The base perimeter is 6 m, so each side of the hexagon is 1 m
Let's concentrate on one of the triangles of the pyramid, it will be isosceles with sides
2 m, 2m and 1 m. Draw a perpendicular and you will have a right-angled triangle, let
the height be h
then h^2 + (1/2)^2 = 2^2
h^2 = 15/4
h = √15/2
So the area of one of those triangles = (1/2)(1)(√15/2) = √15/4
So 6 of them would be 3√15/2 m^2 <---- lateral surface area
No look at the base, it is also made up of 6 isosceles triangle, sketch one of them
the area of one of them = (1/2)(1)(1)sin60° = (1/2)(√3/2) = √3/4
and with 6 of those, we get a base area of 3√3/2 m^2
We need the height of the pyramid, call it k
Back to the equilateral triangle of the base, all sides have 1, so with basic
geometry we see that the "height" in one of these is √3/2
k^2 + (√3/2)^2 = (√15/2)^2
k^2 + 3/4 = 15/4
k^2 = 3
k = √3, <---- nice
volume of pyramid = (1/3)(base area)(height)
= (1/3)(3√3/2)(√3) = 3/2 m^3
better check my arithmetic on that, should have written it out first.
A lateral edge of the hexagonal pyramid is 2.0m long and its base perimeter is 6m determine the volume of the pyramid and its lateral area
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