); showPlots(;0 noAxes0 );

Rice. 1.10: Cuboid

1.3 Properties of parallel sections in a pyramid

1.3.1 Theorems on sections in a pyramid

If the pyramid (1.11) is crossed by a plane parallel to the base, then:

1) side edges and height are divided by this plane into proportional parts;

2) in section, a polygon (abcde) is obtained, similar to the base;

3) the areas of the section and the base are related as the squares of their distances from the top.

1) The lines ab and AB can be considered as the lines of intersection of two parallel planes (base and secant) by the third plane ASB; so abkAB. For the same reason, bckBC, cdkCD.... and amkAM; thereby

aA Sa = bB Sb = cC Sc = ::: = mM Sm :

2) From the similarity of triangles ASB and aSb, then BSC and bSc, etc. we derive:

AB ab = BS bS ; BS bS = BC bc ;

AB ab = BC bc :

BC bc = CS cS ; CS cS = CD cd ;

BC bc = CD cd

We will also prove the proportionality of the remaining sides of the polygons ABCDE and abcde. Since, moreover, these polygons have the corresponding angles (as formed by parallel and equally directed sides), they are similar. The areas of similar polygons are related as the squares of similar sides; that's why

AB ab = AS as = M msS ;

set2D(1; 9; 1; 14);

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;0 dash0 );

			Rice. 1.11: Pyramid
p5 = pointsPlot(

[ 0A 0; 0 B 0; 0 C 0; 0 D 0; 0 E 0; 0 a 0; 0 b 0; 0 c 0; 0d0; 0M0; 0m0; 0S0];

); showPlots(;0 noAxes0 );

1.3.2 Consequence

For a regular truncated pyramid, the upper base is a regular polygon similar to the lower base, and the side faces are equal and equilateral trapezoids (1.11).

The height of any of these trapezoids is called the apothem of a regular truncated pyramid.

1.3.3 Parallel section theorem in a pyramid

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

Let (1.12) B and B1 be the areas of the bases of two pyramids, H the height of each of them, b and b1 the areas of the sections by planes parallel to the bases and at the same distance h from the tops.

According to the previous theorem, we will have:

H2 B1

set2D(2; 36; 2; 23);

23 );

p10 = tableplot(			;0 arrow0 );

p11 = tableplot(			;0 arrow0 );

p12 = tableplot(			;0 arrow0 );

p13 = tableplot(			;0 arrow0 );

p14 = tableplot(				;0 dash0 );

Question:

The pyramid is crossed by a plane parallel to the base. The base area is 1690dm2, and the cross-sectional area is 10dm2. In what ratio, counting from the top, does the section plane divide the height of the pyramid?

Answers:

parallel plane cuts a pyramid similar to this one (h1/h)²=s1/s (h1/h)²=10/1690=1/169 h1/h=√1/169= 1/13 jndtn 1/13

