Dem.—Because the planeAOBcuts the parallel planesABCDE,abcde, the sectionsAB,abare parallel [XI.xvi.] In like mannerBC,bcare parallel. Hence the angleABC=abc[XI.x.]. In like manner the remaining angles of the polygonABCDEare equal to the corresponding angles ofabcde. Again, becauseabis parallel toAB, the trianglesABO,abOare equiangular.
Therefore the polygonsABCDE,abcdeare equiangular, and have the sides about their equal angles proportional.Hence they are similar.
Cor.1.—The edges and the altitude of the pyramid are similarly divided by the parallel plane.
Cor.2.—The areas of parallel sections are in the duplicate ratio of the distances of their planes from the vertex.
Cor.3.—In any two pyramids, sections parallel to their bases, which divide their altitudes in the same ratio, are proportional to their bases.
PROP.V.—Theorem.Pyramids(P–ABCD,p–abc), having equal altitudes(PO,po)and bases(ABCD,abc)of equal areas, have equal volumes.
PROP.V.—Theorem.Pyramids(P–ABCD,p–abc), having equal altitudes(PO,po)and bases(ABCD,abc)of equal areas, have equal volumes.
Dem.—If they be not equal in volume, letabcbe the base of the greater; and letoxbe the altitude of a prism, with an equal base, and whose volume is equal to their difference; then let the equal altitudesPO,pobe divided into such a number of equal parts, by planes parallel to the bases of the pyramids, that each part shall be less thanox. Then [iv.Cor.3] the sections made by these planes will be equal each to each. Now let prisms be constructed on these sections as bases and with the equal parts of the altitudes of the pyramids as altitudes, and let the prisms inP–ABCDbe constructed below the sections, and inp–abc, above. Then it is evident that the sum of the prisms inP–ABCDis less than that pyramid, and the sum of those on the sections ofp–abcgreater thanp–abc. Therefore the difference between the pyramids is less than the difference between the sums of the prisms, that is, less than the lower prism in the pyramidp–abc; but the altitude of this prism is less thanox(const.). Hence the difference between the pyramids is less than the prism whose base is equal to one of the equal bases, and whose altitude is equal toox, and the difference is equal to this prism (hyp.), which is impossible.Therefore the volumes of the pyramids areequal.
Cor.1.—The volume of a triangular pyramidE–ABCis one third the volume ofthe prismABC–DEF, having the same base and altitude.
For, draw the planeEAF, then the pyramidsE–AFC,E–AFDare equal, having equal basesAFC,AFD, and a common altitude; and the pyramidsE–ABC,F–ABCare equal, having a common base and equal altitudes. Hence the pyramidE–ABCis one of three equal pyramids into which the prism is divided.Therefore itis one third of the prism.
Cor.2.—The volume of every pyramid is one-third of the volume of a prismhaving an equal base and altitude.
Because it may be divided into triangular pyramids by planes through the vertex and the diagonals of the base.
PROP.VI.—Theorem.The volume of a cylinderis equal to the product of the area of its base by itsaltitude
PROP.VI.—Theorem.The volume of a cylinderis equal to the product of the area of its base by itsaltitude
.
Dem.—LetObe the centre of its circular base; and take the angleAOBindefinitely small, so that the arcABmay be regarded as a right line. Then planes perpendicular to the base, and cutting it in the linesOA,OB, will be faces of a triangular prism, whose base will be the triangleAOB, and whose altitude will be the altitude of the cylinder. The volume of this prism will be equal to the area of the triangleAOBby the height of the cylinder. Hence, dividing the circle into elementary triangles, the cylinder will be equal to the sum of all the prisms,andtherefore its volume will be equal to the area of the base multiplied by thealtitude.
Cor. 1.—Ifrbe the radius, andhthe height of the cylinder,
vol. of cylinder = πr2h.
Cor. 2.—IfABCDbe a rectangle;Xa line in its plane parallel to the sideAB;Othe middle point of the rectangle; the volume of the solid described by the revolutionofABCDroundXis equal to the area ofABCDmultiplied by the circumference ofthe circle described byO.
Dem.—Produce the linesAD,BCto meetXin the pointsE,F. Then when the rectangle revolves roundX, the rectanglesABFE,DCFEwill describe cylinders whose bases will be circles havingAE,DEas radii, and whose common altitude will beAB. Hence the difference between the volumes of these cylinders will be equal to the differences between the areas of the bases multiplied byAB, that is =π(AE2−DE2).AB. Therefore the volume described byABCD
Hence volume described by the rectangleABCD
= 2π.OG.AB.AD.
= rectangleABCDmultiplied by the circumference of the circle described by its middle pointO.
Observation.—This Cor.is a simple case of Guldinus’s celebrated theorem. By its assistance wegive in the two following corollaries original methods of finding the volumes of the cone and sphere,and it may be applied with equal facility to the solution of several other problems which are usuallydone by the Integral Calculus.
