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perpendicular to OA, and cab perpendicular to oa.

Then angles CAO, BAO, cao, bao

are right angles; and CAB, cab, are the inclination of the planes in question. Now in triangles AOB, aob. we have OA =ca and angles BOA, OAB = angles boa, oab, each to each, .. OB ab (Euclid, I. 26); similarly in triangles

ob and AB =

COA, coa, we have OC oc and AC = ac.

=

sides bo, oc, each to each, = base cb.

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Then in triangles BOC, boc, we have the sides BO, OC : the included angle BOC = boc... (Euclid, I. 4) the base CB Hence, finally, in triangle ABC, abc, we have the sides BA, AC the sides ba, ac each to each, and the base BC base bc... (Euclid, I. 8) CAB : cab. Q.E.D. Cor. 1. Hence (fig. 30), if the solid angle O be superimposed on the solid angle o, so that AO coincides with O oa, and the plane AOB with plane aob, then, because angle AOB angle aob the line OB coincides with ob, and because inclination of plane COA to AOB equals that of coa to cob the plane COA will coincide with coa, and hence OC with oc.

K

B

肉肉

α

e

Fig. 31.

D

C

=

Cor. 2. If ABCDEF and abcdef are

two prisms, the edges of which are equal

each to each, and the plane angles at B equal those at b, the prisms are equal in all respects.

For, if the triangle ABC be applied to abc so that AB coincides with al, and BC with bc, it is plain by the last corollary that BD coincides with bd. And hence the prisms will coincide throughout.

COR. 3.-If ABCD abcd be a parallelepiped * (i.e. a prism on a parallelogram for a base) it can be divided into two equal prisms by a plane ACca passing through the diagonals of its bases; for it is obvious that the edges of these prisms are equal, each to each, and also that the plane angles containing the solid angles at D and B are equal.

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COR. 4.-Again (in fig. 31) if we suppose the base DEF of the one prism to be in all respects equal to that of the other def, and if the angles at D are equal to the angles at d, each to each, then it is plain that if DB is > db the prism BDFE is > prism bdfe, and if DB < db, the prism BDFE < prism bdfe. Also if BD is double of bd, the prism BDFE is double of the prism bdfe, and generally if BD is any multiple of bd, then BDFE is the same multiple of bdfe.

Fig. 32.

Cor. 5.-If we suppose (fig. 31) BD produced to K so that DK is any multiple of DB, the prism KDFE is the same multiple of BDFE; and if we suppose db produced to k so that kd is any multiple of bd, then kdfe is the same multiple of bdfe. But if KD > kd, KDFE > kdfe; if equal, equal; if less, less.. (Euclid, 5. p. 136).

BDFE: bdfe:: BD : bd.

Cor. 6. The results proved in Cors: 4 and 5 to be true of prisms, are manifestly true of paralpps.

* We shall use throughout the abbreviation paralppd for parallelepiped.

(2.) Parallelepipeds on the same Base and of the same Altitude are equal to one another. If the Paralers are on the same base and of the same altitude, it is manifestly that the ends opposite the common base are in the same plane.

(a) Suppose two of the edges of the ends opposite to the base of each figure to be in the same straight line.

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prism eDEF. Hence if we suppose the former prism taken from the whole figure ICDEF, and the latter to be taken from the same figure, the remainders will be equal.. DBEG

=

DBeg.

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.. LKMN is equal to ABCD. Join AL, BM, CN, DK. Then DB MK is a parallelepiped-and by the first part of this proposition, BD MK is equal to each of BDGE, and BDge... BDGE BDge.

=

Q. E. D.

MATHEMATICAL SCIENCES.-No. XIII.

20

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(3.) Solid Parallelepipeds on equal Bases and of the same Altitudes, are equal to each other. (ct). Suppose the edges to be perpendicular to the bases. To avoid a complicated

figure we will only letter the bases, and will call the edges of figure that are perpendicular to the bases by the letters at the angles of the base. Thus the edge B means the edge perpendicular to the base at the point Bi.e., Bb. Let AC, DF be the bases where we suppose the solids to be so placed as to have a common edge D, and the sides AD, DE in the same straight line,

produce CD, and FE to meet in
н
Fig. 35.

H, through G draw LGK parallel
CD and produce BC to meet

KG in L. Now since DH GK, and HE is evidently KF and the angle DHE = GKF :. the base DHE base GKF; and since edges H and K are perpendicular to base, the solid angle at H is contained by plane angles respectively equal to those containing the angle K; and hence the prisms whose bases are DHE and GKF are equal. Add the solid on base DEKG to both, .. paralppd on DK paralppd DF.

Now paralppd BD : paralppd DL :: AD : DG :: BD : DL

and paralppd DL : paralppd DK :: CD : DH :: DL : DK :. (Ex equali).

paralppd BD : paralppd DK :: BD : DK But BD DK, since (Eucl. I. 35) DK DF

.. parasppd BD paralppd DK paralppd DF. (6). If the edges are not perpendicular to the bases, call the paralppd P and Q. If on P's base a paralppd p. is described, whose edges are perpendicular to base and whose altitude is same as P, then p P by last Proposition, and if q be in like manner described on same base as Q, but having its edges perpendicular to the base, then (last Proposition) 9 Q, and by the former part of this Proposition, q = P, ..Q= P.

