10 \sqrt{2}}{\sqrt{4}} &=& a \\ \end{array}\end{align*}

Then, simplify the left side of the equation to solve for the variable.

\begin{align*}\begin{array}{rcl} \frac{10 \sqrt{2}}{\sqrt{4}} &=& a \\ \frac{\overset{5} {\cancel{10}} \sqrt{2} }{{\cancel{2}}} &=& a \\ 5 \sqrt{2} &=& a \\ \end{array}\end{align*}

The answer is \begin{align*}5\sqrt{2}\end{align*}.

The length of a leg is \begin{align*}5 \sqrt{2}\end{align*} meters.

Example 4

The distance between the bases of a baseball diamond is 90 feet. If the umpire runs directly from home plate to second base, how far will he have to run? Round the answer to the nearest tenth.

First, draw and label a diagram to model the problem.

License: CC BY-NC 3.0

First, write down what you know.

\begin{align*}\begin{array}{rcl} AC &=& a \ = \ ? \ m \\ a &=& b \ = \ ? \ m \\ c &=& 10 \ m \ = \ 10 \\ \end{array}\end{align*}

Next, write down an equation to show the relationship between the length of the hypotenuse and the length of a leg.

\begin{align*}C = AC \sqrt{2}\end{align*}

Next, substitute what you know into the equation.

\begin{align*}\begin{array}{rcl} c&=& AC \sqrt{2} \\ 10&=& a \sqrt{2} \\ \end{array}\end{align*}

Then, using the calculator, perform the operations indicated on the right side of the equation.

\begin{align*}\begin{array}{rcl} c&=& 90 \sqrt{2} \\ c &=& 90(1.41) \\ c&=& 126.9 \\ \end{array}\end{align*}The answer is 126.9.

The umpire runs 126.9 feet.

Review

Find the length of the hypotenuse in each of the following \begin{align*}45 ^\circ - 45 ^\circ - 90 ^\circ\end{align*} triangles.

1. \begin{align*}\text{Length of each leg} = 5\end{align*}

2. \begin{align*}\text{Length of each leg} = 4\end{align*}

3. \begin{align*}\text{Length of each leg} = 6\end{align*}

4. \begin{align*}\text{Length of each leg} = 3\end{align*}

5. \begin{align*}\text{Length of each leg} = 7\end{align*}

Use a calculator to figure out the approximate value of each hypotenuse. You may round to the nearest hundredth.

6. \begin{align*} 5\sqrt{2}\end{align*} 

7. \begin{align*} 4\sqrt{2}\end{align*}

8. \begin{align*}6\sqrt{2}\end{align*}

9. \begin{align*} 3\sqrt{2}\end{align*}

10. \begin{align*}7\sqrt{2}\end{align*}

11. \begin{align*} 8\sqrt{2}\end{align*}

12. \begin{align*}10\sqrt{2}\end{align*}

13. \begin{align*} 13\sqrt{2}\end{align*}

14. \begin{align*}21\sqrt{2}\end{align*}

15. \begin{align*} 17\sqrt{2}\end{align*}

Review (Answers)

To see the Review answers, open this PDF file and look for section 7.11.  

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Vocabulary

45-45-90 Theorem

For any isosceles right triangle, if the legs are x units long, the hypotenuse is always x\sqrt{2}.

45-45-90 Triangle

A 45-45-90 triangle is a special right triangle with angles of 45^\circ, 45^\circ, and 90^\circ.

Hypotenuse

The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.

Isosceles Right Triangle

An isosceles right triangle is a triangle with a ninety degree angle and exactly two sides that are the same length.

Legs of a Right Triangle

The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.

Radical

The \sqrt{}, or square root, sign.

Image Attributions

  1. [1]^ License: CC BY-NC 3.0
  2. [2]^ License: CC BY-NC 3.0
  3. [3]^ License: CC BY-NC 3.0
  4. [4]^ License: CC BY-NC 3.0
  5. [5]^ License: CC BY-NC 3.0

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