Pythagorean Converse

[kmcfar]

I have a question not sure how to solve when there is a given
AB =2(BC) and AC =12, find: AB___, BC ____

 

[tkhunny]

So, roughly speaking, \(\displaystyle \overline{AB}^{2} + \overline{AC}^{2} = \overline{BC}^{2}\)?

Then: \(\displaystyle (2\overline{BC})^{2} + 12^{2} = \overline{BC}^{2}\). This appears to lead to a solution for BC.

 

[TchrWill]

I have a question not sure how to solve when there is a given
AB =2(BC) and AC =12, find: AB___, BC ____

It would appear that you have two options that satisfy the given data.

(AB)^2 + (2AB)^2 = 12^2 or (AB)^2 + 12^2 = (2AB)^2 either of which yield you a solution for AB...

 

[tkhunny]

Plus, it's not perfectly clear that we are talking about a RIGHT triangle. There is a good hint in the title, but I would like a more explicit description.

 

[pka]

Here is a LaTeX note.
[TEX]\overline{AB}^~2[/TEX] gives \(\displaystyle \overline{AB}^~2\).

 

[tkhunny]

Excellent. I don't know why I couldn't find that a few days ago.