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gous sides equal and their homologous angles equal, are equal to each other. It is evident, from similar considerations, that—Any plane section LMNOP of the prism, parallel to its base, will be equal to its base ; consequently, plane sections of a prism, parallel to each other, are equal. 240. In any prism, each of the polyedral angles is composed of three plane angles. A prism in which the lateral faces are rectangles, is called a right prism ; all others are called oblique prisms. 241. Suppose, in the prism AI (fig. 130), we have the Fig 130. base given, and the two faces, AG and BII; if we apply the two faces to the base, in the same manner as they are disposed in the figure, to form the triedral angle whose vertex is at B, the magnitude of these angles will determine the inclinations of the planes. These determine the positions of the points F, G and II; the three points H, C, D, determine the plane C.I; and this being a parallelogram fixes the point I. In a similar manner it may be shown that all the other parts of the prism are fixed by the above conditions. We say, therefore— All the parts of any prism are determined by the three planes forming cither of the triedral angles, and the disposition of those planes; therefore—If two prisms have the three planes forming a triedral angle in one, equal respectively to the three planes which form a triedral angle in the other, and disposed in the same order, the prisms are equal. If the planes are respectively cqual, but the arrangement different, the prisms will be symmetrical. Of right prisms, those which are symmetrical are also equal. 242. The prism ABCDEFGH (usually designated by AG) (fig. 131), whose bases are parallelograms, is called Fig.131. a parallelopiped; it has its opposite faces equal and par- " allel. Its base ABCD is a parallelogram, therefore AD is equal and parallel to BC ; ABFE is a parallelogram, consequently AE is equal and parallel to BF ; but when two lines which cut each other are parallel to two other lines which cut each other, the plane determined by the first two is parallel to the plane determined by the second two (211); therefore the plane AEHD, is parallel to BFGC ; and they are evidently equal, their sides and angles being equal. In the same manner we may show that the two other opposite faces are equal and parallel,

We say, therefore—In a parallelopiped any two opposite faces are equal and parallel; and—A prism contained by six planes, of which each is parallel to one of the others, is a parallelopiped. 243. By comparing the plane angles at two opposite vertices in any parallelopiped, the learner will perceive that they are equal respectively, being the corresponding angles in equal parallelograms ; and as each is a triedral angle, the inclination of two faces in the one, must be equal to that of the corresponding faces in the other ; but the plane angles in one are in the inverse order of those of the other ; it will therefore follow, that—In every inclined parallelopiped the opposite triedral angles are symmetrical. 244. A parallelopiped is called rectangular when each of its faces is a rectangle. This will be the case, if, at one of the triedral angles, each of the edges is perpendicular to the plane of the other two. The rectangular parallelopiped of which each of the faces is a square, is called a cube. It is also called a regular heraedron, being contained by six equal regular polygons, and having its diedral angles all equal, and its triedral angles equal. 245. The bases of a parallelopiped are parallelograms; consequently the sum of the squares of their four diagonals, is equal to the sum of all the squares described upon the eight edges at the bases (184). If through opposite diagonals in the bases, and therefore through opposite

vertices of the body (fig. 131), two planes are passed, the

two sections thus made will be two parallelograms, four sides of which are the four diagonals of the bases, and their other sides the four lateral edges of the body. In these two sections there are four diagonals, which are evidently the diagonals of the parallelopiped; and the sum of the squares of these diagonals, must be equal to the slim of the squares of the sides (184), that is, to the sum of the squares of the twelve edges of the parallelopiped. Therefore—In a rectangular parallelopiped, the square of one of the diagonals is equal to the sum of the squares of the three edges which meet at the same verter, And–In the cube, the square of a diagonal is equal to three times the square of one of the edges. *... 246. The cube is distinguished among bodies, as the square among surfaces. All the plane angles in its faces are right-angles; all its diedral angles are right angles;

each of its triedral angles is composed of three rightangles; and all its edges are equal. This character of its angles, and the equality of its sides, give to its form a remarkable simplicity ; on which account it has been selected as a unit in the measure of bodies. ‘s

Of the Volume of Bodies, and its Measure.

247. By the volume of a body, we mean the quantity of space which the body occupies, or the whole amount of space included by its surface. The volume of a body is very distinct from its form. Two bodies may be equal in volume and very different in form. A cubic cistern may be constructed which shall contain the same quantity of water as a hogshead. If the water be drawn from the hogshead into the cistern, the quantity of water remaining the same, the whole amount of space which the water occupies will be the same ; the volume is not changed, though the form is very different. We see, therefore, that two very different geometrical bodies may be equal in volume. If two bodies are equal in volume, they are called equivalent bodies.

248. Suppose we have two parallelopipeds, AG and AL, standing upon the same base ABCD, and having their superior bases in the same plane parallel to ABC Fig.132. (fig. 132). We observe first, that the whole figure con- " sists of the parallelopiped AG, and the triangular prism BFK-CGL, and also of the parallelopiped AL, and the triangular prism AEI-DHM. If, therefore, we can show that these two triangular prisms are equal, as one of them added to one of the parallelopipeds, makes the same body as the other prism added to the other parallelopiped, the two parallelopipeds must be equal in volume.

