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# 8.2: Converse of the Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Understand the converse of the Pythagorean Theorem.
• Identify acute and obtuse triangles from side measures.

## Review Queue

1. Determine if the following sets of numbers are Pythagorean triples.
1. 14, 48, 50
2. 9, 40, 41
3. 12, 43, 44
2. Do the following lengths make a right triangle? How do you know?
1. 5,3,14\begin{align*}\sqrt{5}, 3, \sqrt{14}\end{align*}
2. 6,23,8\begin{align*}6, 2 \sqrt{3}, 8\end{align*}
3. 32,42,52\begin{align*}3 \sqrt{2}, 4 \sqrt{2}, 5 \sqrt{2}\end{align*}

Know What? A friend of yours is designing a building and wantsit to be rectangular. One wall 65 ft. long and the other is 72 ft. long. How can he ensure the walls are going to be perpendicular?

## Converse of the Pythagorean Theorem

In the last lesson, you learned about the Pythagorean Theorem and how it can be used. The converse of the Pythagorean Theorem is also true. We touched on this in the last section with Example 1.

Pythagorean Theorem Converse: If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

With this converse, you can use the Pythagorean Theorem to prove that a triangle is a right triangle, even if you do not know any of the triangle’s angle measurements.

Example 1: Determine if the triangles below are right triangles.

a)

b)

Solution: Check to see if the three lengths satisfy the Pythagorean Theorem. Let the longest sides represent c\begin{align*}c\end{align*}, in the equation.

a) a2+b2=c282+162=?(85)264+256=?645320=320\begin{align*}a^2 + b^2 = c^2\!\\ 8^2 + 16^2 \overset{\underset{{?}}{}}{=} \left ( 8 \sqrt{5} \right )^2\!\\ 64 + 256 \overset{\underset{{?}}{}}{=} 64 \cdot 5\!\\ 320 = 320\end{align*}

The triangle is a right triangle.

b) a2+b2=c2222+242=?262484+576=6761060676\begin{align*}a^2 + b^2 = c^2\!\\ 22^2 + 24^2 \overset{\underset{{?}}{}}{=} 26^2\!\\ 484 + 576 = 676\!\\ 1060 \neq 676\end{align*}

The triangle is not a right triangle.

## Identifying Acute and Obtuse Triangles

We can extend the converse of the Pythagorean Theorem to determine if a triangle has an obtuse angle or is acute. We know that if the sum of the squares of the two smaller sides equals the square of the larger side, then the triangle is right. We can also interpret the outcome if the sum of the squares of the smaller sides does not equal the square of the third.

Theorem 8-3: If the sum of the squares of the two shorter sides in a right triangle is greater than the square of the longest side, then the triangle is acute.

Theorem 8-4: If the sum of the squares of the two shorter sides in a right triangle is less than the square of the longest side, then the triangle is obtuse.

In other words: The sides of a triangle are a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*} and c>b\begin{align*}c > b\end{align*} and c>a\begin{align*}c > a\end{align*}.

If a2+b2>c2\begin{align*}a^2 + b^2 > c^2\end{align*}, then the triangle is acute.

If a2+b2=c2\begin{align*}a^2 + b^2 = c^2\end{align*}, then the triangle is right.

If a2+b2<c2\begin{align*}a^2 + b^2 < c^2\end{align*}, then the triangle is obtuse.

Proof of Theorem 8-3

Given: In \begin{align*}\triangle ABC, a^2 + b^2 > c^2\end{align*}, where \begin{align*}c\end{align*} is the longest side.

In \begin{align*}\triangle LMN, \angle N\end{align*} is a right angle.

