Geometry Help

[matthew042]

In the problem below, I have already solved WY which is the 2 times the square root of six using the pythagorean theorem. I still need help solving YV. Any help would be appreciated. Thanks everyone.
[attachment=0:2h41vq2k]Picture.jpg[/attachment:2h41vq2k]

 

[Mrspi]

In the problem below, I have already solved WY which is the 2 times the square root of six using the pythagorean theorem. I still need help solving YV. Any help would be appreciated. Thanks everyone.
[attachment=0:t28ua83k]Picture.jpg[/attachment:t28ua83k]

I assume that you got 2 sqrt(6) for WY using the fact that the altitude to the hypotenuse of a right triangle is the geometric mean between the two segments of the hypotenuse:

XY/ WY = WY / YZ

4 / WY = WY / 6

(WY)^2 = 24

WY = sqrt(24) = 2 sqrt (6)

Use this same approach to find YV.

You have another right triangle, WZV. YZ is the altitude to the hypotenuse of THAT right triangle, so

WY / YZ = YZ/YV

You know that WY = 2 sqrt(6)
You know that YZ = 6

Let x = YV

2 sqrt(6) / 6 = 6 / x

x(2 sqrt(6)) = 36

You should be able to solve that for x.