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# 1.2: Special Right Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Recognize special right triangles.
• Use the special right triangle ratios to solve special right triangles.

## Special Right Triangle #1: Isosceles Right Triangle

An isosceles right triangle is an isosceles triangle and a right triangle. This means that it has two congruent sides and one right angle. Therefore, the two congruent sides must be the legs.

Because the two legs are congruent, we will call them both $a$ and the hypotenuse $c$. Plugging both letters into the Pythagorean Theorem, we get:

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

From this we can conclude that the hypotenuse length is the length of a leg multiplied by $\sqrt{2}$. Therefore, we only need one of the three lengths to determine the other two lengths of the sides of an isosceles right triangle. The ratio is usually written $x:x:x\sqrt{2}$, where $x$ is the length of the legs and $x\sqrt{2}$ is the length of the hypotenuse.

Example 1: Find the lengths of the other two sides of the isosceles right triangles below.

a.

b.

c.

d.

Solution:

a. If a leg has length 8, by the ratio, the other leg is 8 and the hypotenuse is $8\sqrt{2}$.

b. If the hypotenuse has length $7\sqrt{2}$, then both legs are 7.

c. Because the leg is $10\sqrt{2}$, then so is the other leg. The hypotenuse will be $10\sqrt{2}$ multiplied by an additional $\sqrt{2}$.

$10\sqrt{2} \cdot \sqrt{2} = 10 \cdot 2 = 20$

d. In this problem set $x\sqrt{2} = 9\sqrt{6}$ because $x\sqrt{2}$ is the hypotenuse portion of the ratio.

$x\sqrt{2} & = 9\sqrt{6}\\x & = \frac{9\sqrt{6}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{9\sqrt{12}}{2} = \frac{18\sqrt{3}}{2} = 9\sqrt{3}$

So, the length of each leg is $9\sqrt{3}$.

What are the angle measures in an isosceles right triangle? Recall that the sum of the angles in a triangle is $180^\circ$ and there is one $90^\circ$ angle. Therefore, the other two angles add up to $90^\circ$. Because this is an isosceles triangle, these two angles are equal and $45^\circ$ each. Sometimes an isosceles right triangle is also referred to as a $45-45-90$ triangle.

## Special Right Triangle #2: 30-60-90 Triangle

$30-60-90$ refers to each of the angles in this special right triangle. To understand the ratios of the sides, start with an equilateral triangle with an altitude drawn from one vertex.

Recall from geometry, that an altitude, $h$, cuts the opposite side directly in half. So, we know that one side, the hypotenuse, is $2s$ and the shortest leg is $s$. Also, recall that the altitude is a perpendicular and angle bisector, which is why the angle at the top is split in half. To find the length of the longer leg, use the Pythagorean Theorem:

$s^2 + h^2 & = (2s)^2\\s^2 + h^2 & = 4s^2\\h^2 & = 3s^2\\h & = s\sqrt{3}$

From this we can conclude that the length of the longer leg is the length of the short leg multiplied by $\sqrt{3}$ or $s\sqrt{3}$. Just like the isosceles right triangle, we now only need one side in order to determine the other two in a $30-60-90$ triangle. The ratio of the three sides is written $x:x\sqrt{3}:2x$, where $x$ is the shortest leg, $x\sqrt{3}$ is the longer leg and $2x$ is the hypotenuse.

Notice, that the shortest side is always opposite the smallest angle and the longest side is always opposite $90^\circ$.

If you look back at the Review Questions from the last section we now recognize #8 as a $30-60-90$ triangle.

Example 2: Find the lengths of the two missing sides in the $30-60-90$ triangles.

a.

b.

c.

d.

Solution: Determine which side in the $30-60-90$ ratio is given and solve for the other two.

a. $4\sqrt{3}$ is the longer leg because it is opposite the $60^\circ$. So, in the $x:x\sqrt{3}:2x$ ratio, $4\sqrt{3} = x\sqrt{3}$, therefore $x = 4$ and $2x = 8$. The short leg is 4 and the hypotenuse is 8.

b. 17 is the hypotenuse because it is opposite the right angle. In the $x:x\sqrt{3}:2x$ ratio, $17 = 2x$ and so the short leg is $\frac{17}{2}$ and the long leg is $\frac{17\sqrt{3}}{2}$.

c. 15 is the long leg because it is opposite the $60^\circ$. Even though 15 does not have a radical after it, we can still set it equal to $x\sqrt{3}$.

$x\sqrt{3} & = 15\\x & = \frac{15}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3} \quad \text{So, the short leg is} \ 5\sqrt{3}.$

Multiplying $5\sqrt{3}$ by 2, we get the hypotenuse length, which is $10\sqrt{3}$.

d. $24\sqrt{21}$ is the length of the hypotenuse because it is opposite the right angle. Set it equal to $2x$ and solve for $x$ to get the length of the short leg.

$2x & = 24\sqrt{21}\\x &= 12\sqrt{21}$

To find the length of the longer leg, we need to multiply $12\sqrt{21}$ by $\sqrt{3}$.

