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Let FGH-h be the frustum of a triangular pyramid. Through the three points, F, g, H, pass the plane FgH; it cuts off from the frustum the triangular pyramid g-FGH. This pyramid has for its base the lower base FGH of the frustum; its altitude is equal to that of the frustum, because the vertex g lies in the plane of the upper base fgh.
This pyramid being cut off, there remains the quadrangular pyramid g-fhIIF, whose vertex is g, and base fhHF. Pass the plane gfH through the three points f, g, H; it divides the quadrangular pyramid into two triangular pyramids g-fFH, g-fhH. The latter has for its base the upper base gfh of the frustum; and for its altitude, the altitude of the frustum, because its vertex H lies in the lower base. Thus we already know two of the three pyramids which compose the frustum.
It remains to examine the third pyramid g-FfH. Now, if gK be drawn parallel to fF, and if we conceive a new pyramid K-fFH, having K for its vertex and fFH for its base, these two pyramids have the same base HfF; they also have the same altitude, because their vertices g and K lie in the line gK, parallel to Ff, and consequently, parallel to the plane of the base: hence, these pyramids are equivalent (P. 17, c. 3). But the pyramid K-ƒFH may be regarded as having FKH for its base, and its vertex at f: its altitude is then the same as that of the frustum. We are now to show that the base FKH is a mean proportional between the bases FGH and fgh. The triangles FHK, fgh, have the angle F=f; hence (B. IV., p. 24), FHK fgh :: FKXFH fgfh;
but because of the parallels, FK=f9,
We have also,
FHG FHK :: FG: FK, or fg. But the similar triangles FGH, fgh, give
FG: fg :: FH : fh;
that is, the base FHK is a mean proportional between the two bases FGH, fgh. Hence, the solidity of the frustum of a triangular pyramid is equal to that of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base, and a mean proportional between the two bases.
Similar triangular prisms are to each other as the cubes of their homologous edges.
Let CBD-P, cbd-p, be two similar triangular prisms, and BC, bc, two homologous edges: then will the prism CBD-P be to the prism cbd-p, as BC3 to be.
For, since the prisms are similar, the angles B and b are equal, and the faces which bound them are similar (D. 16). Hence, if these triedral angles be applied, the one to the other, the angles cbd will
coincide with CBD, the edge ba with BA, and the prism cbd-p will take the position Bcd-p. From A draw AH perpendicular to the common base of the prisms: then will the plane BAH be perpendicular to the plane of the common base (B. VI., P. 16). Through a, in the plane BAH, draw ah perpendicular to BH: then will aḥ also be perpendicular to the base BDC (B. VI., P. 17); and AH, ah will be the altitudes of the two prisms.
Since the bases CBD, cbd, are similar, we have (B. IV., p. 25), base CBD : base cbd CB2: cb.
Now, because of the similar triangles ABH, aBh, and of the similar parallelograms AC, ac, we have
AH : ah :: AB : ab :: CB : cb;
hence, multiplying together the corresponding terms, we have base CBDXAH : base cbd Xah :: CB3: cb.
But the solidity of a prism is equal to the base multiplied by the altitude (P. 14); hence,
prism bcd-p :: BC3 : bc3, or as the cubes of any other of their homologous edges. Cor, Whatever be the bases of similar prisms, the prisms are to each other as the cubes of their homologous edges.
For, since the prisms are similar, their bases are similar polygons (D. 16); and these similar polygons may each be divided into the same number of similar triangles, similarly placed (B. IV., P. 26); therefore, each prism may be divided into the same number of triangular prisms, having their faces similar and like placed; hence, their polyedral angles are equal (B. VI., P. 21, s. 2); and consequently, the triangular prisms are similar (D. 16). But these triangular prisms are to each other as the cubes of their homologous edges, and being like parts of the polygonal prisms, their sums, that is, the polygonal prisms, are to each other as the cubes of their homologous edges.
PROPOSITION XX. THEOREM.
Two similar pyramids are to each other as the cubes of their homologous edges.
For, since the pyramids are similar, the homologous polyedral angles at the vertices are equal (D. 16). Hence, the polyedral angles at the vertices may be made to coincide, or the two pyramids may be so placed as to have the polyedral angle S common.
In that position the bases ABCDE, abcde, are parallel; for, the homologous faces being similar, the angle Sab is equal to SAB, and Sbc to SBC; hence, the plane ABC, is parallel to the plane abc (B. VI., P. 13). This being proved, let SO be drawn from the vertex S, perpendicu lar to the plane ABC, and let o, be the point where this perpendicular pierces the plane abc from what has already been
shown, we have (P. 3),
:: SA :
SOS :: AB : ab.
But the bases ABCDE, abcde, being similar figures, we have (B. IV., P. 27),
ABCDE : abcde :: AB2 : ab2 ;
multiply the corresponding terms of these two proportions, there results,
ABCDEXSO: abcdexSo :: AB3
Now, ABCDEXSO measures the solidity of the pyramid S-ABCDE, and abcde×So measures that of the pyramid S-abcde (P. 17); hence, two similar pyramids are to each other as the cubes of their homologous edges.
1. The chief propositions of this Book relating to the solidity of polyedrons, may be expressed in algebraical terins, and so recapitulated in the briefest manner possible. 2. Let B represent the base of a prism; H its altitude: then,
solidity of prism=B × H.
3. Let B represent the base of a pyramid; H its altitude: then,
solidity of pyramid=B×}H.
4. Let H represent the altitude of the frustum of a pyramid, having the parallel bases A and B; VAXB is the mean proportional between those bases; then
solidity of frustum=}H(A+B+√ A×B.)
5. In fine, let P and p represent the solidities of two similar prisms or pyramids; A and a, two homologous edges: then,
THE THREE ROUND BODIES.
1. A CYLINDER is a solid which may be generated by the revolution of a rectangle ABCD, turning about the immovable side AB.
In this movement, the sides AD, BC, continuing always perpendicular to AB, describe the equal circles DHP, CGQ, which are called the bases of the cylinder; the side CD, describing, at the same time, the convex surface.
The immovable line AB is called the axis of the cylinder.
Every section MNKL, made in the
cylinder, by a plane, at right angles to the axis, is a circle equal to either of the bases. For, whilst the rectangle ABCD turns about AB, the line KI, perpendicular to AB, describes a circle, equal to the base, and this circle is nothing else than the section made by a plane, perpendicular to the axis at the point I.
Every section QPHG, made by a plane passing through the axis, is a rectangle double the generating rectangle ABCD.
2. SIMILAR CYLINDERS are those whose axes are proportional to the radii of their bases: hence, they are generated by similar rectangles (B. IV., D. 1).