A cutting plane is passed through a right regular hexagonal pyramid whose measure of one base edge is 8 inches

Question

If the solid plane parallel to and 6 inches from the base of the pyramid makes a section with an area 18.5 square inches. Find the altitude of the pyramid.

Answer 1

S_base = (3sqrt(3)/2) L^2

S_section = S_base * [1-(y/h)]^2

18.65 = (3sqrt(3)/2) 8^2 [1-(6/h)]^2 ==> h= 9.0213 in

Answer 2

Let
h = height pyramid
8 = base edge
s = edge 6 inches above base

Solve for s. For one triangle with edge s:

Area Tri = [(1/2)s] [(√3/2)s] = (√3/4)s²
Area Plane Intersection = 6(√3/4)s² = 18.5 = 37/2
s² = (37/2) / [6(√3/4)] = 37 / (3√3) = 37√3 / 9
s = √(37√3)/3

By similar triangles we have:

(h - 6)/s = h/8

Multiply both sides by 8s to clear the denominators.

8(h - 6) = sh
8h - 48 = sh
8h - sh = 48
(8 - s)h = 48
h = 48/(8 - s)

Plug in the value for s.

h = 48/[8 - √(37√3)/3]
h = 144/[24 - √(37√3)] ≈ 9.0030195 inches