TEXT EXPLANATION OF THE LESSON:

Consider these items:

Building bricks, dice, microwave. These objects are united by a form.

A surface consisting of two equal parallelograms ABCD and A1B1C1D1

and four parallelograms AA1B1B and BB1C1C, CC1D1D, AA1D1D is called a parallelepiped.

The parallelograms that make up the parallelepiped are called faces. Face A1B1C1D1. Face BB1S1S. Edge ABCD.

In this case, the faces ABCD and A1B1C1D1 are more often called bases, and the remaining faces are lateral.

The sides of parallelograms are called edges of the parallelepiped. Edge A1B1. Rib CC1. Edge AD.

The edge CC1 does not belong to the bases, it is called the side edge.

The vertices of parallelograms are called the vertices of the parallelepiped.

Top D1. Pinnacle B. Pinnacle C.

Vertices D1 and B

do not belong to the same face and are called opposite.

The parallelepiped can be drawn in different ways.

The parallelepiped at the base, which is a rhombus, while the images of the faces are parallelograms.

A parallelepiped at the base, which is a square. Invisible edges AA1, AB, AD are shown as dashed lines.

The parallelepiped at the base, which is a square

A parallelepiped at the base of which lies a rectangle or parallelogram

A parallelepiped with all sides square. More often it is called a cube.

All considered parallelepipeds have properties. Let us formulate and prove them.

Property 1. Opposite faces of a parallelepiped are parallel and equal.

Consider the parallelepiped ABCDА1В1С1D1 and prove, for example, that the faces BB1C1C and AA1D1D are parallel and equal.

By the definition of a parallelepiped, the face ABCD is a parallelogram, which means that, by the property of a parallelogram, the edge BC is parallel to the edge AD.

The face ABV1A1 is also a parallelogram, which means that the edges BB1 and AA1 are parallel.

This means that two intersecting lines BC and BB1 of one plane, respectively, are parallel to two lines AD and AA1, respectively, of another plane, which means that the planes ABB1A1 and BCC1D1 are parallel.

All faces of the parallelepiped are parallelograms, which means BC=AD, BB1=AA1.

In this case, the sides of the angles B1BC and A1AD are respectively co-directed, which means they are equal.

Thus, two adjacent sides and the angle between them of the parallelogram ABB1A1 are respectively equal to two adjacent sides and the angle between them of the parallelogram BCC1D1, which means that these parallelograms are equal.

The parallelepiped also has the diagonal property. The diagonal of a parallelepiped is a segment connecting non-neighboring vertices. In the drawing, the dotted line shows the diagonals B1D, BD1, A1C.

So, property 2. The diagonals of the parallelepiped intersect at one point and the intersection point is divided in half.

To prove the property, consider the quadrilateral BB1D1D. Its diagonals В1D, BD1 are the diagonals of the parallelepiped ABCDА1В1С1D1.

In the first property, we have already found out that the edge BB1 ​​is parallel and equal to the edge AA1, but the edge AA1 is parallel and equal to the edge DD1. Hence the edges BB1 and DD1 are parallel and equal, which proves the quadrilateral BB1D1D-parallelogram. And in a parallelogram, according to the property, the diagonals B1D, BD1 intersect at some point O and this point is divided in half.

Quadrilateral BC1D1A is also a parallelogram and its diagonals C1A intersect at one point and bisect this point. The diagonals of the parallelogram C1A, BD1 are the diagonals of the parallelepiped, which means that the stated property is proved.

To consolidate theoretical knowledge about the parallelepiped, consider the problem of proof.

Points L,M,N,P are marked on the edges of the parallelepiped so that BL=CM=A1N=D1P. Prove that ALMDNB1C1P is a parallelepiped.

The face BB1A1A is a parallelogram, which means that the edge BB1 ​​is equal and parallel to the edge AA1, but by condition the segments BL and A1N, which means that the segments LB1 and NA are equal and parallel.

3) Therefore, the quadrilateral LB1NA on the basis of a parallelogram.

