Pyramid concept
Definition 1
A 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.
Types of pyramids
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.
Sample task
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.
Using the midline theorem, 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
When preparing for the Unified State Exam in mathematics, students have to systematize their knowledge of algebra and geometry. I would like to combine all known information, for example, on how to calculate the area of a pyramid. Moreover, starting from the base and side edges to the entire surface area. If the situation with the side faces is clear, since they are triangles, then the base is always different.
How to find the area of the base of the pyramid?
It can be absolutely any figure: from an arbitrary triangle to an n-gon. And this base, in addition to the difference in the number of angles, can be a regular figure or an irregular one. In the Unified State Exam tasks that interest schoolchildren, there are only tasks with correct figures at the base. Therefore, we will talk only about them.
Regular triangle
That is, equilateral. The one in which all sides are equal and are designated by the letter “a”. In this case, the area of the base of the pyramid is calculated by the formula:
S = (a 2 * √3) / 4.
Square
The formula for calculating its area is the simplest, here “a” is again the side:
Arbitrary regular n-gon
The side of a polygon has the same notation. For the number of angles, the Latin letter n is used.
S = (n * a 2) / (4 * tg (180º/n)).
What to do when calculating the lateral and total surface area?
Since the base is a regular figure, all faces of the pyramid are equal. Moreover, each of them is an isosceles triangle, since the side edges are equal. Then, in order to calculate the lateral area of the pyramid, you will need a formula consisting of the sum of identical monomials. The number of terms is determined by the number of sides of the base.
The area of an isosceles triangle is calculated by the formula in which half the product of the base is multiplied by the height. This height in the pyramid is called apothem. Its designation is “A”. The general formula for lateral surface area is:
S = ½ P*A, where P is the perimeter of the base of the pyramid.
There are situations when the sides of the base are not known, but the side edges (c) and the flat angle at its apex (α) are given. Then you need to use the following formula to calculate the lateral area of the pyramid:
S = n/2 * in 2 sin α .
Task No. 1
Condition. Find the total area of the pyramid if its base has a side of 4 cm and the apothem has a value of √3 cm.
Solution. You need to start by calculating the perimeter of the base. Since this is a regular triangle, then P = 3*4 = 12 cm. Since the apothem is known, we can immediately calculate the area of the entire lateral surface: ½*12*√3 = 6√3 cm 2.
For the triangle at the base, you get the following area value: (4 2 *√3) / 4 = 4√3 cm 2.
To determine the entire area, you will need to add the two resulting values: 6√3 + 4√3 = 10√3 cm 2.
Answer. 10√3 cm 2.
Problem No. 2
Condition. There is a regular quadrangular pyramid. The length of the base side is 7 mm, the side edge is 16 mm. It is necessary to find out its surface area.
Solution. Since the polyhedron is quadrangular and regular, its base is a square. Once you know the area of the base and side faces, you will be able to calculate the area of the pyramid. The formula for the square is given above. And for the side faces, all sides of the triangle are known. Therefore, you can use Heron's formula to calculate their areas.
The first calculations are simple and lead to the following number: 49 mm 2. For the second value, you will need to calculate the semi-perimeter: (7 + 16*2): 2 = 19.5 mm. Now you can calculate the area of an isosceles triangle: √(19.5*(19.5-7)*(19.5-16) 2) = √2985.9375 = 54.644 mm 2. There are only four such triangles, so when calculating the final number you will need to multiply it by 4.
It turns out: 49 + 4 * 54.644 = 267.576 mm 2.
Answer. The desired value is 267.576 mm 2.
Task No. 3
Condition. For a regular quadrangular pyramid, you need to calculate the area. The side of the square is known to be 6 cm and the height is 4 cm.
Solution. The easiest way is to use the formula with the product of perimeter and apothem. The first value is easy to find. The second one is a little more complicated.
We will have to remember the Pythagorean theorem and consider It is formed by the height of the pyramid and the apothem, which is the hypotenuse. The second leg is equal to half the side of the square, since the height of the polyhedron falls at its middle.
The required apothem (hypotenuse of a right triangle) is equal to √(3 2 + 4 2) = 5 (cm).
Now you can calculate the required value: ½*(4*6)*5+6 2 = 96 (cm 2).
Answer. 96 cm 2.
Problem No. 4
Condition. The correct side is given. The sides of its base are 22 mm, the side edges are 61 mm. What is the lateral surface area of this polyhedron?
Solution. The reasoning in it is the same as that described in task No. 2. Only there was given a pyramid with a square at the base, and now it is a hexagon.
First of all, the base area is calculated using the above formula: (6*22 2) / (4*tg (180º/6)) = 726/(tg30º) = 726√3 cm 2.
