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Let's determine if each set of numbers can form the sides of a right triangle using the Pythagorean theorem. According to the Pythagorean theorem, for sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], and hypotenuse [tex]\(c\)[/tex], the relationship must satisfy [tex]\(a^2 + b^2 = c^2\)[/tex].
1. For [tex]\(a = 5\)[/tex], [tex]\(b = 12\)[/tex], [tex]\(c = 13\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(5^2 + 12^2 = 25 + 144 = 169\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\(13^2 = 169\)[/tex]
- Since [tex]\(169 = 169\)[/tex], these numbers can form a right triangle.
- Answer: Yes
2. For [tex]\(a = 12\)[/tex], [tex]\(b = 35\)[/tex], [tex]\(c = 20 \sqrt{3}\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(12^2 + 35^2 = 144 + 1225 = 1369\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\((20 \sqrt{3})^2 = 400 \times 3 = 1200\)[/tex]
- Since [tex]\(1369 \neq 1200\)[/tex], these numbers cannot form a right triangle.
- Answer: No
3. For [tex]\(a = 5\)[/tex], [tex]\(b = 10\)[/tex], [tex]\(c = 5 \sqrt{5}\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(5^2 + 10^2 = 25 + 100 = 125\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\((5 \sqrt{5})^2 = 25 \times 5 = 125\)[/tex]
- Since [tex]\(125 = 125\)[/tex], these numbers can form a right triangle.
- Answer: Yes
4. For [tex]\(a = 8\)[/tex], [tex]\(b = 12\)[/tex], [tex]\(c = 15\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(8^2 + 12^2 = 64 + 144 = 208\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\(15^2 = 225\)[/tex]
- Since [tex]\(208 \neq 225\)[/tex], these numbers cannot form a right triangle.
- Answer: No
5. For [tex]\(a = 20\)[/tex], [tex]\(b = 99\)[/tex], [tex]\(c = 101\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(20^2 + 99^2 = 400 + 9801 = 10201\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\(101^2 = 10201\)[/tex]
- Since [tex]\(10201 = 10201\)[/tex], these numbers can form a right triangle.
- Answer: Yes
Thus, we fill out the table as follows:
\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}
1. For [tex]\(a = 5\)[/tex], [tex]\(b = 12\)[/tex], [tex]\(c = 13\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(5^2 + 12^2 = 25 + 144 = 169\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\(13^2 = 169\)[/tex]
- Since [tex]\(169 = 169\)[/tex], these numbers can form a right triangle.
- Answer: Yes
2. For [tex]\(a = 12\)[/tex], [tex]\(b = 35\)[/tex], [tex]\(c = 20 \sqrt{3}\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(12^2 + 35^2 = 144 + 1225 = 1369\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\((20 \sqrt{3})^2 = 400 \times 3 = 1200\)[/tex]
- Since [tex]\(1369 \neq 1200\)[/tex], these numbers cannot form a right triangle.
- Answer: No
3. For [tex]\(a = 5\)[/tex], [tex]\(b = 10\)[/tex], [tex]\(c = 5 \sqrt{5}\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(5^2 + 10^2 = 25 + 100 = 125\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\((5 \sqrt{5})^2 = 25 \times 5 = 125\)[/tex]
- Since [tex]\(125 = 125\)[/tex], these numbers can form a right triangle.
- Answer: Yes
4. For [tex]\(a = 8\)[/tex], [tex]\(b = 12\)[/tex], [tex]\(c = 15\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(8^2 + 12^2 = 64 + 144 = 208\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\(15^2 = 225\)[/tex]
- Since [tex]\(208 \neq 225\)[/tex], these numbers cannot form a right triangle.
- Answer: No
5. For [tex]\(a = 20\)[/tex], [tex]\(b = 99\)[/tex], [tex]\(c = 101\)[/tex]:
- Calculate [tex]\(a^2 + b^2\)[/tex]: [tex]\(20^2 + 99^2 = 400 + 9801 = 10201\)[/tex]
- Calculate [tex]\(c^2\)[/tex]: [tex]\(101^2 = 10201\)[/tex]
- Since [tex]\(10201 = 10201\)[/tex], these numbers can form a right triangle.
- Answer: Yes
Thus, we fill out the table as follows:
\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}
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