(mqs) (pns) (rnq) =

(nqs)3

set2D(0; 20; 4; 20);

;0 dash0 );

;0 dash0 );

;0 dash0 );

dash0);

p8 = pointsPlot(4

[ 0A 0; 0 B 0; 0 C 0; 0 a 0; 0 b 0; 0 c 0; 0 D 0];

showPlots(;0 noAxes0 );

set2D(3; 12; 2; 13);

;0 dash0 );

;0 dash0 );

;0 dash0 );

dash0);

;0 dash0 );

CHAPTER THREE

POLYHEDRALS

II VOLUME OF PRISM AND PYRAMID

82. Key volume assumptions. The size of the part of the space occupied by a geometric body is called the volume of this body.

We set the task - to find an expression for this value in the form of some number that measures this value. In doing so, we will be guided by the following starting points:

1) Equal bodies have equal volumes.

2) The volume of a body(for example, each parallelepiped depicted in Fig. 87), composed of parts(P and Q), is equal to the sum of the volumes of these parts.

Two bodies that have the same volume are called equal.

83. Unit of volume. For a unit of volume, when measuring them, they take the volume of such a cube, in which each edge is equal to a linear unit. So, cubic meters (m 3), cubic centimeters (cm 3), etc. are common.

Volume of the box

84. Theorem.The volume of a rectangular parallelepiped is equal to the product of its three dimensions.

In such a brief expression, this theorem should be understood as follows: the number expressing the volume of a rectangular parallelepiped in a cubic unit is equal to the product of numbers expressing its three dimensions in the corresponding linear unit, i.e., in a unit that is an edge of a cube, the volume of which is taken as a cubic unit . So if X is a number expressing the volume of a cuboid in cubic centimeters, and a, b and With-numbers expressing its three dimensions in linear centimeters, then the theorem states that x=abc.

In the proof, we especially consider the following three cases:

1) Measurements are expressed whole numbers.

Let, for example, the measurements be (Fig. 88): AB = a, BC = b and BD= c,
where a, b and With- some integers (for example, as shown in our drawing: a = 4, b= 2 and With= 5). Then the base of the parallelepiped contains ab such squares, each of which represents the corresponding square unit. On each of these squares, obviously, one cubic unit can be placed. Then you get a layer (shown in the drawing), consisting of ab cubic units. Since the height of this layer is equal to one linear unit, and the height of the entire box contains With such units, then inside the parallelepiped one can put With such layers. Therefore, the volume of this parallelepiped is abc cubic units.

2) Measurements are expressed fractional numbers. Let the dimensions of the box be:

m / n , p / q , r / s

. (Some of these fractions may equal a whole number.) Reducing the fractions to the same denominator, we have:

mqs / nqs , pns / nqs , rnq / nqs

Let's take 1 / nqs fraction of a linear unit for a new (auxiliary) unit of length. Then, in this new unit of measurement of this parallelepiped, they will be expressed as integers, namely: mqs, pns and rnq, and therefore, according to what was proved (in case 1), the volume of the parallelepiped is equal to the product ( mqs) (pns) (rnq) if this volume is measured by a new cubic unit corresponding to the new linear unit. Such cubic units in one cubic unit corresponding to the former linear unit contains ( nqs) 3 ; so the new cubic unit is 1 /( nqs) 3 of the former. Therefore, the volume of the parallelepiped, expressed in the same units, is equal to:

3) Measurements are expressed irrational numbers. Let this parallelepiped (Fig. 89), which for brevity we denote by a single letter Q, have measurements:

AB = α; AC = β; AD = γ,

where all numbers α, β and γ or only some of them are irrational.

Each of the numbers α, β, and γ can be represented as an infinite decimal. Let's take the approximate values ​​of these fractions with P decimal places, first with a deficiency, and then with an excess. Values ​​with deficiency will be denoted by α n , β n , γ n, values ​​with excess α" n , β" n , γ" n. Let us plot on the edge AB, starting from point A, two segments AB 1 = α n and AB 2 \u003d α " n.
On the edge AC from the same point A, we plot the segments AC 1 = β n and AC 2 = β" n and on the edge AD from the same point-segment AD 1 = γ n and AD 2 = γ" n.

