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Let the line AB (fig. 329.) make right angles with the lines AC, AD, it will be perpendicular to the plane which passes through these lines. p

If the line AB were not perpendicular to the FDCG, another plane might be made to pass through the point A, to which the AB would be perpendicular. But this is impossible; for, since the angles BAC, BAD are right angles, this other plane (Defin. 2.) must pass through

the points C, D; it would therefore (Prop. 93.) be the same with the plane FDCG, since these two planes would have three common points 1C. 2=e A, C, D.

975. PRoP. XCV. From a given point in a plane to raise a perpen- F.'”.

dicular to that plane. Let it be required to raise a perpendicular from the point A (fig. 330.) in the plane LM. Form a rectangle CDFG, divide it into two A. rectangles, having a common section AB, and place these rectangles upon the plane LM in such a manner that the bases of the two rectangles AC, AG shall be in the plane LM, and form any angle with each other; the line AB shall be perpendicular to the plane LM. The line AB makes right angles with the two lines AC, AG, which, by supposition, are in the plane LM; it is therefore (Prop. 94.) perpendicular & to the plane LM. Fig. 350. 976. PRoP. XCVI. Two planes cutting each other at right angles, if a right line be drawn in one of the planes perpendicular to their intersection, it will be perpendicular to the other plane. Let the two planes AFBG, ALBM (fig. 331.), cut each other at right angles; if the line LC be perpendicular to their common intersection, it is also per- - i. pendicular to the plane AFB.G. Draw CG perpendicular to AB. F : Because the lines CL, CG are perpendicular to the common in- x : tersection AB, the angle LCG (Defin. 3.) is the angle of inclination \ic of the two planes. Since the two planes cut each other perpendi- A ". p cularly, the angle of inclination LCG is therefore a right angle. - \ And because the line LC is perpendicular to the two lines CA, CG in the plane ABFG, it is (Prop. 94.) perpendicular to the plane AFB.G. 977. PRor. XCVII. If one plane meet another plane, it makes angles with that other plane, which are together equal to two right angles. Let the plane ALBM (fig. 332.) meet the plane AFBG; these planes will make with each other two angles, which will together be equal to two right angles. Through any point C draw the lines FG, LM perpendicular to the line AB. The line CL makes with the line FG two angles together equal to two right angles. But these two angles are (Defin. 3.) the angles of inclination of the two planes. Therefore the two planes make A angles with each other, which are together equal to two right angles. CoRoLLARY. It may be demonstrated in the same manner that planes which intersect each other have their vertical angles equal, that parallel planes have their alternate angles equal, &c. 978. PRor. XCVIII. If two planes be parallel to each other, a right line, which is perpendicular to one of the planes, will be also perpendicular to the other. Let the two planes LM, FG (fig. 33.3.) be parallel. If the line BA 1.Q. be perpendicular to the plane FG, it will also be perpendicular to the plane LM. From any point C in the plane LM draw CD perpendicular to the plane FG, and draw BC, AD. Because the lines BA, CD are perpendicular to the plane FG, the angles A, D are right angles. --Because the planes LM, FG are parallel, the perpendiculars AB, FQ

DC (Defin. 6.) are equal; whence it follows that the lines BC, AD Fig. 335. are parallel. - The line BA, being at right angles to the line AD, will also (Prop. 13.) be a right angles to the parallel line B.C. The line BA is therefore perpendicular to the line BC. ]] In the same manner it may be demonstrated that the line BA is at right angles." " other lines which can be drawn from the point B in the plane LM. Wherefore (De" 2.) the line BA is perpendicular to the plane LM.

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SOLIDs.

