What is the base of a regular triangular pyramid. Formulas and properties of a regular triangular pyramid. Truncated triangular pyramid

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.

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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 the form of triangles, converging at one point.

Relying on modern interpretation, the pyramid is represented as a spatial polyhedron consisting of a certain k-gon and k flat figures triangular shape, having 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:

  1. The basis is a figure of the correct shape.
  2. The edges of the pyramid that limit the side elements have equal numerical values.
  3. The side elements are isosceles triangles.
  4. 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.
  5. All side ribs are inclined to the plane of the base at the same angle.
  6. 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:

  1. In the case when the polygon fits into a circle, the side faces will have the base equal angles.
  2. When describing a circle around a polygon, all edges of the pyramid emanating from the vertex will have equal lengths 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

Correct triangular pyramid– 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 a school geometry course they work with two:

  • axial;
  • parallel to the basis.

An axial section is obtained by intersecting a polyhedron with a plane that 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:

  1. The area of ​​an isosceles 3-gon is Str=1/2(aL), where a is the side of the base, L is the apothem.
  2. 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 is 4a = Rosn, where Rosn is the perimeter of the base. And the expression 1/2*Rosn is its semi-perimeter.
  3. 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 a regular quadrangular pyramid

Pyramid. Truncated pyramid

Pyramid is a polyhedron, one of whose faces is a polygon ( base ), and all other faces are triangles with a common vertex ( side faces ) (Fig. 15). The pyramid is called correct , if its base is a regular polygon and the top of the pyramid is projected into the center of the base (Fig. 16). A triangular pyramid with all edges equal is called tetrahedron .



Lateral rib of a pyramid is the side of the side face that does not belong to the base Height pyramid is the distance from its top to the plane of the base. All lateral edges of a regular pyramid are equal to each other, all lateral faces are equal isosceles triangles. The height of the side face of a regular pyramid drawn from the vertex is called apothem . Diagonal section is called a section of a pyramid by a plane passing through two lateral edges that do not belong to the same face.

Lateral surface area pyramid is the sum of the areas of all lateral faces. Total surface area is called the sum of the areas of all the side faces and the base.

Theorems

1. If in a pyramid all the lateral edges are equally inclined to the plane of the base, then the top of the pyramid is projected into the center of the circle circumscribed near the base.

2. If all the side edges of a pyramid have equal lengths, then the top of the pyramid is projected into the center of a circle circumscribed near the base.

3. If all the faces in a pyramid are equally inclined to the plane of the base, then the top of the pyramid is projected into the center of a circle inscribed in the base.

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

Where V- volume;

S base– base area;

H– height of the pyramid.

For a regular pyramid, the following formulas are correct:

Where p– base perimeter;

h a– apothem;

H- height;

S full

S side

S base– base area;

V– volume of a regular pyramid.

Truncated pyramid called the part of the pyramid enclosed between the base and a cutting plane parallel to the base of the pyramid (Fig. 17). Regular truncated pyramid called the part of a regular pyramid enclosed between the base and a cutting plane parallel to the base of the pyramid.

Reasons truncated pyramid - similar polygons. Side faces – trapezoids. Height of a truncated pyramid is the distance between its bases. Diagonal a truncated pyramid is a segment connecting its vertices that do not lie on the same face. Diagonal section is a section of a truncated pyramid by a plane passing through two lateral edges that do not belong to the same face.


For a truncated pyramid the following formulas are valid:

(4)

Where S 1 , S 2 – areas of the upper and lower bases;

S full– total surface area;

S side– lateral surface area;

H- height;

V– volume of a truncated pyramid.

For a regular truncated pyramid the formula is correct:

Where p 1 , p 2 – perimeters of the bases;

h a– apothem of a regular truncated pyramid.

Example 1. In a regular triangular pyramid, the dihedral angle at the base is 60º. Find the tangent of the angle of inclination of the side edge to the plane of the base.

Solution. Let's make a drawing (Fig. 18).


The pyramid is correct, meaning at the base equilateral triangle and all side faces are equal isosceles triangles. The dihedral angle at the base is the angle of inclination of the side face of the pyramid to the plane of the base. The linear angle is the angle a between two perpendiculars: etc. The top of the pyramid is projected at the center of the triangle (the center of the circumcircle and inscribed circle of the triangle ABC). The angle of inclination of the side edge (for example S.B.) is the angle between the edge itself and its projection onto the plane of the base. For the rib S.B. this angle will be the angle SBD. To find the tangent you need to know the legs SO And O.B.. Let the length of the segment BD equals 3 A. Dot ABOUT line segment BD is divided into parts: and From we find SO: From we find:

Answer:

Example 2. Find the volume of a regular truncated quadrangular pyramid if the diagonals of its bases are equal to cm and cm, and its height is 4 cm.

