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?

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

@hba

- UnkleRhaukus

|dw:1358502234947:dw|

- anonymous

??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

|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

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. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.