The volume of a regular hexagonal pyramid is 6 sides. Volume of a regular hexagonal pyramid

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 a regular hexagonal one.

Pyramid hexagonal

To begin with, let's consider what the figure is, which will be discussed in the article.

Let us 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 having a hexagonal base are shown in the picture 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 one does not intersect the hexagonal base at its geometric center, while the height of the left figure falls exactly into this center. Thanks to this criterion, the left pyramid was called straight, and the right - inclined.

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.

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

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, the following general expression for an n-gon can be used to calculate its area:

S n = n/4 * a 2 * ctg(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:

S 6 \u003d 6/4 * a 2 * ctg (pi / 6) \u003d √3/2 * a 2

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

V 6 \u003d S 6 * h \u003d √3 / 2 * h * a 2

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.

Problem solution example

Let us show how the obtained expression for V 6 can be used to solve the following problem.

It is known that the correct volume is 100 cm 3. 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:

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

V 6 \u003d √3 / 2 * h * (2 * h) 2 \u003d\u003e

h = ∛(V 6 /(2*√3))

To find the value of the height of the figure, it is necessary to take the root of the third degree from the volume, which corresponds to the dimension of length. We substitute the volume value V 6 of the pyramid from the condition of the problem, 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 value found, we obtain the value for it:

a = 2*h = 2*3.0676 = 6.1352 cm

Volume 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 apothem and the length of the side edge.

Problems with pyramids. In this article, we will continue to consider problems with pyramids. They cannot be attributed to any class or type of tasks and give general (algorithms) recommendations for solving. It's just that the rest of the tasks that were not considered earlier are collected here.

I will list the theory that needs to be refreshed in memory before solving: pyramids, similarity properties of figures and bodies, properties of regular pyramids, the Pythagorean theorem, the triangle area formula (it is the second one). Consider the tasks:

From triangular pyramid, whose volume is 80, the triangular pyramid is cut off by a plane passing through the top of the pyramid and the middle line of the base. Find the volume of the cut off triangular pyramid.

The volume of a pyramid is equal to one third of the product of the area of ​​its base and its height:

These pyramids (original and clipped) have a common height, so their volumes are related as the areas of their bases. middle line from the original triangle cuts off a triangle whose area is four times smaller, that is:

You can see more about this here.

This means that the volume of the cut-off pyramid will be four times smaller.

So it will be 20.

Answer: 20

* a similar problem, the formula for the area of ​​a triangle is used.

The volume of a triangular pyramid is 15. The plane passes through the side of the base of this pyramid and intersects the opposite side edge at a point dividing it in a ratio of 1: 2, counting from the top of the pyramid. Find the largest of the volumes of the pyramids into which the plane divides the original pyramid.

Let's build a pyramid, mark the vertices.Mark a point E on the edge AS so that AE is twice as large as ES (in the condition it is said that ES relates to AE as 1 to 2), and construct the indicated plane passing through the edge AC and the point E:

Let's analyze the volume of which pyramid will be larger: EABC or SEBC?

* The volume of a pyramid is equal to one third of the product of the area of ​​its base and its height:

If we consider the two resulting pyramids and take the EBC face as the base in both, then it becomes obvious that the volume of the AEBC pyramid will be greater than the volume of the SEBC pyramid. Why?

The distance from point A to the EBC plane is greater than the distance from point S. And this distance plays the role of height for us.

So, let's find the volume of the EABC pyramid.

The volume of the initial pyramid is given to us, the base of the SABC and EABC pyramids is common. If we establish the ratio of heights, then we can easily determine the volume.

From the ratio of segments ES and AE it follows that AE is equal to two thirds of ES. The heights of the pyramids SABC and EABC are in the same relationship -the height of the pyramid EABC will be equal to 2/3 of the height of the pyramid SABC.

Thus, if

That

Answer: 10

The volume of a regular hexagonal pyramid is 6. The side of the base is 1. Find the side edge.

In a regular pyramid, the top is projected into the center of the base.Let's perform additional constructions:

We can find the side edge from right triangle SOC. To do this, you need to know SO and OS.

SO is the height of the pyramid, we can calculate it using the volume formula:

Calculate the area of ​​the base. this is a regular hexagon with a side equal to 1. The area of ​​a regular hexagon is equal to the area of ​​six equilateral triangles with the same side, more about this (item 6), so:

Means

OS \u003d BC \u003d 1, since in a regular hexagon the segment connecting its center to the vertex is equal to the side of this hexagon.

Thus, according to the Pythagorean theorem:

Answer: 7

VolumeThe size of a tetrahedron is 200. Find the volume of a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.

The volume of the indicated polyhedron is equal to the difference volumes of the initial tetrahedron V 0 and four equal tetrahedra, each of which is obtained by cutting off by a plane passing through the midpoints of edges that have a common vertex:

Let's define what is equal to the volume cut off tetrahedron.