Now you need to find out the semi-perimeter of an isosceles triangle, which is the side face. (22+61*2):2 = 72 cm. All that remains is to use Heron’s formula to calculate the area of each such triangle, and then multiply it by six and add it to the one obtained for the base.
Calculations using Heron's formula: √(72*(72-22)*(72-61) 2)=√435600=660 cm 2. Calculations that will give the lateral surface area: 660 * 6 = 3960 cm 2. It remains to add them up to find out the entire surface: 5217.47≈5217 cm 2.
Answer. The base is 726√3 cm 2, the side surface is 3960 cm 2, the entire area is 5217 cm 2.
Let's consider what properties pyramids have, in which the side faces are perpendicular to the base.
If two adjacent lateral faces of the pyramid are perpendicular to the base, That the common lateral edge of these faces is the height of the pyramid. If the problem says that the edge of a pyramid is its height, then we are talking about this type of pyramid.
The faces of the pyramid perpendicular to the base are right triangles.
If the base of the pyramid is a triangle
In the general case, we look for the lateral surface of such a pyramid as the sum of the areas of all lateral faces.
The base of the pyramid is the orthogonal projection of the face, not perpendicular to the base (in this case, SBC). This means, according to the theorem on the area of orthogonal projection, the area of the base is equal to the product of the area of this face and the cosine of the angle between it and the plane of the base.
If the base of the pyramid is a right triangle
In this case all faces of the pyramid are right triangles.
Triangles SAB and SAC are rectangular, since SA is the height of the pyramid. Triangle ABC is right angled.
The fact that the triangle SBC is right-angled follows from the theorem of three perpendiculars (AB is the projection of the inclined SB onto the plane of the base. Since AB is perpendicular to BC by condition, then SB is perpendicular to BC).
The angle between the side face of SBC and the base in this case is angle ABS.
The lateral surface area is equal to the sum of the areas of right triangles:
Since in this case
If the base of the pyramid is an isosceles triangle
In this case, the angle between the side plane BCS and the base plane is angle AFS, where AF is the altitude, median and bisector of isosceles triangle ABC.
Similarly, if at the base of the pyramid there is an equilateral triangle ABC.
If the base of the pyramid is a parallelogram
In this case, the base of the pyramid is an orthogonal projection of the side faces that are not perpendicular to the base.
If we divide the base into two triangles, then
where α and β are the angles between the ADS and CDS planes and the base plane, respectively.
If BF and BK are the heights of the parallelogram, then angle BFS is the angle of inclination of the side face CDS to the plane of the base, and angle BKS is the angle of inclination of the side ADS.
(the drawing was made for the case when B is an obtuse angle).
If the base of the pyramid is a rhombus ABCD, then the angles BFS and BKS are equal. Triangles ABS and CBS, as well as ADS and CDS, are also equal in this case.
If the base of the pyramid is a rectangle
In this case, the angle between the side face plane SAD and the base plane is angle SAB,
and the angle between the plane of the side face SCD and the plane of the base is angle SCB
(by the theorem of three perpendiculars).
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 complete 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 side 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 lateral surface area 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 theorem 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
- 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. Grades 10-11: Textbook for general education institutions / Sharygin I.F. - M.: Bustard, 1999. - 208 pp.: 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.
Back forward
Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version.
Goals:
- prove the properties of a pyramid with equal edges;
- develop the ability to use this theorem when analyzing the conditions of the problem and constructing a drawing for the problem;
- to develop in students the ability to use this theorem when solving two step problems.
I. Each student receives homework on pre-printed pieces of paper.
Theory: according to the textbook, section 14.2, pp. 110-111, 2) and 3 tasks:
1. In a regular triangular pyramid, the height of the base is h, the side edges are inclined to the plane of the base at an angle?. Find the height of the pyramid.
2. At the base of the pyramid lies a triangle with sides , ,4. The side ribs are inclined to the plane of the base at an angle of 45 0. Find the height of the pyramid.
3. The area of the base of a regular quadrangular pyramid is equal to S. The lateral edges are inclined to the plane of the base at an angle?. Find the height of the pyramid.
II. Oral work based on finished drawings.(Each child receives sheet A-4 with drawings of a triangular pyramid).
2.1. Let us prove 3 (direct) theorems. Given: MABC is a triangular pyramid, MO is the height of the pyramid.
1. Students prove a “simple” theorem from one condition and one conclusion
2. Use the sign of equality of right triangles along the leg and hypotenuse
3. They conclude: from the fact that AO = BO = CO, it follows that O is the center of the circle described around the base.
4. The teacher clarifies the wording of this circumstance: “the base of the pyramid coincides with the center of the circle described around the base” or “the top of the pyramid is projected into the center of the circle described around the base.