In doing so, we will have:

AB 1< АВ < АВ 2 ; АС 1 < АС < АС 2 ; AD 1 < AD < AD 2 .

Let us now construct two auxiliary parallelepipeds; one (let's call it Q 1) with measurements AB 1 , AC 1 and AD 1 and the other (let's call it Q 2 ) with measurements AB 2 , AC 2 and AD 2 . Box Q 1 will all fit inside box Q, and box Q 2 will contain box Q.

By what has been proved (in case 2) we will have:

volume Q 1 \u003d α n β n γ n (1)

volume Q 2 \u003d α " n β" n γ" n (2)

Let's call the volume Q 1< объёма Q 2 .

Let's start increasing the number P. This means that we take approximate values ​​of the numbers α , β , γ with more and more accuracy.

Let's see how the volumes of the parallelepipeds Q 1 and Q 2 change in this case.

With an unlimited increase P the volume Q 1 obviously increases and, by virtue of equality (1), with an infinite increase n has as its limit the limit of the product (α n β n γ n). The volume Q 2 obviously decreases and, by virtue of equality (2), has the limit of the product (α " n β" n γ" n). But it is known from algebra that both products
α n β n γ n and α" n β" n γ" n with unlimited magnification P have a common limit, which is the product of irrational numbers αβγ.

We take this limit as a measure of the volume of the parallelepiped Q: volume Q = αβγ.

It can be proved that the volume thus defined satisfies the conditions established for volume (§ 82). Indeed, with this definition of volume, equal parallelepipeds obviously have equal volumes. Therefore, the first condition (§ 82) is satisfied. Let us now divide this parallelepiped Q into two by a plane parallel to its base: Q 1 and Q 2 (Fig. 90).

Then we will have:

volume Q \u003d AB AC AD,
volume Q 1 \u003d AB AA 1 AD,
volume Q 2 \u003d A 1 B 1 A 1 C A 1 D 1.

Adding term by term the last two equalities and noticing that A 1 B 1 \u003d AB and A 1 D 1 \u003d AD, we get:

volume Q 1 + volume Q 2 \u003d AB AA 1 AD + AB A 1 C AD \u003d AB AD (AA 1 + A 1 C) \u003d AB AD AC, from here we get:

volume Q 1 + volume Q 2 = volume Q.

Consequently, the second condition of § 82 is also satisfied if the parallelepiped is folded from two parts obtained by cutting it with a plane parallel to one of the faces.

85. Consequence. Let the measurements of a rectangular parallelepiped, which serve as the sides of its base, be expressed by numbers a and b, and the third dimension (height) is the number With. Then, denoting its volume in the corresponding cubic units by the letter V, we can write:

V = abs.

Since the work ab expresses the area of ​​the base, then we can say that The volume of a rectangular prism is equal to the product of the area of ​​the base and the height .

Comment. The ratio of two cubic units of different names is equal to the third power of the ratio of those linear units that serve as edges for these cubic units. So, the ratio of a cubic meter to a cubic decimeter is 10 3, i.e. 1000. Therefore, for example, if we have a cube with an edge length a linear units and another cube with an edge of length 3 a linear units, then the ratio of their volumes will be equal to 3 3, i.e. 27, which is clearly seen from drawing 91.

86. Lemma. An inclined prism is equal to such a straight prism, the base of which is equal to the perpendicular section of the inclined prism, and the height is equal to its side edge.

Let an oblique prism ABCDEA 1 B 1 C 1 D 1 E 1 be given (Fig. 92).

Let's continue all its side edges and side faces in the same direction.

Take an arbitrary point on the continuation of one of the edges a and draw a perpendicular section through it abcde. Then, postponing aa 1 \u003d AA 1, let's draw through a 1 perpendicular section a 1 b 1 c 1 d 1 e one . Since the planes of both sections are parallel, then bb 1 = ss 1 = dd 1 = her 1 = aa 1 = AA 1 (§17). As a result, the polyhedron a 1 d, for which the sections we have drawn are taken as bases, is a direct prism, which is mentioned in the theorem.

