If rectangular parallelepipeds have equal bases, then they can be nested one inside the other.
Let AG and AP (Fig.) be two such parallelepipeds. Let's consider two cases.
1. The heights of BF and BN are comparable.
Let the general height measure be contained m times in BF and n times in BN.
Let us draw a series of planes through the division points, parallel to the base.
Then the parallelepiped AG will be divided into m, and the parallelepiped AP into n equal parts.
Thus we get:
\(\frac(BF)(BN)=\frac(m)(n)\) and \(\frac(Volume AG)(Volume AP)=\frac(m)(n) \)
Hence:
\(\frac(Volume AG)(Volume AP)=\frac(BF)(BN) \)
2. The heights of BF and BN are incommensurable.
Divide BN into n equal parts and put one part on BF as many times as possible.
Let the 1/n share of BN be contained in BF more than m times, but less than m+1 times.
Then, drawing as before a series of planes parallel to the base, we divide the par-d AP into n such equal parts that the par-de AG contains more than m, but less than m+1.
Hence:
approx.rel. \(\frac(BF)(BN)=\frac(m)(n)\) and approx.rel. \(\frac(Volume AG)(Volume AP)=\frac(m)(n)\)
Thus, approximate ratios calculated with arbitrary but equal precision are equal. And this is the equality of incommensurable relations.
Lemma 2. The volumes of rectangular parallelepipeds having equal heights are related to the areas of their bases.
Let (Fig.) P and P 1 be two rectangular parallelepipeds. Let us denote the unequal bases of one of them by a and b, and of the other by a 1 and b 1.
Let's take an auxiliary rectangular parallelepiped Q, whose height is the same as those of these bodies, and the base is a rectangle with sides a and b 1.
Parallelepipeds P and Q have equal front faces. If we take these faces as bases, then the heights will be b and b 1, and therefore:
Volume P/Volume Q = b/b1
Parallelepipeds Q and P 1 have equal lateral faces. If we take these faces as bases, then the heights will be a and a 1, and therefore:
Volume Q/Volume P 1 = a/a1
Multiplying the equalities and , we find:
Volume P/Volume P 1 = ab/a 1 b 1
Since ab expresses the area of the base of the par-da P, and a 1 b 1 - the area of the base of the par-da P 1, then the lemma is proven.
Theorem. The volume of a rectangular parallelepiped is equal to the product of the area of the base and the height.
Let (fig.) P be a rectangular parallelepiped, and P 1 some cubic unit.
Let us denote the base area and height of the first by B and H, and of the second by B 1 and H 1.
Let us take an auxiliary rectangular parallelepiped Q, whose base area is B 1 and height H.
Comparing P with Q, and then Q with P 1, we find:
About. P/Vol. Q = B/B1 and vol. Q/rev. P1 = H/H1
Multiplying these equalities, we get:
About. P/Vol. P1 = B/B1 * H/H1
The ratios included in this equality are numbers expressing the volume, base area and height of a given parallelepiped in the corresponding cubic, square and linear units. Therefore, the last equality can be expressed as follows:
The number expressing the volume of a rectangular parallelepiped is equal to the product of the numbers expressing the area of the base and the height in the corresponding units.
This is expressed briefly as follows: the volume of a rectangular parallelepiped is equal to the product of the area of the base and the height, i.e.
where V, B and H are numbers expressing in the corresponding units the volume, base area and height of a rectangular parallelepiped.
Denoting by the letters a, b and c the three dimensions of a rectangular par-da (expressed in numbers), we can write:
because the area of the base is expressed by the product of two of these dimensions, and the height is equal to the third dimension.
Consequences:- The volume of a cube is equal to the third power of its edge.
- The ratio of two cubic units is equal to the third power of the ratio of the corresponding linear units. So, the ratio of m3 to dm3 is 10 3, i.e. 1000.
Volume of any parallelepiped
Lemma. An inclined prism is equal in size to a straight prism whose base is equal to the perpendicular section of the inclined prism, and the height is equal to its side edge.
Through some point a (Fig.) of one of the side edges of the inclined prism A 1 d we draw a perpendicular section abcde. Then we will continue all the side faces down, set aside aa 1 =AA 1 and through point a 1 draw a perpendicular section a 1 b 1 c 1 d 1 e 1 .
Since the planes of the two sections are parallel, the parts of the side ribs enclosed between them are equal, i.e.
bb 1 = ss 1 = dd 1 = ee 1 = aa 1 = AA 1 .
