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lel to the base (30.), that is, to a py- planes VAC, VAD into triangular ramid BCDF, upon the same base pyramids, having the common vertex V; BDF with the last, and having its and the prism, into as many triangular vertex C in the same line CE, parallel prisms, by the planes A C ca, AD da. to the base. Therefore the prism BEF And because the former have the same is equal to the three pyramids A BCD, altitude, and stand upon the same bases BCD E, and BCDF. But of these, with the latter respectively, they are the two last are each of them equal to the equal to their third parts, each of each. first ABCD, because they stand upon Therefore, also, the sum of the former, the same base B C D with it, and their ver- that is, the whole pyramid VABCDE, tices are in the same lines with it, viz., AE, is equal to a third part of the sum of FA, parallel to the base (30.). There. the latter, that is, to a third part of the 'fore, the prism B E F is equal to three whole prism A dc. times the pyramid ABCD; and the py- Therefore, &c. ramid ABCD is equal to a third part Cor. 1. The solid content of every of the prism B EF.

pyramid is equal to one-third of the Therefore, &c.

product of its base and altitude (29. Cor. If a parallelopiped be described, Cor. 1.). having three of its edges the same with

Cor. 2. (Euc. xii. 5, 6.) Pyramids three conterminous

which have equal altitudes are to one edges of a triangular pyramid, the py

another as their bases : and pyramids ramid shall be equal

which have equal bases are to

another as their altitudes: also, any to a sixth part of the parallelopiped; for the latter is divided ratio compounded of the ratios of their

two pyramids are to one another in the by its diagonal plane into two equal bases and altitudes. (29. Cor. 2.). prisms (22.), one of which stands upon the same base, and is of the same altitude

Scholium. with the pyramid, and is, therefore, by

It was observed in the Scholium at i. the proposition, equal to three times the pyramid.

29. that the measurement of any plane Scholium.

rectilineal figure in general depends upon

that of the triangle, because every such This proposition might have been figure may be divided into triangles. In demonstrated somewhat more concisely. like manner, we may here remark, that It will be found, however, that our de- the solid content of the pyramid leads to monstration applies equally to the fol- that of every other polyhedron; and that lowing more general theorem :-" If the the latter may be found by adding togeupper part of a triangular prism (see ther the contents of all the pyramids the figure in the proposition) be cut which constitute the polyhedron. These away by a plane A EF, whether parallel to the base B CD, or otherwise, the may be assumed with a common vertex,

which

may

be either within the solid, remaining solid will be equal to the sum

in which case the several faces will be of three pyramids standing upon the the bases of so many pyramids together same base B C D with the prism, and making up the polyhedron, or at one having for their vertices the upper ex

of the angular points, in which case the tremities A,E,F of the diminished edges." adjoining faces will be divided into sides, PROP. 32.

and the others will be bases of the comEvery pyramid is equal to the third ponent pyramids. part of a prism, having the same base and the same altitude.

PROP. 33. Let V ABCDE be a pyramid, and let abcde be the upper base of a prism

If the upper part of a pyramid be cut standing upon the same

away by a plane parallel to the base; the

remaining frustum shall be equal to the base A B C D E, and having the same altitude

sum of three pyramids, having the same with the pyramid: the py

altitude with the frustum, and for their ramid shall be equal to

bases the bases of the frustum, and a the third part of the prism.

mean proportional between them. Join AC, AD, ac, ad. A

First, let BCD, bcd be the two Then the pyramid VAB

bases of the frustum of a triangular pyCDE is divided by the

ramid : draw cE parallel to b B, to

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mer, that is, the base ABCDE, has the triangles

to the sum of the means the same BCD, bcd are equi

ratio (II. 23. Cor. 1.), viz. the subduangular (12. and 15.),

plicate ratio of ABC to abc, or of and therefore similar,

ABCDE to abcde; and accordBCD is to b c d in

ingly (II. def. 11.) the sum of the the duplicate ratio (II.

means is a mean proportional between 42.) of BC to bc, or

ABCDE and abcde. Therefore, in (I. 22.) BE, that is, in

this case also, the frustum in question is the duplicate ratio of the same BCD equal to the sum of three pyramids, to BED (II. 39.). Therefore the tri- having the same altitude with it, and for angle B E D is a mean proportional be. their bases its two bases, and a mean tween BCD and bcd (il. def. 11.) proportional between them. Now, the frustum is made up of three Therefore, &c. pyramids, namely, cB C D, Dbcd, and cbBD. And, of these the first has the PROP. 34. (Euc. xii. 8. and 8 Cor.) same altitude with the frustum, and stands upon the base BCD; the second Similar pyramids are to one another has likewise the same altitude with the in the triplicate ratio of their homolofrustum, and stands upon the base bcd; gous euges. and the third, because c Eis parallel (10.)