Let us prove that the given oblique prism is equal to this straight line. To do this, we first make sure that the polyhedra a D and a 1 D 1 are equal. their foundations abcde and a 1 b 1 c 1 d 1 e 1 are equal as the bases of the prism a 1 d; on the other hand, adding to both parts of the equality A 1 A = a 1 a along the same line segment A 1 a, we get: a A = a 1 A 1 ; like this b B = b 1 in 1, With C = With 1 C 1, etc. Let us now imagine that the polyhedron a D is embedded in a polyhedron a 1 D 1 so that their bases coincide; then the lateral edges, being perpendicular to the bases and correspondingly equal, will also coincide; so the polyhedron a D is compatible with the polyhedron a 1 D 1 ; so these bodies are equal. Now note that if to a straight prism a 1 d add a polyhedron a D, and add a polyhedron to the inclined prism A 1 D a 1 D 1 equal a D, then we get the same polyhedron a 1 D. It follows that two prisms A 1 D and a 1 d are equal.

87. Theorem. The volume of a parallelepiped is equal to the product of the area of ​​the base and the height.

Earlier we proved this theorem for a right-angled parallelepiped, now we will prove it for a right-angled parallelepiped, and then for an oblique one.

one). Let (Fig. 93) AC 1 be a straight parallelepiped, i.e. one whose base ABCD is some kind of parallelogram, and all side faces are rectangles.

Let's take in it for the base the side face AA 1 B 1 B; then the parallelepiped will be
n a c l o n n y. Considering it as a special case of an inclined prism, we can assert on the basis of the lemma of the previous section that this parallelepiped is equal in size to such a right parallelepiped whose base is the perpendicular section MNPQ and the height is BC. The quadrilateral MNPQ is a rectangle because its angles are the linear angles of the right dihedral angles; therefore, a right parallelepiped having a rectangle MNPQ as its base must be rectangular and, therefore, its volume is equal to the product of its three dimensions, which can be taken as the segments MN, MQ and BC. In this way,

volume AC 1 \u003d MN MQ BC \u003d MN (MQ BC).

But the product MQ BC expresses the area of ​​the parallelogram ABCD, therefore

volume ACX \u003d (area ABCD) MN \u003d (area ABCD) BB 1.

2) Let (Fig. 94) AC 1 be an inclined parallelepiped.

It is equal in size to such a straight line, in which the perpendicular section MNPQ serves as the base (that is, perpendicular to the edges AD, BC, . . .), and the height is the edge BC. But, according to what has been proved, the volume of a right parallelepiped is equal to the product of the base area and the height; means,

volume AC 1 \u003d (area MNPQ) BC.

If RS is the height of the section MNPQ, then the area MNPQ = MQ RS, so

volume AC 1 \u003d MQ RS BC \u003d (BC MQ) RS.

The product BC MQ expresses the area of ​​the parallelogram ABCD; therefore, the volume AC 1 \u003d (area ABCOD) RS.

It remains now to prove that the segment RS is the height of the parallelepiped. Indeed, the section MNPQ, being perpendicular to the edges BC, B 1 C 1 , .. . , must be perpendicular to the faces ABCD, BB 1 C 1 C, .... passing through these edges (§ 43). Therefore, if we set up a perpendicular to the plane ABCD from the point S, then it must lie entirely in the plane MNPQ (§ 44) and, therefore, must merge with the line RS, which lies in this plane and is perpendicular to MQ. Hence, the segment SR is the height of the parallelepiped. Thus, the volume of and an inclined parallelepiped is equal to the product of the base area and the height.

Consequence. If V, B and H are numbers expressing in the appropriate units the volume, base area and height of the parallelepiped, then we can write.

