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Proposition 4. Ratio of Tetrahedrons

244. Theorem. The volumes of two tetrahedrons which have a trihedral angle of one equal to a trihedral angle of the other are to each other as the products of the three edges of these trihedral angles.

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Given S-ABC and S'-A'B'C', two tetrahedrons with trihedral ZS=trihedral ZS'; and V and V', the volumes.

V Prove that

SA · SB · SC

V'S'A'. S'B'. S'C" Proof. Place tetrahedron S'-A'B'C' upon S-ABC so that trihedral ZS shall coincide with its equal, Zs'.

Let CD and C'D' be Is to the plane SAB, and let the plane of CD and C'D' intersect SAB in SD'D.

The faces SAB and SA'B' may be taken as the bases and CD and c'D' as the altitudes of the triangular pyramids C-SAB and C'-SA'B' respectively.

V SABCD SAB CD Then

$123 }'

SA'B'C'D' SA'B' C'D'
SAB SA· SB
But

CD SC
$ ,

($ 24, 1). SA'B' SA': SB'

C'D' SC'
V SA · SB · SC SA · SB · SC

Ax. 5
V SA'. SB': SC S'A': S'B'. S'C''

.

Proposition 5. Ratio of Tetrahedrons

245. Theorem. The volumes of two similar tetrahedrons are to each other as the cubes of any two corresponding edges.

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Given P-ABC and Pl-A'B'C', two similar tetrahedrons; Vand V', the volumes; and PB and P'B', two corresponding edges. Prove that

V

PB
V'

P'B'S

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Given $ 237

$ 244

Proof. Since P-ABC is similar to P'-A'B'C',

the corresponding polyhedral As are equal. Hence

V PB: PC · PA
V' P'B':P'C'. P'A

V PB PC PA or

V'. P'B' P'C' P'A'

PB PC PA But

P'B'

P'C' P'A

PB
Substituting for its equals, we have

P'B'

V PB PB PB
V' P'B' P'B' P'B

V PB or

V

PB8

$ 239

Ax. 5

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Proposition 6. Ratio of Polyhedrons 246. Theorem. The volumes of two similar polyhedrons are to each other as the cubes of any two corresponding edges.

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Given P and P', two similar polyhedrons; V and V', the volumes; and EB and E'B', any two corresponding edges.

Prove that V:V' = EB': E'B''.

Proof. Let P and P' be separated into tetrahedrons similar each to each and similarly placed ($ 238), and let their respective volumes be V1, V2, V2, ..., V1, V, VS, ... Then

V:Vi =EB®: E'B'S,

V.:V, = EB": E'B', and so on. $ 245 ..Vi+Vs+Vs+...:V'+V+V3+... =EB": E'B'*. $ 21,8 But Vi+V2+V3+...=V, and Vi+V:+Vi+.. =V'. :: V:V'=EB": E'B'.

Ax. 5 247. Corollary. The volumes of two similar cylinders, or of two similar cones, are proportional to the cubes of any two corresponding lines.

Consider limits, and apply $ 246.

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248. Cavalieri's Theorem. In connection with the work in mensuration it is desirable to call attention to a theorem which was set forth by an Italian mathematician, Bonaventura Cavalieri (1598-1647), nearly three centuries ago. Since a complete proof requires some knowledge of the calculus, the theorem is here treated informally.

Theorem. If two solids lie between parallel planes, and if sections made by any plane parallel to the given planes are equivalent, the

S

S' solids are equivalent.

That is, if the two solids EB S and S' lie between the parallel planes m and n, and if the planes x, y, ... cut the solids S and s' so that A=A', B=B', ..., the solids S and s' are equivalent.

Let P and P' be the portions lying between x and y, ..., and let the altitude of P and P' be one nth of the altitude h of S and s'. On the bases A, A'; c.

C' B, B';... suppose right cylinders

h or prisms C and c' to stand, each with the altitude h/n. Then C=C', for any value of n.

As n increases indefinitely, the sum of the C's approaches S and the sum of the C's approaches s'. Since each C is always equivalent to each C', we may assume that S=S'.

249. Prismoid. A polyhedron which has for its bases two polygons in parallel planes, and for its lateral faces triangles or trapezoids with one side common to one base and the opposite vertex or side common to the other base, is called a prismoid.

The midsection of such a polyhedron is the section which is parallel to the base and bisects the altitude.

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Proposition 7. Prismoid Formula 250. Theorem. The volume of a prismoid is the product of one sixth the altitude by the sum of the bases and four times the midsection.

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Given CDE...-XY..., a prismoid; V, the volume; B, B', and M, the areas of the bases and midsection; and h, the altitude.

Prove that V=;h(B+B'+4M).

Proof. If any lateral face is a trapezoid, let it be divided into two A by a diagonal, as EY.

Let any point P in the midsection be joined to the vertices of the polyhedron and of the midsection, thus separating the prismoid into pyramids which have their vertices at P, and which have as their respective bases the lower base, the upper base, and the lateral faces of the prismoid.

The pyramid P-XCD, which we may call a lateral pyramid of volume Vp, is composed of the pyramids P-XQR, P-QDR, P-QCD of volumes V1, V2, V, respectively.

Now P-XQR may be regarded as having the vertex X and base PQR, and P-QDR the vertex D and base PQR.

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