TRIG or Physics ppl...Help me please!?

Question

a regular tetrahedron is a three-dimensional object that has four faces, each of which is an equilateral triangle. Each of the edges of such an object has a length L. The height H of a regular tetrahedron is the perpendicular distance from one corner to the center of the opposite triangular face. Show that the ratio between H and L is H/L = sqrt 2/3.
Please help if you can. Answers only are great but they don't help come test time. Can anyone break this down in understndable terms?? All help is appreciated!! Thanks!

Answer 1

Your answer is correct.
You'll have to sketch while reading this description.Imagine the solid resting on a horizontal surface. Lets call the top vertex A. Drop a perpendicular along one face meeting edge CD at E. Also drop a perpendicular from the fourth vertex, F, to the point E.Drop an altitude of the tetrahedron from A meeting FE at G. Triangle ADE is a 30-60-90 with hypotenuse of length L, so AE has length((√3)/2)L. Triangle GCE is also a 30-60-90 with larger leg,CE, of length L/2, so GE has length((√3)/6)L. Now lets look at triangle AGE. It's a right triangle and we know the hypotenuse,AE, and a leg, GE. The other side has length H.
H =√[(((√3)/2)L)^2 - (((√3)/6)L)^2]
H = √[(3/4)L^2 - (3/36)L^2]
H = √[(3/4)L^2 - L/12]
H = √[(9/12)L^2 - (1/12)L^2]
H = √(8/12)L^2
H = (√(2/3))L

Answer 2

well its pretty easy...its just tt i think ur answer is wrong...its should be 2*sqrt 3 not sqrt 2/3

since its all equilateral triangles, (draw a triangle on a piece of paper)

every angle is 60 degrees, drop a vertical line from the tip of one edge to the middle of the opposite side,

thats the height, H and the side which the Height cuts label it L

so using ur trigonometric ratio, tan 60 = H / 0.5 L

its 0.5 L because the height cuts the side at exactly the middle...so its length is half L

tan 60 is a special angle...

tan 60 = sqrt 3 = H/2L

moving the 2 over, we have H/L = 2 sqrt 3

thats all ! ! ! hope u see it...its easy...ur answer is wrong...