In the first place, let ABCD and to the plane B bd, is equal (30.) to the mids, and let the edges BC, bc be ho

abcd be two similar triangular pyrapyramid Eb B D, which has the same altitude, and stands upon the base mologous : the pyramid A B C D shall BED, that is, upon a base which is a mean proportional between B C D and bcd.

Next, let ABCDE, abcde be the two bases of the frustum of any pyramid, having the vertex V. Then, if the planes V AC, VAD be drawn cutting the bases in the straight lines AC, ac, AD, ad,

have to the pyramid abcd the triplicate

ratio of that which B C has to b c. the frustum in question

Because the faces of the two pyramids will be divided into frustums of triangular pyra- A

are similar triangles, it is evident that mids. And each of these,

* If a magnitude A be the first of three propor. by what has been already

tionals A, B, C, and if the same A be the first also of demonstrated, is equal to the sum of than 6, C shall likewise be greater than c.

three other proportionals A, b, c, then if B be greater three pyramids, having the same alti- to c in the ratio which is compounded of the ratios tude with the frustum, and for their of C to B, B to b, and b to c, that is, (because the first

of these ratios is the same with that of B to A, and bases respectively, the two bases of the last the same with that of A to b,) in the ratio the frustum and a mean proportional which is compounded of the ratios of B tu A, A to b, between them. Therefore, their sum,

and B to b, or in the duplicate ratio of B to b. Where

sore because B is greater than b, Cis likewise greater that is, the frustum in question, is equal to three pyramids having the if A°, Bo, Co be likewise proportionals, and if Ahas

Hence it follows, that if A, B, C be proportionals, and same altitude, and for their bases the

a greater ratio to B' than A to B, A' shall also have sum of the lower bases, the sum of

a greater ratio to C'than A to C. For, if b be taken

such that A' may have to b the same ratio as A to the upper bases, and the sum of the B, 6 will be greater than B' (11. 11.Cor. 3.); where. means respectively. Now, the two fore, if c be taken such that À', b, c may bé propor. former sums are the bases ABCDE, tionals, c. will likewise be greater than Ć; and there

fore A' will have to C a greater ratio than it has to abcde. And the sum of the means is a c. But A' has to c the same ratio as A to C(II. 24.). mean between the bases ABCDE, Therefore A' has to Co a greater ratio than À to C.

And hence it is evident, that if any three magni. abcde; for, the triangles ABC, ACD, tudes A, B, C be proportionals, ană likewise any other &c. having to the triangles abc, acd, three, A', B-, C-, and

if A is to C as A'to C', A is to B &c. the same ratio, each to each, viz.

as A' to B'; for if A has not to B the same ratio as

A to B', then neither can A hare to C the same ratio that of V A2 to v a? (II. 42. and II. 32.) as A to C'. This last is the theorem demanded in the must have the same ratio also to their text; and is evidently the converse of 11. 27. Cor. 3.,

from which alone, however, it may be considered as respective means (II. 27. Cor. 3. and sufficiently apparent.

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their corresponding edges are to one an- base D BC to the base DBF (32. Cor. other in the same ratio, so that A B is 2.); that is, as B C to BF (II. 39.), to ab as B C to bc, and as BD to bd. or in the triplicate ratio of BC to bc. And, because the plane and dihedral Therefore, the pyramid ABCD is to angles of the solid angle B are severally the pyramid abcd in the triplicate ratio equal in order to the plane and dihedral of that which B C has to b C* angles of the solid angle b, these solid The proposition is therefore demonangles may be made to coincide. There- strated with regard to triangular pyrafore (27 Cor. 1.) the parallelopipeds, mids. which have the edges B A, B C, BD, Next, let VABCDE, vabcde be and ba, bc, bd, are to one another in the ratio which is compounded of the ratios of B A to ba, B C to bc, and BD to bd, that is, since these ratios are equal to one another, the triplicate ratio (II. 27. Cor. 2.) of that which B C has to.bc. But the pyramids ABCD, abcd are (31. Cor.) the sixth parts of these parallelopipeds, each of each, and therefore have to one another (II. 17.) the same ratio with the parallelopipeds. any two similar pyramids, and let B C,Therefore, the pyramid ABCD has to b c be homologous edges: the pyramid the pyramid abcd the triplicate ratio VABCDE shall be to the pyramid of that which B C has to bc.

vabcde in the triplicate ratio of that Otherwise:

which B C has to bc. Take B E (II. 52.), a third proportional of the solid angle V are severally equal

