Formula for the height of a triangular pyramid. Basics of geometry: a regular pyramid is

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

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 a 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 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

1. 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.
2. Geometry. Grades 10-11: Textbook for general education institutions / Sharygin I.F. - M.: Bustard, 1999. - 208 pp.: 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 of 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.

Video tutorial 2: Pyramid problem. Volume of the pyramid

Video tutorial 3: Pyramid problem. Correct pyramid

Lecture: Pyramid, its base, lateral ribs, height, lateral surface; triangular pyramid; regular pyramid

Pyramid, its properties

Pyramid is a three-dimensional body that has a polygon at its base, and all its faces consist of triangles.

A special case of a pyramid is a cone with a circle at its base.

Let's look at the main elements of the pyramid:

Apothem- this is a segment that connects the top of the pyramid with the middle of the lower edge of the side face. In other words, this is the height of the edge of the pyramid.

In the figure you can see triangles ADS, ABS, BCS, CDS. If you look closely at the names, you can see that each triangle has one common letter in its name - S. That is, this means that all the side faces (triangles) converge at one point, which is called the top of the pyramid.

The segment OS that connects the vertex with the point of intersection of the diagonals of the base (in the case of triangles - at the point of intersection of the heights) is called pyramid height.

A diagonal section is a plane that passes through the top of the pyramid, as well as one of the diagonals of the base.

Since the side surface of the pyramid consists of triangles, to find the total area of ​​the side surface it is necessary to find the area of ​​each face and add them up. The number and shape of faces depends on the shape and size of the sides of the polygon that lies at the base.

The only plane in a pyramid that does not belong to its vertex is called basis pyramids.

In the figure we see that the base is a parallelogram, however, it can be any arbitrary polygon.

Properties:

Consider the first case of a pyramid, in which it has edges of the same length:

• A circle can be drawn around the base of such a pyramid. If you project the top of such a pyramid, then its projection will be located in the center of the circle.
• The angles at the base of the pyramid are the same on each face.
• In this case, a sufficient condition for the fact that a circle can be described around the base of the pyramid, and also that all the edges are of different lengths, can be considered the same angles between the base and each edge of the faces.

If you come across a pyramid in which the angles between the side faces and the base are equal, then the following properties are true:

• You will be able to describe a circle around the base of the pyramid, the apex of which is projected exactly at the center.
• If you draw each side edge of the height to the base, then they will be of equal length.
• To find the lateral surface area of ​​such a pyramid, it is enough to find the perimeter of the base and multiply it by half the length of the height.
• S bp = 0.5P oc H.
• Types of pyramid.
• Depending on which polygon lies at the base of the pyramid, they can be triangular, quadrangular, etc. If at the base of the pyramid there is a regular polygon (with equal sides), then such a pyramid will be called regular.

Regular triangular pyramid

We continue to consider the tasks included in the Unified State Examination in mathematics. We have already studied problems where the condition is given and it is required to find the distance between two given points or an angle.

A pyramid is a polyhedron, the base of which is a polygon, the remaining faces are triangles, and they have a common vertex.

A regular pyramid is a pyramid at the base of which lies a regular polygon, and its vertex is projected into the center of the base.

A regular quadrangular pyramid - the base is a square. The top of the pyramid is projected at the point of intersection of the diagonals of the base (square).

ML - apothem
∠MLO - dihedral angle at the base of the pyramid
∠MCO - angle between the lateral edge and the plane of the base of the pyramid

In this article we will look at problems to solve a regular pyramid. You need to find some element, lateral surface area, volume, height. Of course, you need to know the Pythagorean theorem, the formula for the area of ​​the lateral surface of a pyramid, and the formula for finding the volume of a pyramid.

In the article "" presents the formulas that are necessary to solve problems in stereometry. So, the tasks:

SABCD dot O- center of the base,S vertex, SO = 51, A.C.= 136. Find the side edgeS.C..

In this case, the base is a square. This means that the diagonals AC and BD are equal, they intersect and are bisected by the intersection point. Note that in a regular pyramid the height dropped from its top passes through the center of the base of the pyramid. So SO is the height and the triangleSOCrectangular. Then according to the Pythagorean theorem:

How to extract the root of a large number.

Answer: 85

Decide for yourself:

In a regular quadrangular pyramid SABCD dot O- center of the base, S vertex, SO = 4, A.C.= 6. Find the side edge S.C..

In a regular quadrangular pyramid SABCD dot O- center of the base, S vertex, S.C. = 5, A.C.= 6. Find the length of the segment SO.

In a regular quadrangular pyramid SABCD dot O- center of the base, S vertex, SO = 4, S.C.= 5. Find the length of the segment A.C..

SABC R- middle of the rib B.C., S- top. It is known that AB= 7, a S.R.= 16. Find the lateral surface area.

