anonymous
  • anonymous
AC + BC could equal 15 Always Never Sometimes
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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UsukiDoll
  • UsukiDoll
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
ParthKohli
  • ParthKohli
\[\rm AC = 9\]Now apply the Pythagorean Theorem.\[9^2 + x^2 = 15^2\]Any solutions to the above?

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ParthKohli
  • ParthKohli
Turns out they could never, because the only solution is \(12\) but \(9 + 12 = 21\)
ParthKohli
  • ParthKohli
Also thinking about how AC + BC > 15, we could never get AC + BC = 15.
UsukiDoll
  • UsukiDoll
oh I remember doing that PArth...
ParthKohli
  • ParthKohli
Erm?
UsukiDoll
  • UsukiDoll
the a^2+b^2 = c^2
Directrix
  • Directrix
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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