Note that the original tetrahedron and the "cut off" tetrahedron are similar bodies. It is known that the ratio of the volumes of similar bodies is k 3 , where k is the similarity coefficient. In this case, it is equal to 2 (since all the linear dimensions of the original tetrahedron are twice the corresponding dimensions of the cut one):

Calculate the volume of the cut-off tetrahedron:

Thus, the desired volume will be equal to:

Answer: 100

The surface area of ​​a tetrahedron is 120. Find the surface area of ​​a polyhedron whose vertices are the midpoints of the edges of this tetrahedron.

First way:

The desired surface consists of 8 equilateral triangles with a side half the edge of the original tetrahedron. The surface of the original tetrahedron consists of 16 such triangles (4 triangles on each of the 4 faces of the tetrahedron), so the required area is equal to half the surface area of ​​this tetrahedron and is equal to 60.

Second way:

Since the surface area of ​​the tetrahedron is known, we can find its edge, then determine the length of the edge of the polyhedron, and then calculate its surface area.

Pyramids are: triangular, quadrangular, etc., depending on what the base is - a triangle, a quadrilateral, etc.
The pyramid is called correct ( fig.286,b) if, firstly, its base is a regular polygon, and, secondly, the height passes through the center of this polygon.
Otherwise, the pyramid is called irregular ( Fig.286, in). In a regular pyramid, all side edges are equal to each other (as inclined with equal projections). Therefore, all side faces correct pyramid are equal isosceles triangles.
Analysis of the elements of a regular hexagonal pyramid and their representation in a complex drawing ( fig.287) .

a) Complex drawing of a regular hexagonal pyramid. The base of the pyramid is located on the plane P 1 ; two sides of the base of the pyramid are parallel to the plane of projections П 2 .
b) Base ABCDEF - a hexagon located in the plane of projections П 1 .
c) Lateral face ASF - a triangle located in a plane in general position.
d) Side face FSE - a triangle located in the profile - projecting plane.
e) The edge SE is a segment in general position.
f) Edge SA - frontal segment.
g) The top S of the pyramid is a point in space.
On the ( fig.288 and fig.289) examples of sequential graphic operations are given when performing a complex drawing and visual images (axonometry) of pyramids.

Given:
1. The base is located on the plane P 1.
2. One of the sides of the base is parallel to the x 12 axis.
I. Integrated drawing.
I, a. We design the base of the pyramid - a polygon, according to this condition, lying in the plane П 1 .
We design a vertex - a point located in space. The height of the point S is equal to the height of the pyramid. The horizontal projection S 1 of the point S will be in the center of the projection of the base of the pyramid (by condition).
I, b. We design the edges of the pyramid - segments; to do this, we connect the direct projections of the base vertices ABCDE with the corresponding projections of the top of the pyramid S. The frontal projections S 2 C 2 and S 2 D 2 of the edges of the pyramid are depicted by dashed lines, as invisible, closed by the faces of the pyramid (SBA and SAE).
I, c. The horizontal projection K 1 of the point K on the side face SBA is given, it is required to find its frontal projection. To do this, we draw an auxiliary line S 1 F 1 through the points S 1 and K 1, find its frontal projection and on it, using a vertical line of communication, determine the place of the desired frontal projection K 2 of the point K.
II. The development of the surface of the pyramid is a flat figure consisting of side faces - identical isosceles triangles, one side of which is equal to the side of the base, and the other two - to the side edges, and from a regular polygon - the base.
The natural dimensions of the sides of the base are revealed on its horizontal projection. The natural dimensions of the ribs on the projections were not revealed.
Hypotenuse S 2 ¯A 2 ( fig.288, 1 , b) of a right-angled triangle S 2 O 2 ¯A 2, in which the large leg is equal to the height S 2 O 2 of the pyramid, and the small one is equal to the horizontal projection of the edge S 1 A 1 is the natural size of the edge of the pyramid. The sweep should be built in the following order:
a) from an arbitrary point S (vertex) we draw an arc with a radius R equal to the edge of the pyramid;
b) on the drawn arc, set aside five chords of size R 1 equal to the side of the base;
c) connect the points D, C, B, A, E, D in series with each other and with the point S with straight lines, we get five isosceles equal triangles, which make up the development of the lateral surface of this pyramid, cut along the edge SD ;
d) we attach to any face the base of the pyramid - a pentagon, using the triangulation method, for example, to the face DSE.
The point K is transferred to the sweep using an auxiliary straight line using the size B 1 F 1 taken on the horizontal projection, and the size A 2 K 2 taken on the natural size of the rib.
III. Visual representation of the pyramid in isometry.
III, a. We depict the base of the pyramid, using the coordinates according to ( fig.288, 1 , a).
We depict the top of the pyramid, using the coordinates of ( fig.288, 1 , a).
III, b. We depict the side edges of the pyramid, connecting the top with the tops of the base. The edge S"D" and the sides of the base C"D" and D"E" are shown with dashed lines, as invisible, closed by the faces of the pyramid C"S"B", B"S"A" and A"S"E".
III, e. We determine the point on the surface of the pyramid K, using the dimensions y F and x K. For the dimetric image of the pyramid, the same sequence should be followed.
Image of an irregular triangular pyramid.