Similar questions

• Test on the topic: "Spelling of adverbs" We check the spelling of adverb suffixes, separate and continuous spelling not with adverbs, merged, separate, hyphenation adverbs Option 1. 1. Open the brackets. Mark the "third extra": a) sat (still) motionless; saw (not) hopefully; sang (not) loudly; b) not at all (not) late; not at all (not) beautiful; very (not) decent; c) (not) friendly; (not) in its own way; (not properly; d) (not) ridiculous; (not) bewildering; (not) close, but far; e) extremely (not) coercively; very (un)attractive; not at all (not) threatening; 2. "Not" is written together in all the words of the series: a) (not) true; (not)veve; (not) pleasant; not at all (not) interesting; b) (do not) wonder; (injustice; not at all (not) far; (not) cheerful; c) (not) sincerely; (not) handsome; (not) indignant; (undemanding; d) (ignorance); (not) having arrived; (not) nonsense; (at a wrong time; 3. Select a row with negative adverbs: a) not at all; nobody; nowhere; with no one; b) nowhere nobody; never; nowhere; c) not at all; not at all; nowhere; there is no need; 4. Find the "third extra": a) n ... almost scared; n ... how did not find; n ... how many times; b) n ... where to go; n ... why ask; n ... no matter how envious; c) n ... no matter how upset; n ... when not angry; n ... where to expect; 5. "Нн" is written in all the words of the series: a) beshe ... about spinning; spoke fright...oh; worked desperately...oh; b) suddenly shuddered ... oh; drew qualified ... oh; not working time…oh; c) spoke excitedly ... about; left unexpectedly ... oh; puta answered ... oh; 6. Define the sentence with an adverb: a) The meeting is excited ... about the message. b) Society was excited... oh. c) She spoke excitedly ... oh. In the adverb it is written _____________________________________ 7. Insert the missing letters. Mark the "fourth extra": a) hot ...; fresh…; brilliant ...; good…; b) more ...; melodious ...; viscous ..; sinister...; c) baggage ... m; already ... m; wear ... th; knife ... m; d) belch ... nok; skvorch ... nok; cherry ... nka; hedgehog ... nok; 8. Write down the letters denoting adverbs that are written with suffixes - a and - o: a o a) from afar ...; b) renew ...; c) deaf ...; d) right ...; e) white ...; e) request ...; g) from a young age ...; h) dry ...; i) sons ...; Write down an adverb that does not have suffixes - a and - o: ______________________________ Option 2. 1. Open the brackets. Mark the "third extra": a) not at all (not) interesting; completely (un)interesting; far (not) fun; b) (not) friendly; (not) in our way; (wrong; c) (not) harmonious; (not) friendly; (not) good, but bad; d) read (not) expressively; looked (not) bewilderedly; lived (not) far away; e) very (not) beautiful; it's never too late; extremely (not) thoughtfully; 2. "Not" is written together in all the words of the series: a) (not) a little; (not) ridiculous; (in) intelligible; (not) hiding; b) (not) carelessly; (insincerity; (not) beautiful; (not) thoughtful; c) far (not) fun; (didn't) want; (not) far away; (trouble; d) (not) on time; (fidget; (not) saying; (not) trusting; 3. Highlight a row with negative adverbs: a) nothing; nowhere; nowhere; a lot; b) not at all; there is no need; no way; nowhere; c) nothing; nobody; no one; nobody; 4. Find the "third extra": a) there was no ... where; n…why ask; n ... when he was a coachman; b) did not hurt n ... a little; n ... how much he did not grieve; n…where to stay; c) n ... where I will not go; n ... when I don’t ask; I was n ... when; 5. "N" is written in all the words of the series: a) there is no wind on the street ... o; answering thought ... about; nezhda came ... oh-negada ... oh; b) spoke wisely ... about; entered the wind ... oh; puta said ... oh; c) spun furiously ... oh; sang penetratingly ... oh; worked enthusiastically ... oh; 6. Define the sentence with an adverb: a) His decision will be considered ... oh, professionally. B) He always acts with deliberation…oh. C) Everything was carefully considered ... oh. 7. Insert the missing letters. Mark the "fourth extra": a) speak in general ...; hot…; fresh…; exhausting…; b) friend ... to; strap ... to; cockerel ... to; vish ... nka; c) more ...; protesting...; calling...; sinister...; d) doctor ... m; swift ... m; print…t; save ... t; 8. Write in the boxes letters denoting adverbs that are written with suffixes - a and - o: a o a) first ...; b) from a young age ...; c) light up ...; d) left ...; e) clean ...; e) red-hot ...; g) left ...; h) dark ...; i) for a long time ...; Write down an adverb that does not have suffixes - a and - o: ______________________________

How can you build a pyramid? On surface R construct some polygon, for example, the pentagon ABCDE. Out of plane R take the point S. Connecting the point S with segments to all points of the polygon, we get the pyramid SABCDE (fig.).

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

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

The pyramids are called triangular, quadrangular, n-gonal depending on the number of sides of the base. On fig. images of triangular, quadrangular and hexagonal pyramids are given.