In this lesson we will talk about a rectangular box. Let's recall some of its properties. And then we derive in detail the formulas for calculating the volume of a rectangular parallelepiped. Summary of the lesson "The volume of a cuboid" In this lesson we will talk about a cuboid. Let's recall some of its properties. And then we derive in detail the formulas for calculating the volume of a rectangular parallelepiped. Earlier, we already met with a rectangular parallelepiped. Recall that a box is called rectangular if all six of its faces are rectangles. An idea of ​​the shape of a cuboid is given by a matchbox, a box, a refrigerator, etc. Let's imagine a room that has the shape of a cuboid. If we talk about its dimensions, then the words "length", "width" and "height" are usually used, referring to the lengths of three edges with a common vertex. In geometry, these three quantities are united by a common name: measurements of a rectangular parallelepiped. A rectangular box is shown on the screen as its measurements can be taken, for example, the lengths of the edges, these edges have a common vertex of the box, the width and the box has the following properties: 1) the square of the diagonal of a rectangular box is equal to the sum of the squares of its three dimensions. is the length of the given. Then the edge is its height. . In and, all 2) the volume of a rectangular parallelepiped is equal to the product of its three dimensions. So, the following theorem is true: the volume of a rectangular parallelepiped is equal to the product of its three dimensions. Let's prove this theorem. Let a rectangular parallelepiped be given its dimensions by letters Let's prove that the volume of a rectangular parallelepiped is equal to, and its volume by a letter. Let's denote and. , . Two cases are possible: Consider the first case. Measurements of decimal fractions, in which the number of decimal places does not exceed, are final and (,). In this case, the numbers, and are integers. We divide each edge of the parallelepiped into equal parts of the length. Then, through the split points, we draw planes perpendicular to this edge. Then our box will be divided into equal cubes with the length of each edge. The total number of such cubes will be equal. Since the volume of each such cube is equal, the volume of the entire parallelepiped will be equal. By this we proved that the volume of a rectangular parallelepiped is equal to the product of its three dimensions. Q.E.D. Let's move on to the second case. At least one of the dimensions is an infinite decimal fraction. , and represents Consider the final decimal fractions of numbers c, -th. , which are obtained from, if we discard in each of them all the digits after the decimal point, starting Note that then the inequality is true. Similar inequalities will also hold for numbers where and: . , where, . Let's multiply these inequalities. Then we see that. It is clear from the inequality that a box is a box, and itself is contained in the box. And it says that. Now let's increase indefinitely to become arbitrarily small, and therefore the number differs little from the number. . Then the number will be arbitrarily As a result, they will become equal. Those. . Q.E.D. This theorem has the following corollaries. First consequence. The volume of a rectangular prism is equal to the product of the area of ​​the base and the height. Proof. Let a face with edges of a rectangular parallelepiped. Then the area of ​​the base is the height of the parallelepiped and. is the base, and Then you can see that the formula for calculating the volume of a rectangular parallelepiped is the area of ​​\u200b\u200bthe base, is the height of a rectangular parallelepiped. can be written in the form where Thus, we have proved that the volume of a rectangular parallelepiped is equal to. Q.E.D. Second consequence. The volume of a right prism whose base is a right triangle is equal to the product of the area of ​​the base and the height. Proof. To prove this statement, let's complete a right triangular prism with a parallelepiped base as shown on the screen. Taking into account the first consequence, the volume of this parallelepiped is equal to where is the area of ​​the base) to rectangular (, is the height of the prism. , divides the parallelepiped into two equal straight prisms, one of which is a given plane. These prisms are equal, since they have equal bases and equal heights. Therefore, the volume of this prism is equal, i.e. equal to prove. Remark. Consider a square with side a. As required Based on the Pythagorean theorem, its diagonal is equal. Therefore, the area of ​​the square built on it is twice the area of ​​this square. Thus, not it is difficult to construct a side of a square whose area is twice the area of ​​a given square.Now consider a cube with side a. The question arises: is it possible, using a compass and a straightedge, to construct a side of a cube whose volume is twice the volume of a given cube, i.e. construct a segment, equal? This problem was formulated in ancient times. It has been called the “Doubling the Cube Problem”. Only in 1837 did the French mathematician Pierre Laurent Vanzel prove that such a construction was impossible. At the same time, he proved the unsolvability of another construction problem - the problem of trisection of an angle (an arbitrary given angle is divided into three equal angles). Recall that the problem of squaring a circle (constructing a square whose area is equal to the area of ​​a given circle) also belongs to the class of classical unsolvable construction problems. The impossibility of such a construction was proven in 1882 by the German mathematician Carl Louis Ferdinand Lindemann. Task: find the volume of a cuboid with diagonal sides of the base Solution: write a formula for calculating the volume of a cuboid through its measurements. cm and cm. cm and From the condition of the problem, we know the length, width and diagonal of a rectangular parallelepiped, but its height is unknown. Recall that. We express from this formula the height that the height is (cm). rectangular parallelepiped. We get, and equal to Let's substitute the measurements of our rectangular parallelepiped into the volume formula. Let's count. We get that the volume of the parallelepiped is Let's not forget to write down the answer. (cm3). Task: square. The volume of a cuboid is equal to the height of the cuboid, if the cuboid, the base is cm3. Determine See Solution: In this lesson, we proved that the volume of a cuboid is equal. Express the height from the formula. Hence, the height is equal. Since the base of our rectangular parallelepiped is a square by condition, the area of ​​​​the base is equal to the volume of the rectangular parallelepiped is (cm2). By the condition of the problem, it is also known that. Hence, the height (cm). Let's write down the answer. equals Totals: In this lesson, we remembered the concept of a cuboid. We proved that the volume of a rectangular parallelepiped is equal to the product of its three dimensions. We proved that the volume of a rectangular parallelepiped can be calculated as the product of the base area and the height. They also proved that the volume of a right prism, the base of which is a right triangle, is equal to the product of the area of ​​the base and the height.