As a result, the polyhedron a 1 d is a straight prism, the base of which is a perpendicular section, and the height (or, what is the same, the side edge) is equal to the side edge of the inclined prism.
Let us prove that an inclined prism is equal in size to a straight prism.
To do this, let us first make sure that the polyhedra aD and a 1 D 1 are equal.
Their bases abcde and a 1 b 1 c 1 d 1 e 1 are equal, like the bases of a prism a 1 d.
On the other hand, subtracting from both sides the equalities A 1 A = a 1 a along the same straight line A 1 a , we obtain aA = a 1 A 1 .
Similar to this: bB = b 1 B 1, cC = c 1 C 1, etc.
Let us now imagine that the polyhedron aD is embedded in a 1 D 1 so that their bases coincide. Then the side ribs, being perpendicular to the bases and correspondingly equal, will also coincide.
Therefore, the polyhedron aD is compatible with a 1 D 1. This means that these bodies are equal.
Now note that if we subtract part aD from the whole polyhedron a 1 D , we get a direct prism. And if we subtract part a 1 D 1 from the same polyhedron, we get an inclined prism.
It follows from this that these two prisms are equal in size, since their volumes are the differences in the volumes of equal bodies.
Theorem. The volume of a parallelepiped is equal to the product of the area of the base and the height.
Previously, we proved this theorem for a rectangular parallelepiped, now we will prove it for a straight parallelepiped, and then an inclined one.
1. Let (fig.) AC 1 straight par-d, i.e. one whose base ABCD is some kind of parallelogram, and all the side faces are rectangles.
Let us take the face AA 1 B 1 B as its base. Then the parallelepiped will be inclined.
Considering it as a special case of an inclined prism, we, based on the lemma of the previous paragraph, can assert that this par-d is equivalent to a straight line whose base is a perpendicular section MNPQ and height BC.
The quadrilateral MNPQ is a rectangle because its angles serve as linear angles of right dihedral angles. Therefore, a right parallelepiped having this base must be rectangular, and, therefore, its volume is equal to the product of the area of the base MNPQ and the height BC.
But the area of MNPQ is equal to MN * MQ. Means:
Volume AC1 = MN * MQ * BC
The product MQ * BC expresses the area of parallelogram ABCD. That's why:
Volume AC 1 = (area ABCD) * MN
2. Let (Fig.) AC 1 be inclined. It is equal in size to a straight line whose base is the perpendicular section MNPQ and whose height is the edge BC.
But, according to what has been proven, the volume of a right parallelepiped is equal to the product of the area of the base and the height. Means:
Volume AC 1 = (area MNPQ) * BC
If RS is the height of the section MNPQ, then the area MNPQ = MQ * RS. That's why:
Volume AC1 = MQ * RS * BC
The product BC * MQ expresses the area of parallelogram ABCD. Hence:
Volume AC 1 = (area ABCD) * RS
Those. The volume of any parallelepiped is equal to the product of the area of the base and the height .
Consequence. If V, B and H are numbers expressing in the corresponding units the volume, base area and height of some parallelepiped, then we can write:
Task. The base of a right parallelepiped is a rhombus, the area of which is equal to S. The areas of the diagonal sections are equal to S 1 and S 2. Find the volume of the parallelepiped.
To find the volume of a parallelepiped, you need to find its height H (Fig. 242).
Let us denote the lengths of the diagonals of the base by d 1 and d 2. Then
d 1 H = S 1, d 2 H = S 2, d 1 d 2 = 2S.
From these equations we find
$$ \frac(S_1)(H)\cdot \frac(S_2)(H) = 2S, \;\; H=\sqrt(\frac(S_1 S_2)(2S)) $$
Hence,
$$ V=S\cdot H = S\sqrt(\frac(S_1 S_2)(2S))=\sqrt(\frac(S\cdot S_1\cdot S_2)(2)) $$
CHAPTER THREE
POLYhedra
II VOLUME OF PRISM AND PYRAMID
82. Basic assumptions in volumes. The size of the part of space occupied by a geometric body is called the volume of this body.
We set the task - to find an expression for this quantity in the form of a certain number that measures this quantity. 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 shown in Fig. 87), consisting of parts(P and Q), equal to the sum of the volumes of these parts.
Two bodies having the same volumes are called equal in size.
