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SCHOLIUM 3. The side, x, of a cube equivalent to a given volume, V, is equal to the cube root of V. For x = V; and hence x = ÑV. The problem of the duplication of the cube, famous among the ancient Greek Geometers, consists in determining the side of a cube which shall be double a given cube. Now, if x and c be the sides of the two cubes, then 103 2c", or x = 0 Ņ2, that is, the side of the required cube would have to be to the side of the given cube as the cube root of 2 is to unity. Now, the square root of 2 is easily found by a geometrical construction. But the cube root of .2 cannot be so found, that is, by the simple operations of elementary Geometry, which employ no other lines than straight lines and circumferences. The problem admits of solution, however, by other methods no less rigorous than those of elementary Geometry.

PROPOSITION XVIII.

THEOREM. The volume of a parallelopipedon, and generally of any prism, is equal to the product of its base by its altitude.

First.Any parallelopipedon is equivalent to a rectangular parallelopipedon having the same altitude, and an equivalent base (Prop. XIII.). Now, the volume of the latter is equal to its base multiplied by its altitude; hence the volume of the former is likewise equal to the product of its base by its altitude.

Secondly.--Any triangular prism is half the parallelopipedon so constructed as to have the same altitude and a base twice as great (Prop. X.). But the volume of the latter is equal to its base multiplied by its altitude; hence, that of a triangular prism is equal to the product of its base (half that of the parallelopipedon) multiplied by its altitude.

Thirdly. -Any prism may be divided into triangular prisms having for their bases the different triangles which form the polygon which serves as its base, and for their common altitude the altitude of the prism. But the volume of each triangular prism is equal to its base multiplied by its altitude ; and, since the altitude is the same for all, it follows that the sum of all the partial prisms will be equal to the sum of all the triangles which constitute their bases, multiplied by the common altitude.

Hence, the volume of any polygonal prism is equal to the product of its base by its altitude.

Cor. 1. The volume of any prism is equal to the product of its right section by its lateral edge (Prop. IX.).

Cor. 2. First.-Any two prisms are equivalent when they have equal bases and equal altitudes, or when the products of their bases by their altitudes respectively are equal.

Secondly. Tivo prisms of the same altitude are to each other as their bases.

Thirdly.--Two prisms of equivalent bases are to each other as their altitudes,

PYRAMIDS.

PROPOSITION XIX.

THEOREM. The convex surface of a regular pyramid S-ABCDE is equal to the perimeter of the base, ABCDE, multiplied by half the slant height, SH. For, the convex surface, S, is composed of five

SH isosceles triangles, each equal to SAE

AE X

S

2

C

=

2

D

SH Hence, convex surface S= 5AE X

But 5AE= (AB + BC + CD .. :)

perimeter, ABCDE.

SH Therefore, S= perimeter, ABCDE X

=

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2

SCHOLIUM. The area of the base, ABCDE, being equal to the product of its perimeter by half the apothem, OH (Book IV., Prop. IX.), the whole surface of the pyramid is SH

OH perimeter ABCDE X

+ perimeter ABCDE X

2

2

or

SH + OH perimeter ABCDE X

2

PROPOSITION XX.

THEOREM.

The convex surface of the frustum, AD', of a regular pyramid, is equal to the sum of the perimeters of its bases, ABCDE, abcde, multiplied by half the slant height, Hh, of the frustum.

C

2

For this convex surface, S, is composed of five trapezoids, each equal to AEae, and therefore each

AE + ae measured by

x Hh. Therefore, AE + ae

Hh x Hh = (5AE + 5ae) * But 5AE = perimeter ABCDE, and 5ae = perimeter abcde. Hence, S= (perimeter ABCDE + perim

Нh eter abcde) x

D

S = 5

(***)

2

2

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2

SCHOLIUM. To get the whole surface of the frustum, we add to the convex surface the areas of the two bases, ABCDE, abcde.

PROPOSITION XXI

THEOREM.

