He divided mathematicians who are engaged in the development of the powers of different figures (specks, lines, cuts, two-dimensional and trivi- mer objects), their reconciliation and mutual expansion. For clarity, the calculation of the geometry is subdivided into planometry and stereometry. AT… … Collier Encyclopedia
Geometry of expanse of space, greater than three; the term zastosovuєtsya to quiet expanses, the geometry of such a bula is more or less destined for the vipad of three vimіryuvan and only then it is narrowed down to the number of vimіruvann n>3, the first for all Euclidean expanse, ... Mathematical Encyclopedia
N of the world Euclidean geometry uzgalnennya Euclidean geometry on the expanse of a large number of worlds. If the physical space is trivimirnim, and the human organs are honored for the spying of three vimiriv, N mirna ... Wikipedia
This term may have other meanings, div. Pyramidatsu (meaning). The validity of this section of the article was put under sumniv. It is necessary to verify the accuracy of the facts, which have been shared by whom. On the side of the discussion, you can but ... Wikipedia
- (Constructive Solid Geometry, CSG) technology that wins in the modeling of solid bodies. Constructive block geometry is most often, but not zavzhd, є by the method of modeling in three-dimensional graphics and CAD. Vaughn allows you to create a folding scene chi ... Wikipedia
Constructive Solid Geometry (CSG) is a technology that is used in the modeling of solid bodies. Constructive block geometry is most often, but not zavzhd, є by the method of modeling in three-dimensional graphics and CAD. Vaughn ... ... Wikipedia
This term may have other meanings, div. Obsyag (meaning). It is an additive function in terms of multiplier (setting), which characterizes the space of the area, as it occupies it. On the back of the tongue and zastosovalos without strict ... Wikipedia
Cube Type Regular bagatohedron Face square Vertices Edges Faces ... Wikipedia
It is an additive function in terms of multiplier (setting), which characterizes the space of the area, as it occupies it. On the back of the tongue and zastosovuvalos without a strict designation of the trivi- mer bodies of the trivi- mer Euclidean expanse.
Part of the open space, surrounded by the succes- sion of the final number of flat bollards (div. GEOMETRIYA), closed in such a way that the skin side of any bollard is the side of exactly one other bollard (called ... Collier Encyclopedia
For the help of this video lesson, everyone can independently learn from the topic “Understanding the bagatohedron. Prism. Surface area of the prism. For an hour of employment, the reader will be able to tell the difference between those who have such geometrical positions, like a bagatohedron and prisms, to give specific indications and explain their essence on specific butts.
For the help of this lesson, everyone can learn on their own from the topic “Understanding the bagatohedron. Prism. Surface area of the prism.
Appointment. The surface, which is formed from the bagatokutnikіv and surrounds the deake, is geometrically body, called the rich-faceted surface or the rich-faceted.
Let's take a look at the following:
1. Tetrahedron ABCD- Tse surface, folded from chotiriokh trikutnikov: ABC, adb, bdcі ADC(Fig. 1).
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2. Paralepiped ABCDA 1 B 1 C 1 D 1- Tse surface, folded from six parallelograms (Fig. 2).
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The main elements of a bagatohedron are faces, edges, and vertices.
Borders - tse bagatokutniki, what to make a bagatokonnik.
Edges are the sides of faces.
The vertices are the ends of the ribs.
Look at the tetrahedron ABCD(Fig. 1). Significantly yoga basic elements.
Grani: tricoutniks ABC, ADB, BDC, ADC.
Ribs: AB, AC, ND, DC, AD, BD.
Peaks: A, B, Z, D.
Let's look at the parallelepiped ABCDA 1 B 1 C 1 D 1(Fig. 2).
Grani: parallelograms AA 1 D 1 D, D 1 DCC 1, BB 1 Z 1 Z, AA 1 V 1 V, ABCD, A 1 B 1 C 1 D 1 .
Ribs: AA 1 , BB 1 , SS 1 , DD 1 , AD, A 1 D 1 , B 1 C 1 , BC, AB, A 1 B 1 , D 1 C 1 , DC.
Peaks: A, B, C, D, A1, B1, C1, D1.
Importantly, let's call it a prism.
ABSA 1 V 1 Z 1(Fig. 3).
Rice. 3
Rivnі tricoutniks ABCі A 1 B 1 C 1 spreading in parallel planes α and β so that the ribs AA 1, BB 1, SS 1 parallel.
Tobto ABSA 1 V 1 Z 1- trikutna prism, like:
1) Tricks ABCі A 1 B 1 C 1 equal.
