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

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UnkleRhaukusBest ResponseYou've already chosen the best response.0
dw:1358502234947:dw
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
the sides wont all meet
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
How about 10cm,12cm,20cm? @UnkleRhaukus
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
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.
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
How about the second one... im not sure howt o do the formula though.. @PhoenixFire
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
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.
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
It isn't possible. thats what i got.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
10+12=22 > 20 < true 10+20=30 > 12 < true 12+20=32 > 10 < true The triangle is VALID.
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
I dont understand though.
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
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
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
haha. No problems.
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
I have 2 more questions that i need help on, do you mind giving me a hand? @PhoenixFire
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
Triangle PQR has medians QM and PN that intersect at Z. if Zm=4, what is QZ and QM. @PhoenixFire
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
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.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
It's a property of medians and centroids I believe. I'm not 100% certain on this.
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
Thats what i was thinking too. that it was half of that. so whats QZ and QM then?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
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
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
QZ=8 and QM=12 is this right?
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
okay i have one more.... this one is hard i think lol
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
dw:1358428925101:dw @PhoenixFire
 one year ago

barnoldwrlBest ResponseYou've already chosen the best response.0
you there? @phoenixfire
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
Thinking. Sorry. :P
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
I'm guessing the 2x line goes between midpoints of those sides?
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.2
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. \[4x=x+12\]\[3x=12\]\[x=4\]
 one year ago
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