979. DEFINITIONs:–1. A solid, as we have before observed, is that which has length, breadth, and thickness. 2. A polyhedron is a solid terminated by plane surfaces. 3. A prism is a solid terminated by two identical plane bases parallel to each other, and by surfaces which are parallelograms. (Fig. 334.) 4. A parallelopiped is a prism the bases of which are parallelograms. (Fig. 335.) 5. A cube is a solid terminated by six square surfaces: a die, for example, is a cube. (Fig. 336.) 6. If right lines be raised from every point in the perimeter of *** Fig. 555. any rectilineal figure, and meet in one common point, these lines together with the rectilineal figure inclose a solid which is called a pyramid. (Fig. 337.) 7. A cylinder is a solid terminated by two bases, which are equal and parallel circles, and by a convex surface; or it is a solid formed by the revolution of a parallelogram about one of its sides. (Fig. 838.) 8. If right lines be raised from every point in the circumference of a circle, and meet in one common point, these lines together with the circle inclose a solid, which is called a cone. (Fig. 339.) 9. A semicircle revolving about its diameter forms a solid, which is called a sphere. (Fig. 340.) 10. If from the vertex of a solid a perpendicular be let fall upon the opposite plane, this perpendicular is called the altitude of the solid. In the pyramids ACD, Acd (fig. 341.), AB, ab are their respective altitudes. 11. Solids are said to be equal, if they inclose an equal space: thus a cone and a pyramid are equal solids if the space inclosed within the cone be equal to the space inclosed within the pyramid. 12. Similar solids are such as consist of an equal number of physical points disposed in the same manner. Thus (in the fig. Defin, 10.) the larger pyramid ACD and the smaller pyramid Acd are similar solids if every point in the larger pyramid has a point corresponding to it in the smaller pyramid. A hundred musketballs, and the same number of cannon balls, disposed in the same manner, form two similar solids. 980. Prop. XCIX. The solid content of a cube is equal to the product of one of its sides twice multiplied by itself. Let the lines AB, AD (fig. 342.) be equal. Let the line AD, drawn perpendicular to AB, be supposed to move through the whole length of AB; when it I arrives at BC, and coincides with it, it will have formed the square DABC, # F and will have been multiplied by the line AB. *x Next let the line AF be drawn equal to AD, and perpendicular to the plane DABC; suppose the plane DABC to move perpendicularly through D C: the whole length of the line AF; when it arrives at the plane MFGL, >. and coincides with it, it will have formed the cube AFLC, and will have \ been multiplied by the line AF. Fig. 342. Hence it appears, that to form the cube AFLC, it is necessary, first, to multiply the side AD by the side AB equal to AD; and then to multiply the product, that is, the square of AC, by the side AF equal to AD; that is, it is A to multiply AD by AD, and to multiply the pro- duct again by AD. 981. Prop. C. Similar solids have their homologous lines proportional. Let the two solids A, a (fig. 343.) be similar; and let their homologous lines be AB, ab, BG, bg. AB will be to BG at ab to bg. 6 Because the solids A, a are similar, every point in the solid A has a point corresponding to it, and disposed in the same

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Fig. 336. Fig. 537. Fig. 338.

Fig. 339. Fig. 540. Fig. 341.

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manner, in the solid a. Thus, if the line AB is composed of 20 physical points, and the line BG of 10, the line ab will be composed of 20 corresponding points, and the line by of 10. Now it is evident that 20 is to 10 as 20 is to 10: therefore AB is to BG as ab to by

982. PRor. CI. Similar solids are equiangular.

: the solids (see fig to preced. Prop.) A, a be similar; their corresponding angles are equal.

Because the solids A, a are similar, the surfaces BAF, baf are composed of an equal number of points disposed in the same manner. These surfaces are therefore similar figures, and consequently (Prop. 88.) equiangular. The angles B, A, Fare therefore equal to the angles b, a, f. In the same manner it may be demonstrated that the other correspondent angles are equal.

983. Paor, CII. Solids which have their angles equal and their sides proportional are similar.

If the solids A, a (fig. 844.) have their angles equal and their sides proportional, they are similar.

A. For if the solids A, a were not similar, another solid might be formed upon the line BF similar to the solid a. But this is im- possible; for, in order to form this other solid, some angle or s some side of the solid A must be increased or diminished; and then this new solid would not have all its angles equal and all its sides proportional to those of the solid a, that is (Prop. C c B F & f Fig. 544

100, 101.), would not be similar. 984. PRoP. CIII. Similar solids are to one another as the cubes of their homologous sides. Let A, a (see fig. to preced. Prop.) be two similar solids, the solid A contains the solid a as many times as the cube formed upon the side BF contains the cube formed upon the side bf. i: the solid A is similar to the solid a, every point in the solid A has its corresponding point in the solid a. From whence it follows, that if the side BF is composed, for example, of 50 points, the side bf will also be composed of 50 points: and consequently the cubes formed upon the sides BF, bf will be composed of an equal number of points. Let it then be supposed that the solid A is composed of 4000 points, and the cube of the side BF of 5000 points; the solid A must be composed of 4000 points, and the cube of the side bf of 5000 points. Now it is evident that 4000 is to 5000 as 4000 to 5000. Wherefore the solid A is to the cube of BF as the solid a to the cube of bf; and, alternately, the solid A is to the solid a as the cube of BF to the cube of bf CoRollARY. It may be demonstrated in the same manner that the spheres A, a (fig. 345.), which are similar solids, are to