E

If a pyramid, S-ABCDE, is cut by a plane, abd, parallel to its base,

1. The edges, SA, SB, SC, etc., and the altitude, SO, will be divided proportionally at a, b, c, o.

2. The section abcde will be a polygon similar to the base, ABCDE.

First.Conceive a plane to pass through the vertex, S, parallel to the planes ABCDE, abcde. Then all the edges, SA, SB, etc., and the altitude, SO, being cut by three parallel planes in the points S, A, a, B, 6, O, O, etc., will be divided proportionally (Book V., Prop. XX., Schol.), and we shall have

SA : Sa :: SB : S6 : : SO : So. Secondly. -Since ab is parallel to AB, bc to BC, cd to CD, etc., the angle abc = ABC, bcd = BCD, and so on. Also by reason of the similar triangles SAB, Sab, we have

AB : ab : : SB : Sb, and by reason of the similar triangles SBC, Sbc, we have

SB : Sb : : BC : bc ; hence

AB : ab : : BC : bc.

B

In like manner we may prove

BC : bc : : CD: cd, and so on.

Hence, the polygons ABCDE, abcde have their angles respectively equal, and their homologous sides proportional, hence they are similar,

PROPOSITION XXII.

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THEOREM. If two pyramids, S-ABCDE, S-XYZ, which have the same altitude and the bases in the same plane, be cut by the same plane parallel to the plane of the bases, the sections, abcde, xyz, will be to each other as the bases, ABCDE, XYZ.

For the polygons ABCDE, abcde, being similar (Prop. XXI.), are to each other as the squares of their homologous sides AB, ab (Book III., Prop., XXIX.). But AB : ab : : SA : Sa. Hence

ABCDE : abcde : : SA? : Sa’.
For the same reason

XYZ : xyz : : SX : Sxa.
But since abc and xyz are in one plane, we have likewise

SA : Sa : : SX : Sx (Book V., Prop. XX., Schol.); hence

ABCDE : abcde : : XYZ : XYZ,
abcde : xyz : :

ABCDE : XYZ.
COR. Hence

If the bases, ABCDE, XYZ, are equivalent, the sections made at equal altitudes are also equivalent.

X

Z

B

:

or

PROPOSITION XXIII.

THEOREM.

Two triangular pyramids which have equivalent bases and equal altitudes, are equivalent.

Let S-ABC, s-abc, be the two pyramids. Let their equivalent bases, ABC, abc, be situated in the same plane, and let AT be their common altitude.

If these pyramids are not equivalent, let s-abc be the smaller, and suppose Ax to be the altitude of a prism which, constructed in the base ABC, is equal to their difference.

Divide the common altitude, AT, into equal parts smaller than Ax, and let k be one of these parts ; through the points of division pass planes parallel to the plane of the bases : the sections made in the two pyramids by each of these planes will be equivalent (Prop. XXII., Cor.), namely, DEF to def, GHI to ghi, etc. This being granted, .

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upon the triangles ABC, DEF, GHI, etc., taken as bases, construct exterior prisms, having for edges the parts AD, DG, GK, etc., of the edge SA ; in like manner upon the triangles def, ghi, klm, etc., taken as bases, construct in the second pyramid interior prisms, having for edges the corresponding parts of sa ; all these partial prisms will have the common altitude k.

Now, the sum of the exterior prisms of the pyramid S-ABC is greater than that pyramid, and the sum of the interior prisms of the pyramid s-abc, is less than that pyramid ; hence, the difference between the sum of all the exterior prisms, and the sum of all the interior ones, should be greater than the difference between the two pyramids.

But, beginning with the bases, ABC, abc, the second exterior prism, DEFG, is equivalent to the first interior prism, defa, because their bases, DEF, def, are equivalent, and they have the same altitude, k; for the same reasons, the third exterior prism, GHIK, and the second interior prism, ghid, are equivalent; the fourth exterior and third interior ; and so on to the last of each series.

Hence, all the exterior prisms of the pyramid S-ABC, excepting the first, ABCD,

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