A pyramid with a right triangle at the base. Pyramid Surface area and volume of a truncated pyramid

A three-dimensional figure that often appears in geometric problems is the pyramid. The simplest of all the shapes in this class is triangular. In this article we will analyze in detail the basic formulas and properties of a regular triangular pyramid.

Geometric ideas about the figure

Before moving on to considering the properties of a regular triangular pyramid, let’s take a closer look at what kind of figure we are talking about.

Let's assume that there is an arbitrary triangle in three-dimensional space. Let us select any point in this space that does not lie in the plane of the triangle and connect it with the three vertices of the triangle. We got a triangular pyramid.

It consists of 4 sides, all of which are triangles. The points where three faces meet are called vertices. The figure also has four of them. The lines of intersection of two faces are edges. The pyramid in question has 6 edges. The figure below shows an example of this figure.

Since the figure is formed by four sides, it is also called a tetrahedron.

Correct pyramid

Above we considered an arbitrary figure with a triangular base. Now suppose that we draw a perpendicular segment from the top of the pyramid to its base. This segment is called height. Obviously, you can draw 4 different heights for the figure. If the height intersects the triangular base at the geometric center, then such a pyramid is called straight.

A straight pyramid, the base of which is an equilateral triangle, is called regular. For her, all three triangles forming the lateral surface of the figure are isosceles and equal to each other. A special case of a regular pyramid is the situation when all four sides are equilateral identical triangles.

Let's consider the properties of a regular triangular pyramid and give the corresponding formulas for calculating its parameters.

Base side, height, lateral edge and apothem

Any two of the listed parameters uniquely determine the remaining two characteristics. Let us present formulas that relate these quantities.

Let us assume that the side of the base of a regular triangular pyramid is a. The length of its lateral edge is b. What will be the height of a regular triangular pyramid and its apothem?

For height h we get the expression:

h = √(b 2 - a 2 /3)

This formula follows from the Pythagorean theorem for a right triangle, the sides of which are the side edge, the height and 2/3 of the height of the base.

The apothem of a pyramid is the height for any side triangle. The length of the apothem a b is equal to:

a b = √(b 2 - a 2 /4)

From these formulas it is clear that whatever the side of the base of a triangular regular pyramid and the length of its side edge, the apothem will always be greater than the height of the pyramid.

The two formulas presented contain all four linear characteristics of the figure in question. Therefore, given the known two of them, you can find the rest by solving the system of written equalities.

Figure volume

For absolutely any pyramid (including an inclined one), the value of the volume of space limited by it can be determined by knowing the height of the figure and the area of ​​its base. The corresponding formula is:

Applying this expression to the figure in question, we obtain the following formula:

Where the height of a regular triangular pyramid is h, and its base side is a.

It is not difficult to obtain a formula for the volume of a tetrahedron in which all sides are equal to each other and represent equilateral triangles. In this case, the volume of the figure is determined by the formula:

That is, it is determined uniquely by the length of side a.

Surface area

Let us continue to consider the properties of a regular triangular pyramid. total area of all the faces of a figure is called its surface area. The latter can be conveniently studied by considering the corresponding development. The figure below shows what the development of a regular triangular pyramid looks like.

Let's assume that we know the height h and the side of the base a of the figure. Then the area of ​​its base will be equal to:

Every schoolchild can obtain this expression if he remembers how to find the area of ​​a triangle and also takes into account that the altitude of an equilateral triangle is also a bisector and a median.

The lateral surface area formed by three identical isosceles triangles is:

S b = 3/2*√(a 2 /12+h 2)*a

This equality follows from the expression of the apothem of the pyramid in terms of the height and length of the base.

The total surface area of ​​the figure is:

S = S o + S b = √3/4*a 2 + 3/2*√(a 2 /12+h 2)*a

Note that for a tetrahedron in which all four sides are identical equilateral triangles, the area S will be equal to:

Properties of a regular truncated triangular pyramid

If the top of the considered triangular pyramid is cut off with a plane parallel to the base, then the remaining lower part will be called a truncated pyramid.

In the case of a regular pyramid with a triangular base, the result of the described sectioning method is a new triangle, which is also equilateral, but has a shorter side length than the side of the base. A truncated triangular pyramid is shown below.

We see that this figure is already limited by two triangular bases and three isosceles trapezoids.

