Question
Consider a right-angled triangle with sides a=6
and c=9
. Determine whether these values form a Pythagorean triple. Show your steps and explain your answer. (2 points)
and c=9
. Determine whether these values form a Pythagorean triple. Show your steps and explain your answer. (2 points)
Answers
GPT-4o mini
To determine if the values \( a = 6 \) and \( c = 9 \) form a Pythagorean triple in a right-angled triangle, we need to check if they satisfy the Pythagorean theorem, which states:
\[
a^2 + b^2 = c^2
\]
where \( c \) is the length of the hypotenuse (the longest side), and \( a \) and \( b \) are the lengths of the other two sides.
In this case, since \( c = 9 \) is greater than \( a = 6 \), we identify \( a \) and now we need to find \( b \).
1. We substitute \( a = 6 \) and \( c = 9 \) into the Pythagorean theorem:
\[
6^2 + b^2 = 9^2
\]
2. Calculate \( 6^2 \) and \( 9^2 \):
\[
36 + b^2 = 81
\]
3. Now, isolate \( b^2 \):
\[
b^2 = 81 - 36
\]
\[
b^2 = 45
\]
4. Now, take the square root to find \( b \):
\[
b = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}
\]
Now that we have identified the values of \( a, b, \) and \( c \) as \( 6, 3\sqrt{5}, \) and \( 9 \) respectively, we need to see whether these sides can represent a Pythagorean triple:
For it to be a Pythagorean triple, the values need to be integers. However, \( 3\sqrt{5} \) is not an integer since \( \sqrt{5} \) is an irrational number.
Therefore, while \( a = 6 \), \( b = 3\sqrt{5} \), and \( c = 9 \) satisfy the Pythagorean theorem, they do not form a Pythagorean triple since \( b \) is not an integer.
### Conclusion
The given values \( a = 6 \) and \( c = 9 \) do not form a Pythagorean triple because they result in a non-integer value for the third side \( b \).
\[
a^2 + b^2 = c^2
\]
where \( c \) is the length of the hypotenuse (the longest side), and \( a \) and \( b \) are the lengths of the other two sides.
In this case, since \( c = 9 \) is greater than \( a = 6 \), we identify \( a \) and now we need to find \( b \).
1. We substitute \( a = 6 \) and \( c = 9 \) into the Pythagorean theorem:
\[
6^2 + b^2 = 9^2
\]
2. Calculate \( 6^2 \) and \( 9^2 \):
\[
36 + b^2 = 81
\]
3. Now, isolate \( b^2 \):
\[
b^2 = 81 - 36
\]
\[
b^2 = 45
\]
4. Now, take the square root to find \( b \):
\[
b = \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}
\]
Now that we have identified the values of \( a, b, \) and \( c \) as \( 6, 3\sqrt{5}, \) and \( 9 \) respectively, we need to see whether these sides can represent a Pythagorean triple:
For it to be a Pythagorean triple, the values need to be integers. However, \( 3\sqrt{5} \) is not an integer since \( \sqrt{5} \) is an irrational number.
Therefore, while \( a = 6 \), \( b = 3\sqrt{5} \), and \( c = 9 \) satisfy the Pythagorean theorem, they do not form a Pythagorean triple since \( b \) is not an integer.
### Conclusion
The given values \( a = 6 \) and \( c = 9 \) do not form a Pythagorean triple because they result in a non-integer value for the third side \( b \).