The volume of a regular hexagonal pyramid is 6 sides. Pyramid
Pyramids are: triangular, quadrangular, etc., depending on what the base is - a triangle, a quadrilateral, etc.
A pyramid is called correct (Fig. 286b) 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, c). In a regular pyramid, all side edges are equal to each other (as inclined with equal projections). Therefore, all lateral faces of a regular 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 (Fig.288 and Fig.289) shows examples of sequential graphic operations when performing a complex drawing and visual images (axonometry) of the 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) right 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 leg 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 (
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 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 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 resulting 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
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 apothem and the length of the side edge.
The drawing is the first and very important step in solving a geometric problem. What should be the drawing of a regular pyramid?
Let's remember first parallel design properties:
- parallel segments of the figure are depicted as parallel segments;
- the ratio of the lengths of segments of parallel lines and segments of one straight line is preserved.
Drawing of a regular triangular pyramid
First, draw the base. Since the angles and ratios of the lengths of non-parallel segments are not preserved in parallel design, the regular triangle at the base of the pyramid is represented by an arbitrary triangle.
The center of an equilateral triangle is the intersection point of the medians of the triangle. Since the medians at the intersection point are divided in a ratio of 2: 1, counting from the top, we mentally connect the top of the base with the middle of the opposite side, approximately divide it into three parts, and put a point at a distance of 2 parts from the top. Draw a perpendicular from this point upwards. This is the height of the pyramid. We draw the perpendicular so long that the side edge does not cover the image of the height.
drawing correct quadrangular pyramid
The drawing of a regular quadrangular pyramid also starts from the base. Since the parallelism of the segments is preserved, but the magnitudes of the angles are not, the square at the base is depicted as a parallelogram. Desirable sharp corner make this parallelogram smaller, then the side faces are larger. The center of a square is the intersection point of its diagonals. We draw diagonals, from the point of intersection we restore the perpendicular. This perpendicular is the height of the pyramid. We choose the length of the perpendicular so that the side edges do not merge with each other.
Drawing of a regular hexagonal pyramid
Since parallel projection preserves the parallelism of the segments, the base of a regular hexagonal pyramid - a regular hexagon - is depicted as a hexagon whose opposite sides are parallel and equal. The center of a regular hexagon is the intersection point of its diagonals. In order not to clutter up the drawing, we do not draw diagonals, but we find this point approximately. From it we restore the perpendicular - the height of the pyramid - so that the side edges do not merge with each other.
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 will make another notch with the same radius - we will get point C and connecting it with points B and S we will 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 knows how to 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!
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