C one another as the cubes of their radii AB, C h B." C ab. c 985. PRoP. CIV. The solid content of B a perpendicular prism is equal to the product b | | of its base and height. D : D - 1--The solid content of the perpendicular d ." A. D

prism ABCD (fig. 346.) is equal to the Fig. 345 Fig. 546. product of its base AD, and height AD. If the lower base AD be supposed to move perpendicularly along the height AB till it coincides with the upper base BC, it will have formed the prism ABCD. Now the base AD will have been repeated as many times as there are physical points in the height AB. Therefore the solid content of the prism ABCD is equal to the product of the base multiplied by the height. CoRoLLARY. In the same manner it may be demonstrated that the solid content of the perpendicular cylinder ABCD is equal to the product of its base AD and height AB. 986. Prop. CV. The solid content of an inclined prism is equal to the product of its base and height. Let the inclined prism be CP (fig. 347.), it is equal to the product of its base RP and its height CD. Conceive the base NB of the perpendicular prism NA, and the base RP of the inclined prism PC, to move on in the same time parallel to themselves; when they have reached the points A and C, each of them will have been taken over again the same number of times. But the base NB will have been taken N over again (Prop. 104.) as many times as there are physical points in the height CD. The base RP will therefore have been taken over again as many times as there are physical points in CD. Consequently the solid content of the inclined prism CP is equal to the product of its base RP and the height CD.

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Fig. 347.

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987. Prop. CVI. In a pyramid, a section parallel to the base is similar to the base.

Let the section cd be parallel to the base CD (fig. 348.); this section is a figure similar to the base. Draw AB perpendicular to the base CD; draw also BC, be, BE, be.

Because the planes cd CD are parallel; AB, being perpendicular to the plane CD, will also (Prop. 98.) be perpendicular to the plane cd: whence the triangles Abe, ABC, having the angles b, B right angles, and the angle A common, are equiangular. Therefore (Prop. 61.) Ab is to AB as bc to BC, and as Ac to A.C.

In like manner it may be proved that Ab is to AB as be to BE, and as Ae to A.E. Consequently if Abbe one third part of AB, c be will be one third part of BC, be the same of BE, Ac of AC, and Ae of A.E. Fig. 548.

Again, in the two triangles cAe, CAE, there are about the angle A, common to both, two sides proportional; they are therefore (Prop. 63.) equiangular, and consequently (Prop. 61.) have their other sides proportional. Therefore ce will be one third part of CE.

The two triangles cbe CBE, having their sides proportional, are therefore (Prop. 89.) similar. The same may be demonstrated concerning all the other triangles which form the planes cd, CD. Therefore the section cd is similar to the base CD. REMARK. If the perpendicular AB fall out of the base; by drawing lines from the points b, B, it may be demonstrated in the same manner that the section is similar to the base. 988. Prop. CVII. In a pyramid, sections parallel to the base are to one another as the squares of their heights. Let CD cd (fig. 349.) be parallel sections. From the vertex A draw a perpendicular AB to the plane CD: the plane cd is to the plane CD as the square of the height Ab is to the square of the height AB. Draw BC, bc. A The line AB, being perpendicular to the plane CD, will also (Prop. 98.) be perpendicular to the parallel plane cd: whence the angle Abc is a right angle, and also the angle ABC. Moreover, the angle at A is common to the two triangles Abc, ABC; these two triangles, therefore, are equiangular. Therefore (Prop. 61.) the side cb is to the side CB as the side Ab is to the side AB; and consequently the square of eb is to the square of CB as the square of Ab to the square of AB. The planes cd, CD, being (Prop. 106.) similar figures, are to one C i. another (Prop. 82.) as the squares of the homologous lines cb, CB; Fig. 549. they are therefore also as the squares of the heights Ab, AB. CoRoLLARY. In the same manner it may be demonstrated that in a cone the sections parallel to the base are to one another as the squares of the heights or perpendicular distances from the vertex. 989. Pror. CVIII. Pyramids of the same height are to one another as their bases. Let A, F (fig. 350.) be two pyramids. If the perpendicular AB be equal to the perpendicular FG, the pyramid A is to the pyramid F as the base CD to the base LM. Supposing, for example, the base CD to be triple of the base LM, the pyramid A will be triple of the pyramid F. Two sections cd, lm, being taken at equal heights Ab, Fg, the section cd is (Prop. 107.) to the base CD as the square of the height Ab to the square of the height AB; and the section lm is to the base LM as the square of the Fig. 350. height Fg to the square of the height FG. And because the heights are equal, AB to FG, and Ab to Fg, the section cd is to the base CD as the section lm to the base LM, and, alternately, the section cd is to the section lm as the base CD is to the base LM. But the base CD is triple of the base LM, therefore the section cd is also triple of the section lm. Because the heights AF, FG are equal, it is manifest that the two pyramids are composed of an equal number of physical surfaces placed one upon another. Now it may be demonstrated in the same manner that every surface or section of the pyramid A is triple of the corresponding surface or section of the pyramid F. Therefore the whole pyramid A is triple of the whole pyramid F. CoRollARY. Pyramids of the same height and equal bases are equal, since they are to one another as their bases. 990. Prof. CIX. A pyramid whose base is that of a cube and whose vertex is at the centre of the cube is equal to a third part of the product of its height and base.