83. Unit of volume. When measuring volumes, the volume of a cube in which each edge is equal to a linear unit is taken as a unit of volume. So, cubic meters (m 3), cubic centimeters (cm 3), etc. are used.
Volume of a parallelepiped
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 rectangular parallelepiped 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 will especially consider the following three cases:
1) Measurements are expressed integers.
Let, for example, the measurements be (Fig. 88): AB = A, sun = 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 a corresponding square unit. Each of these squares can obviously accommodate one cubic unit. 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 parallelepiped contains With such units, then inside the parallelepiped we can place With such layers. Therefore, the volume of this parallelepiped is equal to abc cubic units.
2) Measurements are expressed fractional numbers . Let the dimensions of the parallelepiped 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 the share 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 in integers, namely: mqs, pns And rnq, and therefore, according to what has been proven (in case 1), the volume of the parallelepiped is equal to the product ( mqs) (pns) (rnq), if we measure this volume with a new cubic unit, corresponding to a new linear unit. One cubic unit corresponding to the previous linear unit contains ( nqs) 3 ; this means the new cubic unit is 1/( nqs) 3 former. Therefore, the volume of the parallelepiped, expressed in the previous units, is equal to:
3) Measurements are expressed irrational numbers. Let this parallelepiped (Fig. 89), which for brevity we denote by one letter Q, have the dimensions:
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 fraction. Let's take approximate values of these fractions with P in decimal places, first with a deficit and then with an excess. Values with a disadvantage will be denoted by α n , β n , γ n, values with excess α" n , β" n , γ" n. Let us lay down on the edge AB, starting from point A, two segments AB 1 = α n and AB 2 = 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 this case 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. The parallelepiped Q 1 will fit entirely inside the parallelepiped Q, and the parallelepiped Q 2 will contain the parallelepiped Q inside it.
By what has been proven (in case 2) we will have:
volume Q 1 = α n β n γ n (1)
volume Q 2 = α" n β" n γ" n (2)
Let's define the volume Q 1< объёма Q 2 .
Let's now start increasing the number P. This means that we take approximate values of the numbers α, β, γ with an increasingly greater degree of accuracy.
Let's see how the volumes of parallelepipeds Q 1 and Q 2 change.
With unlimited increase P volume Q 1 obviously increases due to 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, due to 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 the irrational numbers αβγ.
We take this limit as a measure of the volume of the parallelepiped Q: volume Q = αβγ.
It can be proven that the volume thus determined satisfies the conditions established for volume (§ 82). In fact, 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 = AB AC AD,
volume Q 1 = AB AA 1 AD,
volume Q 2 = A 1 B 1 A 1 C A 1 D 1.
Adding the last two equalities term by term and noting that A 1 B 1 = AB and A 1 D 1 = AD, we obtain:
volume Q 1 + volume Q 2 = AB AA 1 AD + AB A 1 C AD = AB AD (AA 1 + A 1 C) = 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 dimensions 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 a number With. Then, denoting its volume in the corresponding cubic units by the letter V, we can write:
V= abc.
Since the work ab expresses the area of the base, then we can say that The volume of a rectangular parallelepiped 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 of 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 in size to 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 the inclined 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 one direction.
Let’s take an arbitrary point on the continuation of one edge A and draw a perpendicular section through it abcde. Then, putting aside ahh 1 = AA 1, let's draw through A 1 perpendicular section a 1 b 1 c 1 d 1 e 1 . 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 the base, is a straight prism, which is discussed in the theorem.
Let us prove that this inclined prism is equal in size to this straight line. To do this, let us first make sure that the polyhedra a D and a 1 D 1 are equal. Their reasons abcde And a 1 b 1 c 1 d 1 e 1 are equal as the bases of a prism a 1 d; on the other hand, adding to both sides 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 side ribs, being perpendicular to the bases and correspondingly equal, will also coincide; therefore polyhedron a D will be compatible with the polyhedron a 1 D 1 ; This means that these bodies are equal. Now note that if to a straight prism a 1 d add a polyhedron a D, and to the inclined prism A 1 D we add a polyhedron a 1 D 1 equal a D, then we get the same polyhedron a 1 D. From this it follows that two prisms A 1 D and a 1 d equal in size.
87. Theorem. The volume of a parallelepiped is equal to the product of the area of the base and the height.
Previously, we proved this theorem for a rectangular parallelepiped, now we will prove it for a straight parallelepiped, and then for an inclined one.
1). Let (Fig. 93) AC 1 be a right parallelepiped, i.e., one whose base ABCD is some kind of parallelogram, and all the side faces are rectangles.
