## barnoldwrl 2 years ago Is it possible for a triangle to have sides with the given lengths: 5in, 8in, & 15in?

1. barnoldwrl

@hba

2. UnkleRhaukus

|dw:1358502234947:dw|

3. barnoldwrl

??

4. UnkleRhaukus

the sides wont all meet

5. barnoldwrl

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

6. 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.

7. barnoldwrl

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

8. UnkleRhaukus

Try construction

9. 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.

10. barnoldwrl

It isn't possible. thats what i got.

11. barnoldwrl

@PhoenixFire

12. PhoenixFire

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

13. barnoldwrl

I dont understand though.

14. barnoldwrl

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

15. PhoenixFire

haha. No problems.

16. barnoldwrl

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

17. PhoenixFire

sure.

18. barnoldwrl

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

19. 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.

20. PhoenixFire

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

21. barnoldwrl

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

22. 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

23. barnoldwrl

QZ=8 and QM=12 is this right?

24. PhoenixFire

Yes

25. barnoldwrl

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

26. barnoldwrl

|dw:1358428925101:dw| @PhoenixFire

27. PhoenixFire

find X?

28. barnoldwrl

Yes, sorry.

29. barnoldwrl

you there? @phoenixfire

30. PhoenixFire

Thinking. Sorry. :P

31. barnoldwrl

Its okay.

32. PhoenixFire

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

33. barnoldwrl

Yeah

34. 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$

35. barnoldwrl

Thank you!!!!

36. PhoenixFire

No problem. :)