Students encounter the concept of a pyramid long before studying geometry. The fault lies with the famous great Egyptian wonders of the world. Therefore, when starting to study this wonderful polyhedron, most students already clearly imagine it. All the above-mentioned attractions have the correct shape. What's happened regular pyramid, and what properties it has will be discussed further.
Definition
There are quite a lot of definitions of a pyramid. Since ancient times, it has been very popular.
For example, Euclid defined it as a bodily figure consisting of planes that, starting from one, converge at a certain point.
Heron provided a more precise formulation. He insisted that this was the figure that has a base and planes in in the form of triangles, converging at one point.
Based on modern interpretation, the pyramid is represented as a spatial polyhedron consisting of a certain k-gon and k flat figures triangular in shape with one common point.
Let's look at it in more detail, what elements does it consist of:
- The k-gon is considered the basis of the figure;
- 3-gonal shapes protrude as the edges of the side part;
- the upper part from which the side elements originate is called the apex;
- all segments connecting a vertex are called edges;
- if a straight line is lowered from the vertex to the plane of the figure at an angle of 90 degrees, then its part contained in the internal space is the height of the pyramid;
- in any lateral element, a perpendicular, called an apothem, can be drawn to the side of our polyhedron.
The number of edges is calculated using the formula 2*k, where k is the number of sides of the k-gon. How many faces a polyhedron such as a pyramid has can be determined using the expression k+1.
Important! A pyramid of regular shape is a stereometric figure whose base plane is a k-gon with equal sides.
Basic properties
Correct pyramid has many properties, which are unique to her. Let's list them:
- The basis is a figure of the correct shape.
- The edges of the pyramid that limit the side elements have equal numerical values.
- The side elements are isosceles triangles.
- The base of the height of the figure falls at the center of the polygon, while it is simultaneously the central point of the inscribed and circumscribed.
- All side ribs are inclined to the plane of the base at the same angle.
- All side surfaces have the same angle of inclination relative to the base.
Thanks to all of the listed properties, performing element calculations is much simpler. Based on the above properties, we pay attention to two signs:
- In the case where a polygon fits into a circle, side faces will have with the base equal angles.
- When describing a circle around a polygon, all edges of the pyramid emanating from the vertex will have equal length and equal angles with the base.
The basis is a square
Regular quadrangular pyramid - a polyhedron whose base is a square.
It has four side faces, which are isosceles in appearance.
A square is depicted on a plane, but is based on all the properties of a regular quadrilateral.
For example, if it is necessary to relate the side of a square with its diagonal, then use the following formula: the diagonal is equal to the product of the side of the square and the square root of two.
It is based on a regular triangle
A regular triangular pyramid is a polyhedron whose base is a regular 3-gon.
If the base is a regular triangle and the side edges are equal to the edges of the base, then such a figure called a tetrahedron.
All faces of a tetrahedron are equilateral 3-gons. IN in this case You need to know some points and not waste time on them when calculating:
- the angle of inclination of the ribs to any base is 60 degrees;
- the size of all internal faces is also 60 degrees;
- any face can act as a base;
- , drawn inside the figure, these are equal elements.
Sections of a polyhedron
In any polyhedron there are several types of sections flat. Often in school course geometries work with two:
- axial;
- parallel to the basis.
The axial section is obtained when the plane intersects the polyhedron, which passes through the vertex, side edges and axis. In this case, the axis is the height drawn from the vertex. The cutting plane is limited by the lines of intersection with all faces, resulting in a triangle.
Attention! In a regular pyramid, the axial section is an isosceles triangle.
If the cutting plane runs parallel to the base, then the result is the second option. In this case, we have a cross-sectional figure similar to the base.
For example, if there is a square at the base, then the section parallel to the base will also be a square, only of smaller dimensions.
When solving problems under this condition, they use signs and properties of similarity of figures, based on Thales' theorem. First of all, it is necessary to determine the similarity coefficient.
If the plane is drawn parallel to the base and it cuts off top part polyhedron, then a regular truncated pyramid is obtained in the lower part. Then the bases of a truncated polyhedron are said to be similar polygons. In this case, the side faces are isosceles trapezoids. The axial section is also isosceles.
In order to determine the height of a truncated polyhedron, it is necessary to draw the height in the axial section, that is, in the trapezoid.
Surface areas
The main geometric problems that have to be solved in a school geometry course are finding the surface area and volume of a pyramid.
There are two types of surface area values:
- area of the side elements;
- area of the entire surface.
From the name itself it is clear what we are talking about. The side surface includes only the side elements. It follows from this that to find it, you simply need to add up the areas of the lateral planes, that is, the areas of isosceles 3-gons. Let's try to derive the formula for the area of the side elements:
- The area of an isosceles 3-gon is equal to Str=1/2(aL), where a is the side of the base, L is the apothem.
- The number of lateral planes depends on the type of k-gon at the base. For example, a regular quadrangular pyramid has four lateral planes. Therefore, it is necessary to add the areas of four figures Sside=1/2(aL)+1/2(aL)+1/2(aL)+1/2(aL)=1/2*4a*L. The expression is simplified in this way because the value 4a = Rosn, where Rosn is the perimeter of the base. And the expression 1/2*Rosn is its semi-perimeter.
- So, we conclude that the area of the side elements regular pyramid equal to the product of the semi-perimeter of the base and the apothem: Sside=Rosn*L.
The area of the total surface of the pyramid consists of the sum of the areas of the side planes and the base: Sp.p. = Sside + Sbas.
As for the area of the base, here the formula is used according to the type of polygon.
Volume of a regular pyramid equal to the product of the area of the base plane and the height divided by three: V=1/3*Sbas*H, where H is the height of the polyhedron.
What is a regular pyramid in geometry
Properties of the correct quadrangular pyramid
Hypothesis: we believe that the perfection of the pyramid's shape is due to the mathematical laws inherent in its shape.
Target: having studied the pyramid as geometric body, to explain the perfection of its form.
Tasks:
1. Give a mathematical definition of a pyramid.
2. Study the pyramid as a geometric body.
3. Understand what mathematical knowledge the Egyptians incorporated into their pyramids.
Private questions:
1. What is a pyramid as a geometric body?
2. How can the unique shape of the pyramid be explained from a mathematical point of view?
3. What explains the geometric wonders of the pyramid?
4. What explains the perfection of the pyramid shape?
Definition of a pyramid.