Let us take the side face AA 1 B 1 B as the base; then the parallelepiped will be
slanted. Considering it as a special case of an inclined prism, we can assert, based on the lemma of the previous paragraph, that this parallelepiped is equal in size to a right parallelepiped whose base is a perpendicular section MNPQ and whose height is BC. The quadrilateral MNPQ is a rectangle because its angles serve as linear angles of right dihedral angles; therefore, a right parallelepiped with a base of rectangle MNPQ must be rectangular and, therefore, its volume is equal to the product of its three dimensions, for which the segments MN, MQ and BC can be taken. Thus,
volume AC 1 = MN MQ BC = MN (MQ BC).
But the product MQ BC expresses the area of parallelogram ABCD, therefore
volume ACX = (area ABCD) MN = (area ABCD) BB 1.
2) Let (Fig. 94) AC 1 be an inclined parallelepiped.
It is equal in size to a straight line whose base is the perpendicular section MNPQ (i.e., perpendicular to the edges AD, BC, ...), and the height is the edge BC. But, according to what has been proven, the volume of a right parallelepiped is equal to the product of the area of the base and the height; Means,
volume AC 1 = (area MNPQ) BC.
If RS is the height of the section MNPQ, then the area MNPQ = MQ RS, therefore
volume AC 1 = MQ RS BC = (BC MQ) RS.
The product BC MQ expresses the area of the parallelogram ABCD; therefore, volume AC 1 = (area ABCOD) RS.
It now remains to prove that the segment RS represents 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 construct a perpendicular to the ABCD plane from point S, then it must lie entirely in the MNPQ plane (§ 44) and, therefore, must merge with the straight line RS, which lies in this plane and is perpendicular to MQ. This means that the segment SR is the height of the parallelepiped. Thus, the volume of an inclined parallelepiped is equal to the product of the area of the base and the height.
Consequence. If V, B and H are numbers expressing in the corresponding units the volume, base area and height of the parallelepiped, then we can write.
Any geometric body can be characterized by surface area (S) and volume (V). Area and volume are not the same thing at all. 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, packaging boxes and many other things from everyday life have a similar shape. Rooms in residential and office buildings are usually 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 randomly selected faces are called bases. A parallelepiped can be of several types. The differences are due to the angles between adjacent edges. The formulas for finding the Vs of different polygons are slightly different.
If the 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 find out 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 also of the properties of the figure. The list of basic properties of a rectangular prism is small and very easy to understand:
- The opposite sides of the figure are equal and parallel. This means that the ribs located opposite are the same in length and angle of inclination.
- All lateral faces of a right parallelepiped are rectangles.
- The four main diagonals of a geometric figure intersect at one point and are divided in half by it.
- 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 squares built on the sides of a right triangle is equal to the area of a triangle built on the hypotenuse of the same triangle.
The proof of the last property can be seen in the image below. The process of solving the problem is simple and does not require detailed explanations.
Formula for the volume of a rectangular parallelepiped
The formula for finding for all types of geometric figures is the same: V=S*h, where V is the required 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 dimensions.
Volume is usually expressed in cm3. Knowing all three values of a, b and c, finding the volume of a figure is not at all difficult. The most common type of problem in the Unified State Exam is finding the volume or diagonal of a parallelepiped. Solve many typical Unified State Exam assignments It’s impossible without the 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 determined by multiplying the perimeter of the base by the height of the parallelepiped.
Based on the first property of parallelepipeds AB = A1B1, and face B1D1 = BD. According to corollaries of the Pythagorean theorem, the sum of all angles in right triangle is equal to 180°, and the leg lying opposite the angle of 30° is equal to the hypotenuse. Applying this knowledge to a triangle, we can easily find the length of sides AB and AD. Then we multiply the obtained values and calculate the volume of the parallelepiped.
Formula for finding the volume of an inclined parallelepiped
To find the volume of an inclined parallelepiped, it is necessary to multiply the area of the base of the figure by the height lowered to the given base from the opposite corner.
Thus, the required V can be represented in the form of h - the number of sheets with a base area S, so the volume of the deck consists of the Vs of all cards.
Examples of problem solving
Tasks unified exam must be completed in certain time. Typical tasks, as a rule, do not contain large quantity calculations and complex fractions. Often a student is asked 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 Vs of the component parts.
As you can see from the example in the image above, there is nothing difficult 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 problems in advance.