The triangles AEI, BFK, have their sides AE and BF equal, being opposite sides of the same parallelogram ; for the same reason AI = BK; as these equal lines are also parallel the angles EAI and FBK are equal; the two triangles are therefore equal by the second case ; the faces EC and FD are also equal being opposite faces of the same parallelopiped (242); EF is equal to IK, being each equal to AB; if from each of these we take IF, we have EI = FK; EH is also equal and parallel to FG, therefore the angle IEH is equal to the angle KFG ;

Fig.133.

Fig.134.

and the two parallelograms EM and FL, having two ad-
jacent sides and the contained angle the same, are
equal ; we therefore have the three faces containing the
triedral angle whose vertex is at E, respectively equal to
the three faces which contain the corresponding triedral
angle in the other prism ; the two prisms are therefore
equal, and the parallelopipeds AG and AL are equiv-
alent.
In a similar manner we can show that the perallelo-
piped ABCD-IKLM, is equivalent to ABCD-EFGH
(fig. 133), they being upon the same base ABCD, and
having their superior bases in the same plane. So also
it may be shown that the parallelopiped ABCD-NOPQ,
whose edges are perpendicular to the base is equivalent
to the parallelopiped ABCD-EFGH; therefore the par-
allelopiped ABCD-IKLM is equivalent to ABCD-NOPQ.
But this parallelopiped AP, may be changed into the
equivalent parallelopiped ABRS-NOTU, whose bases
are rectangles (fig. 134) as the triangular prism BCR-
OPT is equal to the prism ADS-NQU.
It is evident that all the faces of this last parallelopi-
ed, are rectangles; and each of these inclined parallelo-
pipeds is equivalent to this rectangular parallelopiped.
We therefore say——Any parallelopiped may be changed
into an equivalent rectangular parallelopiped, having the
same height and an equivalent base.
249. Any parallelopiped may be divided into two
equivalent triangular prisms by a plane passing through
two opposite lateral edges; for each base of the triangular
prism is half the base of the parallelopiped, and their
heights are the same. If, therefore, an inclined paral-
lelopiped may be changed into an equivalent rectangular
parallelopiped of the same base and height; the triangular
prisms which compose the inclined parallelopiped, must
be respectively equivalent to the triangular prisms which
compose the right parallelopiped, as in either case each
prism is one half of the parallelopiped. Whence it ap-
pears that—Any inclined triangular prism may be
changed into an equivalent right prism of the same height
and an equivalent base.
250. Any prism of more than three lateral faces, may
be cut into triangular prisms, by passing planes through
opposite lateral edges. These prisms will have the same
volume when their bases are the same and their height
the same, whether their edges are inclined to the base or
not (249). We say therefore—Any two prisms having
equivalent bases and the same height are equivalent, or
equal in volume.
251. In a comparison of pyramids, we will take the
triangular pyramid, or tetraedron.
Let us suppose that the two pyramids S-ABC, s-a b c,
(fig. 135) have the same height and equivalent bases in
the same plane ; let their height be represented by AH,
perpendicular to the plane of their bases. Divide AH into
any number of equal parts, at the points w, w, y, z, and
through these points, suppose planes to be passed parallel
to the plane of the bases; the corresponding sections of
the two pyramids, made by these planes, will beequiva-
lent (234), namely DEF to def, GHI to g h i, &c.
Upon the upper side of the triangles ABC, DEF, GHI,
&c. as bases, construct prisms, which will be eacterior to
the corresponding segment of the pyramid ; and upon
the lower side of d e f, g h i, k l m, &c., construct prisms
which will be interior to the corresponding segments. It
is manifest that the sum of all the exterior prisms of the
pyramid S-ABC, is greater than the pyramid ; and
that the sum of all the interior prisms in the pyramid
s-a b c is less than the pyramid ; therefore the pyramid
S-ABC minus the pyramid s-a b c, must be less than the
difference between the sum of the interior and the sum
of the exterior prisms. But the prisms all have the same
height; and therefore, those upon equivalent bases are
equivalent; consequently the sum of all the interior
prisms in the pyramid s-a b c, is equal to the sum of all
the exterior prisms about the other pyramid, excepting
the lower prism whose base is ABC. This prism,
therefore, is the difference between the sum of the ex-
terior prisms about the pyramid S-ABC, and the sum
of the interior prisms in the pyramid s-a b c.
If we suppose the pyramids to be cut by planes only
half the distance asunder before supposed, and prisms
constructed as above ; the prism whose base is ABC
will still be the difference between the sums of the two
series of prisms; but having the same base and a less
height, its volume must be less. It is evident that as we
increase the number of prisms in the series, the volume
of that prism which is equal to the difference of the two
series, must be continually diminished; whence we may

Fig.135.

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