The prism is called parallelepiped if its bases are parallelograms. Cm. Fig.1.

Properties of the box:

The opposite faces of the parallelepiped are parallel (i.e. lie in parallel planes) and equal.

The diagonals of the parallelepiped intersect at one point and bisect that point.

Adjacent faces of a box are two faces that have a common edge.

Opposite faces of a parallelepiped– faces that do not have common edges.

Opposite vertices of the box are two vertices that do not belong to the same face.

Diagonal of the box A line segment that connects opposite vertices.

If the lateral edges are perpendicular to the planes of the bases, then the parallelepiped is called direct.

A right parallelepiped whose bases are rectangles is called rectangular. A prism all of whose faces are squares is called cube.

Parallelepiped A prism whose bases are parallelograms.

Right parallelepiped- a parallelepiped whose lateral edges are perpendicular to the plane of the base.

cuboid is a right parallelepiped whose bases are rectangles.

Cube is a rectangular parallelepiped with equal edges.

Parallelepiped a prism is called, the base of which is a parallelogram; thus, the parallelepiped has six faces and all of them are parallelograms.

Opposite faces are pairwise equal and parallel. The parallelepiped has four diagonals; they all intersect at one point and divide in half at it. Any face can be taken as a base; the volume is equal to the product of the base area and the height: V = Sh.

A parallelepiped whose four lateral faces are rectangles is called a right parallelepiped.

A right parallelepiped, in which all six faces are rectangles, is called rectangular. Cm. Fig.2.

The volume (V) of a right parallelepiped is equal to the product of the base area (S) and the height (h): V = Sh .

For a rectangular parallelepiped, in addition, the formula V=abc, where a,b,c are edges.

The diagonal (d) of a cuboid is related to its edges by the relationship d 2 \u003d a 2 + b 2 + c 2 .

cuboid- a parallelepiped whose lateral edges are perpendicular to the bases, and the bases are rectangles.

Properties of a cuboid:

In a cuboid, all six faces are rectangles.

All dihedral angles of a cuboid are right angles.

The square of the diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three dimensions (lengths of three edges that have a common vertex).

The diagonals of a rectangular parallelepiped are equal.

A rectangular parallelepiped, all of whose faces are squares, is called a cube. All edges of a cube are equal; the volume (V) of a cube is expressed by the formula V=a 3, where a is the edge of the cube.

Any geometric body can be characterized by surface area (S) and volume (V). Area and volume are not the same thing. An object can have a relatively small V and a large S, for example, this is how the human brain works. It is much easier to calculate these indicators for simple geometric shapes.