2) Tricks ABCі A 1 B 1 C 1 spreading in parallel planes α and β: ABC║A 1 B 1 C (α ║ β).
3) Ribs AA 1, BB 1, SS 1 parallel.
ABCі A 1 B 1 C 1- Give me a prism.
AA 1, BB 1, SS 1- Bichni prism ribs.
Just from a fair point H 1 one plane (for example, β) lower the perpendicular Mon 1 on the plane α, whose perpendicular is called the height of the prism.
Appointment. If the ribs are perpendicular to the bases, then the prism is called straight, and otherwise - frail.
Let's look at the prism ABSA 1 V 1 Z 1(Fig. 4). The prism is straight. Tobto, її bіchnі ribs are perpendicular to the foundations.
For example, rib AA 1 perpendicular to the plane ABC. Edge AA 1є height tsієї prism.
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Respectfully, what a bіchna line AA 1 V 1 V perpendicular to bases ABCі A 1 B 1 C 1 shards won't pass through the perpendicular. AA 1 to the basics.
Now we can look at the frail prism ABSA 1 V 1 Z 1(Fig. 5). Here the edge is not perpendicular to the plane of the base. How to drop from the point A 1 perpendicular A 1 H on the ABC, whose perpendicular will be the height of the prism. Dear, scho vіdrіzok AN- tse projection vіdrіzka AA 1 on the flat ABC.
Todі kut mіzh straight AA 1 that flat ABC tse kut mizh straight AA 1і її AN projection onto a plane, tobto cut A 1 AH.
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Let's look at a chotiricutnu prism ABCDA 1 B 1 C 1 D 1(Fig. 6). Let's look at how to get out.
1) Chotiriokhkutnik ABCD hello to chotirikutnik A 1 B 1 C 1 D 1: ABCD = A 1 B 1 C 1 D 1.
2) Chotirikutniki ABCDі A 1 B 1 C 1 D 1 ABC║A 1 B 1 C (α ║ β).
3) Chotirikutniki ABCDі A 1 B 1 C 1 D 1 spread out so that the side ribs are parallel, so: AA 1 ║BB 1 ║SS 1 ║DD 1.
Appointment. Diagonal of the prism - tse vіrіzok, scho spoluchaє two vertices of the prism, not overlapping to one facet.
For example, AC 1- diagonal of a chotiricut prism ABCDA 1 B 1 C 1 D 1.
Appointment. Yakshcho bichne rib AA 1 perpendicular to the plane of the base, then such a prism is called a straight line.
Rice. 6
A private view of a chotiric prism is a paralepiped. Paralepiped ABCDA 1 B 1 C 1 D 1 shown in fig. 7.
Let's take a look, like a wine of power:
1) At the base lie equal figures. In this direction - equal parallelograms ABCDі A 1 B 1 C 1 D 1: ABCD = A 1 B 1 C 1 D 1.
2) Parallelograms ABCDі A 1 B 1 C 1 D 1 lie near parallel planes α and β: ABC║A 1 B 1 C 1 (α ║ β).
3) Parallelograms ABCDі A 1 B 1 C 1 D 1 roztashovanі in such a rank that the bіchnі ribs are parallel between themselves: AA 1 ║BB 1 ║SS 1 ║DD 1.
Rice. 7
3 points A 1 omit perpendicular AN on the flat ABC. Vіdrіzok A 1 Hє curls.
We look like a six-cut prism (Fig. 8).
1) At the base lie equal six-pieces ABCDEFі A 1 B 1 C 1 D 1 E 1 F 1: ABCDEF= A 1 B 1 C 1 D 1 E 1 F 1.
2) Squares of shestikutniki ABCDEFі A 1 B 1 C 1 D 1 E 1 F 1 parallel, so the foundations lie at parallel planes: ABC║A 1 B 1 C (α ║ β).
3) Six-piece ABCDEFі A 1 B 1 C 1 D 1 E 1 F 1 spread out so that all the side ribs are parallel to each other: AA 1 ║BB 1 …║FF 1.
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Appointment. If the edge is perpendicular to the plane of the base, then such a six-pointed prism is called a straight line.
Appointment. A straight prism is called correct, because its foundations are correct bagatokutniki.
Let's look at the correct triangular prism ABSA 1 V 1 Z 1.
Rice. nine
trikutna prism ABSA 1 V 1 Z 1- it is correct, tse, that at the bases lie the correct tricots, so that all sides of these trikutniks are equal. So the prism is straight. Also, the rib is perpendicular to the plane of the base. And tse means that all bіchnі faces are equal rectangles.
