x, x, 12; obtuse
x^2 + x^2 > 12^2 2 x^2 > 144 x^2 > 72 --> need to solve this for x.
2.44?
Wouldn't it be a less than instead of greater than since the triangle is obtuse?
2.44 wouldn't actually work because there's an extra restriction now that it's obtuse. You also need \(x+x > 12\) because if that wasn't true, then the two smaller sides wouldn't be able to connect to each other.
@ The King --> I was trying to say that I thought you had the theorem backwards on that other problem. I will look at the solution you cranked out. If the CORRECT theorem is used, there will be no problem.
So it's not 2.44? I am so lost.
The \(x+x > 12\) is merely the triangle inequality theorem: If the sides of your triangle are \(a, b, c\) the the length of \(a+b\) is greater than the length of \(c\). So 2.44 would actually be two small. You need a value greater than 6.
However, you also need to use the theorems that Directrix posted that says \(a^2+b^2 < c^2\) so \(x^2+x^2<12^2 \Longrightarrow 2x^2 < 144\)
Can you solve for \(x\) from there?
I give up. Maybe I should just go for tutoring tomorrow morning.
Just pretend the < sign is an = sign until you solve for x. When you solve for x, merely replace the = sign with < again.
Well, if I could have chance to work through the ENTIRE problem with you without someone else mixing another approach into the work, we might be able to get somewhere. So, I don't know. When the King is finished with this problem, I will try to help.
Go ahead. I'll stay out.
Alright.
x^2 + x^2 > 12^2 2 x^2 > 144 x^2 > 72 --> need to solve this for x. x > (+/-) 6 √ 2 The negative root does not apply here. x > 8.48, say 8.5 Pick a number x such that x > 8.48.
Post it here.
8.48
Is 8.48 GREATER THAN 8.48 ? (Pick a number x such that x > 8.48.) This is from above.
Pick a number > 8.48. I need to find the instructions for what we are doing.
Any number??
YES
9 would be good
I cannot find the instructions for the original problem?
Whatever they are, 9, 9, 12 could be the sides of this obtuse triangle. There are others.
this is so confusing for me..
Could I just try one last one?
Let's start in a new thread with the instructions. I don't know what we do after getting the sides. Plus, you owe me a meY@l. lol
hahah :P
What about that other thread tonight where I was working for you lol
I fell asleep on my desk lol. The other night I was up till 3 studying for a different test.
Not what I mean. The first problem from tonight about the acute triangle.
I'd like to point out that a 9, 9, 12 triangle is an acute triangle, where the measure of angle ABC is approximately 84 degrees if the length of AB=BC=9.
new question posted!
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