Q.E.D. COR. 1.-Hence if there be two paralppds the same altitude, but the base of the one is double of the base of the other, the former is double of the latter, and generally if the base of the one is any multiple of the base of the other, the former paralppd is the same multiple of the latter; and hence, by reasoning similar to that employed in Cor. 5. Prop. 1. of the present article, it appears that paralppds of equal altitudes are as their bases. And again, if they have equal bases, they are to one another as their altitudes.

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COR. 2.—Let there be two paralppds P, P', the base and height of one of which are A and h, and one of the other A' and h', where Ah A'k' are in numbers. And suppose S to be a third paralppd on the base A' and the height h.

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COR. 3.-If we suppose P' to be a cube which has one of its edges equal to unity, then A'

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1. and h

= 1.

.. P: P' :: Ah: 1.

hence if we consider P' to be the unit of solid measure (a cubic inch, or foot, for instance) then Ah.

Р

Hence the volume of a paralppd is found by multiplying the area of one face by the distance between that face and the opposite one.

=

=

COR. 4. We have seen that a prism on a triangular base is half of a parald of the same altitude and on a base which is double of the triangle. Hence if A is the area of the triangle, and h the height of the prism, the volume of this paraloo1 2A x h, and.. the volume of the prism Ax h; or the volume of a prism on a triangular base is found by multiplying the area of the base by the altitude. COR. 5. It is plain that a prism on a polygonal base can be divided by planes passing through one edge of the prism into a number of prisms on triangular bases, each having the same altitude as the original prism.

...

Let A, A2 A3. be the areas of these triangles, and h the common altitude, the volumes of all these prisms

=

(A, + A2 + A ̧ + ) h.

3

...

But the polygon is equal to all the triangles; hence if A is the area of the polygon, A1 + A2 + A ̧ +

A =

1

3

.. The volume of the prism on a polygonal base

multiplied by the altitude.

=

Ah, or is equal to the base

Cor. 6.-It is plain that Cor. 5 is true of a regular polygon of any number of sides, and therefore is true in limit; now when the number of sides of the polygon is increased, its limit is the circumscribing circle, and the limit of the corresponding prism is the circumscribing cylinder. Hence the volume of a cylinder base x height.

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N.B. When we speak of the volume of a solid, a prism for instance, being equal to the base multiplied by the height, it is of course understood that all the measurements are referred to the same unit. Thus, if we were asked what is the volume of a prism whose base is 2 square feet, and height 18 inches, the answer is,

not 2 x 18, but 2 ×

18
12

=

3, and the 3 is in CUBIT FEET. The cubit foot and

the square foot being the unit of content and of area corresponding to the linear

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Since the plane EDF is parallel to ABC, ED is parallel to AB. Now AP is bisected in D.

is bisected in E.

.. PB

Fig. 36.

Similarly AC is bisected in H. Also since plane DEF is parallel

to ABC. ;. ED is parallel to KH, and . KH is parallel to AB. Draw CN perpendicular to AB, meeting KH in M, then since CH : HA::CM: MN (Euclid VI. — 2), and CH HA ::. CM MN. Hence, if from N and C perpendiculars are drawn on the plane, DEKH, they are equal. Hence paralppds on the base DEKH, with these perpendiculars respectively for altitudes, are equal. Now the prism BEKHDG is half the former, and the prism KCHDFE is half the latter. Hence the prisms are equal, and they are together double of one of them. But the two prisms make up the figure EDGHCB, which is, therefore, double of the prism EDFHKC.

(5.) Pyramids on triangular Bases and of equal Altitudes are to one another as

their Bases.

Let P, ABC, P, abc be the two pyramids, and let them be divided by planes as in the last proposition.

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Now efd : bca in the duplicate ratio of fd : ca ; but fd:ca ::1:2:. efd : oca ::1:4. Similarly EFD : BCA ::1:4.. efd : bca : : EFD : BCA or efd : EFD :: bca : BCA. But the prism efdkch : prism EFDKCH :: efd : EFD :: abc : ABC. .: (by last proposition)

The double prism edghcb : EDGHCB : : abc : ABC. Now if we suppose pd and PD, and also da and DA figures will be formed p,dfe, and P,DFE of the same kind as in p,abc and P,ABC, and also in d,agh and D,AGH, and these figures will have to each other the ratios efd : EFD and agh : AGH respectively; which ratios are each equal to the ratio abc : ABC. And the same will be true, however often we continue to bisect the bisections of the sides, and hence

All the double prisms in p,abc : all those in P,ABC :: abc : ABC. And the number of successive bisections being as great as we please, this is true in the limit. Now the pyramids are the limits of the sum of these double prisms. Hence p,abc : P,ABC :: abc : ABC.

Q. E. D. Cor : Hence pyramids on equal bases and of the same altitude are equal.

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