Otzhe, yakscho trikutna prism ABSA 1 V 1 Z 1 is correct, then:
1) The side edge is perpendicular to the plane of the base, tobto є in height: AA 1 ⊥ ABC.
2) It is based on the correct tricot: ∆ ABC- correct.
Appointment. The total area of the surface of the prism is the sum of the areas of її faces. be appointed S renew.
Appointment. The area of the beaded surface is the sum of the areas of the mustaches of the beetle faces. be appointed S bik.
The prism may have two supports. Todi total surface area of the prism:
S surf = S bik + 2S main.
The square of the square of the surface of the straight prism is more advanced than the perimeter of the base and the height of the prism.
The proof will be carried out with the butt of a triangular prism.
Given: ABSA 1 V 1 Z 1- Direct prism, tobto. AA 1 ⊥ ABC.
AA1 = h.
Bring: S bik \u003d R main ∙ h.
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proof.
trikutna prism ABSA 1 V 1 Z 1- Straight, that means AA 1 B 1 B, AA 1 C 1 C, BB 1 C 1 C - rectangles.
We know the area of \u200b\u200bbіchnoi surface like the sum of the squares of rectangle AA 1 V 1 V, AA 1 Z 1 Z, BB 1 Z 1 Z:
S bіk \u003d AB ∙h + BC ∙h + CA ∙h \u003d (AB + BC + CA) ∙h \u003d P main ∙h.
We take S bik \u003d R main ∙ h, what it was necessary to bring.
We got acquainted with the rich-faceted, prism, її different types. They brought the theorem about the bіchnіy surface of the prism. On the approaching urn, mi virishuvatimemo zavdannya on the prism.
- Geometry. Grade 10-11: a teacher for students of zagalnosvitnіkh installations (basic and profile level) / I. M. Smirnova, V. A. Smirnov. - 5th edition, corrected and supplemented - M.: Mnemozina, 2008. - 288 p. : il.
- Geometry. Grade 10-11: Handyman for the primordial lighting of the primary mortgages / Sharigin I. F. - M.: Bustard, 1999. - 208 p.: il.
- Geometry. Grade 10: A handyman for sacral and enlightening mortgages with destruction and profile studies of mathematics /Є. V. Potoskuev, L. I. Zvalich. - 6th sightings, stereotype. - M.: Bustard, 008. - 233 p. :il.
- Yaclas().
- Shkolo.ru ().
- old school ().
- wikihow().
- What is the minimum number of faces possible for a prism? How many vertices, edges does such a prism have?
- What is a prism, how can it be exactly 100 ribs?
- The side rib is heeled to the surface under the apex of 60°. To know the height of the prism, as if the rib is healthy 6 div.
- A straight triangular prism has equal ribs. The area of \u200b\u200bb_chnї surface becomes 27 cm2. Know the area of the surface of the prism again.
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Description of the presentation with four slides:
1 slide
Description of the slide:
2 slide
Description of the slide:
Appointment 1. A bagatohedron, two faces of which are the same bagatokniki, which lie near parallel planes, and be it two edges, which do not lie near these planes, are parallel, are called a prism. The term "prism" of the Greek pohodzhennya and literally means "vіdpilane" (body). Bagatokutniki, which are near parallel planes, are called prism supports, and other faces - beech faces. On top of the prism, in such a rank, it is composed of two equal bagatokutnikiv (podstav) and parallelograms (bіchnih faces). Distinguish prisms trikutnі, chotirikutnі, p'yatikutnі thinly. fallow in view of the number of tops of the base.
3 slide
Description of the slide:
Usі prismi podіlyayutsya on the straight line and pohili. (Fig. 2) If the edge of the prism is perpendicular to the plane of the її base, then such a prism is called a straight line; if the edge of the prism is perpendicular to the plane of the її base, then such a prism is called frail. At a straight prism, the bichnі faces are rectangular. Perpendicular to the planes of the substav, the kіnci of which lie on these planes, is called the height of the prism.
4 slide
Description of the slide:
Prism power. 1. Submit prisms with equal bagatokutniks. 2. Bichni faces of the prism are parallelograms. 3. Bichni ribs of the prism are even.
5 slide
Description of the slide:
The area of the surface of the prism is the area of the surface of the prism. The surface of the bagatohedron is formed from the final number of bagatokutnikov (facets). The area of the surface of the bagatohedron is the sum of the areas of all its faces. The area of the surface prisms (Spr) is equal to the sum of the areas of the side faces (the area of the side surfaces Sside) and the area of the two bases (2Sosn) - equal bagatokutnikov: Spov=Sside+2Sosn. Theorem. The area of the flank surface of the prism is equal to the perimeter of the perimeter perpendicular to the cut and the back of the flank rib.
