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8.4: Special Right Triangles

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Learn and use the 45-45-90 triangle ratio.
  • Learn and use the 30-60-90 triangle ratio.

Review Queue

Find the value of the missing variables. Simplify all radicals.

  1. Is 9, 12, and 15 a right triangle?
  2. Is 3, \begin{align*}3 \sqrt{3}\end{align*}, and 6 a right triangle?

Know What? A baseball diamond is a square with sides that are 90 feet long. Each base is a corner of the square. What is the length between \begin{align*}1^{st}\end{align*} and \begin{align*}3^{rd}\end{align*} base and between \begin{align*}2^{nd}\end{align*} base and home plate? (the red dotted lines in the diagram).

Isosceles Right Triangles

There are two special right triangles. The first is an isosceles right triangle.

Isosceles Right Triangle: A right triangle with congruent legs and acute angles. This triangle is also called a 45-45-90 triangle (after the angle measures).

\begin{align*}\triangle ABC\end{align*} is a right triangle with:

\begin{align*}m \angle A &= 90^\circ\\ \overline {AB} & \cong \overline{AC}\\ m \angle B &= m \angle C = 45^\circ\end{align*}

Investigation 8-2: Properties of an Isosceles Right Triangle

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

1. Draw an isosceles right triangle with 2 inch legs and the \begin{align*}90^\circ\end{align*} angle between them.

2. Find the measure of the hypotenuse, using the Pythagorean Theorem. Simplify the radical.

\begin{align*}2^2 + 2^2 &= c^2\\ 8 &= c^2\\ c &= \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}\end{align*}

What do you notice about the length of the legs and hypotenuse?

3. 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. How is it 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 Theorem: If a right triangle is isosceles, then its sides are \begin{align*}x:x:x \sqrt{2}\end{align*}.

For any isosceles right triangle, the legs are \begin{align*}x\end{align*} and the hypotenuse is always \begin{align*}x \sqrt{2}\end{align*}. Because the three angles are always \begin{align*}45^\circ, 45^\circ,\end{align*} and \begin{align*}90^\circ\end{align*}, all isosceles right triangles are similar.

Example 1: Find the length of the missing sides.

a)

b)

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

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

b) \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*}.

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

a)

b)

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

a) \begin{align*}12 \sqrt{2}\end{align*} is the diagonal of the square. Remember that the diagonal of a square bisects each angle, so it splits the square into two 45-45-90 triangles. \begin{align*}12 \sqrt{2}\end{align*} would be the hypotenuse, or equal to \begin{align*}x \sqrt{2}\end{align*}.

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

b) Here, we are given the hypotenuse. 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}} = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}\end{align*}

In part b, we rationalized the denominator which we learned in the first section.

30-60-90 Triangles

The second special right triangle is called a 30-60-90 triangle, after the three angles. To draw a 30-60-90 triangle, start with an equilateral triangle.

Investigation 8-3: Properties of a 30-60-90 Triangle

Tools Needed: Pencil, paper, ruler, compass

1. Construct an equilateral triangle with 2 inch sides.

http://www.mathsisfun.com/geometry/construct-equitriangle.html

2. Draw or construct the altitude from the top vertex to form two congruent triangles.

3. Find the measure of the two angles at the top vertex and the length of the shorter leg.

The top angles are each \begin{align*}30^\circ\end{align*} and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector.

4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical.

\begin{align*}1^2 + b^2 &= 2^2\\ 1+b^2 &= 4\\ b^2 &= 3\\ b &= \sqrt{3}\end{align*}

5. Now, let’s say the shorter leg is length \begin{align*}x\end{align*} and the hypotenuse is \begin{align*}2x\end{align*}. Use the Pythagorean Theorem to find the longer leg. How is this similar to your answer in #4?

\begin{align*}x^2 + b^2 &= (2x)^2\\ x^2+b^2 &= 4x^2\\ b^2 &= 3x^2\\ b &= x \sqrt{3}\end{align*}

30-60-90 Theorem: If a triangle has angle measures \begin{align*}30^\circ, 60^\circ\end{align*} and \begin{align*}90^\circ\end{align*}, then the sides are \begin{align*}x:x \sqrt{3}:2x\end{align*}.

The shortest leg is always \begin{align*}x\end{align*}, the longest leg is always \begin{align*}x \sqrt{3}\end{align*}, and the hypotenuse is always \begin{align*}2x\end{align*}. If you ever forget these theorems, you can still use the Pythagorean Theorem.