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Let the cube AM and the pyramid C (fig. 351.) have the same base AD, and let the vertex of the pyramid be at the centre of the cube C; this pyramid N M is equal to a third part of the product of its height and base. - i Conceive right lines drawn from the centre of the cube to its eight angles A, B, D, F, N, G, L, M, the cube will be divided into six equal pyramids, each of which has one surface of the cube for its base, and half the height of the cube for its height; for F example, the pyramid CABDF. Three of these pyramids will therefore be equal to half the cube. Now the solid content of half the cube is (Prop. 99.) equal to the product of the base and half the height. Each pyramid, therefore, will be equal to one third part of the product of the base, and half the height of the cube; that is. the whole height of the pyramid. 991. Prof. CX. The solid content of a pyramid is equal to a third part of the product of its height and base. Let RPS (fig. 352.) be a pyramid, its solid content is equal to a third part of the product of its height and its base R.S. Form a cube the height of which BL is double of the height N M of the pyramid RPS. A pyramid the base of which is that of this cube and the vertex of which is C, the centre of the cube,

Fig. 551.

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will be equal to a third part of the product of its base and C P height. - L- : The pyramids C and Phave the same height; they are there- r:#2× *@ fore (Corol. to Prop. 108.) to one another as their bases. If the 2–: -base AFDB is double of the base RS, the pyramid C will there- Fig.552.

fore be double of the pyramid P. But the pyramid C is equal to a third part of the product of its height and base. The pyramid P will therefore be equal to a third part of the product of the same height, and half the base AFDB, or, which is the same thing, the whole base RS. 992. PRor. CXI. The solid content of a come is equal to the third part of the product of its height and base. For the base of a cone may be considered as a polygon composed of exceedingly small sides, and consequently the cone may be considered as a pyramid having a great number of exceedingly small surfaces; whence its solid contents will be equal (Prop. 110.) to one third part of the product of its height and base. 993. Pror. CXII. The solid content of a cone is a third part of the solid content of a cylinder described about it. Let the cone BAC and the cylinder BDFC (fig. 353.) have the same height and base, the cone is a third part of the cylinder. For the cylinder is equal to the product of its height and base, and the cone is equal to a third part of this product. Therefore the cone is a third part of the cylinder. 994. PRor. CXIII. The solid content of a sphere is equal to a third part of the product of its radius and surface. Two points not being sufficient to make a curve line, three points will not be sufficient to make a curve surface. If therefore, all the physical Fig. 355. points which compose the surface of the sphere C (fig. 354.) be taken three by three, the whole surface will be divided into exceedingly small plane surfaces; and radii

being drawn to each of these points, the sphere will be divided into small pyramids, which have their vertex at the centre, and have plane bases. /* C t # /

The solid contents of all these small pyramids will be equal (Prop. 110.) to a third part of the product of the height and bases. Therefore the solid content of the whole sphere will be equal to a third part of the product of the height and all the bases, that is, of its radius and surface. 995. Pror. CXIV. The surface of a sphere is equal to four of its great circles. Fig. 354. If a plane bisect a sphere, the section will pass through the centre, and it is called a great circle of the sphere. Let ABCD (fig. 355.) be a square; describe the fourth part of the circumference of a circle BLD; draw the diagonal AC, the right line FM, parallel to AD, a and the right line AL. In the triangle ABC, on account of the equal sides AB, BC, the angles A and C are (Prop. 4.) equal; therefore, since the angle B is a right angle, the angles A and C are each half a right angle. Again, in the triangle AFG, because the angle F is a right angle, and the angle A half a right |A angle, the angle G is also half a right angle; therefore (Prop. 26.) A F is equal to FG.

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