Let us assume that the height of the resulting figure is equal to h, the lengths of the sides of the lower and upper bases are a 1 and a 2, respectively, and the apothem (height of the trapezoid) is equal to a b. Then the surface area of ​​the truncated pyramid can be calculated using the formula:

S = 3/2*(a 1 +a 2)*a b + √3/4*(a 1 2 + a 2 2)

Here the first term is the area of ​​the lateral surface, the second term is the area of ​​the triangular bases.

The volume of the figure is calculated as follows:

V = √3/12*h*(a 1 2 + a 2 2 + a 1 *a 2)

To unambiguously determine the characteristics of a truncated pyramid, it is necessary to know its three parameters, as demonstrated by the given formulas.

Formulas and properties of a regular triangular pyramid. Truncated triangular pyramid - all interesting facts and achievements of science and education on the site

S floor = S base + S side.

Stage III: Virtual trip into the world of pyramids - student presentation

Stage IV - Study of a new topic - accompanied by a multimedia presentation

Concepts studied:

Write down the topic.

House. task: No. 70.

Stage VI Reflection.

Control questions

7. What is the height of the pyramid?

House task: No. 75

Grading

Annex 1

Appendix 2

The magical properties of the pyramids

Another way to achieve the effect is to put pure spring water in the pyramid, let it sit for 24 hours, and then rub it into the scalp before going to bed. It takes longer, but is more practical.

For example, if you breed fish in a glass pyramid-aquarium, the result is amazing: the water purifies itself! There are no signs of rotting, no mud deposits on the bottom, the glass does not turn green and there is no need to spend money on buying aquarium filters - the pyramid cleans everything itself. The geometry of the pyramid structures water molecules in a special way, setting a program to suppress rotting inside the aquarium.

The pyramid is a radiation damper. If you place it on a computer and correctly orient it to the cardinal points, the pyramid will create a more beneficial field. How larger pyramid, the greater her goodness factor. All negative impacts will either be extinguished or redistributed into something neutral.

View document contents
"Frustum Pyramids"

Lesson topic: Truncated pyramid, its main elements.

Lesson objectives:

Educational: familiarize students with the concept of a truncated pyramid, its elements and formulas for calculating the areas of the lateral and total surfaces;

Educational: develop students’ spatial imagination, the ability to depict pyramids and recognize them among other spatial figures;

Educators: This topic contributes to the development of curiosity, intelligence, attentiveness and the development of interest in mathematics, the formation of accuracy in the construction of mathematical figures.

Lesson type: familiarization with new material

TOOL of the lesson: interactive whiteboard, computer, presentations “Pyramids”, “Truncated Pyramids”, “Virtual Journey to the World of Pyramids”.

Lesson steps:

Stage I: organizational

Stage II Knowledge updating

1) oral survey using slides

List of questions:

(slide 2) - among the figures depicted, name the numbers of those that are pyramids.

Among the models, also select pyramids.

Which polyhedron is called a pyramid? Name and show their main elements, show them on models. (Slides 3,4)

Types of pyramids. (slides 5-7)

Make a drawing of a triangular and quadrangular pyramid.

What does the full surface of the pyramid consist of? (slide 8)

Properties of the lateral edges and lateral faces of a regular pyramid. (Slide 9)

Formulas for calculating the surface areas of pyramids (write on the board, check on the screen) (slide 10-11)

2) solve a problem from a textbook using ready-made drawings

S floor = S base + S side.

Stage III: Virtual trip to the world of pyramids – presentation of students

Stage IV – Study of a new topic – accompanied by a multimedia presentation

Concepts studied:

Truncated pyramid (definition);

Elements of a truncated pyramid;

Regular truncated pyramid;

The lateral surface area of ​​a truncated pyramid;

The total surface area of ​​a truncated pyramid.

Write down the topic.

Draw a random pyramid in your notebook.

Draw a plane parallel to the base.

This plane divides the pyramid into two parts. What can you say about them?

Define a truncated pyramid.

Name the main elements of a truncated pyramid.

What can you say about the side edges?

Which truncated pyramid is called regular? What can you say about its side faces?

What does the full surface of a truncated pyramid consist of?

Write a formula to calculate its total surface.

What is the side surface made of?