Because the plane and dihedral angles to BC, bc; and, again, BF a third proportional to bc, BE; so that B C to the plane and dihedral angles of the may have to BF the triplicate ratio of solid angle v, these solid angles may be that which it has to bc (11. def. u.): made to coincide ; and, if this be done, join D Е, DF, A F; and, from A, a, let the bases ABCDE, abcde will be AP, ap be drawn perpendicular to the parallel (15.), because the faces V AB bases BCD, bed of the pyramids, similar, their sides A B and ab, BC

and vab, VBC and vbc, &c. being each to each. Then, because the solid angles B, 6 may be made to coincide, and b c, &c. will be parallel (1. 15.). and that in this case (5. Cor.), the tri- and vac, VAD and vad make equal

Hence it appears that the planes VAC angles A B P,* abp will be in the same plane, and on account of the parallelism angles with the adjoining faces of of A'P, ap (5.) will be equiangular va Čand va'c, as also VA D and vad,

the pyramids; and that the triangles (I. 15.), A Pis to ap (II.31.) as A B to ab, or as B C to bc. Again, because, that the triangles A B C, A CD, ADE

are similar. And it is easy to show in the triangles D B C, dbc, D B is to db as BC to bc; that is, as bc to BE;

are respectively similar to the triangles the triangles D BE, dbc are equal to the pyramids in question are divided by

e one another (II. 41.) But D BE is to D BF (II. 39.) as B E to BF; that ramids V ABC and vabc, VACD

those planes into similar triangular pyis, as B C to bc. Therefore, dbc is to DBF as B C to bc. But B C is to bc

and vacd, VADE and vad e (def. 10.). as A P to ap; therefore, dbc is to And, since the homologous edges of DB Fas A P to ap, and consequently cd, DÉ and de, are to one another in

these pyramids, B C and bc, C D and (II. 11. Cor. 2.) the ratio, which is compounded of the ratios of dbc to DBF, the same ratio ; and, by what has been and ap to A P is a ratio of equality. * These demonstrations are respectively analogons But the pyramid abcd is to the pyra

to those given of 11. 42. in page 66. The second, it is mid A BFD in a ratio which is com

obvious, admits of considerable abridgment, by refer

ence to 32. Cor. 2.; for, since the pyramids are to pounded of the ratios of dbc to DBF, one another in the ratio which is compounded of the and ap to AP (32. Cor. 2.); therefore, ratios of their bases and altitudes, and that the bases

are to one another in the duplicate ratio of two homo. the pyramid AB FD is equal to the py- logous edges (II. 42.), and the altitudes have to one ramid abcd. Now, the pyramid AB .

another the same ratio with those edges, it follows

that the pyramids are to one another in the triplicate C D is to the pyramid A BFD, as the ratio of their homologous edges. And in this form

the demonstration applies equally to the general case * The lines B P, bp are omitted in the figure,

in which the bases are similar polygons.

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already demonstrated, the pyramids to points A, a, with all the angular points one another in the triplicate ratios of of their respective figures, to which they their homologous edges, that is likewise are not already joined by the edges (II. 27. Cor. 3.) in the same ratio ; the terminating in 'A, a. Then, because whole pyramid V ABCDE is (II. 23. the triangles 'ABC, abc, are similar Cor. 1.) to the whole pyramid v abcde (11. 32.) the angles BAC, bac are equal in the latter ratio, i. e. in the triplicate to one another, but the planes B AC, ratio of that which B C has to b c. bac are parallel, and A B is parallel to Therefore, &c.

ab; therefore, also (15.), AC is parallel Cor. Since the triplicate ratio of two to ac: and, for the like reason, AH is straight lines is the same with the ratio parallel to ah. Again, because A F and of their cubes (27. Cor. 2.), similar py- FK are parallel respectively, and in the ramids are to one another as the cubes same ratio, to af and fk, the triangles of their homologous edges.

AFK, af k are similar (15. and II. 32.),

and their planes parallel; wherefore, as PROP. 35.

before, AK is parallei to ak; and for the Similar polyhedrons are to one ano- like reason the triangles ADL, adl are ther in the triplicate ratio of their ho. similar and in parallel planes, and AL mologous edges.

is parallel to al. And because the triLet A CG, acg be two similar poly- angles A CL, acl have the three sides hedrons, and let B C, bc be any two of the one parallel to the three sides of

the other, each to each, they are equiangular (15.), and lie in parallel planes: the same also may be said of the triangles AKL, akl, and AHK, ahk.

