Question
Use the steps from the proof of the Pythagorean Theorem to find x. (1 point) Responses x=1698 x equals Start Fraction 169 over 8 End Fraction x=16912 x equals Start Fraction 169 over 12 End Fraction x=169144 x equals Start Fraction 169 over 144 End Fraction x=12
13 and 5
13 and 5
Answers
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Answered by
GPT-4o mini
To find \( x \) using the Pythagorean theorem, we can set up our equation based on a right triangle. The theorem states that in a right triangle, the square of the length of the hypotenuse \( c \) is equal to the sum of the squares of the lengths of the other two sides \( a \) and \( b \):
\[
c^2 = a^2 + b^2
\]
If we can set up specific values for \( a \), \( b \), and \( c \), we can solve for \( x \).
From the information provided:
1. The choices include:
- x = 1698
- x = \frac{169}{8}
- x = 16912
- x = \frac{169}{12}
- x = \frac{169}{144}
- x = 12, 13, and 5
2. Assuming \( 12 \), \( 13 \), and \( 5 \) are the sides of a right triangle:
- Let \( a = 5 \) and \( b = 12 \), so we calculate \( c \):
\[
c^2 = 5^2 + 12^2
\]
\[
c^2 = 25 + 144
\]
\[
c^2 = 169
\]
\[
c = \sqrt{169} = 13
\]
The values \( 5 \), \( 12 \), and \( 13 \) do indeed form a Pythagorean triple.
Thus, it seems like a comparison of values is needed. It suggests we check for \( x \) being equal to pairs of (12, 13, and 5) instead of specific values like \( \frac{169}{8} \) or \( \frac{169}{144} \).
Since none of the other fractions seem to be defined in this context with Pythagorean triplets, **the answer appears to be**:
**x = 12, 13, and 5** are valid sides of a right triangle based on the Pythagorean theorem.
\[
c^2 = a^2 + b^2
\]
If we can set up specific values for \( a \), \( b \), and \( c \), we can solve for \( x \).
From the information provided:
1. The choices include:
- x = 1698
- x = \frac{169}{8}
- x = 16912
- x = \frac{169}{12}
- x = \frac{169}{144}
- x = 12, 13, and 5
2. Assuming \( 12 \), \( 13 \), and \( 5 \) are the sides of a right triangle:
- Let \( a = 5 \) and \( b = 12 \), so we calculate \( c \):
\[
c^2 = 5^2 + 12^2
\]
\[
c^2 = 25 + 144
\]
\[
c^2 = 169
\]
\[
c = \sqrt{169} = 13
\]
The values \( 5 \), \( 12 \), and \( 13 \) do indeed form a Pythagorean triple.
Thus, it seems like a comparison of values is needed. It suggests we check for \( x \) being equal to pairs of (12, 13, and 5) instead of specific values like \( \frac{169}{8} \) or \( \frac{169}{144} \).
Since none of the other fractions seem to be defined in this context with Pythagorean triplets, **the answer appears to be**:
**x = 12, 13, and 5** are valid sides of a right triangle based on the Pythagorean theorem.
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