Prove: \begin{align*}\triangle ABC\end{align*} is an acute triangle. (all angles are less than \begin{align*}90^\circ\end{align*})

Statement Reason
1. In \begin{align*}\triangle ABC, a^2 + b^2 > c^2\end{align*}, and \begin{align*}c\end{align*} is the longest side. In \begin{align*}\triangle LMN, \angle N\end{align*} is a right angle. Given
2. \begin{align*}a^2 + b^2 = h^2\end{align*} Pythagorean Theorem
3. \begin{align*}c^2 < h^2\end{align*} Transitive PoE
4. \begin{align*}c < h\end{align*} Take the square root of both sides
5. \begin{align*}\angle C\end{align*} is the largest angle in \begin{align*}\triangle ABC\end{align*}. The largest angle is opposite the longest side.
6. \begin{align*}m \angle N = 90^\circ\end{align*} Definition of a right angle
7. \begin{align*}m \angle C < m \angle N\end{align*} SSS Inequality Theorem
8. \begin{align*}m \angle C < 90^\circ\end{align*} Transitive PoE
9. \begin{align*}\angle C\end{align*} is an acute angle. Definition of an acute angle
10. \begin{align*}\triangle ABC\end{align*} is an acute triangle. If the largest angle is less than \begin{align*}90^\circ\end{align*}, then all the angles are less than \begin{align*}90^\circ\end{align*}.

The proof of Theorem 8-4 is very similar and is in the review questions.

Example 2: Determine if the following triangles are acute, right or obtuse.

a)

b)

Solution: Set the shorter sides in each triangle equal to \begin{align*} a \end{align*} and \begin{align*} b \end{align*} and the longest side equal to \begin{align*}c\end{align*}.

a) \begin{align*}6^2 + (3 \sqrt{5})^2 \ ? \ 8^2\!\\ 36 + 45 \ ? \ 64\!\\ 81 > 64\end{align*}

The triangle is acute.

b) \begin{align*}15^2 + 14^2 \ ? \ 21^2\!\\ 225 + 196 \ ? \ 441\!\\ 421 < 441\end{align*}

The triangle is obtuse.

Example 3: Graph \begin{align*}A(-4, 1), B(3, 8)\end{align*}, and \begin{align*}C(9, 6)\end{align*}. Determine if \begin{align*}\triangle ABC\end{align*} is acute, obtuse, or right.

Solution: This looks like an obtuse triangle, but we need proof to draw the correct conclusion. Use the distance formula to find the length of each side.

\begin{align*}AB & = \sqrt{(-4-3)^2 + (1 - 8)^2} = \sqrt{49 + 49} = \sqrt{98} = 7 \sqrt{2}\\ BC & = \sqrt{(3 - 9)^2 + (8 - 6)^2} = \sqrt{36 + 4 } = \sqrt{40} = 2 \sqrt{10}\\ AC & = \sqrt{(-4-9)^2 + (1-6)^2} = \sqrt{169 + 25} = \sqrt{194}\end{align*}

Now, let’s plug these lengths into the Pythagorean Theorem.

\begin{align*}\left ( \sqrt{98} \right )^2 + \left ( \sqrt{40} \right )^2 & \ ? \ \left ( \sqrt{194} \right )^2\\ 98 + 40 & \ ? \ 194\\ 138 & < 194\end{align*}

\begin{align*}\triangle ABC\end{align*} is an obtuse triangle.

Know What? Revisited To make the walls perpendicular, find the length of the diagonal.

\begin{align*}65^2 + 72^2 & = c^2\\ 4225 + 5184 & = c^2\\ 9409 & = c^2\\ 97 & = c\end{align*}

In order to make the building rectangular, both diagonals must be 97 feet.

## Review Questions

1. The two shorter sides of a triangle are 9 and 12.
1. What would be the length of the third side to make the triangle a right triangle?
2. What is a possible length of the third side to make the triangle acute?
3. What is a possible length of the third side to make the triangle obtuse?
2. The two longer sides of a triangle are 24 and 25.
1. What would be the length of the third side to make the triangle a right triangle?
2. What is a possible length of the third side to make the triangle acute?
3. What is a possible length of the third side to make the triangle obtuse?
3. The lengths of the sides of a triangle are \begin{align*}8x, 15x,\end{align*} and \begin{align*}17x\end{align*}. Determine if the triangle is acute, right, or obtuse.

Determine if the following lengths make a right triangle.

1. 15, 20, 25
2. 20, 25, 30
3. \begin{align*}8 \sqrt{3}, 6, 2 \sqrt{39}\end{align*}

Determine if the following triangles are acute, right or obtuse.

1. 7, 8, 9
2. 14, 48, 50
3. 5, 12, 15
4. 13, 84, 85
5. 20, 20, 24
6. 35, 40, 51
7. 39, 80, 89
8. 20, 21, 38
9. 48, 55, 76

Graph each set of points and determine if \begin{align*}\triangle ABC\end{align*} is acute, right, or obtuse.