PYRAMID (from Greek pyramis, gen. pyramidos) - a polyhedron whose base is a polygon, and the remaining faces are triangles having a common vertex (drawing). Based on the number of base angles, pyramids are classified as triangular, quadrangular, etc.
PYRAMID - a monumental structure that has the geometric shape of a pyramid (sometimes also stepped or tower-shaped). Pyramids are the name given to the giant tombs of the ancient Egyptian pharaohs of the 3rd-2nd millennium BC. e., as well as ancient American temple pedestals (in Mexico, Guatemala, Honduras, Peru), associated with cosmological cults.
It is possible that the Greek word “pyramid” comes from the Egyptian expression per-em-us, i.e., from a term meaning the height of the pyramid. The outstanding Russian Egyptologist V. Struve believed that the Greek “puram...j” comes from the ancient Egyptian “p"-mr".
From history. Having studied the material in the textbook “Geometry” by the authors of Atanasyan. Butuzov and others, we learned that: A polyhedron composed of a n-gon A1A2A3 ... An and n triangles PA1A2, PA2A3, ..., PAnA1 is called a pyramid. Polygon A1A2A3...An is the base of the pyramid, and triangles PA1A2, PA2A3,..., PAnA1 are the side faces of the pyramid, P is the top of the pyramid, segments PA1, PA2,..., PAn are lateral ribs.
However, this definition of a pyramid did not always exist. For example, the ancient Greek mathematician, the author of theoretical treatises on mathematics that have come down to us, Euclid, defines a pyramid as a solid figure bounded by planes that converge from one plane to one point.
But this definition was criticized already in ancient times. So Heron proposed the following definition of a pyramid: “It is a figure bounded by triangles converging at one point and the base of which is a polygon.”
Our group, having compared these definitions, came to the conclusion that they do not have a clear formulation of the concept of “foundation”.
We examined these definitions and found the definition of Adrien Marie Legendre, who in 1794 in his work “Elements of Geometry” defines a pyramid as follows: “A pyramid is a solid figure formed by triangles converging at one point and ending on different sides of a flat base.”
It seems to us that the last definition gives a clear idea of the pyramid, since it talks about the fact that the base is flat. Another definition of a pyramid appeared in a 19th-century textbook: “a pyramid is a solid angle intersected by a plane.”
Pyramid as a geometric body.
That. A pyramid is a polyhedron, one of whose faces (base) is a polygon, the remaining faces (sides) are triangles that have one common vertex (the vertex of the pyramid).
The perpendicular drawn from the top of the pyramid to the plane of the base is called heighth pyramids.
In addition to the arbitrary pyramid, there are correct pyramid at the base of which is a regular polygon and truncated pyramid.
In the figure there is a pyramid PABCD, ABCD is its base, PO is its height.
Total surface area pyramid is the sum of the areas of all its faces.
Sfull = Sside + Smain, Where Side– the sum of the areas of the side faces.
Volume of the pyramid is found by the formula:
V=1/3Sbas. h, where Sbas. - base area, h- height.
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Apothem ST is the height of the side face of a regular pyramid.
The area of the lateral face of a regular pyramid is expressed as follows: Sside. =1/2P h, where P is the perimeter of the base, h- height of the side face (apothem of a regular pyramid). If the pyramid is intersected by plane A’B’C’D’, parallel to the base, then:
1) the side ribs and height are divided by this plane into proportional parts;
2) in cross-section a polygon A’B’C’D’ is obtained, similar to the base;
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Bases of a truncated pyramid– similar polygons ABCD and A`B`C`D`, side faces are trapezoids.
Height truncated pyramid - the distance between the bases.
Truncated volume pyramid is found by the formula:
V=1/3 h(S + https://pandia.ru/text/78/390/images/image019_2.png" align="left" width="91" height="96"> The lateral surface area of a regular truncated pyramid is expressed as follows: Sside. = ½(P+P') h, where P and P’ are the perimeters of the bases, h- height of the side face (apothem of a regular truncated piri
Sections of a pyramid.
Sections of a pyramid by planes passing through its apex are triangles.
A section passing through two non-adjacent lateral edges of a pyramid is called diagonal section.
If the section passes through a point on the side edge and the side of the base, then its trace to the plane of the base of the pyramid will be this side.
A section passing through a point lying on the face of the pyramid and a given section trace on the base plane, then the construction should be carried out as follows:
· find the point of intersection of the plane of a given face and the trace of the section of the pyramid and designate it;
construct a straight line passing through given point and the resulting intersection point;
· repeat these steps for the next faces.
, which corresponds to the ratio of the legs right triangle 4:3. This ratio of the legs corresponds to the well-known right triangle with sides 3:4:5, which is called the “perfect”, “sacred” or “Egyptian” triangle. According to historians, the “Egyptian” triangle was given a magical meaning. Plutarch wrote that the Egyptians compared the nature of the universe to a “sacred” triangle; they symbolically likened the vertical leg to the husband, the base to the wife, and the hypotenuse to that which is born from both.
For a triangle 3:4:5, the equality is true: 32 + 42 = 52, which expresses the Pythagorean theorem. Was it not this theorem that the Egyptian priests wanted to perpetuate by erecting a pyramid based on the triangle 3:4:5? It is difficult to find a more successful example to illustrate the Pythagorean theorem, which was known to the Egyptians long before its discovery by Pythagoras.
Thus, the brilliant creators Egyptian pyramids sought to amaze distant descendants with the depth of their knowledge, and they achieved this by choosing the “golden” right triangle as the “main geometric idea” for the Cheops pyramid, and the “sacred” or “Egyptian” triangle for the Khafre pyramid.
Very often in their research, scientists use the properties of pyramids with Golden Ratio proportions.
In mathematics encyclopedic dictionary The following definition of the Golden Section is given - this is a harmonic division, division in extreme and average ratio - dividing the segment AB into two parts in such a way that its larger part AC is the average proportional between the entire segment AB and its smaller part NE.
Algebraic determination of the Golden section of a segment AB = a reduces to solving the equation a: x = x: (a – x), from which x is approximately equal to 0.62a. The ratio x can be expressed as fractions 2/3, 3/5, 5/8, 8/13, 13/21...= 0.618, where 2, 3, 5, 8, 13, 21 are Fibonacci numbers.