Name objects that have the shape of a truncated pyramid. (slide)

Stage V Solving problems - No. 71, 77 from the textbook Geometry 7-11 A.V. Pogorelov.

Solving problems in pairs. (Annex 1)

House. task: No. 70.

Stage VI Reflection.

Control questions

1. Which polyhedron is called a pyramid?

2. Which pyramid is called triangular?

3. Which pyramid is called correct?

4. What is the apothem of a regular pyramid?

5 Which pyramid is called a tetrahedron?

6. Which pyramid is called truncated?

7. What is the height of the pyramid?

8. What is the lateral surface area of ​​a regular pyramid?

9. What is the lateral surface area of ​​the truncated pyramid?

House task: No. 75

Grading

Annex 1

Solving free choice problems - pairs choose a problem and solve it.

1. The base of the pyramid is a rectangle with sides 6 cm and 8 cm. Each edge of the pyramid is 13 cm. Calculate the height of the pyramid.

2. The base of the pyramid is a right triangle with legs 6 cm and 8 cm. All dihedral angles at the base of the pyramid are 60°. Find the height of the pyramid.

3. For a quadrangular truncated pyramid, the sides of one base are 6, 7, 8, 9 cm, and the smaller side of the other base is 5 cm. Find the remaining sides of this base.

4. In a regular triangular pyramid with height h, a plane is drawn through the side of the base a, intersecting the opposite side edge at a right angle. Find the cross-sectional area.

5. The base side of a regular hexagonal pyramid is a, and the dihedral angle at the base is 45°. Find the volume of the pyramid.

6. In a regular truncated quadrangular pyramid, the sides of the lower and upper bases are equal to a and b, and the dihedral angle at the edge of the lower base is equal to a. Find the volume of the pyramid.

7. Construct a section of the pyramid with a plane passing through the top of the pyramid and two given points on its base.

8. In a regular quadrangular truncated pyramid, the height is 2 cm, and the sides of the bases are 3 cm and 5 cm. Find the diagonal of this pyramid.

Appendix 2

The magical properties of the pyramids

The term "pyramid" is borrowed from the Greek "pyramis" or "pyramidos". The Greeks, in turn, borrowed this word from the Egyptian language. Others believe that the term originates from the shape of the bread in Ancient Greece(“piros” - rye). Due to the fact that the shape of the flame resembles the image of a pyramid, some scientists believed that the term comes from the Greek word “pyre” - which means fire, and fire, as is known, is a symbol of the life of all creatures.

The pyramids can be considered one of the most mysterious on the planet.

It has now been proven that the pyramid concentrates high-quality energy that is useful for humans. It has been established that objects in the shape of a pyramid have an effect on environment positive impact.

Czech engineer Karel Duban, a specialist in radio waves, argued. that the pyramids concentrate cosmic energy, which is the “actor” in them.

He discovered a connection between the shape of the pyramid space and the biological and physicochemical processes occurring in this space.

It turned out that the energy of the pyramid shape “can do” a lot: instant coffee, after standing over the pyramid, acquires a natural taste; cheap wines significantly improve their taste; water acquires properties to promote healing, tones the body, reduces the inflammatory reaction after bites, burns and acts as a natural aid to improve digestion; meat, fish, eggs, vegetables, fruits are mummified, but do not spoil; milk does not sour for a long time; the cheese does not mold. If you sit under the pyramid, the process of meditation improves, the intensity of headaches and toothaches decreases, and the healing of wounds and ulcers accelerates. Pyramids eliminate geopathogenic influences around them and harmonize the interior space of premises. Dutch pyramid researcher Paul Likens experimented with a variety of materials: with seeds of garden crops (radish grew 2 times larger in size than the control from the same set of seeds), herbs - they remain green and continue to carry their energy charge, the healing power increases significantly.

If you place a pyramid with certain parameters in an apartment, cockroaches leave the room.

By placing a model of a pyramidal design on the head of a bald person and orienting it to the cardinal points, the effect of stimulating the hair follicles is achieved. The harmonious radiation generated by the pyramid model penetrates sufficiently into the skin structure and contributes to the effect of a gentle massage of the hair follicles.

Another way to achieve the effect is to put pure spring water in the pyramid, let it sit for 24 hours, and then rub it into the scalp before going to bed. It takes longer, but is more practical.

