Calculating the volumes of spatial figures is one of the important tasks stereometry. In this article we will consider the issue of determining the volume of such a polyhedron as a pyramid, and also give a hexagonal regular one.
Hexagonal pyramid
First, let’s look at what the figure that will be discussed in the article is.
Let us have an arbitrary hexagon, the sides of which are not necessarily equal to each other. Let's also assume that we have chosen a point in space that is not located 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 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 connecting point of which is called the vertex. The difference between the pyramids depicted is that the height h of the right one does not intersect the hexagonal base at its geometric center, and the height of the left figure falls exactly in this center. Thanks to this criterion, the left pyramid was called straight, and the right pyramid was called inclined.
Since the base of the left figure in the figure is formed by a hexagon with equal sides and angles, it is called regular. Further in the article we will talk only about this pyramid.
To calculate the volume of an arbitrary pyramid, we have following formula:
Here h is the length of the height of the figure, S o is the area of its base. Let's use this expression to determine the volume of a hexagonal regular pyramid.
Since the base of the figure in question is an equilateral hexagon, to calculate its area you can use the following general expression for n-gon:
S n = n/4 * a 2 * ctg(pi/n)
Here n is an integer equal to the number of sides (angles) 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 = 6/4 * a 2 * ctg(pi/6) = √3/2 * a 2
Now all that remains is to substitute this expression into general formula for volume V:
V 6 = S 6 * h = √3/2 * h * a 2
Thus, to calculate the volume of the pyramid in question, it is necessary to know its two linear parameter: length of the side of the base and height of the figure.
Example of problem solution
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 the formula for volume includes only a and h, you can substitute any of these parameters into it, expressed in terms of the other. For example, substituting a, we get:
V 6 = √3/2*h*(2*h) 2 =>
h = ∛(V 6 /(2*√3))
To find the height of a figure, you need to take the third root of the volume, which corresponds to the dimension of length. We substitute the value of the volume V 6 of the pyramid from the problem conditions, 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 as large as the found value, 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.
Date: 2015-01-19
If you need step by step instructions How to build a pyramid scan, then I ask you to join our lesson. First, evaluate whether your pyramid is deployed in a similar way as in Figure 1.
If you have it rotated 90 degrees, then the edge marked in the figure as “known real values” in your case can be found on the profile projection that you will need to construct. In my case, this is not required; we already have all the quantities necessary for construction. It is important not to forget that in this drawing only the edges SA and SD in the front 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 at full size. Based on this, let's proceed.
1. For greater beauty, let’s draw the first line horizontally (Figure 1). Then, let's draw a wide arc with radius R=a, i.e. radius equal to length lateral edge of the pyramid. Let's get point A. Using a compass, we'll make a notch on the arc from it, with radius r=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 required quantity times (it all depends on how many faces your pyramid has) we will get a fan like this (Figure 3). If constructed correctly, you should get all the base points, and the extreme ones should be repeated.
4. This is not always required, but it is still necessary: add the base of the pyramid to the development of the side surface. I believe everyone who has read this far 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 needs to be drawn in the right place and at the right angle. We draw an axis through the middle of any face. From the point of intersection with the straight line of the base, we plot the distance m, as shown in Figure 4. 
By drawing a perpendicular through this point, we obtain the axes of the future hexagon. From the resulting center we draw a circle, as you did when constructing the top view. Please note that the circle must pass through two points on the side face (in my case these are F and A)
5. Figure 5 shows the final view of the development of a hexagonal prism. 
This completes the construction of the pyramid. Build your developments, learn to find solutions, be meticulous and never give up. Thank you for stopping by. Don't forget to recommend us to your friends:) All the best!
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Pyramids are: triangular, quadrangular, etc., depending on what is the base - triangle, quadrangle, etc.
A pyramid is called regular (Fig. 286, b) if, firstly, its base is a regular polygon, and, secondly, its height passes through the center of this polygon.
Otherwise, the pyramid is called irregular (Fig. 286, c). Everything is in the right pyramid lateral ribs equal to each other (as inclined with equal projections). Therefore everything side faces regular pyramid there are equal isosceles triangles.
Analysis of the elements of a regular hexagonal pyramid and their depiction in a complex drawing (Fig. 287).
A) Complex drawing 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 projection plane P 2.
b) The base ABCDEF is a hexagon located in the projection plane P 1.
c) The lateral face of ASF is a triangle located in the general plane.
d) The side face of FSE is a triangle located in the profile-projecting plane.
e) Edge SE is a segment in general position.
f) Rib SA - frontal segment.
g) The top S of the pyramid is a point in space.
Figures 288 and 289 show 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-axis 12.
I. Complex drawing.
I, a. We design the base of the pyramid - a polygon, according to this condition lying in the plane P1.
