What is the volume of a hexagonal base pyramid? ?

Refer to the figure below.

The hexagonal base can be divided into six congruent equilateral triangles.
Height of each equilateral triangle
= 1 × sin60°
= (√3)/2

* Area of an equilateral triangle
= (1/2) × base of triangle × height of triangle
= (1/2) × 1 × (√3)/2
= (√3)/4

Area of the hexagonal base
= (Area of an equilateral triangle) × 6
= [(√3)/4] × 6
= 3(√3)/2

Volume of the pyramid
= (1/3) × base area × height of pyramid
= (1/3) × [3(√3/2)] × 10
= 5√3 cubic units
≈ 8.66 cubic units

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* The area of an equilateral triangle can also be calculated by either of the following ways:
Area = (1/2) × 1² × sin(60°) = (1/2) × (√3/2) = (√3)/4
Area = √{s[s - 1]³} = √{(3/2)[(3/2) - 1]³ = √(3/16) = (√3)/4

Volume (V)  =  (1/3) * ( Area of the base ) * Height of the pyramid.

As it is clear from the following figure, Area (A) of the base = 6 * (area of the triangle)

A  =  6 * 0.5 * √(3)/2 = 2.598 sq.unit

Hence Volume  =  (1/3) * 2.598 * 10  

=  8.66 cub units  .............................. Answer

Any pyramid with a constant shape cross section essentially has the same formula for volume, which is 1/3 of the volume of the prism it would sit in, ie 
  
Volume = base area x height x 1/3

The volume of a pyramid is 1/3 * height * area of base.
If it's a regular hexagon,  google "area of a regular hexagon"
If not, figure out what the shape is and do some 2D geometry.