Here you can find basic information about pyramids and related formulas and concepts. All of them are studied with a mathematics tutor in preparation for the Unified State Exam.
Consider a plane, a polygon 
, lying in it and a point S, not lying in it. Let's connect S to all the vertices of the polygon. The resulting polyhedron is called a pyramid. The segments are called side ribs. 
The polygon is called the base, and point S is the top of the pyramid. Depending on the number n, the pyramid is called triangular (n=3), quadrangular (n=4), pentagonal (n=5) and so on. An alternative name for a triangular pyramid is tetrahedron. The height of a pyramid is the perpendicular descending from its top to the plane of the base. 
A pyramid is called regular if a regular polygon, and the base of the pyramid's altitude (the base of the perpendicular) is its center.
Tutor's comment:
Do not confuse the concepts of “regular pyramid” and “regular tetrahedron”. In a regular pyramid, the side edges are not necessarily equal to the edges of the base, but in a regular tetrahedron, all 6 edges are equal. This is his definition. It is easy to prove that the equality implies that the center P of the polygon coincides with a base height, so a regular tetrahedron is a regular pyramid.
What is an apothem?
The apothem of a pyramid is the height of its side face. If the pyramid is regular, then all its apothems are equal. The reverse is not true. 
A mathematics tutor about his terminology: 80% of work with pyramids is built through two types of triangles:
1) Containing apothem SK and height SP
2) Containing the lateral edge SA and its projection PA 
To simplify references to these triangles, it is more convenient for a math tutor to call the first of them apothemal, and second costal. Unfortunately, you will not find this terminology in any of the textbooks, and the teacher has to introduce it unilaterally.
Pyramid volume formula:
1) 
, where is the area of the base of the pyramid, and is the height of the pyramid
2) , where is the radius of the inscribed sphere, and is the area of the total surface of the pyramid.
3) 
, where MN is the distance between any two crossing edges, and is the area of the parallelogram formed by the midpoints of the four remaining edges.
Point P (see figure) coincides with the center of the inscribed circle at the base of the pyramid if one of the following conditions is met:
1) All apothems are equal
2) All side faces are equally inclined to the base
3) All apothems are equally inclined to the height of the pyramid
4) The height of the pyramid is equally inclined to all side faces
Math tutor's comment: Please note that all points have one thing in common general property: one way or another, lateral faces are involved everywhere (apothems are their elements). Therefore, the tutor can offer a less precise, but more convenient for learning, formulation: point P coincides with the center of the inscribed circle, the base of the pyramid, if there is any equal information about its lateral faces. To prove it, it is enough to show that all apothem triangles are equal.
Point P coincides with the center of a circle circumscribed near the base of the pyramid if one of three conditions is true:
1) All side edges are equal
2) All side ribs are equally inclined to the base
3) All side ribs are equally inclined to the height
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 vertices 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
The full surface of the pyramid consists of the 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:
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 - 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 an equilateral triangle ABC. notice, 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 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 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 lateral surface area 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 of a regular triangle (m), we 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.
Bibliography
Homework
1. When all side edges are of the same size, then:
2. When the side faces have an angle of inclination to the plane of the base of the same value, then:
3. A sphere can be described around a pyramid if at the base of the pyramid there is a polygon around which a circle can be described (a necessary and sufficient condition). The center of the sphere will be the point of intersection of the planes that pass through the middles of the edges of the pyramid perpendicular to them. From this theorem we conclude that a sphere can be described both around any triangular and around any regular pyramid.
4. A sphere can be inscribed into a pyramid if the bisector planes of the internal dihedral angles of the pyramid intersect at the 1st point (a necessary and sufficient condition). This point will become the center of the sphere.
The simplest pyramid.
Based on the number of angles, the base of the pyramid is divided into triangular, quadrangular, and so on.
There will be a pyramid triangular, quadrangular, and so on, when the base of the pyramid is a triangle, a quadrangle, and so on. A triangular pyramid is a tetrahedron - a tetrahedron. Quadrangular - pentagonal and so on.
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 \((a)\) implies \((b)\) . 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 \(PA_1=PA_2=PA_3=...=PA_n\) . So, \(A_1H=A_2H=...=A_nH\) . 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 from \((b)\) it follows \((d)\) .
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 a 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. U 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.
Pyramid concept
Definition 1
Geometric figure, formed by a polygon and a point not lying in the plane containing this polygon, connected to all the vertices of the polygon is called a pyramid (Fig. 1).
The polygon from which the pyramid is made is called the base of the pyramid; the resulting triangles, when connected to a point, are the side faces of the pyramid, the sides of the triangles are the sides of the pyramid, and the point common to all triangles is the top of the pyramid.
Depending on the number of angles at the base of the pyramid, it can be called triangular, quadrangular, and so on (Fig. 2).
Figure 2.
Another type of pyramid is the regular pyramid.
Let us introduce and prove the property of a regular pyramid.
Theorem 1
All lateral faces of a regular pyramid are isosceles triangles that are equal to each other.
Proof.
Consider a regular $n-$gonal pyramid with vertex $S$ of height $h=SO$. Let us draw a circle around the base (Fig. 4).
Figure 4.
Consider the triangle $SOA$. According to the Pythagorean theorem, we get
Obviously, any side edge will be defined this way. Consequently, all side edges are equal to each other, that is, all side faces are isosceles triangles. Let us prove that they are equal to each other. Since the base is a regular polygon, the bases of all side faces are equal to each other. Consequently, all lateral faces are equal according to the III criterion of equality of triangles.
The theorem has been proven.
Let us now introduce the following definition related to the concept of a regular pyramid.
Definition 3
The apothem of a regular pyramid is the height of its side face.
Obviously, by Theorem One, all apothems are equal to each other.
Theorem 2
The lateral surface area of a regular pyramid is determined as the product of the semi-perimeter of the base and the apothem.
Proof.
Let us denote the side of the base of the $n-$gonal pyramid by $a$, and the apothem by $d$. Therefore, the area of the side face is equal to
Since, according to Theorem 1, all sides are equal, then
The theorem has been proven.
Another type of pyramid is a truncated pyramid.
Definition 4
If a plane parallel to its base is drawn through an ordinary pyramid, then the figure formed between this plane and the plane of the base is called a truncated pyramid (Fig. 5).
Figure 5. Truncated pyramid
The lateral faces of the truncated pyramid are trapezoids.
Theorem 3
The lateral surface area of a regular truncated pyramid is determined as the product of the sum of the semi-perimeters of the bases and the apothem.
Proof.
Let us denote the sides of the bases of the $n-$gonal pyramid by $a\ and\ b$, respectively, and the apothem by $d$. Therefore, the area of the side face is equal to
Since all sides are equal, then
The theorem has been proven.
Example 1
Find the area of the lateral surface of a truncated triangular pyramid if it is obtained from a regular pyramid with base side 4 and apothem 5 by cutting off a plane passing through the midline of the side faces.
Solution.
By the theorem about midline we find that the upper base of the truncated pyramid is equal to $4\cdot \frac(1)(2)=2$, and the apothem is equal to $5\cdot \frac(1)(2)=2.5$.
Then, by Theorem 3, we get