We design a vertex - a point located in space. The height of point S is equal to the height of the pyramid. The horizontal projection S 1 of 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 projections of the vertices of the base ABCDE with the corresponding projections of the vertex of the pyramid S by straight lines. We depict the frontal projections S 2 C 2 and S 2 D 2 of the edges of the pyramid with dashed lines, as invisible, closed by the edges of the pyramid (SА and SAE).
I, c. Given a horizontal projection K 1 of point K on the side face of SBA, you need to find its frontal projection. To do this, we draw an auxiliary straight line S 1 F 1 through points S 1 and K 1, find its frontal projection and on it using vertical line connection, we determine the location of the desired frontal projection K 2 of point K.
II. Development of the surface of the pyramid - 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 are equal to the side edges, and from regular polygon- grounds.
The natural dimensions of the sides of the base are revealed on its horizontal projection. The natural dimensions of the ribs were not revealed on the projections.
Hypotenuse S 2 ¯A 2 (Fig. 288, 1
, b) right triangle S 2 O 2 ¯A 2, which has a large leg equal to height S 2 O 2 of the pyramid, and the small one is the horizontal projection of the edge S 1 A 1 is the natural size of the edge of the pyramid. The construction of the sweep should be performed in the following order:
a) from arbitrary point S (vertices) draw an arc of radius R, equal to the edge pyramids;
b) on the drawn arc we plot five chords of size R 1 equal to side grounds;
c) connect points D, C, B, A, E, D with straight lines in series with each other and with point S, we get five isosceles equal triangles, constituting the development of the lateral surface of this pyramid, cut along the edge SD;
d) we attach the base of the pyramid - a pentagon - to any face using the triangulation method, for example to the DSE face.
The transfer of point K to the scan is carried out by an auxiliary straight line using the dimension B 1 F 1 taken on the horizontal projection and the dimension A 2 K 2 taken on the natural size of the rib.
III. A visual representation of a 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 according to (Fig. 288, 1
, A).
III, b. We depict the side edges of the pyramid, connecting the top with the vertices of the base. The edge S"D" and the sides of the base C"D" and D"E" are depicted with dashed lines, as invisible, closed by the edges of the pyramid C"S"B", B"S"A" and A"S"E".
III, e. We determine point K on the surface of the pyramid using the dimensions y F and x K. For a dimetric image of a 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. Complex drawing
I, a. Designing the base of the pyramid - isosceles triangle, lying in the plane P 1, and the vertex S is 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 straight lines of the same-name projections of the base vertices with the same-name projections of the pyramid's apex. We depict the horizontal projection of the side of the base of the aircraft with a dashed line, as invisible, covered by two faces of the pyramid ABS, ACS.
I, c. On the frontal projection A 2 C 2 S 2 of the side face, a projection D 2 of point D is given. You need to find its horizontal projection. To do this, through point D 2 we draw an auxiliary 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 connection line, we determine the location of the desired horizontal projection D 1 of point D.
II. Constructing a pyramid scan.
The natural dimensions of the sides of the base are revealed on the horizontal projection. The natural size of rib AS was revealed on the frontal projection; there are no natural size edges BS and CS in the projections; the size of these edges is revealed by rotating them around the i axis perpendicular to the plane P1 passing through the top of the pyramid S. New frontal projection¯C 2 S 2 is the natural value of the edge CS.
The sequence of constructing the 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 rib SC;
b) we attach 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 CBA of the pyramid, as a result we obtain a complete development of the surface of this pyramid.
Transferring point D to the scan is carried out in the following order: first, on the scan of the side face ASC, draw a horizontal line using size R 1 and then determine the location of point D on the horizontal line using size R 2.
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 coordinates according to (
A drawing is the first and very important step in solving geometric problem. What should the drawing of a regular pyramid look like?
First let's remember parallel design properties:
- parallel segments of the figure are depicted 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 we draw the base. Since when parallel design angles and length ratios are not parallel segments are not saved, the regular triangle at the base of the pyramid is depicted as an arbitrary triangle.
Center regular triangle is the point of intersection of the medians of the triangle. Since the medians at the intersection point are divided in a ratio of 2:1, counting from the vertex, we mentally connect the vertex of the base with the middle of the opposite side, approximately divide it into three parts, and place a point at a distance of 2 parts from the vertex. From this point upward we draw a perpendicular. This is the height of the pyramid. We draw a perpendicular of such length that the side edge does not cover the image of the height.
Drawing correct quadrangular pyramid
We also start drawing a regular quadrangular pyramid 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. Preferably acute angle make this parallelogram smaller, then the side faces will be larger. The center of a square is the point of intersection of its diagonals. We draw diagonals and restore a perpendicular from the intersection point. This perpendicular is the height of the pyramid. We choose the length of the perpendicular so that the side ribs do not merge with each other.
Drawing of a regular hexagonal pyramid
Since during parallel design the parallelism of the segments is preserved, 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 point of intersection of its diagonals. In order not to clutter the drawing, we do not draw diagonals, but find this point approximately. From it we restore the perpendicular - the height of the pyramid - so that the side ribs do not merge with each other.
