barnoldwrl
Is it possible for a triangle to have sides with the given lengths: 5in, 8in, & 15in?
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barnoldwrl
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@hba
UnkleRhaukus
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|dw:1358502234947:dw|
barnoldwrl
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??
UnkleRhaukus
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the sides wont all meet
barnoldwrl
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How about 10cm,12cm,20cm? @UnkleRhaukus
PhoenixFire
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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.
barnoldwrl
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How about the second one... im not sure howt o do the formula though.. @PhoenixFire
UnkleRhaukus
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Try construction
PhoenixFire
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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.
barnoldwrl
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It isn't possible. thats what i got.
barnoldwrl
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@PhoenixFire
PhoenixFire
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10+12=22 > 20 <---- true
10+20=30 > 12 <---- true
12+20=32 > 10 <---- true
The triangle is VALID.
barnoldwrl
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I dont understand though.
barnoldwrl
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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
PhoenixFire
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haha. No problems.
barnoldwrl
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I have 2 more questions that i need help on, do you mind giving me a hand? @PhoenixFire
PhoenixFire
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sure.
barnoldwrl
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Triangle PQR has medians QM and PN that intersect at Z. if Zm=4, what is QZ and QM. @PhoenixFire
PhoenixFire
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|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.
PhoenixFire
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It's a property of medians and centroids I believe. I'm not 100% certain on this.
barnoldwrl
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Thats what i was thinking too. that it was half of that. so whats QZ and QM then?
PhoenixFire
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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
barnoldwrl
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QZ=8 and QM=12 is this right?
PhoenixFire
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Yes
barnoldwrl
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okay i have one more.... this one is hard i think lol
barnoldwrl
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|dw:1358428925101:dw| @PhoenixFire
PhoenixFire
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find X?
barnoldwrl
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Yes, sorry.
barnoldwrl
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you there? @phoenixfire
PhoenixFire
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Thinking. Sorry. :P
barnoldwrl
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Its okay.
PhoenixFire
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I'm guessing the 2x line goes between midpoints of those sides?
barnoldwrl
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Yeah
PhoenixFire
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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.
\[4x=x+12\]\[3x=12\]\[x=4\]
barnoldwrl
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Thank you!!!!
PhoenixFire
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No problem. :)