Solution. To find the volume of a truncated pyramid, we use formula (4). To find the area of ​​the bases, you need to find the sides of the base squares, knowing their diagonals. The sides of the bases are equal to 2 cm and 8 cm, respectively. This means the areas of the bases and Substituting all the data into the formula, we calculate the volume of the truncated pyramid:

Answer: 112 cm 3.

Example 3. Find the area of ​​the lateral face of a regular triangular truncated pyramid, the sides of the bases of which are 10 cm and 4 cm, and the height of the pyramid is 2 cm.

Solution. Let's make a drawing (Fig. 19).


The side face of this pyramid is an isosceles trapezoid. To calculate the area of ​​a trapezoid, you need to know the base and height. The bases are given according to the condition, only the height remains unknown. We'll find her from where A 1 E perpendicular from a point A 1 on the plane of the lower base, A 1 D– perpendicular from A 1 per AC. A 1 E= 2 cm, since this is the height of the pyramid. To find DE Let's make an additional drawing showing the top view (Fig. 20). Dot ABOUT– projection of the centers of the upper and lower bases. since (see Fig. 20) and On the other hand OK– radius inscribed in the circle and OM– radius inscribed in a circle:

MK = DE.

According to the Pythagorean theorem from

Side face area:


Answer:

Example 4. At the base of the pyramid lies an isosceles trapezoid, the bases of which A And b (a> b). Each side face forms an angle equal to the plane of the base of the pyramid j. Find the total surface area of ​​the pyramid.

Solution. Let's make a drawing (Fig. 21). Total surface area of ​​the pyramid SABCD equal to the sum of the areas and the area of ​​the trapezoid ABCD.

Let us use the statement that if all the faces of the pyramid are equally inclined to the plane of the base, then the vertex is projected into the center of the circle inscribed in the base. Dot ABOUT– vertex projection S at the base of the pyramid. Triangle SOD is the orthogonal projection of the triangle CSD to the plane of the base. Using the theorem on the area of ​​the orthogonal projection of a plane figure, we obtain:


Likewise it means Thus, the problem was reduced to finding the area of ​​the trapezoid ABCD. Let's draw a trapezoid ABCD separately (Fig. 22). Dot ABOUT– the center of a circle inscribed in a trapezoid.


Since a circle can be inscribed in a trapezoid, then or From the Pythagorean theorem we have

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 the 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:

  • its base is a regular polygon;
  • the segment connecting the top of the pyramid to 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 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 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 right triangle ROM.

Now we can find the lateral surface of the pyramid:

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 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.

Bibliography

  1. Geometry. Grades 10-11: textbook for students educational institutions(basic and profile levels) / I. M. Smirnova, V. A. Smirnov. - 5th ed., rev. and additional - M.: Mnemosyne, 2008. - 288 p.: ill.
  2. Geometry. 10-11 grade: Textbook for general education educational institutions/ Sharygin I.F. - M.: Bustard, 1999. - 208 p.: ill.
  3. 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.
  1. Internet portal "Yaklass" ()
  2. Internet portal "Festival pedagogical ideas"First of September" ()
  3. Internet portal “Slideshare.net” ()

Homework

  1. Can a regular polygon be the base of an irregular pyramid?
  2. Prove that disjoint edges of a regular pyramid are perpendicular.
  3. 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.
  4. RAVS- regular triangular pyramid. Construct the linear angle of the dihedral angle at the base of the pyramid.
  • apothem- the height of the side face of a regular pyramid, which is drawn from its vertex (in addition, the apothem is the length of the perpendicular, which is lowered from the middle of the regular polygon to one of its sides);
  • side faces (ASB, BSC, CSD, DSA) - triangles that meet at the vertex;
  • lateral ribs ( AS , B.S. , C.S. , D.S. ) — common aspects side edges;
  • top of the pyramid (t. S) - a point that connects the side ribs and which does not lie in the plane of the base;
  • height ( SO ) - a perpendicular segment drawn through the top of the pyramid to the plane of its base (the ends of such a segment will be the top of the pyramid and the base of the perpendicular);
  • diagonal section of the pyramid- a section of the pyramid that passes through the top and the diagonal of the base;
  • base (ABCD) - a polygon that does not belong to the vertex of the pyramid.

Properties of the pyramid.

1. When all the side edges have the same size, then:

  • it is easy to describe a circle near the base of the pyramid, and the top of the pyramid will be projected into the center of this circle;
  • the side ribs form equal angles with the plane of the base;
  • Moreover, the opposite is also true, i.e. when the side ribs form equal angles with the plane of the base, or when a circle can be described around the base of the pyramid and the top of the pyramid will be projected into the center of this circle, it means that all the side edges of the pyramid are the same size.

2. When the side faces have an angle of inclination to the plane of the base of the same value, then:

  • it is easy to describe a circle near the base of the pyramid, and the top of the pyramid will be projected into the center of this circle;
  • the heights of the side faces are of equal length;
  • the area of ​​the side surface is equal to ½ the product of the perimeter of the base and the height of the side face.

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 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 both along the leg and acute angle. 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 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. 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.

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