The formula for the volume of a hexagonal pyramid: an example of solving a problem

2024 Author: Angel Austin | [email protected]. Last modified: 2024-01-02 17:39

Calculation of volumes of spatial figures is one of the important tasks of stereometry. In this article, we will consider the issue of determining the volume of such a polyhedron as a pyramid, and also give the formula for the volume of a regular hexagonal pyramid.

hexagonal pyramid

First, let's look at what the figure is, which will be discussed in the article.

Let's have an arbitrary hexagon whose sides are not necessarily equal to each other. Also suppose that we have chosen a point in space that is not in the plane of the hexagon. By connecting all the corners of the latter with the selected point, we get a pyramid. Two different pyramids with a hexagonal base are shown in the figure below.

It can be seen that in addition to the hexagon, the figure consists of six triangles, the connection point of which is called the vertex. The difference between the depicted pyramids is that the height h of the right of them does not intersect the hexagonal base in its geometric center, and the height of the left figure fallsright in that center. Thanks to this criterion, the left pyramid was called straight, and the right - oblique.

Since the base of the left figure in the figure is formed by a hexagon with equal sides and angles, it is called correct. Further in the article we will talk only about this pyramid.

Volume of the hexagonal pyramid

To calculate the volume of an arbitrary pyramid, the following formula is valid:

V=1/3hS_o

Here h is the length of the figure's height, S_{o is the area of its base. Let's use this expression to determine the volume of a regular hexagonal pyramid.}

Since the figure under consideration is based on an equilateral hexagon, to calculate its area, you can use the following general expression for an n-gon:

S=n/4a2ctg(pi/n)

Here n is an integer equal to the number of sides (corners) of the polygon, a is the length of its side, the cotangent function is calculated using the appropriate tables.

Applying the expression for n=6, we get:

S₆=6/4a² ctg(pi/6)=√3/2a ²

Now it remains to substitute this expression into the general formula for the volume V:

V₆=S₆h=√3/2ha²

Thus, to calculate the volume of the pyramid under consideration, it is necessary to know its two linear parameters: the length of the side of the base and the height of the figure.

Example of problem solving

Let's show how the obtained expression for V_{6 can be used to solve the following problem.}

It is known that the volume of a regular hexagonal pyramid is 100 cm3. It is necessary to determine the side of the base and the height of the figure, if it is known that they are related to each other by the following equality:

a=2h

Since only a and h are included in the formula for volume, any of these parameters can be substituted into it, expressed in terms of the other. For example, substitute a, we get:

V₆=√3/2h(2h)^2=>

h=∛(V_6/(2√3))

To find the value of the height of a figure, you need to take the root of the third degree from the volume, which corresponds to the dimension of length. We substitute the volume value V_{6of the pyramid from the problem statement, we get the height:}

h=∛(100/(2√3)) ≈ 3.0676 cm

Since the side of the base, in accordance with the condition of the problem, is twice the found value, we get the value for it:

a=2h=23, 0676=6, 1352cm

The volume of a hexagonal pyramid can be found not only through the height of the figure and the value of the side of its base. It is enough to know two different linear parameters of the pyramid to calculate it, for example, the apotema and the length of the side edge.