Question

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

Answer 1

@hba

Answer 2

??

Answer 3

the sides wont all meet

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

Try construction

Answer 8

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.

Answer 9

It isn't possible. thats what i got.

Answer 10

@PhoenixFire

Answer 11

10+12=22 > 20 <---- true
10+20=30 > 12 <---- true
12+20=32 > 10 <---- true

The triangle is VALID.

Answer 12

I dont understand though.

Answer 13

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

Answer 14

haha. No problems.

Answer 15

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

Answer 16

sure.

Answer 17

Triangle PQR has medians QM and PN that intersect at Z. if Zm=4, what is QZ and QM. @PhoenixFire

Answer 18

Supposedly ZM is half the length of QZ:


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.