Cor.3.—The volume of a cone is one-third the volume of a cylinder having thesame base and altitude.
Dem.—LetABCDbe a rectangle whose diagonal isAC. The triangleABCwill describe a cone, and the rectangle a cylinder by revolving roundAB. Take two pointsE,Finfinitely near each other inAC, and form two rectangles,EH,EK, by drawing lines parallel toAD,AB. Now ifO,O′be the middle points of these rectangles, it is evident that, when the whole figure revolves roundAB, the circumference of the circle described byO′will ultimately be twice the circumference of the circle described byO; and since the parallelogramEKis equal toEH[I.xliii.], the solid described byEK(Cor. 1) will be equal to twice the solid described byEH. Hence, ifACbe divided into an indefinite number of equal parts, and rectangles corresponding toEH,EKbe inscribed in the trianglesABC,ADC, the sum of the solids described by the rectangles in the triangleADCis equal to twice the sum of the solids described by the rectangles in the triangleABC—that is, the difference between the cylinder and cone is equal to twice the cone. Hence the cylinder is equal to three times the cone.
Or thus:We may regard the cone and the cylinder as limiting cases of a pyramid and prismhaving the same base and altitude; and since (v.Cor.2) the volume of a pyramid is one-third ofthe volume of a prism, having the same base and altitude, the volume of the cone is one-third of thevolume of the cylinder.
Cor.4.—Ifrbe the radius of the base of a cone, andhits height,
vol. of cone = πr2h-. 3
Cor.5.—The volume of a sphereis two-thirds of the volume of a circumscribedcylinder.
Dem.—LetABbe the diameter of the semicircle which describes the sphere;ABCDthe rectangle which describes the cylinder. Take two pointsE,Findefinitely near each other in the semicircle. JoinEF, which will be a tangent, and produce it to meet the diameterPQperpendicular toABinN. LetRbe the centre. JoinRE; drawEG,FH,NLparallel toAB; andEI,FKparallel toPQ; and produce to meetLNinMandL; and letO,O′be the middle points of the rectanglesEH,EK.
Now the rectangleNG.GR=PG.GQ, because each is equal toGE2. HenceNG:GP::GQ:GR, orME:IE::RP+RG:RG. Now, denoting the radii of the circles described by the pointsO,O′byρ,ρ′respectively, we have ultimatelyρ=GRandρ′=1 2(RP+RG). HenceME:IE:: 2ρ′:ρ; butME:IE:: rectangleEL: rectangleEK:: [I.xliii.]EH:EK;
Hence the solid described byEHequal twice the solid described byEK. Therefore we infer, as in the last Cor., that the whole volume of the sphere is equal to twice the difference between the cylinder and sphere.Therefore the sphere is two-thirds of thecylinder.
Cor.6.—Ifrbe the radius of a sphere,
vol. = 4πr3. 3
PROP.VII.—Theorem.The surface of a sphereis equal to the convex surface of the circumscribedcylinder.
PROP.VII.—Theorem.The surface of a sphereis equal to the convex surface of the circumscribedcylinder.
Dem.—LetABbe the diameter of the semicircle which describes the sphere. Take two points,E,F, indefinitely near each other in the semicircle. JoinEF, and produce to meet the tangentCDparallel toABinN. DrawEI,FKparallel toPQ. ProduceEIto meetABinG. LetObe the centre. JoinOE.
because the trianglesENIandOEGare similar.
HenceEF:IK::IG:EG; andIG:EG:: circumference of circle described by the pointI: circumference of circle described by the pointE. Hence the rectangle contained byEF, and circumference of circle described byEis equal to the rectangle contained byIK, and circumference of circle described byI—that is, the portion of the spherical surfacedescribed byEFis equal to the portion of the cylindrical surface described byIK. Hence it is evident, if planes be drawn perpendicular to the diameterAB—that the portions of cylindrical and spherical surface between any two of them are equal.Hencethe whole spherical surface is equal to the cylindrical surface described byCD.
Or thus:Conceive the whole surface of the sphere divided into an indefinitely great number ofequal parts, then it is evident that each of these may be regarded as the base of a pyramid havingthe centre of the sphere as a common vertex. Therefore the volume of the sphere is equal to thewhole area of the surface multiplied by one-third of the radius. Hence ifSdenote the surface, wehave
That is, the area of the surface of a sphere is equal four times the area of one of its great circles.
Exercises.
Exercises.
1.The convex surface of a cone is equal to half the rectangle contained by the circumference ofthe base and the slant height.
2.The convex surface of a right cylinder is equal to the rectangle contained by thecircumference of the base and the altitude.
3.IfPbe a point in the baseABCof a triangular pyramidO–ABC, and if parallels tothe edgesOA,OB,OC, throughP, meet the faces in the pointsa,b,c, the sum of theratios
P a Pb Pc OA-, OB, OC-= 1.