Example 3: Find the length of the missing sides.

a)

b)

Solution: In part a, we are given the shortest leg and in part b, we are given the hypotenuse.

a) If \begin{align*}x=5\end{align*}, then the longer leg, \begin{align*}b=5 \sqrt{3}\end{align*}, and the hypotenuse, \begin{align*}c=2(5)=10\end{align*}.

b) Now, \begin{align*}2x=20\end{align*}, so the shorter leg, \begin{align*}f = \frac{20}{2} = 10\end{align*}, and the longer leg, \begin{align*}g=10 \sqrt{3}\end{align*}.

Example 4: Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

a)

b)

Solution: In part a, we are given the longer leg and in part b, we are given the hypotenuse.

a) \begin{align*}x \sqrt{3} = 12\!\\ x = \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12 \sqrt{3}}{3} = 4 \sqrt{3}\!\\ \text{The hypotenuse is}\\ y = 2(4 \sqrt{3}) = 8 \sqrt{3}\end{align*}

b) \begin{align*}2x=16\!\\ x = 8\!\\ \text{The longer leg is}\!\\ y = 8 \cdot \sqrt 3 = 8 \sqrt{3}\end{align*}

Example 5: A rectangle has sides 4 and \begin{align*}4 \sqrt{3}\end{align*}. What is the length of the diagonal?

Solution: If you are not given a picture, draw one.

The two lengths are \begin{align*}x, x \sqrt{3}\end{align*}, so the diagonal would be \begin{align*}2x\end{align*}, or \begin{align*}2(4) = 8\end{align*}.

If you did not recognize this is a 30-60-90 triangle, you can use the Pythagorean Theorem too.

\begin{align*}4^2 + \left( 4 \sqrt{3} \right )^2 &= d^2\\ 16 + 48 &= d^2\\ d &= \sqrt{64} = 8\end{align*}

Example 6: A square has a diagonal with length 10, what are the sides?

Solution: 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*}

Know What? Revisited The distance between \begin{align*}1^{st}\end{align*} and \begin{align*}3^{rd}\end{align*} base is one of the diagonals of the square. So, it would be the same as the hypotenuse of a 45-45-90 triangle. Using our ratios, the distance is \begin{align*}90 \sqrt{2} \approx 127.3 \ ft\end{align*}. The distance between \begin{align*}2^{nd}\end{align*} base and home plate is the same length.

Review Questions

  • Questions 1-4 are similar to Example 1-4.
  • Questions 5-8 are similar to Examples 5 and 6.
  • Questions 9-23 are similar to Examples 1-4.
  • Questions 24 and 25 are a challenge.
  1. In an isosceles right triangle, if a leg is 4, then the hypotenuse is __________.
  2. In a 30-60-90 triangle, if the shorter leg is 5, then the longer leg is __________ and the hypotenuse is ___________.
  3. In an isosceles right triangle, if a leg is \begin{align*}x\end{align*}, then the hypotenuse is __________.
  4. In a 30-60-90 triangle, if the shorter leg is \begin{align*}x\end{align*}, then the longer leg is __________ and the hypotenuse is ___________.
  5. A square has sides of length 15. What is the length of the diagonal?
  6. A square’s diagonal is 22. What is the length of each side?
  7. A rectangle has sides of length 6 and \begin{align*}6 \sqrt{3}\end{align*}. What is the length of the diagonal?
  8. Two (opposite) sides of a rectangle are 10 and the diagonal is 20. What is the length of the other two sides?

For questions 9-23, find the lengths of the missing sides. Simplify all radicals.

Challenge For 24 and 25, find the value of \begin{align*}y\end{align*}. You may need to draw in additional lines. Round all answers to the nearest hundredth.

Review Queue Answers

  1. \begin{align*}4^2+4^2 = x^2\!\\ {\;} \quad \ \ 32 = x^2\!\\ {\;} \qquad \ x = 4 \sqrt{2}\end{align*}
  2. \begin{align*}3^2+y^2 = 6^2\!\\ {\;} \qquad y^2 = 27\!\\ {\;} \qquad \ y = 3 \sqrt{3}\end{align*}
  3. \begin{align*}x^2 + x^2 = \left ( 10 \sqrt{2} \right )^2\!\\ {\;} \quad \ 2x^2 = 200\!\\ {\;} \qquad x^2 = 100\!\\ {\;} \qquad \ x = 10\end{align*}
  4. Yes, \begin{align*}9^2 + 12^2 = 15^2 \rightarrow 81+144 = 225\end{align*}
  5. Yes, \begin{align*}3^2 + \left( 3 \sqrt{3} \right )^2 = 6^2 \rightarrow 9+27 = 36\end{align*}

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