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To determine which three pipes can form a right triangle, we can use the Pythagorean theorem which states that for a right triangle with sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] (where [tex]\(c\)[/tex] is the hypotenuse), the following relationship holds:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
We need to check all combinations of three pipes from the given lengths: [tex]\(6 \, \text{cm}\)[/tex], [tex]\(4 \, \text{cm}\)[/tex], [tex]\(8 \, \text{cm}\)[/tex], [tex]\(10 \, \text{cm}\)[/tex], and [tex]\(24 \, \text{cm}\)[/tex] to see if they satisfy this condition.
We'll go through the combinations step by step and find their squares.
1. Combination: 6 cm, 4 cm, 8 cm
[tex]\[ 6^2 = 36, \quad 4^2 = 16, \quad 8^2 = 64 \][/tex]
[tex]\[ 36 + 16 = 52 \, \text{(not equal to 64)} \][/tex]
2. Combination: 6 cm, 4 cm, 10 cm
[tex]\[ 6^2 = 36, \quad 4^2 = 16, \quad 10^2 = 100 \][/tex]
[tex]\[ 36 + 16 = 52 \, \text{(not equal to 100)} \][/tex]
3. Combination: 6 cm, 4 cm, 24 cm
[tex]\[ 6^2 = 36, \quad 4^2 = 16, \quad 24^2 = 576 \][/tex]
[tex]\[ 36 + 16 = 52 \, \text{(not equal to 576)} \][/tex]
4. Combination: 6 cm, 8 cm, 10 cm
[tex]\[ 6^2 = 36, \quad 8^2 = 64, \quad 10^2 = 100 \][/tex]
[tex]\[ 36 + 64 = 100 \, \text{(equal to 100)} \][/tex]
5. Combination: 6 cm, 8 cm, 24 cm
[tex]\[ 6^2 = 36, \quad 8^2 = 64, \quad 24^2 = 576 \][/tex]
[tex]\[ 36 + 64 = 100 \, \text{(not equal to 576)} \][/tex]
6. Combination: 6 cm, 10 cm, 24 cm
[tex]\[ 6^2 = 36, \quad 10^2 = 100, \quad 24^2 = 576 \][/tex]
[tex]\[ 36 + 100 = 136 \, \text{(not equal to 576)} \][/tex]
7. Combination: 4 cm, 8 cm, 10 cm
[tex]\[ 4^2 = 16, \quad 8^2 = 64, \quad 10^2 = 100 \][/tex]
[tex]\[ 16 + 64 = 80 \, \text{(not equal to 100)} \][/tex]
8. Combination: 4 cm, 8 cm, 24 cm
[tex]\[ 4^2 = 16, \quad 8^2 = 64, \quad 24^2 = 576 \][/tex]
[tex]\[ 16 + 64 = 80 \, \text{(not equal to 576)} \][/tex]
9. Combination: 4 cm, 10 cm, 24 cm
[tex]\[ 4^2 = 16, \quad 10^2 = 100, \quad 24^2 = 576 \][/tex]
[tex]\[ 16 + 100 = 116 \, \text{(not equal to 576)} \][/tex]
10. Combination: 8 cm, 10 cm, 24 cm
[tex]\[ 8^2 = 64, \quad 10^2 = 100, \quad 24^2 = 576 \][/tex]
[tex]\[ 64 + 100 = 164 \, \text{(not equal to 576)} \][/tex]
From the combinations above, we see only one set satisfies the Pythagorean theorem:
[tex]\[ \boxed{(6 \, \text{cm}, 8 \, \text{cm}, 10 \, \text{cm})} \][/tex]
Therefore, the three pipes that can be used to form a right triangle are the pipes measuring [tex]\(6 \, \text{cm}\)[/tex], [tex]\(8 \, \text{cm}\)[/tex], and [tex]\(10 \, \text{cm}\)[/tex].
[tex]\[ a^2 + b^2 = c^2 \][/tex]
We need to check all combinations of three pipes from the given lengths: [tex]\(6 \, \text{cm}\)[/tex], [tex]\(4 \, \text{cm}\)[/tex], [tex]\(8 \, \text{cm}\)[/tex], [tex]\(10 \, \text{cm}\)[/tex], and [tex]\(24 \, \text{cm}\)[/tex] to see if they satisfy this condition.
We'll go through the combinations step by step and find their squares.
1. Combination: 6 cm, 4 cm, 8 cm
[tex]\[ 6^2 = 36, \quad 4^2 = 16, \quad 8^2 = 64 \][/tex]
[tex]\[ 36 + 16 = 52 \, \text{(not equal to 64)} \][/tex]
2. Combination: 6 cm, 4 cm, 10 cm
[tex]\[ 6^2 = 36, \quad 4^2 = 16, \quad 10^2 = 100 \][/tex]
[tex]\[ 36 + 16 = 52 \, \text{(not equal to 100)} \][/tex]
3. Combination: 6 cm, 4 cm, 24 cm
[tex]\[ 6^2 = 36, \quad 4^2 = 16, \quad 24^2 = 576 \][/tex]
[tex]\[ 36 + 16 = 52 \, \text{(not equal to 576)} \][/tex]
4. Combination: 6 cm, 8 cm, 10 cm
[tex]\[ 6^2 = 36, \quad 8^2 = 64, \quad 10^2 = 100 \][/tex]
[tex]\[ 36 + 64 = 100 \, \text{(equal to 100)} \][/tex]
5. Combination: 6 cm, 8 cm, 24 cm
[tex]\[ 6^2 = 36, \quad 8^2 = 64, \quad 24^2 = 576 \][/tex]
[tex]\[ 36 + 64 = 100 \, \text{(not equal to 576)} \][/tex]
6. Combination: 6 cm, 10 cm, 24 cm
[tex]\[ 6^2 = 36, \quad 10^2 = 100, \quad 24^2 = 576 \][/tex]
[tex]\[ 36 + 100 = 136 \, \text{(not equal to 576)} \][/tex]
7. Combination: 4 cm, 8 cm, 10 cm
[tex]\[ 4^2 = 16, \quad 8^2 = 64, \quad 10^2 = 100 \][/tex]
[tex]\[ 16 + 64 = 80 \, \text{(not equal to 100)} \][/tex]
8. Combination: 4 cm, 8 cm, 24 cm
[tex]\[ 4^2 = 16, \quad 8^2 = 64, \quad 24^2 = 576 \][/tex]
[tex]\[ 16 + 64 = 80 \, \text{(not equal to 576)} \][/tex]
9. Combination: 4 cm, 10 cm, 24 cm
[tex]\[ 4^2 = 16, \quad 10^2 = 100, \quad 24^2 = 576 \][/tex]
[tex]\[ 16 + 100 = 116 \, \text{(not equal to 576)} \][/tex]
10. Combination: 8 cm, 10 cm, 24 cm
[tex]\[ 8^2 = 64, \quad 10^2 = 100, \quad 24^2 = 576 \][/tex]
[tex]\[ 64 + 100 = 164 \, \text{(not equal to 576)} \][/tex]
From the combinations above, we see only one set satisfies the Pythagorean theorem:
[tex]\[ \boxed{(6 \, \text{cm}, 8 \, \text{cm}, 10 \, \text{cm})} \][/tex]
Therefore, the three pipes that can be used to form a right triangle are the pipes measuring [tex]\(6 \, \text{cm}\)[/tex], [tex]\(8 \, \text{cm}\)[/tex], and [tex]\(10 \, \text{cm}\)[/tex].
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