In this lesson we will talk about a rectangular parallelepiped. Let's recall some of its properties. And then we will derive in detail the formulas for calculating the volume of a rectangular parallelepiped. Lesson summary "Volume of a rectangular parallelepiped" In this lesson we will talk about a rectangular parallelepiped. Let's recall some of its properties. And then we will derive in detail the formulas for calculating the volume of a rectangular parallelepiped. Earlier we already got acquainted with a rectangular parallelepiped. Recall that a parallelepiped is called rectangular if all its six faces are rectangles. An idea of the shape of a rectangular parallelepiped is given by a matchbox, box, refrigerator, etc. Let's imagine a room that has the shape of a rectangular parallelepiped. If we talk about its dimensions, we usually use the words “length”, “width” and “height”, meaning 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 parallelepiped is depicted on the screen. As its dimensions, you can take, for example, the lengths of the edges, these edges have a common vertex of the parallelepiped, - the width and The rectangular parallelepiped has the following properties: 1) the square of the diagonal of the cuboid is equal to the sum of the squares of its three dimensions. – this 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 with its dimensions in letters. Let us prove that the volume of a rectangular parallelepiped is equal, and its volume in letters. Let's denote and. , . There are two possible cases: Let's consider the first case. Measurements of decimal fractions, in which the number of decimal places does not exceed, are finite and (,). In this case, the numbers and are integers. Let us divide each edge of the parallelepiped into equal parts of length. Then, through the splitting points, we draw planes perpendicular to this edge. Then our parallelepiped 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 measurements is an infinite decimal fraction. , and represents Consider the finite decimal fractions of numbers c, -oi. , 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 valid. Similar inequalities will hold for numbers, where and: . , Where, . Let's multiply these inequalities. Then we see that. From the inequality it is clear that a parallelepiped is a parallelepiped, and itself is contained in a parallelepiped. And this suggests that. Now let's increase without limit to become as small as desired, and therefore the number will differ little from the number. . Then the number will be arbitrary. In the end, they will become equal. Those. . Q.E.D. The following corollaries are true from this theorem. First consequence. The volume of a rectangular parallelepiped is equal to the product of the area of the base and the height. Proof. Let the face with the 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 notice that the formula for calculating the volume of a cuboid is the area of the base, and is the height of the cuboid. can be written in the form where Thus, we have proven 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, we will complete a straight triangular prism with the base of a parallelepiped 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 the 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... Which is what was required. Based on the Pythagorean theorem, its diagonal is equal. Therefore, the area of the square constructed on it is twice the area of the given 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 ruler, to construct the side of a cube whose volume is twice the volume given cube, i.e. construct a segment equal to? This task was formulated in ancient times. It was called the “doubling the cube problem.” Only in 1837 did the French mathematician Pierre Laurent Wantzel 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 (dividing an arbitrary given angle into three equal angles). Let us recall that the classic unsolvable construction problems also include the problem of squaring a circle (construct a square whose area is equal to the area of the given circle). The impossibility of such a construction was proven in 1882 by the German mathematician Karl Louis Ferdinand Lindemann. Problem: find the volume of a rectangular parallelepiped with diagonal sides of the base. Solution: write a formula for calculating the volume of a rectangular parallelepiped through its measurements. cm and cm cm and From the conditions of the problem we know the length, width and diagonal of the rectangular parallelepiped, but its height is unknown. Let us remind you that. Let us express from this formula the height that the height is equal to (cm). rectangular parallelepiped. We get, and is equal to Substitute the measurements of our rectangular parallelepiped into the volume formula. Let's do the math. We find that the volume of the parallelepiped is equal to. Don’t forget to write down the answer. (cm3). Problem: square. The volume of a rectangular parallelepiped is equal to the height of the rectangular parallelepiped, if the base is cm3. Determine see. Solution: In this lesson we proved that the volume of a cuboid is equal to. Let's express the height from the formula. Hence, the height is equal. Since at the base of our rectangular parallelepiped there is a square, according to the condition, the area of the base is equal to the volume of the rectangular parallelepiped (cm2). According to the conditions of the problem, it is also known that. Hence, height (cm). Let's write down the answer. equal to Results: In this lesson we remembered the concept of a rectangular parallelepiped. They proved that the volume of a rectangular parallelepiped is equal to the product of its three dimensions. They proved that the volume of a rectangular parallelepiped can be calculated as the product of the area of the base 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.
