[tex]\mathbb{SOLUTION:}[/tex]
[tex]\begin{gathered}\begin{array}{l}{A. \: \rm In \: the \: △ABC, Using \: Pythagorean \: Theorem} \\ \sf h = \sqrt{800^{2} - 400^{2}} \\ \sf h = \sqrt{480000} \\ \sf h = 400\sqrt{3} \\ \\ \rm Answer: \small \boxed{\sf The \: height \: of \: each \: triangle \: is \: 400\sqrt{3} \: ft} \\ \ \ \\ {B. \: \sf The \: solid \: figure \: shown \: to \: the \: attached \: picture} \\ \textrm{To find the point G is the center of square base.} \\ \\ So, \\ {{\rm{ FG = \frac{1}{2}AD}}}\Longrightarrow {\bold {\boxed {\frac{1}{2} × 800}}} \Longrightarrow {\bold {\boxed{400}}} \\ \\ \sf{According \: to \: A, we \: can \: know \: EF = 400\sqrt{3}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{In \: △FGE}, \\ GE = \sqrt{EF^{2} - FG^{2}} \\ GE = \sqrt{(400\sqrt{3})^{2} - 400^{2}} \\ GE = \sqrt{320000} \\ GE = 400\sqrt{2} \\ \\ \rm Answer: \small \boxed{\sf The \: height \: of \: the \: pyramid \: is \: 400\sqrt{2} \: ft}\end{array} \end{gathered}[/tex]
