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AC + BC could equal 15 Always Never Sometimes

Mathematics
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Well one side is 9. This can get tricky since if A = 1, B = 1, and C = 9 you'll have 18. And you can't have a zero side or a negative side. That's impossible
\[\rm AC = 9\]Now apply the Pythagorean Theorem.\[9^2 + x^2 = 15^2\]Any solutions to the above?

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Other answers:

Turns out they could never, because the only solution is \(12\) but \(9 + 12 = 21\)
Also thinking about how AC + BC > 15, we could never get AC + BC = 15.
oh I remember doing that PArth...
Erm?
the a^2+b^2 = c^2
Triangle Inequality Theorem: One side of a triangle is less than the sum of the other two. Therefore, 15 < AC + BC which is equivalent to AC + BC > 15. AC + BC could never be equal to 15. There would be no triangle. @daustin15

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