The geometric construction of the Golden Section of the segment AB is carried out as follows: at point B, a perpendicular to AB is restored, the segment BE = 1/2 AB is laid out on it, A and E are connected, DE = BE is laid off and, finally, AC = AD, then the equality AB is satisfied: CB = 2:3.
Golden ratio often used in works of art, architecture, and found in nature. Vivid examples are the sculpture of Apollo Belvedere, the Parthenon. During the construction of the Parthenon, the ratio of the height of the building to its length was used and this ratio is 0.618. Objects around us also provide examples of the Golden Ratio, for example, the bindings of many books have a width-to-length ratio close to 0.618. Considering the arrangement of leaves on the common stem of plants, you can notice that between every two pairs of leaves the third is located at the Golden Ratio (slides). Each of us “carries” the Golden Ratio with us “in our hands” - this is the ratio of the phalanges of the fingers.
Thanks to the discovery of several mathematical papyri, Egyptologists have learned something about the ancient Egyptian systems of calculation and measurement. The tasks contained in them were solved by scribes. One of the most famous is the Rhind Mathematical Papyrus. By studying these problems, Egyptologists learned how the ancient Egyptians dealt with the various quantities that arose when calculating measures of weight, length, and volume, which often involved fractions, as well as how they handled angles.
The ancient Egyptians used a method of calculating angles based on the ratio of the height to the base of a right triangle. They expressed any angle in the language of a gradient. The slope gradient was expressed as a whole number ratio called "seced". In Mathematics in the Age of the Pharaohs, Richard Pillins explains: “The seked of a regular pyramid is the inclination of any of the four triangular faces to the plane of the base, measured by the nth number of horizontal units per vertical unit of rise. Thus, this unit of measurement is equivalent to our modern cotangent of the angle of inclination. Therefore, the Egyptian word "seced" is related to our modern word"gradient"".
The numerical key to the pyramids lies in the ratio of their height to the base. In practical terms, this is the easiest way to make the templates necessary to constantly check the correct angle of inclination throughout the construction of the pyramid.
Egyptologists would be happy to convince us that each pharaoh longed to express his individuality, hence the differences in the angles of inclination for each pyramid. But there could be another reason. Perhaps they all wanted to embody different symbolic associations, hidden in different proportions. However, the angle of Khafre's pyramid (based on the triangle (3:4:5) appears in the three problems presented by the pyramids in the Rhind Mathematical Papyrus). So this attitude was well known to the ancient Egyptians.
To be fair to Egyptologists who claim that the ancient Egyptians were not aware of the 3:4:5 triangle, the length of the hypotenuse 5 was never mentioned. But math problems questions concerning pyramids are always decided on the basis of the second angle - the ratio of the height to the base. Since the length of the hypotenuse was never mentioned, it was concluded that the Egyptians never calculated the length of the third side.
The height-to-base ratios used in the Giza pyramids were undoubtedly known to the ancient Egyptians. It is possible that these relationships for each pyramid were chosen arbitrarily. However, this contradicts the importance attached to number symbolism in all types of Egyptian fine art. It is very likely that such relationships were significant because they expressed specific religious ideas. In other words, the entire Giza complex was subordinated to a coherent design designed to reflect a certain divine theme. This would explain why the designers chose different angles the inclination of the three pyramids.
In The Mystery of Orion, Bauval and Gilbert presented convincing evidence linking the pyramids of Giza with the constellation Orion, in particular with the stars of Orion's Belt. The same constellation is present in the myth of Isis and Osiris, and there is reason to view each pyramid as a representation of one of the three main deities - Osiris, Isis and Horus.
"GEOMETRICAL" MIRACLES.
Among the grandiose pyramids of Egypt, it occupies a special place Great Pyramid of Pharaoh Cheops (Khufu). Before we begin to analyze the shape and size of the Cheops pyramid, we should remember what system of measures the Egyptians used. The Egyptians had three units of length: a "cubit" (466 mm), which was equal to seven "palms" (66.5 mm), which, in turn, was equal to four "fingers" (16.6 mm).
Let us analyze the dimensions of the Cheops pyramid (Fig. 2), following the arguments given in the wonderful book of the Ukrainian scientist Nikolai Vasyutinsky " Golden ratio" (1990).
Most researchers agree that the length of the side of the base of the pyramid, for example, GF equal to L= 233.16 m. This value corresponds almost exactly to 500 “elbows”. Full compliance with 500 “elbows” will occur if the length of the “elbow” is considered equal to 0.4663 m.
Height of the pyramid ( H) is estimated by researchers variously from 146.6 to 148.2 m. And depending on the accepted height of the pyramid, all its ratios change geometric elements. What is the reason for the differences in estimates of the height of the pyramid? The fact is that, strictly speaking, the Cheops pyramid is truncated. Its upper platform today measures approximately 10 ´ 10 m, but a century ago it was 6 ´ 6 m. Obviously, the top of the pyramid was dismantled, and it does not correspond to the original one.
When assessing the height of the pyramid, it is necessary to take into account such a physical factor as the “draft” of the structure. For long time under the influence of colossal pressure (reaching 500 tons per 1 m2 of the lower surface), the height of the pyramid decreased compared to its original height.
What was the original height of the pyramid? This height can be recreated by finding the basic "geometric idea" of the pyramid.
Figure 2.
In 1837, the English Colonel G. Wise measured the angle of inclination of the faces of the pyramid: it turned out to be equal a= 51°51". This value is still recognized by most researchers today. The specified angle value corresponds to the tangent (tg a), equal to 1.27306. This value corresponds to the ratio of the height of the pyramid AC to half its base C.B.(Fig.2), that is A.C. / C.B. = H / (L / 2) = 2H / L.
And here the researchers were in for a big surprise!.png" width="25" height="24">= 1.272. Comparing this value with the tg value a= 1.27306, we see that these values are very close to each other. If we take the angle a= 51°50", that is, reduce it by only one arc minute, then the value a will become equal to 1.272, that is, it will coincide with the value. It should be noted that in 1840 G. Wise repeated his measurements and clarified that the value of the angle a=51°50".
These measurements led the researchers to the following very interesting hypothesis: the triangle ACB of the Cheops pyramid was based on the relation AC / C.B. = = 1,272!