The use of this method is relevant in conditions of increased radiation, when many children lose their hair. This is a drug-free method that does not require large financial costs, easy to use.

According to a number of testers, ordinary water perfectly captures the energy of the pyramids and exhibits new properties: it acquires the taste of pure spring water, has a healing effect, stimulates plant growth, it is also known that the use of such water is effective for strengthening hair, removing dandruff, softening the skin and smoothing out wrinkles, getting rid of from sweaty feet, etc.

For example, if you breed fish in a glass pyramid-aquarium, the result is amazing: the water purifies itself! There are no signs of rotting, no mud deposits on the bottom, the glass does not turn green and there is no need to spend money on buying aquarium filters - the pyramid cleans everything itself. The geometry of the pyramid structures water molecules in a special way, setting a program to suppress rotting inside the aquarium.

Another example. FAMOUS GENETICIST GENNADY BERDISHEV says: “THE MEAT IN MY PYRAMID CAN EVEN LIE WITHOUT A REFRIGERATOR FOR A WHOLE WEEK IN THE HEAT!”

Having built a pyramid at his dacha, a famous scientist says that in it he loses years.

The pyramid is a radiation absorber. If you place it on a computer and correctly orient it to the cardinal points, the pyramid will create a more beneficial field. The larger the pyramid, the greater its goodness factor. All negative impacts will either be extinguished or redistributed into something neutral.

And many such examples can be given.

The pyramid, provided that it is oriented with the edges of the base towards the cardinal points, turns into an accumulator of cosmic energy. Therefore, in recent years, all sorts of souvenirs in the shape of pyramids have been in fashion: it is believed that they cleanse space and emit positive energy.

Task

Solution.

At the base of the pyramid lies a right triangle. The center of the circumscribed circle of a right triangle lies on its hypotenuse. Accordingly, AB = 10 cm, AO = 5 cm.

Since the height ON = 12 cm, the size of the ribs AN and NB is equal
AN 2 = AO 2 + ON 2
AN 2 = 5 2 + 12 2
AN = √169
AN=13

Since we know the value AO = OB = 5 cm and the size of one of the legs of the base (8 cm), then the height lowered to the hypotenuse will be equal to
CB 2 = CO 2 + OB 2
64 = CO 2 + 25
CO2 = 39
CO = √39

Accordingly, the size of the edge CN will be equal to
CN 2 = CO 2 + NO 2
CN 2 = 39 + 144
CN = √183

Answer: 13, 13 , √183

Task

Solution.
We find the volume of the pyramid using the formula:
V = 1/3 Sh

We find the area of ​​the base using the formula for finding the area of ​​a right triangle:
S = ab/2 = 8 * 6 / 2 = 24
where
V = 1/3 * 24 *10 = 80 cm 3.

MUNICIPAL EDUCATIONAL INSTITUTION
"SCHOOL No. 2" OF THE CITY OF ALUSHTA

LESSON PLAN

Problem solving.

Pyramid. Truncated pyramid

Mathematic teacher

Pikhidchuk Irina Anatolevna

2016 G.

LESSON

Geometry. Grade 11.

The lesson lasts 3 hours. It is recommended to carry out general repetition.

SUBJECT: Pyramid. Truncated pyramid. Problem solving.

THE MAIN TASK: Preparing for test work(identify problems; systematize and adjust knowledge on the topic).

GOALS: 1) Check your knowledge of definitions: the angle between a straight line and a plane; linear dihedral angle (construction); correct pyramid.

Repeat the formulas: volume of the pyramid; radii of the inscribed and circumscribed circle of the polygon;

test your drawing skills; the ability to justify the angles between the side edge and the plane of the base, between the side edge and the plane of the base.

strengthen computing skills.

DURING THE CLASSES:

Organizing time. Communicating the goals and objectives of the lesson.

Repetition.

Drawings on the folding board:

Assignment for the drawings: formulate the definition of the angle between a straight line and a plane. Show the angle in the pictures and justify it.

Main board

Show the angle between the side edge and the plane of the base of a regular triangular pyramid. Calculate the volume of the pyramid if the side of the base is equal to a, the angle between the side edge and the plane of the base is equal to a.

Find the volume of each of the given regular pyramids

CONCLUSION: 1) The angle between the side edge and the plane of the base is the angle between the side edge and the radius of the circle circumscribed near the base;

2) The angle between the side face and the plane of the base of the pyramid is the angle between the apothem and the radius of the circle inscribed in the base.

