Geometry

[vampirewitchreine]

Okay, so basically, I've completely forgotten how to do this. The question from the textbook is "Can the lengths 2, 6, and 5 make a triangle?" My first thought was to kinda do it like Pythagorean theorem, but I'm unsure if this would be correct.

This is what I got when I did that
2^2*6^2(=?)5^2
4*36(=?)25
40(=\)25

Is this wrong? I just want to know how to do this right so that I know for future reference. Please help.

 

[royhaas]

Think about the triangle inequality.

 

[lookagain]

... from the textbook is "Can the lengths 2, 6, and 5 make a triangle?"
My first thought was to kinda do it like Pythagorean theorem, but I'm unsure if this would be correct.

This is what I got when I did that
2^2*6^2(=?)5^2
4*36(=?)25
40(=\)25

Is this wrong? I just want to know how to do this right so that I know for future reference. Please help.

\(\displaystyle \text{ Please refer to the reply of royhaas regarding your specific question.}\)

---------------------------------------------------------------------------------------------------------


What if instead you're asked if those could be the side lengths of a right triangle. ***

Choose the side length which is longest to be c (the hypotenuse),
because the hypotenuse is always the longest side of a right triangle.


\(\displaystyle a^2 \ + \ b^2 \ = \ c^2\)

\(\displaystyle 2^2 \ + \ 5^2 \ = \ ? \ 6^2\)

\(\displaystyle 4 \ + \ 25 \ = \ ? \ 36\)

\(\displaystyle 29 \ \ne \ 36\)

So, for this different question, it would not be a right triangle.
But you wouldn't bother to use this method if you found out
at first that those sides couldn't even form *any* triangle at all.



\(\displaystyle \text{*** However, if the question were different}\)
\(\displaystyle \text{and had asked if those sides could}\)
\(\displaystyle \text{be those of a right triangle, then yes,}\)
\(\displaystyle \text{you could the Pythagorean Theorem.}\)

 

[vampirewitchreine]

So then I would do it like this?

2+5 (7)> 6
5+6 (11) > 2
6+2 (8) > 5

Making the answer yes, right?

 

[mmm4444bot]

Yes. One confirmation is good enough.

 

[Deleted member 4993]

Yes. One confirmation is good enough.

I would suggest that 3 confirmations are needed.

Suppose the sides were 3,2,1

then

3 + 2> 1

3 + 1> 2

but third inequality fails.

However, one failure is good enough to reject "triangularity".

 

[mmm4444bot]

Ahh, yes. I see now that I should have elaborated. I assumed that everyone picks the two smallest sides to see if their sum exceeds the longest side. That's what I do, along with a graphical animation in my mind.

It's better to be thorough.

PS: Are you sure that's not "triangularitylessness"? :lol:

 

[vampirewitchreine]

I would suggest that 3 confirmations are needed.

Suppose the sides were 3,2,1

then

3 + 2> 1

3 + 1> 2

but third inequality fails.

However, one failure is good enough to reject "triangularity".

I appreciate the thought of testing the sums with other numbers, but my book was specifically asking if 2, 5 and 6 could form a triangle. (Also, I totally found the answer in the selected answers, and thanks to lookagain's helpful comment, this was correct.)