Consider now the right triangle ABC, in which the ratio of the legs A.C. / C.B.= (Fig. 2). If now the lengths of the sides of the rectangle ABC designate by x, y, z, and also take into account that the ratio y/x= , then in accordance with the Pythagorean theorem, the length z can be calculated using the formula:
If we accept x = 1, y= https://pandia.ru/text/78/390/images/image027_1.png" width="143" height="27">
Figure 3."Golden" right triangle.
A right triangle in which the sides are related as t:golden" right triangle.
Then, if we take as a basis the hypothesis that the main “geometric idea” of the Cheops pyramid is a “golden” right triangle, then from here we can easily calculate the “design” height of the Cheops pyramid. It is equal to:
H = (L/2) ´ = 148.28 m.
Let us now derive some other relations for the Cheops pyramid, which follow from the “golden” hypothesis. In particular, we will find the ratio of the outer area of the pyramid to the area of its base. To do this, we take the length of the leg C.B. per unit, that is: C.B.= 1. But then the length of the side of the base of the pyramid GF= 2, and the area of the base EFGH will be equal SEFGH = 4.
Let us now calculate the area of the side face of the Cheops pyramid SD. Since the height AB triangle AEF equal to t, then the area of the side face will be equal to SD = t. Then the total area of all four lateral faces of the pyramid will be equal to 4 t, and the ratio of the total outer area of the pyramid to the area of the base will be equal to the golden ratio! This is it - the main geometric mystery of the Cheops pyramid!
The group of “geometric miracles” of the Cheops pyramid includes real and far-fetched properties of the relationships between various dimensions in the pyramid.
As a rule, they are obtained in search of certain “constants”, in particular, the number “pi” (Ludolfo’s number), equal to 3.14159...; grounds natural logarithms"e" (Neper's number), equal to 2.71828...; the number "F", the number of the "golden section", equal to, for example, 0.618... etc.
You can name, for example: 1) Property of Herodotus: (Height)2 = 0.5 art. basic x Apothem; 2) Property of V. Price: Height: 0.5 art. base = Square root of "F"; 3) Property of M. Eist: Perimeter of the base: 2 Height = "Pi"; in a different interpretation - 2 tbsp. basic : Height = "Pi"; 4) Property of G. Edge: Radius of the inscribed circle: 0.5 art. basic = "F"; 5) Property of K. Kleppisch: (Art. main.)2: 2(Art. main. x Apothem) = (Art. main. W. Apothema) = 2(Art. main. x Apothem) : ((2 art. main X Apothem) + (v. main)2). And so on. You can come up with many such properties, especially if you connect two adjacent pyramids. For example, as “Properties of A. Arefyev” it can be mentioned that the difference in the volumes of the pyramid of Cheops and the pyramid of Khafre is equal to twice the volume of the pyramid of Mikerin...
Many interesting provisions, in particular on the construction of pyramids according to the “golden ratio”, are set out in the books by D. Hambidge “Dynamic symmetry in architecture” and M. Gick “Aesthetics of proportion in nature and art”. Let us recall that the “golden ratio” is the division of a segment in such a ratio that part A is as many times greater than part B, how many times A is smaller than the entire segment A + B. The ratio A/B in this case is equal to the number “F” == 1.618. .. The use of the “golden ratio” is indicated not only in individual pyramids, but also in the entire complex of pyramids at Giza.
The most curious thing, however, is that one and the same Cheops pyramid simply “cannot” contain so many wonderful properties. Taking a certain property one by one, it can be “fitted”, but all of them do not fit at once - they do not coincide, they contradict each other. Therefore, if, for example, when checking all properties, we initially take the same side of the base of the pyramid (233 m), then the heights of pyramids with different properties will also be different. In other words, there is a certain “family” of pyramids that are externally similar to Cheops, but have different properties. Note that there is nothing particularly miraculous in the “geometric” properties - much arises purely automatically, from the properties of the figure itself. A “miracle” should only be considered something that was clearly impossible for the ancient Egyptians. This, in particular, includes “cosmic” miracles, in which the measurements of the Cheops pyramid or the pyramid complex at Giza are compared with some astronomical measurements and “even” numbers are indicated: a million times less, a billion times less, and so on. Let's consider some "cosmic" relationships.
One of the statements is: “if you divide the side of the base of the pyramid by the exact length of the year, you get exactly 10 millionths of the earth’s axis.” Calculate: divide 233 by 365, we get 0.638. The radius of the Earth is 6378 km.
Another statement is actually the opposite of the previous one. F. Noetling pointed out that if we use the “Egyptian cubit” he himself invented, then the side of the pyramid will correspond to “the most accurate duration of the solar year, expressed to the nearest one billionth of a day” - 365.540.903.777.
P. Smith's statement: "The height of the pyramid is exactly one billionth of the distance from the Earth to the Sun." Although the height usually taken is 146.6 m, Smith took it as 148.2 m. According to modern radar measurements, the semi-major axis of the earth's orbit is 149,597,870 + 1.6 km. This is the average distance from the Earth to the Sun, but at perihelion it is 5,000,000 kilometers less than at aphelion.
One last interesting statement:
“How can we explain that the masses of the pyramids of Cheops, Khafre and Mykerinus relate to each other, like the masses of the planets Earth, Venus, Mars?” Let's calculate. The masses of the three pyramids are: Khafre - 0.835; Cheops - 1,000; Mikerin - 0.0915. The ratios of the masses of the three planets: Venus - 0.815; Earth - 1,000; Mars - 0.108.
So, despite skepticism, we note the well-known harmony of the construction of statements: 1) the height of the pyramid, like a line “going into space”, corresponds to the distance from the Earth to the Sun; 2) the side of the base of the pyramid, closest “to the substrate,” that is, to the Earth, is responsible for the earth’s radius and earth’s circulation; 3) the volumes of the pyramid (read - masses) correspond to the ratio of the masses of the planets closest to the Earth. A similar “cipher” can be traced, for example, in the bee language analyzed by Karl von Frisch. However, we will refrain from commenting on this matter for now.
PYRAMID SHAPE
The famous tetrahedral shape of the pyramids did not arise immediately. The Scythians made burials in the form of earthen hills - mounds. The Egyptians built "hills" of stone - pyramids. This first happened after the unification of Upper and Lower Egypt, in the 28th century BC, when the founder of the Third Dynasty, Pharaoh Djoser (Zoser), was faced with the task of strengthening the unity of the country.
