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Mathematics 24 Online
OpenStudy (anonymous):

x, x, 12; obtuse

Directrix (directrix):

x^2 + x^2 > 12^2 2 x^2 > 144 x^2 > 72 --> need to solve this for x.

OpenStudy (anonymous):

2.44?

OpenStudy (kinggeorge):

Wouldn't it be a less than instead of greater than since the triangle is obtuse?

Directrix (directrix):

OpenStudy (kinggeorge):

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.

Directrix (directrix):

@ 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.

OpenStudy (anonymous):

So it's not 2.44? I am so lost.

OpenStudy (kinggeorge):

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.

OpenStudy (kinggeorge):

However, you also need to use the theorems that Directrix posted that says \(a^2+b^2 < c^2\) so \(x^2+x^2<12^2 \Longrightarrow 2x^2 < 144\)

OpenStudy (kinggeorge):

Can you solve for \(x\) from there?

OpenStudy (anonymous):

I give up. Maybe I should just go for tutoring tomorrow morning.

OpenStudy (kinggeorge):

Just pretend the < sign is an = sign until you solve for x. When you solve for x, merely replace the = sign with < again.

Directrix (directrix):

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.

OpenStudy (kinggeorge):

Go ahead. I'll stay out.

OpenStudy (anonymous):

Alright.

Directrix (directrix):

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.

Directrix (directrix):

Post it here.

OpenStudy (anonymous):

8.48

Directrix (directrix):

Is 8.48 GREATER THAN 8.48 ? (Pick a number x such that x > 8.48.) This is from above.

Directrix (directrix):

Pick a number > 8.48. I need to find the instructions for what we are doing.

OpenStudy (anonymous):

Any number??

Directrix (directrix):

YES

Directrix (directrix):

9 would be good

Directrix (directrix):

I cannot find the instructions for the original problem?

Directrix (directrix):

Whatever they are, 9, 9, 12 could be the sides of this obtuse triangle. There are others.

OpenStudy (anonymous):

this is so confusing for me..

OpenStudy (anonymous):

Could I just try one last one?

Directrix (directrix):

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

OpenStudy (anonymous):

hahah :P

Directrix (directrix):

What about that other thread tonight where I was working for you lol

OpenStudy (anonymous):

I fell asleep on my desk lol. The other night I was up till 3 studying for a different test.

Directrix (directrix):

Not what I mean. The first problem from tonight about the acute triangle.

OpenStudy (kinggeorge):

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.

OpenStudy (anonymous):

new question posted!

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