6 slide
Description of the slide:
Proof. The bichni faces of the straight prism are rectangular, the bases of which are the sides of the base of the prism, and the heights are equal to the heights of the prism. Sbіk surface prisms are more expensive sums of S assigned trikutnikіv, tobto. dorivnyuє sum tvorіv storіn foundation height h. Winning the multiplier h for the arms, take away the sum of the sides of the arms, substituting the prism, tobto. perimeter P. Later, Sside = Ph. The theorem has been completed. Last. The square of the square surface of the straight prism is more advanced than the perimeter and the base of the height. Indeed, in a straight prism, the base can be seen as perpendicular to the ribs, and the edge is the height.
7 slide
Description of the slide:
Pererіz prism 1. Peretin prism with a plane parallel to the base. At the peretina, the bagatokutnik is established, equal to the bagatokutnik, which lies on the base. 2. Peretin prism with a flatness to pass through two non-superficial ribs. At the peritoneum, a parallelogram is established. Such an overcut is called a diagonal overlay of a prism. Some vipadkas can have a rhombus, a rectangle or a square.
8 slide
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9 slide
Description of the slide:
Appointment 2. A straight prism, the basis of which is a regular bagatokutnik, is called a regular prism. The power of the correct prism 1. Zasnuvannya correct prism є correct bagatokutnikami. 2. Bichni faces of regular prism and equal rectangles. 3. Bichni ribs of the right prism are equal.
10 slide
Description of the slide:
Resection of the correct prism. 1. Retin correct prism with a plane parallel to the base. At the perimeter, the correct bagatokutnik is established, equal to the bagatokutnik, who lies at the base. 2. Peretin of the correct prism with a plane to pass through two non-adherent lateral ribs. At the peretina, a straight cut is established. Some vipadkas may have a square.
11 slide
Description of the slide:
Symmetry of a regular prism 1. The center of symmetry with a paired number of sides of the base is the cross point of the diagonals of a regular prism (Fig. 6)
Diagonal cross-sections A prism rib with a plane that passes through the diagonal of the base and two side ribs that adjoin to it is called a diagonal prism rib. The crossbar of the pyramid with a plane that passes through the diagonal of the base and the top is called the diagonal crossbar of the pyramid. Let the plane cross the pyramid and be parallel to the її base. Part of the pyramid, laid between the flat and the base, is called a truncated pyramid. Peretin pyramid is also called the basis of the truncated pyramid.
Pobudova perezіv When pbudovі pererіzіv bahatohedral, basic є pobudovy points of the line of the straight line and the plane, as well as the line of the line of the two planes. If two points A and B are given on the straight line and in their projections A' and B' on the plane, the crossing point of these lines and the plane will be the crossing point of the straight lines AB and A'B' In this case, three points A, B, C of the plane are given in the projections A', B', C' to the other plane, then the significant line of the line of these planes is to find the points P and Q of the line of lines AB and AC to the other plane. The straight line PQ will be a line of flats.
To the right 1 Make the cross section of the cube a plane that passes through points E, F, that lies on the edges of the cube and vertex B. Solution. To encourage the cut of the cube, to pass through points E, F and vertex B, it is necessary to cross points E and B, F and B. Through points E and F we draw a straight line, parallel to BF and BE, obviously. Taking away the parallelogram BFGE will be a shukani peretin.
To the right 2 Make the cross section of the cube flat, which will pass through the points E, F, G, but lie on the edges of the cube. Solution. To induce a cross section of the cube to pass through the points E, F, G, we draw a straight line EF i and signifi cantly її a cross point in AD. Let Q be the intersection point of the lines PG and AB. Z'ednaёmo points E і Q, F і G. Otriman's trapezium EFGQ will be a shukanim peretina.
To the right 3 Make the floor of the cube flat, which will pass through the points E, F, G, but lie on the edges of the cube. Solution. To induce a cross section of the cube to pass through the points E, F, G, we draw a straight line EF i and signifi cantly її a cross point in AD. Significantly Q, R are the breakpoints of the line PG from AB and DC. The crossing point FR c СС 1 is significant. Three points E і Q, G і S.
To the right 4 Make the cross section of the cube flat, which will pass through the points E, F, G, but lie on the edges of the cube. Solution. To induce the cross section of the cube to pass through the points E, F, G, we know the point P, the cross section of the straight line EF and the plane of the face ABCD. Significantly Q, R are the crossing points of the line PG 3 AB and CD. Draw the line RF and signifi- cantly S, T її the breakpoints in CC 1 and DD 1. Draw the line TE and signifi- cant U six-piece EUFSGQ will be shukanim peretin.
