SHSAT Geometry queation

[wind tinsel]

If you look at triangles HEH' and H'E'E, you'll notice they're right triangles so you can use Pythagoras' Theorem.

We need to find HE.

HE is the hypotenuse of HEH' and equal to √ (HH'^2+HE'^2).
HH' is 4 but we don't know H'E.
H'E is part of H'E'E though so it can be found using
H'E^2=H'E^2+E'E^2
=4^2+(2*4)^2
H'E^2 =80
Use this value to finally obtain
HE=√ (HH'^2+HE'^2)
=√ (4^2+80)
=√ (96)
=4√ 6