Homework on cards (assignment attached).

Geometry 11th grade, (continued)

PROBLEM SOLVING: Pyramid. Truncated pyramid.

Problem No. 1. At the base of the pyramid lies a right triangle. Two faces containing legs are perpendicular to the plane of the base. Show the angles between the side ribs and the plane of the base. Will they be equal if the triangle is isosceles?

Problem No. 2. At the base of the pyramid lies an isosceles triangle. The lateral ribs are inclined to the base plane at one angle. Construct the height of the pyramid and the angles between the side edges and the plane of the base (justify the construction)

CONCLUSION: The height of the pyramid is projected into the center of the circumcircle if: the side edges are equal; the side ribs are inclined to the plane of the base at one angle; The pyramid is correct.

Homework. In a regular pyramid (triangular, quadrangular, hexagonal), construct an angle between the side face and the plane of the base. Justify the construction.

How can you build a pyramid? On surface R Let's construct a polygon, for example the pentagon ABCDE. Out of plane R Let's take point S. By connecting point S with segments to all points of the polygon, we get the SABCDE pyramid (Fig.).

Point S is called top, and the polygon ABCDE is basis this pyramid. Thus, a pyramid with top S and base ABCDE is the union of all segments where M ∈ ABCDE.

Triangles SAB, SBC, SCD, SDE, SEA are called side faces pyramids, common sides of the lateral faces SA, SB, SC, SD, SE - lateral ribs.

The pyramids are called triangular, quadrangular, p-angular depending on the number of sides of the base. In Fig. Images of triangular, quadrangular and hexagonal pyramids are given.

The plane passing through the top of the pyramid and the diagonal of the base is called diagonal, and the resulting section is diagonal. In Fig. 186 one of the diagonal sections of the hexagonal pyramid is shaded.

The perpendicular segment drawn through the top of the pyramid to the plane of its base is called the height of the pyramid (the ends of this segment are the top of the pyramid and the base of the perpendicular).

The pyramid is called correct, if the base of the pyramid is a regular polygon and the vertex of the pyramid is projected at its center.

All lateral faces of a regular pyramid are congruent isosceles triangles. In a regular pyramid, all lateral edges are congruent.

The height of the side face of a regular pyramid drawn from its vertex is called apothem pyramids. All apothems of a regular pyramid are congruent.

If we designate the side of the base as A, and the apothem through h, then the area of ​​one side face of the pyramid is 1/2 ah.

The sum of the areas of all the lateral faces of the pyramid is called lateral surface area pyramid and is designated by S side.

Since the lateral surface of a regular pyramid consists of n congruent faces, then

S side = 1/2 ahn= P h / 2 ,

where P is the perimeter of the base of the pyramid. Hence,

S side = P h / 2

i.e. 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.

The total surface area of ​​the pyramid is calculated by the formula

S = S ocn. + S side. .

The volume of the pyramid is equal to one third of the product of the area of ​​its base S ocn. to height H:

V = 1 / 3 S main. N.

The derivation of this and some other formulas will be given in one of the subsequent chapters.

Let's now build a pyramid in a different way. Let a polyhedral angle be given, for example, pentahedral, with vertex S (Fig.).

Let's draw a plane R so that it intersects all the edges of a given polyhedral angle at different points A, B, C, D, E (Fig.). Then the SABCDE pyramid can be considered as the intersection of a polyhedral angle and a half-space with the boundary R, in which the vertex S lies.

Obviously, the number of all faces of the pyramid can be arbitrary, but not less than four. When a trihedral angle intersects with a plane, a triangular pyramid is obtained, which has four sides. Any triangular pyramid is sometimes called tetrahedron, which means tetrahedron.

Truncated pyramid can be obtained if the pyramid is intersected by a plane parallel to the plane of the base.

In Fig. An image of a quadrangular truncated pyramid is given.

Truncated pyramids are also called triangular, quadrangular, n-gonal depending on the number of sides of the base. From the construction of a truncated pyramid it follows that it has two bases: upper and lower. The bases of a truncated pyramid are two polygons, the sides of which are parallel in pairs. The lateral faces of the truncated pyramid are trapezoids.

