Geometry (B): READ: Special Right Triangles Reading

Special Right Triangles

Learning Objectives

Identify and use the ratios involved with right isosceles triangles.
Identify and use the ratios involved with $30^\circ-60^\circ-90^\circ$ triangles.
Identify and use ratios involved with equilateral triangles.
Employ right triangle ratios when solving real-world problems.

Introduction

What happens when you cut an equilateral triangle in half using an altitude? You get two right triangles. What about a square? If you draw a diagonal across a square you also get two right triangles. These two right triangles are special special right triangles called the $30^\circ-60^\circ-90^\circ$ and the $45^\circ-45^\circ-90^\circ$ right triangles. They have unique properties and if you understand the relationships between the sides and angles in these triangles, you will do well in geometry, trigonometry, and beyond.

Right Isosceles Triangles

The first type of right triangle to examine is isosceles. As you know, isosceles triangles have two sides that are the same length. Additionally, the base angles of an isosceles triangle are congruent as well. An isosceles right triangle will always have base angles that each measure $45^\circ$ and a vertex angle that measures $90^\circ$ .

Don’t forget that the base angles are the angles across from the congruent sides. They don’t have to be on the bottom of the figure.

Because the angles of all $45^\circ-45^\circ-90^\circ$ triangles will, by definition, remain the same, all $45^\circ-45^\circ-90^\circ$ triangles are similar, so their sides will always be proportional. To find the relationship between the sides, use the Pythagorean Theorem.

Example 1

The isosceles right triangle below has legs measuring $1\;\mathrm{centimeter}$ .

Use the Pythagorean Theorem to find the length of the hypotenuse.

Since the legs are $1\;\mathrm{centimeter}$ each, substitute $1$ for both $a$ and $b$ , and solve for $c$ :

$a^2 + b^2 &= c^2\\ 1^2+1^2&=c^2\\ 1+1&=c^2\\ 2&=c^2\\ \sqrt{2}&=\sqrt{c^2}\\ c&=\sqrt{2}$

In this example $c = \sqrt{2}\;\mathrm{cm}$ .

What if each leg in the example above was $5\;\mathrm{cm}$ ? Then we would have

$a^2 + b^2 &= c^2\\ 5^2+5^2&=c^2\\ 25+25&=c^2\\ 50&=c^2\\ \sqrt{50}&=\sqrt{c^2}\\ c&=5\sqrt{2}$

If each leg is $5\;\mathrm{cm}$ , then the hypotenuse is $5\sqrt{2}\;\mathrm{cm}$ .

When the length of each leg was $1$ , the hypotenuse was $1\sqrt{2}$ . When the length of each leg was $5$ , the hypotenuse was $5\sqrt{2}$ . Is this a coincidence? No. Recall that the legs of all $45^\circ-45^\circ-90^\circ$ triangles are proportional. The hypotenuse of an isosceles right triangle will always equal the product of the length of one leg and $\sqrt{2}$ . Use this information to solve the problem in example 2.

Example 2

What is the length of the hypotenuse in the triangle below?

Since the length of the hypotenuse is the product of one leg and $\sqrt{2}$ , you can easily calculate this length. One leg is $4\;\mathrm{inches}$ , so the hypotenuse will be $4\sqrt{2}\;\mathrm{inches}$ , or about $5.66\;\mathrm{inches}$ .

Equilateral Triangles

Remember that an equilateral triangle has sides that all have the same length. Equilateral triangles are also equiangular—all angles have the same measure. In an equilateral triangle, all angles measure exactly $60^\circ$ .

Notice what happens when you divide an equilateral triangle in half.

When an equilateral triangle is divided into two equal parts using an altitude, each resulting right triangle is a $30^\circ-60^\circ-90^\circ$ triangle. The hypotenuse of the resulting triangle was the side of the original, and the shorter leg is half of an original side. This is why the hypotenuse is always twice the length of the shorter leg in a $30^\circ-60^\circ-90^\circ$ triangle. You can use this information to solve problems about equilateral triangles.

30º-60º-90º Triangles

Another important type of right triangle has angles measuring $30^\circ$ , $60^\circ$ , and $90^\circ$ . Just as you found a constant ratio between the sides of an isosceles right triangle, you can find constant ratios here as well. Use the Pythagorean Theorem to discover these important relationships.

Example 3

Find the length of the missing leg in the following triangle. Use the Pythagorean Theorem to find your answer.

Just like you did for $45^\circ-45^\circ-90^\circ$ triangles, use the Pythagorean theorem to find the missing side. In this diagram, you are given two measurements: the hypotenuse $(c)$ is $2\;\mathrm{cm}$ and the shorter leg $(a)$ is $1\;\mathrm{cm}$ . Find the length of the missing leg $(b)$ .

$a^2 + b^2 &= c^2\\ 1^2+b^2&=2^2 \\ 1+b^2&=4 \\ b^2&=3 \\ b&= \sqrt{3}$

You can leave the answer in radical form as shown, or use your calculator to find the approximate value of $b \approx 1.732\;\mathrm{cm}$ .

On your own, try this again using a hypotenuse of $6\;\mathrm{feet}$ . Recall that since the $30^\circ-60^\circ-90^\circ$ triangle comes from an equilateral triangle, you know that the length of the shorter leg is half the length of the hypotenuse.

Now you should be able to identify the constant ratios in $30^\circ-60^\circ-90^\circ$ triangles. The hypotenuse will always be twice the length of the shorter leg, and the longer leg is always the product of the length of the shorter leg and $\sqrt{3}$ . In ratio form, the sides, in order from shortest to longest are in the ratio $x:x\sqrt{3}:2x$ .

