Let's determine if each set of numbers can be the lengths of the sides of a right triangle by verifying whether they satisfy the Pythagorean theorem [tex]\(a^2 + b^2 = c^2\)[/tex].
1. For sides [tex]\(a = 5\)[/tex], [tex]\(b = 12\)[/tex], and [tex]\(c = 13\)[/tex]:
[tex]\[
5^2 + 12^2 = 25 + 144 = 169 = 13^2
\][/tex]
This set does satisfy the Pythagorean theorem, so it forms a right triangle.
- Pythagorean triple? Yes
2. For sides [tex]\(a = 12\)[/tex], [tex]\(b = 35\)[/tex], and [tex]\(c = 20\sqrt{3}\)[/tex]:
[tex]\[
12^2 + 35^2 = 144 + 1225 = 1369
\][/tex]
[tex]\[
(20\sqrt{3})^2 = 1200
\][/tex]
This set does not satisfy the Pythagorean theorem, so it does not form a right triangle.
- Pythagorean triple? No
3. For sides [tex]\(a = 5\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 5\sqrt{5}\)[/tex]:
[tex]\[
5^2 + 10^2 = 25 + 100 = 125
\][/tex]
[tex]\[
(5\sqrt{5})^2 = 125
\][/tex]
This set does satisfy the Pythagorean theorem, so it forms a right triangle.
- Pythagorean triple? Yes
4. For sides [tex]\(a = 8\)[/tex], [tex]\(b = 12\)[/tex], and [tex]\(c = 15\)[/tex]:
[tex]\[
8^2 + 12^2 = 64 + 144 = 208
\][/tex]
[tex]\[
15^2 = 225
\][/tex]
This set does not satisfy the Pythagorean theorem, so it does not form a right triangle.
- Pythagorean triple? No
5. For sides [tex]\(a = 20\)[/tex], [tex]\(b = 99\)[/tex], and [tex]\(c = 101\)[/tex]:
[tex]\[
20^2 + 99^2 = 400 + 9801 = 10201
\][/tex]
[tex]\[
101^2 = 10201
\][/tex]
This set does satisfy the Pythagorean theorem, so it forms a right triangle.
- Pythagorean triple? Yes
Here is the final table with the correct answers:
\begin{tabular}{|c|c|c|c|}
\hline
[tex]$a$[/tex] & [tex]$b$[/tex] & [tex]$c$[/tex] & Pythagorean triple? \\
\hline
5 & 12 & 13 & Yes \\
\hline
12 & 35 & [tex]$20 \sqrt{3}$[/tex] & No \\
\hline
5 & 10 & [tex]$5 \sqrt{5}$[/tex] & Yes \\
\hline
8 & 12 & 15 & No \\
\hline
20 & 99 & 101 & Yes \\
\hline
\end{tabular}