The plane passing through the top of the pyramid and the diagonal of the base is called diagonal, and the resulting cross section - diagonal. On fig. 186 one of the diagonal sections hexagonal pyramid shaded.

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

The pyramid is called correct if the base of the pyramid is a regular polygon and the top of the pyramid is projected into its center.

All side edges correct pyramid are congruent isosceles triangles. In a regular pyramid, all side edges are congruent.

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

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

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

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

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

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

S side = P h / 2

i.e. the area of ​​the lateral surface of a regular pyramid is equal to half the product of the perimeter of the base and the apothem.

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

S = S ocn. + S side. .

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

V = 1 / 3 S ocn. N.

The derivation of this and some other formulas will be given in a later chapter.

Now let's build a pyramid in a different way. Let a polyhedral angle be given, for example, a five-sided one, with a vertex S (fig.).

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

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

truncated pyramid can be obtained if the pyramid is crossed by a plane parallel to the plane of the base.

On fig. the image of a quadrangular truncated pyramid is given.

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

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

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

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

If in the correct truncated n- coal pyramid through a and b n denote the lengths of the sides of the upper and lower bases, and through h- the length of the apothem, then the area of ​​\u200b\u200beach side face of the pyramid is

1 / 2 (a + b n) h

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

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

Because pa= P and nb n\u003d P 1 - the perimeters of the bases of the truncated pyramid, then

S side \u003d 1 / 2 (P + P 1) h ,

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

Section parallel to the base of the pyramid

Theorem. If the pyramid is crossed by a plane parallel to the base, then:

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

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

3) the areas of the section and the base are related as the squares of their distances from the top.

It suffices to prove the theorem for a triangular pyramid.

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

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

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

Therefore, ΔSAB ~ ΔSA 1 B 1 and

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

∆SBC ~ ∆SB 1 C 1 and

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

In this way,

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

Corresponding angles of triangles ABC and A 1 B 1 C 1 are congruent, like angles with parallel and equally directed sides. That's why

∆ABC ~ ∆A 1 B 1 C 1

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

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

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

Consequently,

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

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

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

According to the previous theorem, we will have:

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

Consequence. If B \u003d B 1, then and b = b 1 , i.e. if two pyramids with equal heights have equal bases, then the sections equidistant from the top are also equal.

Other materials

CHAPTER THREE

POLYHEDRALS

1. PARALLELEPIPED AND PYRAMID

Properties of parallel sections in a pyramid

74. Theorem. If the pyramid (dev. 83) crossed by a plane parallel to the base, then:

1) side edges and height are divided by this plane into proportional parts;

2) the cross section is a polygon (abcde ), ground-like;

3) the areas of the section and the base are related as the squares of their distances from the top.

1) Direct ab and AB can be considered as the lines of intersection of two parallel planes (base and secant) by the third plane ASB; that's why ab||AB (§ 16). For the same reason bc||BC, cd||CD, ... and at||AM; thereby

S a / a A=S b / b B=S c / c C=...=S m / m M

2) From the similarity of triangles ASB and a S b, then BSC and b S c etc. output:

AB / ab= BS / bs; BS / bs= BC / bc ,

AB / ab= BC / bc

BC / bc= CS / cs; CS / cs= CD / cd whence BC / bc= CD / cd .

We will also prove the proportionality of the remaining sides of the polygons ABCDE and abcde. Since, moreover, these polygons have equal corresponding angles (as formed by parallel and equally directed sides), they are similar.

3) The areas of similarities of polygons are related as squares of similar sides; that's why

75. Consequence. The correct truncated pyramid the upper base is a regular polygon similar to the lower base, and the side faces are equal and isosceles trapezoids(dev. 83).

The height of any of these trapezoids is called apothema regular truncated pyramid.

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

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

According to the previous theorem, we will have:

77. Consequence. If B \u003d B 1, then and b = b 1 , i.e. if two pyramids with equal heights have equal bases, then the sections equidistant from the top are also equal.