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Is it possible for a triangle to have sides with the given lengths: 5in, 8in, & 15in?

Mathematics
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Other answers:

the sides wont all meet
How about 10cm,12cm,20cm? @UnkleRhaukus
Three things must be true for the sides: Two sides added together must be larger than the third side. a+b>c, a+c>b, b+c>a So, for your first one: a=5, b=8, c=15 5+8=13 > 15 <---- false 8+15=23 > 5 <---- true 5+15=20 > 8 <---- true Because there is one false, the sides are invalid.
How about the second one... im not sure howt o do the formula though.. @PhoenixFire
Try construction
a=10 b=12 c=20 Check the three inequalities I gave you and see if any result in a false. If so, the triangle is invalid. If they are all true, then it is a valid triangle.
It isn't possible. thats what i got.
10+12=22 > 20 <---- true 10+20=30 > 12 <---- true 12+20=32 > 10 <---- true The triangle is VALID.
I dont understand though.
Nevermind! i understand it now.. it has to be greater then the number. thats the part i wasn't understanding. i was doing it the opposite way thats why it through me off . thank you so much!! :D
haha. No problems.
I have 2 more questions that i need help on, do you mind giving me a hand? @PhoenixFire
sure.
Triangle PQR has medians QM and PN that intersect at Z. if Zm=4, what is QZ and QM. @PhoenixFire
|dw:1358503491884:dw| Supposedly ZM is half the length of QZ: |dw:1358503968318:dw| So if ZM=4 then QZ=8 which means QM=ZM+QZ=4+8=12 @UnkleRhaukus If you can validate this. I'm not a triangle expert.
It's a property of medians and centroids I believe. I'm not 100% certain on this.
Thats what i was thinking too. that it was half of that. so whats QZ and QM then?
I gave the answer in my post. ZM=Centroid to midpoint=a=4 QZ=vertex to centroid=2a=8 QM=vertex to midpoint=2a+a=12
QZ=8 and QM=12 is this right?
Yes
okay i have one more.... this one is hard i think lol
|dw:1358428925101:dw| @PhoenixFire
find X?
Yes, sorry.
you there? @phoenixfire
Thinking. Sorry. :P
Its okay.
I'm guessing the 2x line goes between midpoints of those sides?
Yeah
So what I've gathered is that the line connecting two midpoints is half the size of the third side. \[2x={1 \over 2}(x+12)\] So, solve for x I guess. \[4x=x+12\]\[3x=12\]\[x=4\]
Thank you!!!!
No problem. :)

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