Right 5 Make the edge of the cube flat enough to pass through points E, F, G, so that the faces BB 1 C 1 C, CC 1 D 1 D, AA 1 B 1 B lie perfectly. Solution. From these points, let's drop the perpendiculars EE', FF', GG' to the plane of the face ABCD, and we know the points I and H of the span of the straight lines FE and FG with the plane. IH will be the line of the line of the shukano plane and the plane of the facet ABCD. Significantly Q, R are the cross-points of the straight line ї IH z AB and BC. Draw lines PG and QE and signifi- cantly R, S їx crossing points AA 1 and CC 1. Draw lines SU, UV and RV, parallel to PR, PQ and QS. Taking off the six-piece RPQSUV will be a shukani peretin.
To the right 6 Make the cube's span a plane that passes through points E, F, that lies on the edges of the cube, parallel to the diagonal BD. Solution. Draw straight lines FG and EH, parallel to BD. Draw a straight line FP, parallel to EG, and draw points P and G. Draw points E and G, F and H.
Try to cross the prism ABCA 1 B 1 C 1 with a plane to pass through points E, F, G. Right 8 Solutions. Draw the line E and F. Draw the line FG and її the line point with CC 1 signifi- cantly H. Draw the line EH and її the line point with A 1 C 1 signifi- cantly I. Draw the point I and G. .
Try to cross the prism ABCA 1 B 1 C 1 with a plane, so as to pass through points E, F, G. Right 9 Solutions. Draw a straight line EG і signifi- cantly H and I її crossing point s CC 1 and AC. We draw the line IF and її the line point with AB is signifi- cantly K. We draw the line FH and її the line point with B 1 C 1 is L significantly.
Try to cross the prism ABCA 1 B 1 C 1 with a plane parallel to AC 1, so as to pass through the points D 1. Right 10 Solutions. Through the point D we draw a line parallel to AC 1 and it is signifi- cantly E її the crossing point with the line BC 1. This point lies on the plane of the face ADD 1 A 1. Draw a line through point D parallel to line FD and signifi- cantly G point її traverse with edge A 1 C 1 H – point її traverse with line A 1 B 1. Draw line DH і significantly P її point of traverse with the edge AA 1. With the edge of the point P and G.
Encourage the plane to cross the prism ABCA 1 B 1 C 1 through points E on the edge BC, F on the edge ABB 1 A 1 and G on the edge ACC 1 A 1. Right 11 Solutions. Let us draw the line GF and find the point H її over the line with the plane ABC. Draw the line EH, and it is significant P and I її crossing points over AC and AB. Draw a straight line PG and IF, that is significant S, R and Q їx the points of the break s A 1 C 1, A 1 B 1 and BB 1. We draw the points E and Q, S and R. .
Encourage the perimeter of a regular six-curve prism with a plane to pass through points A, B, D 1. Right 12 Solutions. Respectfully, we pass through the point E 1. Draw the line AB and find її points of the line K and L with lines CD and FE. Let's draw lines KD 1, LE 1 and know їx points of the line P, Q іz lines CC 1 and FF 1. The six-curve line ABPD 1 E 1 Q will be the line of the line.
Induce a cross section of a regular six-curve prism with a plane to pass through points A, B', F'. Right 13 Decision. Let's draw AB' and AF'. Draw a straight line through the point B', parallel to AF' and її the crossing point from EE 1 is signifi- cantly E'. Draw a straight line through the point F', parallel to AB' and її the crossing point in CC 1 is signifi- cantly C'. Through the points E' and C' we draw a straight line parallel to AB' and AF', and the crossing points D 1 E 1 i C 1 D 1 are signifi- cantly D', D”. We need points B', C'; D', D"; F', E'. Otrimany seven-piece AB'C'D'D'E'F' will be a shukani peretin.
Encourage the perimeter of a regular six-cut prism with a plane, like passing through the points F', B', D'. Right 14 Decision. Let's draw a straight line F'B' and F'D' and find the crossing points P and Q with the area ABC. Draw a straight line PQ. Significantly R is the break point PQ and FC. The break point F'R and CC 1 is meaningfully C'. We need points B', C' and C', D'. Draw a straight line through the point F', parallel to C'D' and B'C'; We need points A', B' and E', D'. Taking off the six-piece A'B'C'D'E'F' will be a shukani peretin.