Therefore, the pyramids into which the solid ACG is divided by the planes

ACL, ADLAFK, AHK, AKL are simihomologous edges: the polyhedronlar to the pyramids into which the solid ACG shall be to the polyhedron acg acg is divided by the planes a cl, adl, in the triplicate ratio of that which afk, ahk, akl, each to each : for it BC has to bc.

has been shown that the faces in these In the first place, it is evident that planes are similar, each to each, and the corresponding edges of the polyhe- their other faces are similar, because drons are to one another each to each in they are similar faces, or similar parts the same ratio, because they are homo- of similar faces, of the polyhedrons; also logous sides of the similar faces of the the dihedral angles made by any two polyhedrons. Let the polyhedrons be similar and adjoining faces are equal so placed as to have two corresponding (17. Scholium, 3.), because their planes edges A B, ab parallel to one ano- are parallel. And the former pyramids ther, and two corresponding faces are to the latter (34.) in the triplicate ABCD, abcd adjoining to those ratios of their homologous edges, each edges likewise parallel. Then, be- to each; that is, in the triplicate ratio cause the angles ABC, abc ai of B C to bc, because their edges have equal to one another, B C is likewise to one another the common ratio of parallel to bc (15.); and for the like BC to be. Therefore, also, the sum reason CD, D A are parallel to cd, da of the former pyramids, that is, the porespectively. Again, since A B is pa- lyhedron A CG, is (II. 23. Cor. 1.) to rallel to ab, a plane may be made to the sum of the latter, that is, to the pass through A B parallel to the plane polyhedron a cg, in the same ratio, viz., abhg (15.), and that plane is no other the triplicate ratio of BC to b c. than A BHG, because the dihedral an- Therefore, &c. gles at A B and ab are equal of one ano- Cor. Similar polyhedrons are to one ther(17. Scholium, 3.and 11.Cor.1.); and another as the cubes of their homolofor the like reason the other similar faces gous edges (27. Cor. 2.). of the two polyhedrons lie in parallel planes. Therefore, also, any two corresponding edges being the common

SECTION 4.- Problems. sections of such planes, are parallel In the following problems it is as(12. Cor.). Join AC and ac, AH and ah, AK from any centre, and at any distance

sumed that a sphere may be described and ak, A L and al; that is, join the rom that centre: also, a plane is considered to be drawn, when two straight are therefore (9. Cor.) right angles ; lines are drawn which lie in that plane. and every other point of the perpendi

cular on the same side of AB is at a PROP. 36. Prob 1.

greater or a less distance from D than To draw a straight line perpendicu. Gis (I.12. Cor. 2.). Reversely, therefore, lar to a given plane A B, from a given from the centres D, E, F, with any the point C without it.

same radius greater than D C, describe From C, to the plane, draw

three spheres, and let G be the point

any straight line CD; from the centre c,

common to the surfaces of these
spheres; join CG; CG will be the
perpendicular required.
Therefore, &c.
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Cor. It appears from the demonstration, that, if a point be equally distant from the three angles of a triangle, it

must lie in a perpendicular to the plane with the radius C D, describe a sphere of the triangle, which passes through DEF. If this sphere touch the plane the centre of the circumscribing circle. in D, Ć D is the perpendicular required

Scholium. (8. Cor.); if not, take K, the centre of the circle DGH in which it cuts the

Had the intersection of the spherical plane, and join CK; C K will be the surfaces, in this problem, been consiperpendicular required (8. Cor.).

dered at a greater length, the analogy of Therefore, &c.

the construction to that employed in the Another method has been seen in first method of Book i. prop. 44, might

have been lost sight of in the detail.

The subject is, however, of sufficient inPROP. 37. Prob. 2.

terest to merit the attention of the stuTo draw a straight line perpendicular dent; and, as it has not hitherto found to a given plane A B. from a given point a place in this treatise, the present is, C in the same.

perhaps, the most proper for its consiFrom the centre C, with any radius deration. CD, describe, · in

1. Two spheres will, 1°, cut one anothe plane A B, the

ther, or 20, touch one another, or 3°, circle DEF. In the circumference of this circle, take any three points D, E, F; and join A CD, CE, CF.

Now, let C G be the perpendicular required. From the centre D, with any radius greater than D C, describe a sphere, and let its surface cut the perpendicular C G in G; and join GD, GE, GF. Then, because GE and GF are each of them equal to GD (8.), if spheres be described from one of them fall wholly without the the points E and F, each with a radius other, according as the distance beequal to DG, their surfaces will like. tween their centres is, 1°, less than the wise each of them pass through the sum, and greater than the difference point G. And G is the only point on of their radii, or, 2°, equal to the that side of the plane A B, which is sum, or to the difference of their rafound at the same time in the surface of dii, or, 3°, greater than the sum, or less each of the three spheres. For, every than the difference of their radii. point, as G, which is at equal distances This is easily inferred from what has from the three points D, E, F, must lie been already demonstrated with regard in the perpendicular which passes to the intersection of circles; for it through the centre of the circle D EF, is evident that the sections of two because (I. 7.) the angles G CD, GCE, spheres, made by a plane passing G C F are equal to one another, and through their centre and through any

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