4) Since CC1D1D is a parallelogram, it means that the edge CC1 is equal and parallel to the edge D1D, and CM is equal to D1P by condition, so the segments MC1 and DP are equal and parallel

Therefore, quadrilateral MC1PD is also a parallelogram.

5) Angles LB1N and MC1P are equal as angles with respectively parallel and equally directed sides.

6) We have obtained that the parallelograms and MC1PD have the corresponding sides equal and the angles between them are equal, so the parallelograms are equal.

7) The segments are equal by condition, so BLMC is a parallelogram and the side BC is parallel to the side LM is parallel to the side B1C1.

8) Similarly, from the parallelogram NA1D1P it follows that the side A1D1 is parallel to the side NP and parallel to the side AD.

9) The opposite faces ABB1A1 and DCC1D1 of the parallelepiped are parallel by property, and the segments of parallel straight lines enclosed between parallel planes are equal, which means that the segments B1C1, LM, AD, NP are equal.

It is obtained that in the quadrilaterals ANPD, NB1C1P, LB1C1M, ALMD two sides are parallel and equal, which means they are parallelograms. Then our surface ALMDNB1C1P consists of six parallelograms, two of which are equal, and by definition it is a parallelepiped.

Or (equivalently) a polyhedron with six faces and each of them - parallelogram.

Types of box

There are several types of parallelepipeds:

  • A cuboid is a cuboid whose faces are all rectangles.
  • A right parallelepiped is a parallelepiped with 4 side faces that are rectangles.
  • An oblique box is a box whose side faces are not perpendicular to the bases.

Essential elements

Two faces of a parallelepiped that do not have a common edge are called opposite, and those that have a common edge are called adjacent. Two vertices of a parallelepiped that do not belong to the same face are called opposite. The line segment connecting opposite vertices is called the diagonal of the parallelepiped. The lengths of three edges of a cuboid that have a common vertex are called its dimensions.

Properties

  • The parallelepiped is symmetrical about the midpoint of its diagonal.
  • Any segment with ends belonging to the surface of the parallelepiped and passing through the middle of its diagonal is divided by it in half; in particular, all the diagonals of the parallelepiped intersect at one point and bisect it.
  • Opposite faces of a parallelepiped are parallel and equal.
  • The square of the length of the diagonal of a cuboid is equal to the sum of the squares of its three dimensions.

Basic Formulas

Right parallelepiped

Lateral surface area S b \u003d R o * h, where R o is the perimeter of the base, h is the height

Total surface area S p \u003d S b + 2S o, where S o is the area of ​​\u200b\u200bthe base

Volume V=S o *h

cuboid

Lateral surface area S b \u003d 2c (a + b), where a, b are the sides of the base, c is the side edge of the rectangular parallelepiped

Total surface area S p \u003d 2 (ab + bc + ac)

Volume V=abc, where a, b, c are the dimensions of the cuboid.

Cube

Surface area: S=6a^2
Volume: V=a^3, Where a- the edge of the cube.

Arbitrary box

The volume and ratios in a skew box are often defined using vector algebra. The volume of a parallelepiped is equal to the absolute value of the mixed product of three vectors defined by the three sides of the parallelepiped emanating from one vertex. The ratio between the lengths of the sides of the parallelepiped and the angles between them gives the statement that the Gram determinant of these three vectors is equal to the square of their mixed product: 215 .

In mathematical analysis

In mathematical analysis, under an n-dimensional rectangular parallelepiped B understand many points x = (x_1,\ldots,x_n) kind B = \(x|a_1\leqslant x_1\leqslant b_1,\ldots,a_n\leqslant x_n\leqslant b_n\)

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An excerpt characterizing the Parallelepiped