And here, according to historians, an important role in strengthening central government played by the “new concept of deification” of the king. Although the royal burials were distinguished by greater splendor, they, in principle, did not differ from the tombs of court nobles; they were the same structures - mastabas. Above the chamber with the sarcophagus containing the mummy, a rectangular hill of small stones was poured, where a small building made of large stone blocks was then erected - a “mastaba” (in Arabic - “bench”). Pharaoh Djoser erected the first pyramid on the site of the mastaba of his predecessor, Sanakht. It was stepped and was a visible transitional stage from one architectural form to another, from a mastaba to a pyramid.
In this way, the sage and architect Imhotep, who was later considered a wizard and identified by the Greeks with the god Asclepius, “raised” the pharaoh. It was as if six mastabas were erected in a row. Moreover, the first pyramid occupied an area of 1125 x 115 meters, with an estimated height of 66 meters (according to Egyptian standards - 1000 “palms”). At first, the architect planned to build a mastaba, but not oblong, but square in plan. Later it was expanded, but since the extension was made lower, it seemed like there were two steps.
This situation did not satisfy the architect, and on the upper platform of the huge flat mastaba, Imhotep placed three more, gradually decreasing towards the top. The tomb was located under the pyramid.
Several more step pyramids are known, but later the builders moved on to building tetrahedral pyramids that are more familiar to us. Why, however, not triangular or, say, octagonal? An indirect answer is given by the fact that almost all pyramids are perfectly oriented along the four cardinal directions, and therefore have four sides. In addition, the pyramid was a “house”, the shell of a quadrangular burial chamber.
But what determined the angle of inclination of the faces? In the book “The Principle of Proportions” an entire chapter is devoted to this: “What could have determined the angles of inclination of the pyramids.” In particular, it is indicated that “the image to which the great pyramids of the Old Kingdom gravitate is a triangle with a right angle at the apex.
In space it is a semi-octahedron: a pyramid in which the edges and sides of the base are equal, the edges are equilateral triangles." Certain considerations are given on this subject in the books of Hambidge, Gick and others.
What is the advantage of the semi-octahedron angle? According to descriptions by archaeologists and historians, some pyramids collapsed under their own weight. What was needed was a “durability angle,” an angle that was the most energetically reliable. Purely empirically, this angle can be taken from the vertex angle in a pile of crumbling dry sand. But to get accurate data, you need to use a model. Taking four firmly fixed balls, you need to place a fifth one on them and measure the angles of inclination. However, you can make a mistake here, so a theoretical calculation helps: you should connect the centers of the balls with lines (mentally). The base will be a square with a side equal to twice the radius. The square will be just the base of the pyramid, the length of the edges of which will also be equal to twice the radius.
Thus, a close packing of balls like 1:4 will give us a regular semi-octahedron.
However, why do many pyramids, gravitating towards a similar shape, nevertheless not retain it? The pyramids are probably aging. Contrary to the famous saying:
“Everything in the world is afraid of time, and time is afraid of pyramids,” the buildings of the pyramids must age, not only processes of external weathering can and should occur in them, but also processes of internal “shrinkage,” which may cause the pyramids to become lower. Shrinkage is also possible because, as revealed by the work of D. Davidovits, the ancient Egyptians used the technology of making blocks from lime chips, in other words, from “concrete”. It is precisely similar processes that could explain the reason for the destruction of the Medum Pyramid, located 50 km south of Cairo. It is 4600 years old, the dimensions of the base are 146 x 146 m, the height is 118 m. “Why is it so disfigured?” asks V. Zamarovsky. “The usual references to the destructive effects of time and the “use of stone for other buildings” are not suitable here.
After all, most of its blocks and facing slabs have remained in place to this day, in ruins at its foot." As we will see, a number of provisions even make us think that the famous Pyramid of Cheops also "shrivelled." In any case, in all ancient images the pyramids are pointed ...
The shape of the pyramids could also have been generated by imitation: some natural samples, “miracle perfection,” say, some crystals in the form of an octahedron.
Similar crystals could be diamond and gold crystals. Characteristic large number"overlapping" signs for such concepts as Pharaoh, Sun, Gold, Diamond. Everywhere - noble, brilliant (brilliant), great, impeccable, and so on. The similarities are not accidental.
The solar cult, as is known, formed an important part of the religion Ancient Egypt. “No matter how we translate the name of the greatest of the pyramids,” notes one of modern aids- “The firmament of Khufu” or “The firmament of Khufu”, it meant that the king is the sun.” If Khufu, in the brilliance of his power, imagined himself to be the second sun, then his son Djedef-Ra became the first of the Egyptian kings to call himself “the son of Ra ", that is, the son of the Sun. The Sun of almost all peoples was symbolized by the "solar metal", gold. "A large disk of bright gold" - that’s what the Egyptians called our daylight. The Egyptians knew gold perfectly, they knew its native forms, where gold crystals can appear in the form of octahedrons.
The “sun stone”—diamond—is also interesting here as a “sample of forms.” The name of the diamond came precisely from the Arab world, “almas” - the hardest, most hard, indestructible. The ancient Egyptians knew diamond and its properties quite well. According to some authors, they even used bronze tubes with diamond cutters for drilling.
Nowadays the main supplier of diamonds is South Africa, but Western Africa is also rich in diamonds. The territory of the Republic of Mali is even called the “Diamond Land”. Meanwhile, it is on the territory of Mali that the Dogon live, with whom supporters of the paleo-visit hypothesis pin many hopes (see below). Diamonds could not have been the reason for the contacts of the ancient Egyptians with this region. However, one way or another, it is possible that precisely by copying the octahedrons of diamond and gold crystals, the ancient Egyptians thereby deified the pharaohs, “indestructible” like diamond and “brilliant” like gold, the sons of the Sun, comparable only to the most wonderful creations of nature.
Conclusion:
Having studied the pyramid as a geometric body, becoming acquainted with its elements and properties, we were convinced of the validity of the opinion about the beauty of the shape of the pyramid.
As a result of our research, we came to the conclusion that the Egyptians, having collected the most valuable mathematical knowledge, embodied it in a pyramid. Therefore, the pyramid is truly the most perfect creation of nature and man.
