## Liv1234 one year ago I need help with a question.

1. Liv1234

Finding the labeled lengths of a triangle. |dw:1435941173879:dw|

2. Liv1234

3. anonymous

Do you have any better pictures?

4. Liv1234

No, but I can try and draw it better.

5. anonymous

I would appreciate that

6. Liv1234

Okay, just a sec.

7. Liv1234

|dw:1435941385187:dw|

8. Liv1234

Is that better? Cause I can try again.

9. anonymous

There is a tool in the draw section that allows you to make a bar, ya know

10. Liv1234

Oh okay thanks. I'm fixing it now

11. Liv1234

|dw:1435941515188:dw|

12. Liv1234

I'm trying to find the labeled lengths.

13. anonymous

And to make sure, the lengths are 4, 4, x for the first triangle and y, 1, 2 for the second?

14. Liv1234

Yes.

15. Liv1234

So, how would I determine the labeled lengths? That's what I'm confused about.

16. Liv1234

@rvc

17. rvc

as legend said

18. Liv1234

Yes, those are the lengths like legend asked.

19. anonymous

I feel like this is about the properties of 45-45-90 triangles and 30-60-90 triangles. The first is definitely a 45-45-90 triangle though.

20. rvc

|dw:1435941970291:dw|

21. Liv1234

So, how would I solve the labeled lengths though? And thanks those pictures are a lot better then mine.

22. rvc

correct?

23. Liv1234

And yeah, Legend is definitely correct.

24. rvc

25. anonymous

But for the first one, to find x, just apply the properties of a 45-45-90 triangle. $\large Hypotenuse~=~Leg\sqrt{2}$

26. Liv1234

So, the square root of 2?

27. anonymous

Should make it a bit clearer |dw:1435942323278:dw|

28. Liv1234

Ohh okay, so 4*4=16 and then the square root of 16 which would be 4?

29. anonymous

No, just $\large x~=~4\sqrt{2}$

30. Liv1234

5.65?

31. Liv1234

or 5.7 rounded?

32. anonymous

Yeah, but it depends on what form they want you to leave it in. I usually would just leave it as $$4\sqrt{2}$$

33. Liv1234

Okay, I have another question.

34. Liv1234

Want me to open another post?

35. anonymous

Are we not gonna solve for y?