Example 4

What is the length of the missing leg in the triangle below?

Since the length of the longer leg is the product of the shorter leg and $\sqrt{3}$ , you can easily calculate this length. The short leg is $8\;\mathrm{inches}$ , so the longer leg will be $8\sqrt{3}\;\mathrm{inches}$ , or about $13.86\;\mathrm{inches}$ .

Example 5

What is $AC$ below?

To find the length of segment $\overline {AC}$ , identify its relationship to the rest of the triangle. Since it is an altitude, it forms two congruent triangles with angles measuring $30^\circ$ , $60^\circ$ , and $90^\circ$ . So, $AC$ will be the product of $BC$ (the shorter leg) and $\sqrt {3}$ .

$AC &= BC \sqrt{3} \\ &=4 \sqrt{3}$

$AC=4\sqrt{3}\;\mathrm{yards}$ , or approximately $6.93\;\mathrm{yards}$ .

Special Right Triangles in the Real World

You can use special right triangles in many real-world contexts. Many real-life applications of geometry rely on special right triangles, so being able to recall and use these ratios is a way to save time when solving problems.

Example 6

The diagram below shows the shadow a flagpole casts at a certain time of day.

If the length of the shadow cast by the flagpole is $13\;\mathrm{m}$ , what is the height of the flagpole and the length of the hypotenuse of the right triangle shown?

The wording in this problem is complicated, but you only need to notice a few things. You can tell in the picture that this triangle has angles of $30^\circ , 60^\circ$ , and $90^\circ$ (This assumes that the flagpole is perpendicular to the ground, but that is a safe assumption). The height of the flagpole is the longer leg in the triangle, so use the special right triangle ratios to find the length of the hypotenuse.

The longer leg is the product of the shorter leg and $\sqrt{3}$ . The length of the shorter leg is given as $13\;\mathrm{meters}$ , so the height of the flagpole is $13\sqrt{3}\;\mathrm{m}$ .

The length of the hypotenuse is the hypotenuse of a $30^\circ-60^\circ-90^\circ$ triangle. It will always be twice the length of the shorter leg, so it will equal $13 \cdot 2$ , or $26\;\mathrm{meters}$ .

Example 7

Antonio built a square patio in his backyard.

He wants to make a water pipe for flowers that goes from one corner to another, diagonally. How long will that pipe be?

The first step in a word problem of this nature is to add important information to the drawing. Because the problem asks you to find the length from one corner to another, you should draw that segment in.

Once you draw the diagonal path, you can see how triangles help answer this question. Because both legs of the triangle have the same measurement $(17\;\mathrm{feet})$ , this is an isosceles right triangle. The angles in an isosceles right triangle are $45^\circ , 45^\circ$ , and $90^\circ$ .

In an isosceles right triangle, the hypotenuse is always equal to the product of the length of one leg and $\sqrt{2}$ . So, the length of Antonio’s water pipe will be the product of $17$ and $\sqrt{2}$ , or $17 \sqrt{2} \approx 17(1.414)\;\mathrm{feet}$ . This value is approximately equal to $24.04\;\mathrm{feet}$ .

Lesson Summary

In this lesson, we explored how to work with different radicals both in theory and in practical situations. Specifically, we have learned:

How to identify and use the ratios involved with right isosceles triangles.
How to identify and use the ratios involved with $30^\circ-60^\circ-90^\circ$ triangles.
How to identify and use ratios involved with equilateral triangles.
How to employ right triangle ratios when solving real-world problems.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

The following questions are for your own review. The answers are listed below for you to check your work and understanding.

Review Questions

Mildred had a piece of scrap wood cut into an equilateral triangle. She wants to cut it into two smaller congruent triangles. What will be the angle measurement of the triangles that result?
Roberto has a square pizza. He wants to cut two congruent triangles out of the pizza without leaving any leftovers. What will be the angle measurements of the triangles that result?
What is the length of the hypotenuse in the triangle below?
What is the length of the hypotenuse in the triangle below?
What is the length of the longer leg in the triangle below?
What is the length of one of the legs in the triangle below?
What is the length of the shorter leg in the triangle below?
A square window has a diagonal of $5\sqrt2\;\mathrm{feet}$ . What is the length of one of its sides?
A square block of foam is cut into two congruent wedges. If a side of the original block was $3\;\mathrm{feet}$ , how long is the diagonal cut?
They wants to find the area of an equilateral triangle but only knows that the length of one side is $6\;\mathrm{inches}$ . What is the height of Thuy’s triangle? What is the area of the triangle?

Review Answers

$30^\circ , 60^\circ$ , and $90^\circ$
$45^\circ , 45^\circ$ , and $90^\circ$
$10$
$11\sqrt2\;\mathrm{cm}$ or approx. $15.56\;\mathrm{cm}$
$6\sqrt3\;\mathrm{miles}$ or approx. $10.39\;\mathrm{miles}$
$3\;\mathrm{mm}$
$14\;\mathrm{feet}$
$5\;\mathrm{feet}$
$3\sqrt2\;\mathrm{feet}$ or approx. $4.24\;\mathrm{feet}$
$3\sqrt3\;\mathrm{inches}$ or approx. $5.2\;\mathrm{in}$ . The area is $9\sqrt3\approx 15.59\; \mathrm{inches}^2$

Last modified: Tuesday, June 29, 2010, 10:31 AM