LIST OF REFERENCES USED
"Geometry: Textbook. for 7 – 9 grades. general education institutions\, etc. - 9th ed. - M.: Education, 1999
History of mathematics in school, M: “Prosveshchenie”, 1982.
Geometry 10-11 grades, M: “Enlightenment”, 2000
Peter Tompkins “Secrets of the Great Pyramid of Cheops”, M: “Tsentropoligraf”, 2005.
Internet resources
http://veka-i-mig. *****/
http://tambov. *****/vjpusk/vjp025/rabot/33/index2.htm
http://www. *****/enc/54373.html
Definition
Pyramid is a polyhedron composed of a polygon \(A_1A_2...A_n\) and \(n\) triangles with a common vertex \(P\) (not lying in the plane of the polygon) and sides opposite it, coinciding with the sides of the polygon.
Designation: \(PA_1A_2...A_n\) .
Example: pentagonal pyramid \(PA_1A_2A_3A_4A_5\) .
Triangles \(PA_1A_2, \PA_2A_3\), etc. are called side faces pyramids, segments \(PA_1, PA_2\), etc. – lateral ribs, polygon \(A_1A_2A_3A_4A_5\) – basis, point \(P\) – top.
Height pyramids are a perpendicular descended from the top of the pyramid to the plane of the base.
A pyramid with a triangle at its base is called tetrahedron.
The pyramid is called correct, if its base is a regular polygon and one of the following conditions is met:
\((a)\) the lateral edges of the pyramid are equal;
\((b)\) the height of the pyramid passes through the center of the circle circumscribed near the base;
\((c)\) the side ribs are inclined to the plane of the base at the same angle.
\((d)\) the side faces are inclined to the plane of the base at the same angle.
Regular tetrahedron is a triangular pyramid, all of whose faces are equal equilateral triangles.
Theorem
Conditions \((a), (b), (c), (d)\) are equivalent.
Proof
Let's find the height of the pyramid \(PH\) . Let \(\alpha\) be the plane of the base of the pyramid.
1) Let us prove that from \((a)\) it follows \((b)\) . Let \(PA_1=PA_2=PA_3=...=PA_n\) .
Because \(PH\perp \alpha\), then \(PH\) is perpendicular to any line lying in this plane, which means the triangles are right-angled. This means that these triangles are equal in common leg \(PH\) and hypotenuse \(PA_1=PA_2=PA_3=...=PA_n\) . This means \(A_1H=A_2H=...=A_nH\) . This means that the points \(A_1, A_2, ..., A_n\) are at the same distance from the point \(H\), therefore, they lie on the same circle with the radius \(A_1H\) . This circle, by definition, is circumscribed about the polygon \(A_1A_2...A_n\) .
2) Let us prove that \((b)\) implies \((c)\) .
\(PA_1H, PA_2H, PA_3H,..., PA_nH\) rectangular and equal on two legs. This means that their angles are also equal, therefore, \(\angle PA_1H=\angle PA_2H=...=\angle PA_nH\).
3) Let us prove that \((c)\) implies \((a)\) .
Similar to the first point, triangles \(PA_1H, PA_2H, PA_3H,..., PA_nH\) rectangular and along the leg and sharp corner. This means that their hypotenuses are also equal, that is, \(PA_1=PA_2=PA_3=...=PA_n\) .
4) Let us prove that \((b)\) implies \((d)\) .
Because in a regular polygon the centers of the circumscribed and inscribed circles coincide (generally speaking, this point is called the center of a regular polygon), then \(H\) is the center of the inscribed circle. Let's draw perpendiculars from the point \(H\) to the sides of the base: \(HK_1, HK_2\), etc. These are the radii of the inscribed circle (by definition). Then according to TTP (\(PH\) is perpendicular to the plane, \(HK_1, HK_2\), etc. are projections perpendicular to the sides) inclined \(PK_1, PK_2\), etc. perpendicular to the sides \(A_1A_2, A_2A_3\), etc. respectively. So, by definition \(\angle PK_1H, \angle PK_2H\) equal to the angles between the side faces and the base. Because triangles \(PK_1H, PK_2H, ...\) are equal (as rectangular on two sides), then the angles \(\angle PK_1H, \angle PK_2H, ...\) are equal.
5) Let us prove that \((d)\) implies \((b)\) .
Similar to the fourth point, the triangles \(PK_1H, PK_2H, ...\) are equal (as rectangular along the leg and acute angle), which means the segments \(HK_1=HK_2=...=HK_n\) are equal. This means, by definition, \(H\) is the center of a circle inscribed in the base. But because For regular polygons, the centers of the inscribed and circumscribed circles coincide, then \(H\) is the center of the circumscribed circle. Chtd.
Consequence
The lateral faces of a regular pyramid are equal isosceles triangles.
Definition
The height of the lateral face of a regular pyramid drawn from its vertex is called apothem.
The apothems of all lateral faces of a regular pyramid are equal to each other and are also medians and bisectors.
Important Notes
1. The height of a regular triangular pyramid falls at the point of intersection of the heights (or bisectors, or medians) of the base (the base is a regular triangle).
2. The height of a regular quadrangular pyramid falls at the point of intersection of the diagonals of the base (the base is a square).
3. The height of a regular hexagonal pyramid falls at the point of intersection of the diagonals of the base (the base is a regular hexagon).
4. The height of the pyramid is perpendicular to any straight line lying at the base.
Definition
The pyramid is called rectangular, if one of its side edges is perpendicular to the plane of the base.
Important Notes
1. In a rectangular pyramid, the edge perpendicular to the base is the height of the pyramid. That is, \(SR\) is the height.
2. Because \(SR\) is perpendicular to any line from the base, then \(\triangle SRM, \triangle SRP\)– right triangles.
3. Triangles \(\triangle SRN, \triangle SRK\)- also rectangular.
That is, any triangle formed by this edge and the diagonal emerging from the vertex of this edge lying at the base will be rectangular.
\[(\Large(\text(Volume and surface area of the pyramid)))\]
Theorem
The volume of the pyramid is equal to one third of the product of the area of the base and the height of the pyramid: \
Consequences
Let \(a\) be the side of the base, \(h\) be the height of the pyramid.
