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Let the parallelograms AC and GK be the bases as before, and let AE and GM be the altitudes of two parallelopipeds Y and Z on these bases. Then, if the upright parallelopipeds AF and GO be constituted on the bases AC and GK, with the altitudes AE and GM, they will be equal to the parallelopipeds Y and Z (7. 3. Sup.). Now, the solids AF and GO, by the first case, are in the ratio compounded of the ratios of the bases AC and GK, and of the altitudes AE and GM; therefore also the solids Y and Z have to one another a ratio that is compounded of the same ratios.
COR. 1. Hence, two straight lines may be found having the same ratio with the two parallelopipeds AF and GO. To AB, one of the sides of the parallelogram AC, apply the parallelogram BV equal to GK, having an angle equal to the angle BAD (Prob. 15. 1.); and as AE to GM, so let AV be to AX (12. 6.), then AD is to AX as the solid AF to the solid GO. For the ratio of AD to AX is compounded of the ratios (def. 10. 5.) of AD to AV, and of AV to AX; but the ratio of AD to AV is the same with that of the parallelogram AC to the parallelogram BV (1. 6.) or GK; and the ratio of AV to AX is the same with that of AE to GM; therefore the ratio of AD to AX is compounded of the ratios of AC to GK, and of AE to GM (E. 5.). But the ratio of the solid AF to the solid GO is compounded of the same ratios; therefore, as AD to AX, so is the solid AF to the solid GO.
COR. 2. If AF and GO are two parallelopipeds, and if to AB, to the perpendicular from A upon DC, and to the altitude of the parallelopiped AF, the numbers L, M, N, be proportional: and if to AB, to GH, to the perpendicular from G on LK, and to the altitude of the parallelopiped GO, the numbers L, l, m, n, be proportional; the solid AF is to the solid GO as LX MXN to lxmxn.
For it may be proved, as in the 7th of the 1st of the Sup. that LMX N is to xmxn in the ratio compounded of the ratio of LX M to lxm, and of the ratio of N to n. Now the ratio of LX M to 1x m is that of the area of the parallelogram AC to that of the parallelogram GK; and the ratio of N to n is the ratio of the altitudes of the parallelopipeds, by hypothesis, therefore, the ratio of LX MXN to 1x mxn is compounded of the ratio of the areas of the bases, and of the ratio of the altitudes of the parallelopipeds AF and GO; and the ratio of the parallelopipeds themselves is shewn, in this proposition, to be compounded of the same ratios; therefore it is the same with that of the product LX MXN to the product l×m×n.
COR. 3. Hence all prisms are to one another in the ratio compounded of the ratios of their bases, and of their altitudes. For every prism is equal to a parallelopiped of the same altitude with it, and of an equal base (2. Cor. 8. 3. Sup.).
PROP. X. THEOR.
Solid parallelopipeds, which have their bases and altitudes reciprocally proportional, are equal; and parallelopipeds which are equal, have their bases and altitudes reciprocally proportional.
Let AG and KQ be two solid parallelopipeds, of which the bases are
AC and KM, and the altitudes AE and KO, and let AC be to KM as KO to AE; the solids AG and KQ are equal.
As the base AC to the base KM, so let the straight line KO be to the straight line S. Then, since AC is to KM as KO to S, and also by hypothesis, AC to KM as KO to AE, KO has the same ratio to S that it has to AE (11. 5.); wherefore AF is equal to S (9. 5.). But the solid AG is
to the solid KQ, in the ratio compounded of the ratios of AE to KO, and of AC to KM (9. 3. Sup.), that is, in the ratio compounded of the ratios of AE to KO, and of KO to S. And the ratio of AE to S is also compounded of the same ratios (def. 10. 5.); therefore, the solid AG has to the solid KQ the same ratio that AE has to S. But AE was proved to be equal to S, therefore AG is equal to KQ.
equal, the base AC is to the base Take S, so that AC may be
Again, if the solids AG and KQ be KM as the altitude KO to the altitude AE. to KM as KO to S, and it will be shewn, as was done above, that the solid AG is to the solid KQ as AE to S; now, the solid AG is, by hypothesis, equal to the solid KQ therefore, AE is equal to S; but, by construction, AC is to KM, as KO is to S; therefore, AC is to KM as KO to AE.
COR. In the same manner, it may be demonstrated, that equal prisms have their bases and altitudes reciprocally proportional, and conversely.
PROP. XI. THEOR.
Similar solid parallelopipeds are to one another in the triplicate ratio of their homologous sides.
Let AG, KQ be two similar parallelopipeds, of which AB and KL are two homologous sides; the ratio of the solid AG to the solid KQ is triplicate of the ratio of AB to KL.
Because the solids are similar, the parallelograms AF, KB are similar (def. 2. 3. Sup.), as also the parallelograms AH, KR; therefore, the ratios of AB to KL, of AE to KO, and of AD to KN are all equal (def. 1. 6.). But the ratio of the solid AG to the solid KQ is compounded of the ratios of AC to KM, and of AE to KO. they are equiangular parallelograms, of AB to KL, and of AD to KN.
Now, the ratio of AC to KM, because is compounded (23. 6.) of the ratios Wherefore, the ratio of AG to KQ is
compounded of the three ratios of AB to KL, AD to KN, and AE to KO: and the three ratios have already been proved to be equal; therefore, the ratio that is compounded of them, viz. the ratio of the solid AG to the solid KQ, is triplicate of any of them (def. 12. 5.): it is therefore triplicate of the ratio of AB to KL.