Parallelepiped: definition, types and properties

A parallelepiped is a quadrangular prism with a parallelogram at its base. Why might you need a formula for finding the volume of a figure? Books, packing boxes and many other things from everyday life have a similar shape. Rooms in residential and office buildings, as a rule, are rectangular parallelepipeds. To install ventilation, air conditioning and determine the number of heating elements in a room, it is necessary to calculate the volume of the room.

The figure has 6 faces - parallelograms and 12 edges, two arbitrarily chosen faces are called bases. The parallelepiped can be of several types. The differences are due to the angles between adjacent edges. The formulas for finding the V-s of various polygons are slightly different.

If 6 faces of a geometric figure are rectangles, then it is also called rectangular. A cube is a special case of a parallelepiped in which all 6 faces are equal squares. In this case, to find V, you need to know the length of only one side and raise it to the third power.

To solve problems, you will need knowledge not only of ready-made formulas, but of the properties of the figure. The list of basic properties of a rectangular prism is small and very easy to understand:

1. Opposite faces of the figure are equal and parallel. This means that the ribs located opposite are the same in length and angle of inclination.
2. All side faces of a right parallelepiped are rectangles.
3. The four main diagonals of a geometric figure intersect at one point, and divide it in half.
4. The square of the diagonal of a parallelepiped is equal to the sum of the squares of the dimensions of the figure (follows from the Pythagorean theorem).

Pythagorean theorem states that the sum of the areas of the squares built on the legs of a right triangle is equal to the area of ​​the triangle built on the hypotenuse of the same triangle.

The proof of the last property can be seen in the image below. The course of solving the problem is simple and does not require detailed explanations.

The formula for the volume of a rectangular parallelepiped

The formula for finding for all types of geometric shapes is the same: V=S*h, where V is the desired volume, S is the area of ​​the base of the parallelepiped, h is the height lowered from the opposite vertex and perpendicular to the base. In a rectangle, h coincides with one of the sides of the figure, so to find the volume of a rectangular prism, you need to multiply three measurements.

The volume is usually expressed in cm3. Knowing all three values ​​a, b and c, finding the volume of the figure is not at all difficult. The most common type of problem in the USE is the search for the volume or diagonal of a parallelepiped. It is impossible to solve many typical USE tasks without a formula for the volume of a rectangle. An example of a task and the design of its solution is shown in the figure below.

Note 1. The surface area of ​​a rectangular prism can be found by multiplying by 2 the sum of the areas of the three faces of the figure: the base (ab) and two adjacent side faces (bc + ac).

Note 2. The surface area of ​​the side faces can be easily found by multiplying the perimeter of the base by the height of the parallelepiped.

Based on the first property of parallelepipeds, AB = A1B1, and the face B1D1 = BD. According to the consequences of the Pythagorean theorem, the sum of all angles in a right triangle is equal to 180 °, and the leg opposite the angle of 30 ° is equal to the hypotenuse. Applying this knowledge for a triangle, we can easily find the length of the sides AB and AD. Then we multiply the obtained values ​​​​and calculate the volume of the parallelepiped.

The formula for finding the volume of a slanted box

To find the volume of an inclined parallelepiped, it is necessary to multiply the area of ​​\u200b\u200bthe base of the figure by the height lowered to this base from the opposite angle.

Thus, the desired V can be represented as h - the number of sheets with an area S of the base, so the volume of the deck is made up of the Vs of all cards.

Examples of problem solving

The tasks of the single exam must be completed within a certain time. Typical tasks, as a rule, do not contain a large number of calculations and complex fractions. Often a student is offered how to find the volume of an irregular geometric figure. In such cases, you should remember the simple rule that the total volume is equal to the sum of the V-s of the constituent parts.

As you can see from the example in the image above, there is nothing complicated in solving such problems. Tasks from more complex sections require knowledge of the Pythagorean theorem and its consequences, as well as the formula for the length of the diagonal of a figure. To successfully solve test tasks, it is enough to familiarize yourself with samples of typical tasks in advance.