1. The volume of a regular triangular pyramid is \(V_(\text(right triangle.pir.))=\dfrac(\sqrt3)(12)a^2h\),
2. The volume of a regular quadrangular pyramid is \(V_(\text(right.four.pir.))=\dfrac13a^2h\).
3. The volume of a regular hexagonal pyramid is \(V_(\text(right.six.pir.))=\dfrac(\sqrt3)(2)a^2h\).
4. The volume of a regular tetrahedron is \(V_(\text(right tetr.))=\dfrac(\sqrt3)(12)a^3\).
Theorem
The area of the lateral surface of a regular pyramid is equal to the half product of the perimeter of the base and the apothem.
\[(\Large(\text(Frustum)))\]
Definition
Consider an arbitrary pyramid \(PA_1A_2A_3...A_n\) . Let us draw a plane parallel to the base of the pyramid through a certain point lying on the side edge of the pyramid. This plane will split the pyramid into two polyhedra, one of which is a pyramid (\(PB_1B_2...B_n\)), and the other is called truncated pyramid(\(A_1A_2...A_nB_1B_2...B_n\) ).
The truncated pyramid has two bases - polygons \(A_1A_2...A_n\) and \(B_1B_2...B_n\) which are similar to each other.
The height of a truncated pyramid is a perpendicular drawn from some point of the upper base to the plane of the lower base.
Important Notes
1. All lateral faces of a truncated pyramid are trapezoids.
2. The segment connecting the centers of the bases of a regular truncated pyramid (that is, a pyramid obtained by cross-section of a regular pyramid) is the height.
A triangular pyramid is a pyramid that has a triangle at its base. The height of this pyramid is the perpendicular that is lowered from the top of the pyramid to its base.
Finding the height of a pyramid
How to find the height of a pyramid? Very simple! To find the height of any triangular pyramid, you can use the volume formula: V = (1/3)Sh, where S is the area of the base, V is the volume of the pyramid, h is its height. From this formula, derive the height formula: to find the height of a triangular pyramid, you need to multiply the volume of the pyramid by 3, and then divide the resulting value by the area of the base, it will be: h = (3V)/S. Since the base of a triangular pyramid is a triangle, you can use the formula to calculate the area of a triangle. If we know: the area of the triangle S and its side z, then according to the area formula S=(1/2)γh: h = (2S)/γ, where h is the height of the pyramid, γ is the edge of the triangle; the angle between the sides of the triangle and the two sides themselves, then using the following formula: S = (1/2)γφsinQ, where γ, φ are the sides of the triangle, we find the area of the triangle. The value of the sine of angle Q needs to be looked at in the table of sines, which is available on the Internet. Next, we substitute the area value into the height formula: h = (2S)/γ. If the task requires calculating the height of a triangular pyramid, then the volume of the pyramid is already known.
Regular triangular pyramid
Find the height of a regular triangular pyramid, that is, a pyramid in which all faces are equilateral triangles, knowing the edge size γ. In this case, the edges of the pyramid are the sides of equilateral triangles. The height of a regular triangular pyramid will be: h = γ√(2/3), where γ is the edge of the equilateral triangle, h is the height of the pyramid. If the area of the base (S) is unknown, and only the length of the edge (γ) and the volume (V) of the polyhedron are given, then the necessary variable in the formula from the previous step must be replaced by its equivalent, which is expressed in terms of the length of the edge. The area of a triangle (regular) is equal to 1/4 of the product of the side length of this triangle squared by the square root of 3. We substitute this formula instead of the area of the base in the previous formula, and we obtain the following formula: h = 3V4/(γ 2 √3) = 12V/(γ 2 √3). The volume of a tetrahedron can be expressed through the length of its edge, then from the formula for calculating the height of a figure, you can remove all variables and leave only the side triangular face figures. The volume of such a pyramid can be calculated by dividing by 12 from the product the cubed length of its face by the square root of 2.
Substituting this expression into the previous formula, we obtain the following formula for calculation: h = 12(γ 3 √2/12)/(γ 2 √3) = (γ 3 √2)/(γ 2 √3) = γ√(2 /3) = (1/3)γ√6. Also correct triangular prism can be inscribed in a sphere, and knowing only the radius of the sphere (R) one can find the height of the tetrahedron itself. The length of the tetrahedron edge is: γ = 4R/√6. We replace the variable γ with this expression in the previous formula and get the formula: h = (1/3)√6(4R)/√6 = (4R)/3. The same formula can be obtained by knowing the radius (R) of a circle inscribed in a tetrahedron. In this case, the length of the edge of the triangle will be equal to 12 ratios between square root of 6 and radius. We substitute this expression into the previous formula and we have: h = (1/3)γ√6 = (1/3)√6(12R)/√6 = 4R.
How to find the height of a regular quadrangular pyramid
To answer the question of how to find the length of the height of a pyramid, you need to know what a regular pyramid is. A quadrangular pyramid is a pyramid that has a quadrangle at its base. If in the conditions of the problem we have: volume (V) and area of the base (S) of the pyramid, then the formula for calculating the height of the polyhedron (h) will be as follows - divide the volume multiplied by 3 by the area S: h = (3V)/S. Given a square base of a pyramid with a given volume (V) and side length γ, replace the area (S) in the previous formula with the square of the side length: S = γ 2 ; H = 3V/γ2. The height of a regular pyramid h = SO passes exactly through the center of the circle that is circumscribed near the base. Since the base of this pyramid is a square, point O is the intersection point of diagonals AD and BC. We have: OC = (1/2)BC = (1/2)AB√6. Next, in the right triangle SOC we find (using the Pythagorean theorem): SO = √(SC 2 -OC 2). Now you know how to find the height of a regular pyramid.
This video tutorial will help users get an idea of the Pyramid theme. Correct pyramid. In this lesson we will get acquainted with the concept of a pyramid and give it a definition. Let's consider what a regular pyramid is and what properties it has. Then we prove the theorem about the lateral surface of a regular pyramid.
In this lesson we will get acquainted with the concept of a pyramid and give it a definition.
Consider a polygon A 1 A 2...A n, which lies in the α plane, and the point P, which does not lie in the α plane (Fig. 1). Let's connect the dots P with peaks A 1, A 2, A 3, … A n. We get n triangles: A 1 A 2 R, A 2 A 3 R and so on.
Definition. Polyhedron RA 1 A 2 ...A n, made up of n-square A 1 A 2...A n And n triangles RA 1 A 2, RA 2 A 3 …RA n A n-1 is called n-coal pyramid. Rice. 1.