COR. 1. If as AB to KL, so KL to m, and as KL to m, so is m to n, then AB is to n as the solid AG to the solid KQ. For the ratio of AB to n is triplicate of the ratio of AB to KL (def. 12. 5.), and is therefore equal to that of the solid AG to the solid KQ.
COR. 2. As cubes are similar solids, therefore the cube on AB is to the cube on KL in the triplicate ratio of AB to KL, that is in the same ratio with the solid AG, to the solid KQ. Similar solid parallelopipeds are therefore to one another as the cubes on their homologous sides.
COR. 3. In the same manner it is proved, that similar prisms are to one another in the triplicate ratio, or in the ratio of the cubes of their homologous sides.
PROP. XII. THEOR.
If two triangular pyramids, which have equal bases and altitudes, be cut by planes that are parallel to the bases, and at equal distances from them, the sections - are equal to one another.
Let ABCD and EFGH be two pyramids, having equal bases BDC and FGH, and equal altitudes, viz. the perpendiculars AQ, and ES drawn from A and E upon the planes BDC and FGH: and let them be cut by planes parallel to BDC and FGH, and at equal altitudes QR and ST above those planes, and let the sections be the triangles KLM, NOP; KLM and NOP are equal to one another.
Because the plane ABD cuts the parallel planes BDC, KLM, the common sections BD and KM are parallel (14.2. Sup.). For the same reason, DC and ML are parallel. Since therefore KM and ML are parallel to BD and DC, each to each, though not in the same plane with them, the angle KLM is equal to the angle BDC (9. 2. Sup.). In like manner the other angles of these triangles are proved to be equal; therefore, the triangles are equiangular, and consequently similar; and the same is true of the triangles NOP, FGH.
Now, since the straight lines ARQ, AKB meet the parallel planes BDC
and KML, they are cut by them proportionally (16. 2. Sup.), or QR: RA :: BK: KA; and AQ : AR::AB: AK (18. 5.), for the same reason, ES ET EF: EN; therefore AB: AK :: EF: EN, because AQ is equal to ES, and AR to ET. Again, because the triangles ABC, AKL are similar,
AB: AK:: BC: KL; and for the same reason
EF EN FG: NO; therefore,
BC: KL:: FG: NO. And, when four straight lines are proportionals, the similar figures described on them are proportionals (22. 6.); therefore the triangle BCD is to the triangle KLM as the triangle FGH to the triangle NOP; but the triangle BDC, FGH are equal; therefore, the triangle KLM is also equal to the triangle NOP (1. 5.).
COR. 1. Because it has been shewn that the triangle KLM is similar to the base BCD; therefore, any section of a triangular pyramid parallel to the base, is a triangle similar to the base. And in the same manner it is shewn, that the sections parallel to the base of a polygonal pyramid are similar to the base.
COR. 2. Hence also, in polygonal pyramids of equal bases and altitudes, the sections parallel to the bases, and at equal distances from them, are equal to one another.
PROP. XIII. THEOR.
A series of prisms of the same altitude may be circumscribed about any pyramid, such that the sum of the prisms shall exceed the pyramid by a solid less than any given solid.
Let ABCD be a pyramid, and Z* a given solid; a series of prisms having all the same altitude, may be circumscribed about the pyramid ABCD, so that their sum shall exceed ABCD, by a solid less than Z.
*The solid Z is not represented in the figure of this, or the following Proposition.
Let Z be equal to a prism standing on the same base with the pyramid, viz. the triangle BCD, and having for its altitude the perpendicular drawn from a certain point E in the line AC upon the plane BCD. It is evident, that
CE multiplied by a certain number m will be greater than AC; divide CA into as many equal parts as there are units in m, and let these be CF, FG, GH, HA, each of which will be less than CE. Through each of the points F, G, H, let planes be made to pass parallel to the plane BCD, making with the sides of the pyramid the sections FPQ, GRS, HTU, which will be all similar to one another, and to the base BCD (1. cor. 12. 3. Sup.). From the point B draw in the plane of the triangle ABC, the straight line BK parallel to CF meeting FP produced in K. In like manner, from D draw DL parallel to CF, meeting FQ in L: Join KL, and it is plain, that the solid KBCDLF is a prism (def. 4. 3. Sup.). By the same construction, let the prisms PM, RO, TV
be described. Also, let the straight line IP, which is in the plane of the triangle ABC, be produced till it meet BC in h; and let the line MQ be produced till it meet DC in g: Join hg; then hC gQFP is a prism, and is equal to the prism PM (1. Cor. 8. 3. Sup.). In the same manner is described the prism MS equal to the prism RO, and the prism qU equal to the prism TV. The sum, therefore, of all the inscribed prisms hQ, mS, and qU is equal to the sum of the prisms PM, RO and TV, that is, to the sum of all the circumscribed prisms except the prism BL; wherefore, BL is the excess of the prism circumscribed about the pyramid ABCD above the prisms inscribed within it. But the prism BL is less than the prism which has the triangle BCD for its base, and for its altitude the perpendicular from E upon the plane BCD; and the prism which has BCD for its base, and the perpendicular from E for its altitude, is by hypothesis equal to the given solid Z; therefore the excess of the circumscribed, above the inscribed prisms, is less than the given solid Z. But the excess of the circumscribed prisms above the inscribed is greater than their excess above the pyramid ABCD, because ABCD is greater than the sum of the inscribed prisms. Much more, therefore, is the excess of the circumscribed prisms above the pyramid, less than the solid Z. A series of prisms of the same altitude has therefore been circumscribed about the pyramid ABCD, exceeding it by a solid less than the given solid Z.
PROP. XIV. THEOR
Pyramids that have equal bases and altitudes are equal to one another.
Let ABCD, EFGH, be two pyramids that have equal bases BCD, FGH