Question

In a regular hexagonal pyramid, each lateral face is inclined at an angle of 60 degrees with the base. if the are of the base is 200 sqrt of 3 ft^2. Find the volume of the pyramid.

Answer 1

Area of a regular polygon is defined by the formula
\[A = \frac{ 1 }{ 2 }nR^2\sin \theta\]
where n is the number of sides
R is the radius of the circle that can be circumscribing the polygon
theta is 360/n
\[A = 200\sqrt{3}\]
\[n=6\]
\[\theta = \frac{ 360^o }{ 6 } \rightarrow 60^o\]
\[200\sqrt{3} = \frac{ 1 }{ 2 }6R^2\sin60^o\]
\[R^2 = \frac{ 200\sqrt{3(2)} }{ 6\sin60^o }\]
\[R^2 = \frac{ 400 }{ 3 }\]
\[R = \frac{ 20\sqrt{3} }{ 3 }\]

Answer 2

internet connection's annoying -_-

Answer 3

The main purpose of getting R is to find 'x' and eventually 'h' or the altitude of the pyramid. Hence the formula for the volume of pyramid is\[V = \frac{ 1 }{ 3 } Bh\]
where B is the area of the base and h is the altitude.
Area is given, so all we have to know is the height.

Answer 4

find 'x' using trigonometry
\[\cos30^o = \frac{ x }{ R }\]
\[x = Rcos30^o\]
\[x = \frac{ 20\sqrt{3} }{ 3 }\cos30^o\]
\[x = 10\]
Since the lateral faces are inclined at 60 degrees then the height can be solved
\[\tan60^o = \frac{ h }{ x }\]
\[h = xtan60^o\]
\[h = 10\tan60^o\]
\[h = 10\sqrt3\]
Volume then would be:
\[V = \frac{ 1 }{ 3 }Bh\]
\[V = \frac{ 1 }{ 3 } (200\sqrt{3}) (10\sqrt{3})\]
\[V = 2,000 ft^3\]