AC + BC could equal 15

Always
Never
Sometimes

Question

AC + BC could equal 15

Always
Never
Sometimes

anonymous · Answer

attachment

usukidoll · Answer

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 · Answer

$$\rm AC = 9$$Now apply the Pythagorean Theorem.$$9^2 + x^2 = 15^2$$Any solutions to the above?

parthkohli · Answer

Turns out they could never, because the only solution is $12$ but $9 + 12 = 21$

parthkohli · Answer

Also thinking about how AC + BC > 15, we could never get AC + BC = 15.

usukidoll · Answer

oh I remember doing that PArth...

parthkohli · Answer

Erm?

usukidoll · Answer

the a^2+b^2 = c^2

directrix · Answer

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