- On dit que les rivaux se sont reconcilies grace a l "angine ... [They say that the rivals reconciled thanks to this illness.]
The word angine was repeated with great pleasure.
- Le vieux comte est touchant a ce qu "on dit. Il a pleure comme un enfant quand le medecin lui a dit que le cas etait dangereux. [The old count is very touching, they say. He cried like a child when the doctor said that dangerous case.]
Oh, ce serait une perte terrible. C "est une femme ravissante. [Oh, that would be a great loss. Such a lovely woman.]
“Vous parlez de la pauvre comtesse,” said Anna Pavlovna, coming up. - J "ai envoye savoir de ses nouvelles. On m" a dit qu "elle allait un peu mieux. Oh, sans doute, c" est la plus charmante femme du monde, - said Anna Pavlovna with a smile over her enthusiasm. - Nous appartenons a des camps differents, mais cela ne m "empeche pas de l" estimer, comme elle le merite. Elle est bien malheureuse, [You are talking about the poor countess... I sent to find out about her health. I was told that she was a little better. Oh, without a doubt, this is the most beautiful woman in the world. We belong to different camps, but this does not prevent me from respecting her according to her merits. She is so unhappy.] Anna Pavlovna added.
Believing that with these words Anna Pavlovna slightly lifted the veil of secrecy over the countess's illness, one careless young man allowed himself to express surprise that famous doctors were not called, but a charlatan who could give dangerous means was treating the countess.
“Vos informations peuvent etre meilleures que les miennes,” Anna Pavlovna suddenly lashed out venomously at the inexperienced young man. Mais je sais de bonne source que ce medecin est un homme tres savant et tres habile. C "est le medecin intime de la Reine d" Espagne. [Your news may be more accurate than mine... but I know from good sources that this doctor is a very learned and skillful person. This is the life physician of the Queen of Spain.] - And thus destroying the young man, Anna Pavlovna turned to Bilibin, who in another circle, picking up the skin and, apparently, about to dissolve it, to say un mot, spoke about the Austrians.
- Je trouve que c "est charmant! [I find it charming!] - he said about a diplomatic paper, under which the Austrian banners taken by Wittgenstein were sent to Vienna, le heros de Petropol [the hero of Petropolis] (as he was called in Petersburg).
- How, how is it? Anna Pavlovna turned to him, rousing silence to hear mot, which she already knew.
And Bilibin repeated the following authentic words of the diplomatic dispatch he had compiled:
- L "Empereur renvoie les drapeaux Autrichiens," Bilibin said, "drapeaux amis et egares qu" il a trouve hors de la route, [The Emperor sends the Austrian banners, friendly and misguided banners that he found off the real road.] - finished Bilibin loosening the skin.
- Charmant, charmant, [Charming, charming,] - said Prince Vasily.
- C "est la route de Varsovie peut etre, [This is the Warsaw road, maybe.] - Prince Hippolyte said loudly and unexpectedly. Everyone looked at him, not understanding what he wanted to say with this. Prince Hippolyte also looked around with cheerful surprise around him. He, like others, did not understand what the words he said meant. During his diplomatic career, he noticed more than once that words suddenly spoken in this way turned out to be very witty, and just in case, he said these words, "Maybe it will turn out very well," he thought, "but if it doesn't, they'll be able to arrange it there." Anna Pavlovna, and she, smiling and shaking her finger at Ippolit, invited Prince Vasily to the table, and, bringing him two candles and a manuscript, asked him to begin.

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Theorem. In any parallelepiped, opposite faces are equal and parallel.

So, the faces (Fig.) BB 1 C 1 C and AA 1 D 1 D are parallel, because two intersecting lines BB 1 and B 1 C 1 of one face are parallel to two intersecting lines AA 1 and A 1 D 1 of the other. These faces are equal, since B 1 C 1 =A 1 D 1 , B 1 B=A 1 A (as opposite sides of parallelograms) and ∠BB 1 C 1 = ∠AA 1 D 1 .

Theorem. In any parallelepiped, all four diagonals intersect at one point and are divided in half at it.

Take (fig.) in a parallelepiped any two diagonals, for example, AC 1 and DB 1, and draw straight lines AB 1 and DC 1.


Since the edges AD and B 1 C 1 are respectively equal and parallel to the edge BC, they are equal and parallel to each other.