The area of ​​the lateral surface of a regular triangular pyramid is equal to half the product of the perimeter of the base and the apothem (apothem is the height of the lateral face of a regular pyramid drawn from its vertex):

Or we can say this: the area of ​​the lateral surface of the pyramid is equal to the sum of the areas of the three lateral faces. The lateral faces in a regular triangular pyramid are triangles of equal area. In this case:

Answer: 168

Decide for yourself:

In a regular triangular pyramid SABC R- middle of the rib B.C., S- top. It is known that AB= 1, a S.R.= 2. Find the lateral surface area.

In a regular triangular pyramid SABC R- middle of the rib B.C., S- top. It is known that AB= 1, and the area of ​​the lateral surface is 3. Find the length of the segment S.R..

In a regular triangular pyramid SABC L- middle of the rib B.C., S- top. It is known that SL= 2, and the area of ​​the lateral surface is 3. Find the length of the segment AB.

In a regular triangular pyramid SABC M. Area of ​​a triangle ABC is 25, the volume of the pyramid is 100. Find the length of the segment MS.

The base of the pyramid is an equilateral triangle. That's why Mis the center of the base, andMS- height of a regular pyramidSABC. Volume of the pyramid SABC equals: view solution

In a regular triangular pyramid SABC the medians of the base intersect at the point M. Area of ​​a triangle ABC equals 3, MS= 1. Find the volume of the pyramid.

In a regular triangular pyramid SABC the medians of the base intersect at the point M. The volume of the pyramid is 1, MS= 1. Find the area of ​​the triangle ABC.

Let's finish here. As you can see, problems are solved in one or two steps. In the future, we will consider other problems from this part, where bodies of revolution are given, don’t miss it!

I wish you success!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

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.

A pyramid is a polyhedron with a polygon at its base. All faces, in turn, form triangles that converge at one vertex. Pyramids are triangular, quadrangular, and so on. In order to determine which pyramid is in front of you, it is enough to count the number of angles at its base. The definition of “height of a pyramid” is very often found in geometry problems in the school curriculum. In this article we will try to look at different ways to find it.

Parts of the pyramid

Each pyramid consists of the following elements:

• side faces, which have three corners and converge at the apex;
• the apothem represents the height that descends from its apex;
• the top of the pyramid is a point that connects the side ribs, but does not lie in the plane of the base;
• the base is a polygon on which the vertex does not lie;
• the height of a pyramid is a segment that intersects the top of the pyramid and forms a right angle with its base.

How to find the height of a pyramid if its volume is known

Through the formula V = (S*h)/3 (in the formula V is the volume, S is the area of ​​the base, h is the height of the pyramid) we find that h = (3*V)/S. To consolidate the material, let's immediately solve the problem. The triangular base is 50 cm 2 , whereas its volume is 125 cm 3 . The height of the triangular pyramid is unknown, which is what we need to find. Everything is simple here: we insert the data into our formula. We get h = (3*125)/50 = 7.5 cm.

How to find the height of a pyramid if the length of the diagonal and its edges are known

As we remember, the height of the pyramid forms a right angle with its base. This means that the height, edge and half of the diagonal together form Many, of course, remember the Pythagorean theorem. Knowing two dimensions, it will not be difficult to find the third quantity. Let us recall the well-known theorem a² = b² + c², where a is the hypotenuse, and in our case the edge of the pyramid; b - the first leg or half of the diagonal and c - respectively, the second leg, or the height of the pyramid. From this formula c² = a² - b².

Now the problem: in a regular pyramid the diagonal is 20 cm, when the length of the edge is 30 cm. You need to find the height. We solve: c² = 30² - 20² = 900-400 = 500. Hence c = √ 500 = about 22.4.

How to find the height of a truncated pyramid

It is a polygon with a cross section parallel to its base. The height of a truncated pyramid is the segment that connects its two bases. The height can be found for a regular pyramid if the lengths of the diagonals of both bases, as well as the edge of the pyramid, are known. Let the diagonal of the larger base be d1, while the diagonal of the smaller base is d2, and the edge has length l. To find the height, you can lower the heights from the two upper opposite points of the diagram to its base. We see that we have two right triangles; all that remains is to find the lengths of their legs. To do this, subtract the smaller one from the larger diagonal and divide by 2. So we will find one leg: a = (d1-d2)/2. After which, according to the Pythagorean theorem, all we have to do is find the second leg, which is the height of the pyramid.

Now let's look at this whole thing in practice. We have a task ahead of us. A truncated pyramid has a square at the base, the diagonal length of the larger base is 10 cm, while the smaller one is 6 cm, and the edge is 4 cm. You need to find the height. First, we find one leg: a = (10-6)/2 = 2 cm. One leg is equal to 2 cm, and the hypotenuse is 4 cm. It turns out that the second leg or height will be equal to 16-4 = 12, that is, h = √12 = about 3.5 cm.