Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![### Problem: Find the Missing Side Length
Given the right triangles below, find the missing side length and determine if the side lengths form a Pythagorean triple. Explain your reasoning.
#### Step-by-Step Solutions:
**1. Problem 6:**
**Triangle with sides 4 and 5**
- The triangle has one leg of length 4 and the other leg of length 5.
- To find the hypotenuse \(c\), use the Pythagorean theorem: \(a^2 + b^2 = c^2\).
\[
4^2 + 5^2 = c^2
\]
\[
16 + 25 = c^2
\]
\[
41 = c^2
\]
\[
c = \sqrt{41}
\]
- Since \( \sqrt{41} \) is not an integer, the side lengths do not form a Pythagorean triple.
**2. Problem 7:**
**Triangle with sides 12 and 20**
- The triangle has legs of length 12 and 20.
- To find the hypotenuse \(c\), use the Pythagorean theorem: \(a^2 + b^2 = c^2\).
\[
12^2 + 20^2 = c^2
\]
\[
144 + 400 = c^2
\]
\[
544 = c^2
\]
\[
c = \sqrt{544} = 4\sqrt{34}
\]
- Since \(4\sqrt{34}\) is not an integer, these side lengths do not form a Pythagorean triple.
**3. Problem 8:**
**Triangle with legs 1.5 and 1.7**
- The triangle has one leg of length 1.5 and one leg of length 1.7.
- To find the hypotenuse \(c\), use the Pythagorean theorem: \(a^2 + b^2 = c^2\).
\[
1.5^2 + 1.7^2 = c^2
\]
\[
2.25 + 2.89 = c^2
\]
\[
5.14 = c^2
\]
\[
c = \sqrt{5.14}
\]
- Since \(\sqrt{5.14}\) is not](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c2d1f60-2481-46b6-b554-6dc9ed915f51%2F301f0a2f-983d-47f5-8d62-2b269285af04%2Fmntqf4u_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem: Find the Missing Side Length
Given the right triangles below, find the missing side length and determine if the side lengths form a Pythagorean triple. Explain your reasoning.
#### Step-by-Step Solutions:
**1. Problem 6:**
**Triangle with sides 4 and 5**
- The triangle has one leg of length 4 and the other leg of length 5.
- To find the hypotenuse \(c\), use the Pythagorean theorem: \(a^2 + b^2 = c^2\).
\[
4^2 + 5^2 = c^2
\]
\[
16 + 25 = c^2
\]
\[
41 = c^2
\]
\[
c = \sqrt{41}
\]
- Since \( \sqrt{41} \) is not an integer, the side lengths do not form a Pythagorean triple.
**2. Problem 7:**
**Triangle with sides 12 and 20**
- The triangle has legs of length 12 and 20.
- To find the hypotenuse \(c\), use the Pythagorean theorem: \(a^2 + b^2 = c^2\).
\[
12^2 + 20^2 = c^2
\]
\[
144 + 400 = c^2
\]
\[
544 = c^2
\]
\[
c = \sqrt{544} = 4\sqrt{34}
\]
- Since \(4\sqrt{34}\) is not an integer, these side lengths do not form a Pythagorean triple.
**3. Problem 8:**
**Triangle with legs 1.5 and 1.7**
- The triangle has one leg of length 1.5 and one leg of length 1.7.
- To find the hypotenuse \(c\), use the Pythagorean theorem: \(a^2 + b^2 = c^2\).
\[
1.5^2 + 1.7^2 = c^2
\]
\[
2.25 + 2.89 = c^2
\]
\[
5.14 = c^2
\]
\[
c = \sqrt{5.14}
\]
- Since \(\sqrt{5.14}\) is not
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