4.The volume of the frustum of a cone, made by a plane parallel to the base, is equal to thesum of the three cones whose bases are the two ends of the frustum, and the circle whose diameter isa mean proportional between the end diameters, and whose common altitude is equal to one-third ofthe altitude of the frustum.
5.If a pointPbe joined to the angular pointsA,B,C,Dof a tetrahedron, and thejoining lines, produced if necessary, meet the opposite faces ina,b,c,d, the sum of theratios
Pa P b Pc Pd Aa, Bb , Cc, Dd-= 1.
6.The surface of a sphere is equal to the rectangle by its diameter, and the circumference of agreat circle.
7.The surface of a sphere is two thirds of the whole surface of its circumscribedcylinder.
8.If the four diagonals of a quadrangular prism be concurrent, it is a parallelopiped.
9.If the slant height of a right cone be equal to the diameter of its base, its total surface is tothe surface of the inscribed sphere as 9 : 4.
10.The middle points of two pairs of opposite edges of a triangular pyramid are coplanar, andform a parallelogram.
11.If the four perpendiculars from the vertices on the opposite faces of a pyramidABCDbeconcurrent, then
AB2 +CD2 = BC2 +AD2 = CA2+ BD2.
12.Every section of a sphere by a plane is a circle.
13.The locus of the centres of parallel sections is a diameter of the sphere.
14.If any number of lines in space pass through a fixed point, the feet of the perpendiculars onthem from another fixed point are homospheric.
15.Extend the property of Ex.4 to the pyramid.
16.The volume of the ring described by a circle which revolves round a line in its plane is equalto the area of the circle, multiplied by the circumference of the circle described by itscentre.
17.Any plane bisecting two opposite edges of a tetrahedron bisects its volume.
18.Planes which bisect the dihedral angles of a tetrahedron meet in a point.
19.Planes which bisect perpendicularly the edges of a tetrahedron meet in a point.
20.The volumes of two triangular pyramids, having a common solid angle, are proportional tothe rectangles contained by the edges terminating in that angle.
21.A plane bisecting a dihedral angle of a tetrahedron divides the opposite edge into portionsproportional to the areas containing that edge.
22.The volume of a sphere: the volume of the circumscribed cube asπ: 6.
23.Ifhbe the height, andρthe radius of a segment of a sphere, its volume isπh 6(h2+ 3ρ2).
24.Ifhbe the perpendicular distance between two parallel planes, which cut a sphere insections whose radii areρ1,ρ2, the volume of the frustum isπh 6-{h2+ 3(ρ12+ρ22)}.
25.Ifδbe the distance of a pointPfrom the centre of a sphere whose radius isR, thesum of the squares of the six segments at three rectangular chords passing throughPis= 6R2−2δ2.
26.The volume of a sphere : the volume of an inscribed cube asπ: 2.
27.IfO–ABCbe a tetrahedron whose anglesAOB,BOC,COAare right, the square of thearea of the triangleABCis equal to the sum of the squares of the three other triangularfaces.
28.In the same case, ifpbe the perpendicular fromOon the faceABC,
12 =-1-2 +-12-+ -12. p OA OB OC
29.Ifhbe the height of an æronaut, andRthe radius of the earth, the extent of surface visible=2πR2h R-+-h.
30.If the four sides of a gauche quadrilateral touch a sphere, the points of contact areconcyclic.
NOTES._____NOTE A.MODERN THEORY OF PARALLEL LINES.
NOTES.
_____
NOTE A.
MODERN THEORY OF PARALLEL LINES.
In every plane there is one special line called theline at infinity. The point where any other linein the plane cuts the line at infinity is called thepoint at infinityin that line. All other points in theline are calledfinite points. Two lines in the plane which meet the line at infinity in the same pointare said to have thesame direction, and two lines which meet it in different points to havedifferent directions. Two lines which have the same direction cannot meet in any finite point[I.Axiomx.], and are parallel. Two lines which have different directions must intersect in somefinite point, since, if produced, they meet the line at infinity in different points. This is afundamental conception in Geometry, it is self-evident, and may be assumed as an Axiom(seeObservations on the Axioms, BookI.). Hence we may infer the following generalproposition:—“Any two lines in the same plane must meet in some point in that plane; thatis—(1)at infinity, when the lines have the same direction;(2)in some finite point,when they have different directions.”—SeePoncelet,Propriétés Projectives, page52.
________________NOTE B.legendre’sandhamilton’sproofsofeuclid, I.xxxii.
________________
NOTE B.
legendre’sandhamilton’sproofsofeuclid, I.xxxii.
The discovery of the Proposition that “the sum of the three angles of a triangle is equal to tworight angles” is attributed to Pythagoras. Until modern times no proof of it, independent of thetheory of parallels, was known. We shall give here two demonstrations, each independent of thattheory. These are due to two of the greatest mathematicians of modern times—one, thefounder of the Theory of Elliptic Functions; the other, the discoverer of the Calculus ofQuaternions.