As a result, the figure ADC 1 B 1 is a parallelogram in which C 1 A and DB 1 are diagonals, and in the parallelogram the diagonals intersect in half.

This proof can be repeated for every two diagonals.

Therefore, the diagonal AC 1 intersects with BD 1 in half, the diagonal BD 1 with A 1 C in half.

Thus, all diagonals intersect in half and, therefore, at one point.

Theorem. In a cuboid, the square of any diagonal is equal to the sum of the squares of its three dimensions.

Let (fig.) AC 1 be some diagonal of a rectangular parallelepiped.


After drawing AC, we get two triangles: AC 1 C and ACB. Both are rectangular.


the first because the box is straight, and therefore the edge CC 1 is perpendicular to the base,

the second is because the parallelepiped is rectangular, which means that it has a rectangle at its base.

From these triangles we find:

AC 2 1 = AC 2 + CC 2 1 and AC 2 = AB 2 + BC 2


Therefore, AC 2 1 = AB 2 + BC 2 + СС 2 1 = AB 2 + AD 2 + AA 2 1

Consequence. In a cuboid, all diagonals are equal.

A parallelepiped is a prism whose bases are parallelograms. In this case, all edges will parallelograms.
Each parallelepiped can be considered as a prism in three different ways, since every two opposite faces can be taken as bases (in Fig. 5 faces ABCD and A "B" C "D", or ABA "B" and CDC "D", or BC "C" and ADA "D").
The body under consideration has twelve edges, four equal and parallel to each other.
Theorem 3 . The diagonals of the parallelepiped intersect at one point, coinciding with the midpoint of each of them.
The parallelepiped ABCDA"B"C"D" (Fig. 5) has four diagonals AC", BD", CA", DB". We must prove that the midpoints of any two of them, for example, AC and BD, coincide. This follows from the fact that the figure ABC "D", which has equal and parallel sides AB and C "D", is a parallelogram.
Definition 7 . A right parallelepiped is a parallelepiped that is also a straight prism, that is, a parallelepiped whose side edges are perpendicular to the base plane.
Definition 8 . A rectangular parallelepiped is a right parallelepiped whose base is a rectangle. In this case, all its faces will be rectangles.
A rectangular parallelepiped is a right prism, no matter which of its faces we take as the base, since each of its edges is perpendicular to the edges coming out of the same vertex with it, and will therefore be perpendicular to the planes of the faces defined by these edges. In contrast, a straight, but not rectangular, box can be viewed as a right prism in only one way.
Definition 9 . The lengths of three edges of a cuboid, of which no two are parallel to each other (for example, three edges coming out of the same vertex), are called its dimensions. Two |rectangular parallelepipeds having correspondingly equal dimensions are obviously equal to each other.
Definition 10 A cube is a rectangular parallelepiped, all three dimensions of which are equal to each other, so that all its faces are squares. Two cubes whose edges are equal are equal.
Definition 11 . An inclined parallelepiped in which all edges are equal and the angles of all faces are equal or complementary is called a rhombohedron.
All faces of a rhombohedron are equal rhombuses. (The shape of a rhombohedron has some crystals of great importance, for example, crystals of Iceland spar.) In a rhombohedron, one can find such a vertex (and even two opposite vertices) that all angles adjacent to it are equal to each other.
Theorem 4 . The diagonals of a rectangular parallelepiped are equal to each other. The square of the diagonal is equal to the sum of the squares of three dimensions.
In a rectangular parallelepiped ABCDA "B" C "D" (Fig. 6), the diagonals AC "and BD" are equal, since the quadrilateral ABC "D" is a rectangle (line AB is perpendicular to the plane BC "C", in which lies BC ") .
In addition, AC" 2 =BD" 2 = AB2+AD" 2 based on the hypotenuse square theorem. But based on the same theorem AD" 2 = AA" 2 + +A"D" 2; hence we have:
AC "2 \u003d AB 2 + AA" 2 + A "D" 2 \u003d AB 2 + AA "2 + AD 2.