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

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358502234947:dw

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0the sides wont all meet

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How about 10cm,12cm,20cm? @UnkleRhaukus

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2Three 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.

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

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2a=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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It isn't possible. thats what i got.

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.210+12=22 > 20 < true 10+20=30 > 12 < true 12+20=32 > 10 < true The triangle is VALID.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I dont understand though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Nevermind! 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

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

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

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2dw: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.

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

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

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0QZ=8 and QM=12 is this right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay i have one more.... this one is hard i think lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358428925101:dw @PhoenixFire

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you there? @phoenixfire

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2Thinking. Sorry. :P

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2I'm guessing the 2x line goes between midpoints of those sides?

PhoenixFire
 3 years ago
Best ResponseYou've already chosen the best response.2So 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. \[4x=x+12\]\[3x=12\]\[x=4\]
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