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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.
• Determine if a triangle is acute or obtuse 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
4. 12, 35, 37
1. $\left( 5 \sqrt{12} \right )^2$
2. $\frac{14}{\sqrt 2}$
3. $\frac{18}{\sqrt 3}$

Know What? A friend of yours is designing a building and wants it 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

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.

If $a^2 + b^2 = c^2$, then $\triangle ABC$ 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 angle measures.

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 side represent $c$.

a) $a^2 + b^2 = c^2\!\\8^2 + 16^2 \overset{?}= \left( 8 \sqrt{5} \right )^2\!\\64 + 256 \overset{?}= 64 \cdot 5\!\\320 = 320 \qquad \text{Yes}$

b) $a^2+b^2 = c^2\!\\22^2 + 24^2 \overset{?}= 26^2\!\\484 + 576 \overset{?}= 676\!\\1060 \neq 676 \qquad \text{No}$

Example 2: Do the following lengths make a right triangle?

a) $\sqrt{5}, 3, \sqrt{14}$

b) $6, 2 \sqrt{3}, 8$

c) $3 \sqrt{2}, 4 \sqrt{2}, 5\sqrt{2}$

Solution: Even though there is no picture, you can still use the Pythagorean Theorem. Again, the longest length will be $c$.

a) $\left( \sqrt{5} \right )^2 + 3^2 = \sqrt{14}^2\!\\5+9=14\!\\\text{Yes}$

b) $6^2 + \left( 2 \sqrt{3} \right )^2 = 8^2\!\\36+(4 \cdot 3) = 64\!\\36+12 \neq 64$

c) This is a multiple of $\sqrt{2}$ of a 3, 4, 5 right triangle. Yes, this is a right triangle.

## Identifying Acute and Obtuse Triangles

We can extend the converse of the Pythagorean Theorem to determine if a triangle is an obtuse or acute triangle.

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.

$b < c$ and $a < c$

If $a^2 + b^2 > c^2$, 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.

$b < c$ and $a < c$

If $a^2+b^2, then the triangle is obtuse.

Example 3: Determine if the following triangles are acute, right or obtuse.

a)

b)

Solution: Set the longest side equal to $c$.

a) $6^2 + \left( 3 \sqrt{5} \right)^2 \ ? \ 8^2\!\\36 + 45 \ ? \ 64\!\\81 > 64$

The triangle is acute.

b) $15^2 + 14^2 \ ? \ 21^2\!\\225 + 196 \ ? \ 441\!\\421 < 441$

The triangle is obtuse.

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

Solution: Use the distance formula to find the length of each side.

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

Plug these lengths into the Pythagorean Theorem.

$\left( \sqrt{98} \right )^2 + \left( \sqrt{40} \right)^2 & \ ? \ \left ( \sqrt{194} \right )^2\\98 + 40 & \ ? \ 194\\138 & < 194$

$\triangle ABC$ is an obtuse triangle.

Know What? Revisited Find the length of the diagonal.

$65^2 + 72^2 &= c^2\\4225 + 5184 &= c^2\\9409 &= c^2\\97 &= c && \text{To make the building rectangular, both diagonals must be 97 feet.}$

## Review Questions

• Questions 1-6 are similar to Examples 1 and 2.
• Questions 7-15 are similar to Example 3.
• Questions 16-20 are similar to Example 4.
• Questions 21-24 use the Pythagorean Theorem.
• Question 25 uses the definition of similar triangles.

Determine if the following lengths make a right triangle.

1. 7, 24, 25
2. $\sqrt{5}, 2 \sqrt{10}, 3 \sqrt{5}$
3. $2 \sqrt{3}, \sqrt{6}, 8$
4. 15, 20, 25
5. 20, 25, 30
6. $8 \sqrt{3}, 6, 2 \sqrt{39}$

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 $\triangle ABC$ is acute, right, or obtuse, using the distance formula.

1. $A(3, -5), B(-5, -8), C(-2, 7)$
2. $A(5, 3), B(2, -7), C(-1, 5)$
3. $A(1, 6) , B(5, 2), C(-2, 3)$
4. $A(-6, 1), B(-4, -5), C(5, -2)$
5. Show that #18 is a right triangle by using the slopes of the sides of the 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 faces are perpendicular.
6. Find $c$.
7. Find $d$.

Now, the figure is a cube, where all the sides are squares. If all the sides have length 4, find:

1. Find $c$.
2. Find $d$.
3. Writing Explain why $m \angle A = 90^{\circ}$.

1. Yes
2. Yes
3. No
4. Yes
1. $\left ( 5 \sqrt{12} \right )^2 = 5^2 \cdot 12 = 25 \cdot 12=300$
2. $\frac{14}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{14 \sqrt{2}}{2} = 7 \sqrt{2}$
3. $\frac{18}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}= \frac{18 \sqrt{3}}{3} = 6 \sqrt{3}$

8 , 9 , 10

Feb 22, 2012

Feb 03, 2016