Given:
1. The base is located on the plane P 1.
2. Side BC of the base is perpendicular to the X axis.
I. Integrated drawing
I, a. We design the base of the pyramid - an isosceles triangle lying in the plane P 1, and the top S - a point located in space, the height of which is equal to the height of the pyramid.
I, b. We design the edges of the pyramid - segments, for which we connect the same-named projections of the vertices of the base with the same-named projections of the top of the pyramid with straight lines. We depict the horizontal projection of the side of the base of the aircraft with a dashed line, as an invisible one, closed by two faces of the pyramid ABS, ACS.
I, c. On the frontal projection A 2 C 2 S 2 of the side face, the projection D 2 of the point D is given. It is required to find its horizontal projection. To do this, through point D 2 we draw an auxiliary straight line parallel to the x 12 axis - the frontal projection of the horizontal, then we find its horizontal projection and on it, using a vertical line of communication, we determine the location of the desired horizontal projection D 1 of point D.
II. Construction of a pyramid sweep.
The natural dimensions of the sides of the base are revealed in the horizontal projection. The natural size of the rib AS is revealed in the frontal projection; there are no natural size of the ribs BS and CS in the projections, the size of these ribs is revealed by rotating them around the i axis, perpendicular to the plane P 1 passing through the top of the pyramid S. The new frontal projection ¯C 2 S 2 is the natural value of the edge CS .
The sequence of constructing a development of the surface of the pyramid:
a) draw an isosceles triangle - face CSB, the base of which is equal to the side of the base of the pyramid CB, and sides- natural size of the rib SC ;
b) we add two triangles to the sides SC and SB of the constructed triangle - the faces of the pyramid CSA and BSA, and to the base CB of the constructed triangle - the base of the CBA pyramid, as a result we obtain a complete unfolding of the surface of this pyramid.
The transfer of point D to the development is carried out in the following order: first, draw a horizontal line on the ASC side face development using the R 1 dimension, and then determine the location of the point D on the horizontal line using the R 2 dimension.
III. A visual representation of the pyramid e frontal dimetric projection
III, a. We depict the base A "B" C and the top S "of the pyramid, using the coordinates according to (

Date: 2015-01-19

If you need step-by-step instruction how to build a pyramid sweep, then I ask for our lesson. First of all, evaluate whether your pyramid is unfolded in the same way as in Figure 1.

If you have it turned at 90 degrees, then the edge marked in the figure as "known real values" in your case can be found on the profile projection, which you will need to build. In my case, this is not required, we already have all the quantities necessary for constructing. It is important not to forget that in this drawing only the edges SA and SD on the frontal projection are displayed in full size. All others are projected with length distortion. In addition, in the top view, all sides of the hexagon are also projected in full size. Based on this, let's get started.

1. For greater beauty, let's draw the first line horizontally (Figure 1). Then, we will draw a wide arc with a radius R=a, i.e. with a radius equal to the length of the lateral edge of the pyramid. We get point A. From it we make a notch on the arc with a compass, with a radius r \u003d b (the length of the side of the base of the pyramid). Let's get point B. We already have the first face of the pyramid!

2. From point B we make another notch with the same radius - we get point C and connecting it with points B and S we get the second side face of the pyramid (Figure 2).

3. Repeating these steps the required number of times (it all depends on how many faces your pyramid has) we will get such a fan (Figure 3). With the correct construction, you should get all the points of the base, and the extreme ones should be repeated.

4. This is not always required, but still it is necessary: ​​add the base of the pyramid to the development of the side surface. I believe that everyone who has read up to this point can draw a six-eight-pentagon (how to draw a pentagon is described in detail in the lesson) The difficulty lies in the fact that the figure must be drawn in right place and at the right angle. Draw an axis through the middle of any face. From the point of intersection with the line of the base, we plot the distance m, as shown in Figure 4.


Drawing a perpendicular through this point, we get the axes of the future hexagon. From the resulting center we draw a circle, as you did when building a top view. Please note that the circle must pass through two points of the side face (in my case, these are F and A)

5. Figure 5 shows the final unfolded view of the hexagonal prism.


This completes the construction of the pyramid sweep. Build your sweeps, learn to find solutions, be corrosive and never give up. Thanks for stopping by. Don't forget to recommend us to your friends :) All the best!

or write down our phone number and tell your friends about us - someone is probably looking for a way to make drawings

or create a note about our lessons on your page or blog - and someone else will be able to master the drawing.