1. \begin{align*}A(3, -5), B(-5, -8), C(-2, 7)\end{align*}
2. \begin{align*}A(5, 3), B(2, -7), C(-1, 5)\end{align*}
3. Writing Explain the two different ways you can show that a triangle in the coordinate plane is a right triangle.

The figure to the right is a rectangular prism. All sides (or faces) are either squares (the front and back) or rectangles (the four around the middle). All sides are perpendicular.

1. Find \begin{align*}c\end{align*}.
2. Find \begin{align*}d\end{align*}.

1. Writing Explain why \begin{align*}m \angle A = 90^\circ\end{align*}.
2. Fill in the blanks for the proof of Theorem 8-4. Given: In \begin{align*}\triangle ABC,a^2 + b^2 < c^2\end{align*}, where \begin{align*}c\end{align*} is the longest side. In \begin{align*}\triangle LMN, \angle N\end{align*} is a right angle. Prove: \begin{align*}\triangle ABC\end{align*} is an obtuse triangle. (one angle is greater than \begin{align*}90^\circ\end{align*})
Statement Reason
1. In \begin{align*}\triangle ABC, a^2 + b^2 < c^2\end{align*}, and \begin{align*}c\end{align*} is the longest side. In \begin{align*}\triangle LMN, \angle N\end{align*} is a right angle.
2. \begin{align*}a^2 + b^2 = h^2\end{align*}
3. \begin{align*}c^2 > h^2\end{align*}
4.
5. \begin{align*}\angle C\end{align*} is the largest angle in \begin{align*}\triangle ABC\end{align*}.
6. \begin{align*}m \angle N = 90^\circ\end{align*}
7. \begin{align*}m \angle C > m \angle N\end{align*}
8. Transitive PoE
9. \begin{align*}\angle C\end{align*} is an obtuse angle.
10. \begin{align*}\triangle ABC\end{align*} is an obtuse triangle.

Given \begin{align*}\overline{AB}\end{align*}, with \begin{align*}A(3, 3)\end{align*} and \begin{align*}B(2, -3)\end{align*} determine whether the given point \begin{align*}C\end{align*} in problems 23-25 makes an acute, right or obtuse triangle.

1. \begin{align*}C(3, -3)\end{align*}
2. \begin{align*}C(4, -1)\end{align*}
3. \begin{align*}C(5, -2)\end{align*}

Given \begin{align*}\overline{AB}\end{align*}, with \begin{align*}A(-2, 5)\end{align*} and \begin{align*}B(1, -3)\end{align*} find at least two possible points, \begin{align*}C\end{align*}, such that \begin{align*}\triangle ABC\end{align*} is

1. right, with right \begin{align*}\angle C\end{align*}.
2. acute, with acute \begin{align*}\angle C\end{align*}.
3. obtuse, with obtuse \begin{align*}\angle C\end{align*}.
4. Construction
1. Draw \begin{align*}\overline{AB}\end{align*}, such that \begin{align*}AB = 3 \ in\end{align*}.
2. Draw \begin{align*}\overrightarrow{AD}\end{align*} such that \begin{align*}\angle BAD < 90^\circ\end{align*}.
3. Construct a line through \begin{align*} B \end{align*} which is perpendicular to \begin{align*}\overrightarrow{AD}\end{align*}, label the intersection \begin{align*}C\end{align*}.
4. \begin{align*}\triangle ABC\end{align*} is a right triangle with right \begin{align*}\angle C\end{align*}.
5. Is the triangle you made unique? In other words, could you have multiple different outcomes with the same \begin{align*}AB\end{align*}? Why or why not? You may wish to experiment to find out.
6. Why do the instructions specifically require that \begin{align*}\angle BAD < 90^\circ\end{align*}?
7. Describe how this construction could be changed so that \begin{align*}\angle B\end{align*} is the right angle in the triangle.

## Review Queue Answers

1. Yes
2. Yes
3. No
1. Yes
2. No
3. Yes

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Feb 22, 2012
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CK.MAT.ENG.SE.1.Geometry.8.2