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

- anonymous

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

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- anonymous

@hba

- UnkleRhaukus

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- anonymous

??

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## More answers

- UnkleRhaukus

the sides wont all meet

- anonymous

How about 10cm,12cm,20cm? @UnkleRhaukus

- PhoenixFire

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.

- anonymous

How about the second one... im not sure howt o do the formula though.. @PhoenixFire

- UnkleRhaukus

Try construction

- PhoenixFire

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.

- anonymous

It isn't possible. thats what i got.

- anonymous

@PhoenixFire

- PhoenixFire

10+12=22 > 20 <---- true
10+20=30 > 12 <---- true
12+20=32 > 10 <---- true
The triangle is VALID.

- anonymous

I dont understand though.

- anonymous

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

haha. No problems.

- anonymous

I have 2 more questions that i need help on, do you mind giving me a hand? @PhoenixFire

- PhoenixFire

sure.

- anonymous

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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Supposedly ZM is half the length of QZ:
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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

It's a property of medians and centroids I believe. I'm not 100% certain on this.

- anonymous

Thats what i was thinking too. that it was half of that. so whats QZ and QM then?

- PhoenixFire

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

- anonymous

QZ=8 and QM=12 is this right?

- PhoenixFire

Yes

- anonymous

okay i have one more.... this one is hard i think lol

- anonymous

|dw:1358428925101:dw| @PhoenixFire

- PhoenixFire

find X?

- anonymous

Yes, sorry.

- anonymous

you there? @phoenixfire

- PhoenixFire

Thinking. Sorry. :P

- anonymous

Its okay.

- PhoenixFire

I'm guessing the 2x line goes between midpoints of those sides?

- anonymous

Yeah

- PhoenixFire

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.
\[4x=x+12\]\[3x=12\]\[x=4\]

- anonymous

Thank you!!!!

- PhoenixFire

No problem. :)

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