$12\sqrt{21} \cdot \sqrt{3} = 12\sqrt{3 \cdot 3 \cdot 7} = 36\sqrt{7}$

The length of the longer leg is $36\sqrt{7}$.

Be careful when doing these problems. You can always check your answers by finding the decimal approximations of each side. For example, in 2d, short leg $= 12\sqrt{21} \approx 54.99$, long leg $= 36\sqrt{7} \approx 95.25$ and the hypotenuse $= 24\sqrt{21} \approx 109.98$. This is an easy way to double-check your work and verify that the hypotenuse is the longest side.

## Using Special Right Triangle Ratios

Special right triangles are the basis of trigonometry. The angles $30^\circ, \ 45^\circ, \ 60^\circ$ and their multiples have special properties and significance in the unit circle (sections 1.5 and 1.6). Students are usually required to memorize these two ratios because of their importance.

First, let’s compare the two ratios, so that we can better distinguish the difference between the two. For a $45-45-90$ triangle the ratio is $x:x:x\sqrt{2}$ and for a $30-60-90$ triangle the ratio is $x:x\sqrt{3}:2x$. An easy way to tell the difference between these two ratios is the isosceles right triangle has two congruent sides, so its ratio has the $\sqrt{2}$, whereas the $30-60-90$ angles are all divisible by 3, so that ratio includes the $\sqrt{3}$. Also, if you are ever in doubt or forget the ratios, you can always use the Pythagorean Theorem. The ratios are considered a short cut.

Example 3: Determine if the sets of lengths below represent special right triangles. If so, which one?

a. $8\sqrt{3}:24:16\sqrt{3}$

b. $\sqrt{5}:\sqrt{5}:\sqrt{10}$

c. $6\sqrt{7}:6\sqrt{21}:12$

Solution:

a. Yes, this is a $30-60-90$ triangle. If the short leg is $x = 8\sqrt{3}$, then the long leg is $8\sqrt{3} \cdot \sqrt{3} = 8 \cdot 3 = 24$ and the hypotenuse is $2 \cdot 8\sqrt{3} = 16\sqrt{3}$.

b. Yes, this is a $45-45-90$ triangle. The two legs are equal and $\sqrt{5} \cdot \sqrt{2} = \sqrt{10}$, which would be the length of the hypotenuse.

c. No, this is not a special right triangle, nor a right triangle. The hypotenuse should be $12\sqrt{7}$ in order to be a $30-60-90$ triangle.

## Points to Consider

• What is the difference between Pythagorean triples and special right triangle ratios?
• Why are these two ratios considered “special”?

## Review Questions

Solve each triangle using the special right triangle ratios.

1. A square window has a diagonal of 6 ft. To the nearest hundredth, what is the height of the window?
2. Pablo has a rectangular yard with dimensions 10 ft by 20 ft. He is decorating the yard for a party and wants to hang lights along both diagonals of his yard. How many feet of lights does he need? Round your answer to the nearest foot.
3. Can $2:2:2\sqrt{3}$ be the sides of a right triangle? If so, is it a special right triangle?
4. Can $\sqrt{5}:\sqrt{15}:2\sqrt{5}$ be the sides of a right triangle? If so, is it a special right triangle?

1. Each leg is $\frac{16}{\sqrt{2}} = \frac{16}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}$.
2. Short leg is $\frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac{6}{3}} = \sqrt{2}$ and hypotenuse is $2\sqrt{2}$.
3. Short leg is $\frac{12}{\sqrt{3}} = \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$ and hypotenuse is $8\sqrt{3}$.
4. The hypotenuse is $4\sqrt{10} \cdot \sqrt{2} = 4\sqrt{20} = 8\sqrt{5}$.
5. Each leg is $\frac{5\sqrt{2}}{\sqrt{2}} = 5$.
6. The short leg is $\frac{15}{2}$ and the long leg is $\frac{15\sqrt{3}}{2}$.
7. If the diagonal of a square is 6 ft, then each side of the square is $\frac{6}{\sqrt{2}}$ or $3\sqrt{2} \approx 4.24 \ ft$.
8. These are not dimensions for a special right triangle, so to find the diagonal (both are the same length) do the Pythagorean Theorem: $10^2 + 20^2 & = d^2\\100 + 400 & = d^2\\\sqrt{500} & = d\\10\sqrt{5} & = d$ So, if each diagonal is $10\sqrt{5}$, two diagonals would be $20\sqrt{5} \approx 45 \ ft.$ Pablo needs 45 ft of lights for his yard.
9. $2:2:2\sqrt{3}$ does not fit into either ratio, so it is not a special right triangle. To see if it is a right triangle, plug these values into the Pythagorean Theorem: $2^2 + 2^2 & = \left ( 2\sqrt{3} \right )^2\\4 + 4 & = 12\\8 & < 12$ this is not a right triangle, it is an obtuse triangle.
10. $\sqrt{5}:\sqrt{15}:2\sqrt{5}$ is a $30-60-90$ triangle. The long leg is $\sqrt{5} \cdot \sqrt{3} = \sqrt{15}$ and the hypotenuse is $2\sqrt{5}$.

## Date Created:

Feb 23, 2012

Dec 16, 2014
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