Height a truncated pyramid is a perpendicular segment drawn from any point of the upper base to the plane of the lower one.

Regular truncated pyramid called the part of a regular pyramid enclosed between the base and a section plane parallel to the base. The height of the side face of a regular truncated pyramid (trapezoid) is called apothem.

It can be proven that a regular truncated pyramid has congruent lateral edges, all lateral faces are congruent, and all apothems are congruent.

If in the correct truncated n-coal pyramid through A And b n indicate the lengths of the sides of the upper and lower bases, and through h is the length of the apothem, then the area of ​​each side face of the pyramid is equal to

1 / 2 (A + b n) h

The sum of the areas of all the lateral faces of the pyramid is called the area of ​​its lateral surface and is designated S side. . Obviously, for a correct truncated n-coal pyramid

S side = n 1 / 2 (A + b n) h.

Because pa= P and nb n= P 1 - the perimeters of the bases of the truncated pyramid, then

S side = 1 / 2 (P + P 1) h,

that is, the area of ​​the lateral surface of a regular truncated pyramid is equal to half the product of the sum of the perimeters of its bases and the apothem.

Section parallel to the base of the pyramid

1) the side ribs and height will be divided into proportional parts;

2) in cross-section you will get a polygon similar to the base;

3) the cross-sectional areas and bases are related as the squares of their distances from the top.

It is enough to prove the theorem for a triangular pyramid.

Since parallel planes are intersected by a third plane along parallel lines, then (AB) || (A 1 B 1), (BC) ||(B 1 C 1), (AC) || (A 1 C 1) (fig.).

Parallel lines cut the sides of an angle into proportional parts, and therefore

$$ \frac(\left|(SA)\right|)(\left|(SA_1)\right|)=\frac(\left|(SB)\right|)(\left|(SB_1)\right| )=\frac(\left|(SC)\right|)(\left|(SC_1)\right|) $$

Therefore, ΔSAB ~ ΔSA 1 B 1 and

$$ \frac(\left|(AB)\right|)(\left|(A_(1)B_1)\right|)=\frac(\left|(SB)\right|)(\left|(SB_1 )\right|) $$

ΔSBC ~ ΔSB 1 C 1 and

$$ \frac(\left|(BC)\right|)(\left|(B_(1)C_1)\right|)=\frac(\left|(SB)\right|)(\left|(SB_1 )\right|)=\frac(\left|(SC)\right|)(\left|(SC_1)\right|) $$

Thus,

$$ \frac(\left|(AB)\right|)(\left|(A_(1)B_1)\right|)=\frac(\left|(BC)\right|)(\left|(B_ (1)C_1)\right|)=\frac(\left|(AC)\right|)(\left|(A_(1)C_1)\right|) $$

The corresponding angles of triangles ABC and A 1 B 1 C 1 are congruent, like angles with parallel and identical sides. That's why

ΔABC ~ ΔA 1 B 1 C 1

The areas of similar triangles are related as the squares of the corresponding sides:

$$ \frac(S_(ABC))(S_(A_1 B_1 C_1))=\frac(\left|(AB)\right|^2)(\left|(A_(1)B_1)\right|^2 ) $$

$$ \frac(\left|(AB)\right|)(\left|(A_(1)B_1)\right|)=\frac(\left|(SH)\right|)(\left|(SH_1 )\right|) $$

Hence,

$$ \frac(S_(ABC))(S_(A_1 B_1 C_1))=\frac(\left|(SH)\right|^2)(\left|(SH_1)\right|^2) $$

Theorem. If two pyramids with equal heights are cut at the same distance from the top by planes parallel to the bases, then the areas of the sections are proportional to the areas of the bases.

Let (Fig. 84) B and B 1 be the areas of the bases of two pyramids, H be the height of each of them, b And b 1 - sectional areas by planes parallel to the bases and removed from the vertices at the same distance h.

According to the previous theorem we will have:

$$ \frac(b)(B)=\frac(h^2)(H^2)\: and \: \frac(b_1)(B_1)=\frac(h^2)(H^2) $ $
where
$$ \frac(b)(B)=\frac(b_1)(B_1)\: or \: \frac(b)(b_1)=\frac(B)(B_1) $$

Consequence. If B = B 1, then b = b 1, i.e. If two pyramids with equal heights have equal bases, then the sections equally spaced from the top are also equal.