Rice. 1
Consider a quadrangular pyramid PABCD(Fig. 2).
R- the top of the pyramid.
ABCD- the base of the pyramid.
RA- side rib.
AB- base rib.
From point R let's drop the perpendicular RN to the base plane ABCD. The perpendicular drawn is the height of the pyramid.
Rice. 2
Full surface The pyramid consists of a lateral surface, that is, the area of all lateral faces, and the area of the base:
S full = S side + S main
A pyramid is called correct if:
- its base is a regular polygon;
- the segment connecting the top of the pyramid with the center of the base is its height.
Explanation using the example of a regular quadrangular pyramid
Consider a regular quadrangular pyramid PABCD(Fig. 3).
R- the top of the pyramid. Base of the pyramid ABCD- a regular quadrilateral, that is, a square. Dot ABOUT, the point of intersection of the diagonals, is the center of the square. Means, RO is the height of the pyramid.
Rice. 3
Explanation: in the correct n In a triangle, the center of the inscribed circle and the center of the circumcircle coincide. This center is called the center of the polygon. Sometimes they say that the vertex is projected into the center.
The height of the lateral face of a regular pyramid drawn from its vertex is called apothem and is designated h a.
1. all lateral edges of a regular pyramid are equal;
2. The side faces are equal isosceles triangles.
We will give a proof of these properties using the example of a regular quadrangular pyramid.
Given: PABCD- regular quadrangular pyramid,
ABCD- square,
RO- height of the pyramid.
Prove:
1. RA = PB = RS = PD
2.∆ABP = ∆BCP =∆CDP =∆DAP See Fig. 4.
Rice. 4
Proof.
RO- height of the pyramid. That is, straight RO perpendicular to the plane ABC, and therefore direct JSC, VO, SO And DO lying in it. So triangles ROA, ROV, ROS, ROD- rectangular.
Consider a square ABCD. From the properties of a square it follows that AO = VO = CO = DO.
Then the right triangles ROA, ROV, ROS, ROD leg RO- general and legs JSC, VO, SO And DO are equal, which means that these triangles are equal on two sides. From the equality of triangles follows the equality of segments, RA = PB = RS = PD. Point 1 has been proven.
Segments AB And Sun are equal because they are sides of the same square, RA = PB = RS. So triangles AVR And VSR - isosceles and equal on three sides.
In a similar way we find that triangles ABP, VCP, CDP, DAP are isosceles and equal, as required to be proved in paragraph 2.
The area of the lateral surface of a regular pyramid is equal to half the product of the perimeter of the base and the apothem:
To prove this, let’s choose a regular triangular pyramid.
Given: RAVS- regular triangular pyramid.
AB = BC = AC.
RO- height.
Prove: 
. See Fig. 5.
Rice. 5
Proof.
RAVS- regular triangular pyramid. That is AB= AC = BC. Let ABOUT- center of the triangle ABC, Then RO is the height of the pyramid. At the base of the pyramid lies equilateral triangle ABC. Note that 
.
Triangles RAV, RVS, RSA- equal isosceles triangles (by property). A triangular pyramid has three side faces: RAV, RVS, RSA. This means that the area of the lateral surface of the pyramid is:
S side = 3S RAW
The theorem has been proven.
The radius of a circle inscribed at the base of a regular quadrangular pyramid is 3 m, the height of the pyramid is 4 m. Find the area of the lateral surface of the pyramid.
Given: regular quadrangular pyramid ABCD,
ABCD- square,
r= 3 m,
RO- height of the pyramid,
RO= 4 m.
Find: S side. See Fig. 6.
Rice. 6
Solution.
According to the proven theorem, .
Let's first find the side of the base AB. We know that the radius of a circle inscribed at the base of a regular quadrangular pyramid is 3 m.
Then, m.
Find the perimeter of the square ABCD with a side of 6 m:
Consider a triangle BCD. Let M- middle of the side DC. Because ABOUT- middle BD, That 
(m).
Triangle DPC- isosceles. M- middle DC. That is, RM- median, and therefore the height in the triangle DPC. Then RM- apothem of the pyramid.
RO- height of the pyramid. Then, straight RO perpendicular to the plane ABC, and therefore direct OM, lying in it. Let's find the apothem RM from a right triangle ROM.
Now we can find lateral surface pyramids:
Answer: 60 m2.
The radius of the circle circumscribed around the base of a regular triangular pyramid is equal to m. The area of the lateral surface is 18 m 2. Find the length of the apothem.
Given: ABCP- regular triangular pyramid,
AB = BC = SA,
R= m,
S side = 18 m2.
Find: . See Fig. 7.
Rice. 7
Solution.
In a right triangle ABC The radius of the circumscribed circle is given. Let's find a side AB this triangle using the law of sines.
Knowing the side regular triangle(m), let's find its perimeter.
By the theorem on the lateral surface area of a regular pyramid, where h a- apothem of the pyramid. Then:
Answer: 4 m.
So, we looked at what a pyramid is, what a regular pyramid is, and we proved the theorem about the lateral surface of a regular pyramid. In the next lesson we will get acquainted with the truncated pyramid.
References
- Geometry. Grades 10-11: textbook for students of general education institutions (basic and specialized levels) / I. M. Smirnova, V. A. Smirnov. - 5th ed., rev. and additional - M.: Mnemosyne, 2008. - 288 p.: ill.
- Geometry. 10-11 grade: Textbook for general education educational institutions/ Sharygin I.F. - M.: Bustard, 1999. - 208 p.: ill.
- Geometry. Grade 10: Textbook for general education institutions with in-depth and specialized study of mathematics /E. V. Potoskuev, L. I. Zvalich. - 6th ed., stereotype. - M.: Bustard, 008. - 233 p.: ill.
- Internet portal "Yaklass" ()
- Internet portal “Festival of pedagogical ideas “First of September” ()
- Internet portal “Slideshare.net” ()
Homework
- Can a regular polygon be the base of an irregular pyramid?
- Prove that disjoint edges of a regular pyramid are perpendicular.
- Find the value of the dihedral angle at the side of the base of a regular quadrangular pyramid if the apothem of the pyramid is equal to the side of its base.
- RAVS- regular triangular pyramid. Construct the linear angle of the dihedral angle at the base of the pyramid.