(to Fig. 2,3). Replace the conditions of the theorem and preserve its conclusion. Based on the signs of equality of right triangles, students come to the conclusion that they can require equality of angles between the side edges and the plane of the base or equality of angles between the side edges and the height of the pyramid.
So, from what conditions can we conclude that the base of the height of the pyramid coincides with the center of the circle described around the base?
2.2. Let us formulate the converse statements. Are these statements true?
Students, using the signs of equality of right triangles, prove converse statements. Given: MABC is a triangular pyramid, MO is the height of the pyramid, O is the center of the circle circumscribed around the base, AO=BO=CO.
2.3. Statement of the theorem for an n-gonal pyramid.
Statement of the problem: is this statement true for an n-gonal pyramid? Students are asked to prove three direct statements by analogy.
Theorem. In an n-gonal pyramid with equal lateral edges, the base of the height coincides with the center of the circle circumscribed around the base; the height is equal angles with the side ribs; the side ribs make equal angles with the plane of the base.
Figure 7.
2.4. Work after the proof of the theorem (look back).
| A – The lateral edges of the pyramid are equal B – The lateral edges of the pyramid make equal angles with the plane of the base C – The lateral edges of the pyramid make equal angles with the height of the pyramid M - The base of the pyramid coincides with the center of the circle described around the base |
Given all 6 simple theorems, students are led to the conclusion 2. The teacher shows statements A(B, C, M), the student formulates 3 simple theorems. |
III. Formulation of the lesson topic.(Properties of a pyramid with equal lateral edges).
What is the topic of today's lesson? (Any of statements A, B, C, M can be taken as the topic of the lesson).
IV. Drawing up an algorithm
| Given: triangular pyramid MAVS, MO – height of the pyramid. Determine the height of the pyramid. Algorithm for solving two step problems. 1. The presence in the problem statement of one of the conditions (A, B, C,). From these conditions follows M. 2. Solve the base (find the radius of the circle described around the base). 3. Solve a right triangle, for example MOA. |
1. Drawing up an algorithm. 2. Updating knowledge: a) the center of the circle described near the base - the point of intersection of the bisector perpendiculars to the sides of the triangle; b) the location of the center of the circumscribed circle in acute, obtuse, and right triangles; c) formula S = . |
V. Application of the properties of a pyramid with equal lateral edges to solving problems.
| Task 1. At the base of the pyramid lies an isosceles right triangle with a leg equal to 2. The lateral edges are inclined to the plane of the base at an angle of 60 0. Find the height of the pyramid.
Figure 8 |
1.Each student receives a sheet with the conditions of the problems to solve 2. We do not do stereometric drawings. Presence of condition “B” We make a drawing of the base. O - middle of the hypotenuse, AB = 4, R = 2 We construct triangle AMO, find MO = 6 Answer: 6 |
| Task 2. The base of the pyramid is a triangle, two sides of which are 2 and form an angle of 45 0. Each side edge is equal to .
Find the height of the pyramid. |
Figure 9 Solution. We work according to the algorithm: 1. Presence of condition “A”. 2. We make a drawing of the base. Using the cosine theorem, we find the third side (), which means the triangle is isosceles and right-angled. O - the middle of the hypotenuse. Hypotenuse is 2, R = 1 |
| 3. Construct triangle AMO, find MO = 3 Answer: 3
Problem 3 At the base of the pyramid lies a triangle with sides 5, 12, 13. The angle between the height and each side edge is 45 0. Find the height of the pyramid. |
Figure 9 Figure 10 1. Presence of condition “C” 2. We make a drawing of the base. According to the theorem inverse to the Pythagorean theorem, we find out that the triangle is right-angled, O is the middle of the hypotenuse, AB = 13, R = 6.5 |
| 3. We construct an isosceles triangle AMO, find MO = 6.5 Answer: 6.5
Problem 4 The base of the pyramid is an isosceles triangle, the sides of which are equal and form an angle of 120 0. Each side edge is equal to . Find the height of the pyramid. |
Figure 11 Solution. We work according to the algorithm: 1. Presence of condition “A”. 2. We make a drawing of the base. angle A is obtuse, O - outside the triangle, AO is the perpendicular bisector to BC, triangle AOC is equilateral, AB =, |
3.Construct triangle AMO, MO = = 6 Answer: 6
VI. Summarize the lesson when solving problems:
1. At the base of the pyramid lies a trapezoid, the side ribs are equal. Determine the type of trapezoid (isosceles).
2. At the base of the pyramid lies a parallelogram, the angles between the side edges and the plane of the base are equal. Determine the type of parallelogram (rectangle).
3. At the base of the pyramid is a rhombus. The angles between the side ribs and the height of the pyramid are equal. Find the angles of the rhombus. (90 o).