manner, in the solid a. Thus, if the line AB is composed of 20 physical points, and t line BG of 10, the line ab will be composed of 20 corresponding points, and the line of 10. Now it is evident that 20 is to 10 as 20 is to 10: therefore AB is to BG as ab to 982. PROP. CI. Similar solids are equiangular. Let the solids (see fig. to preced. Prop.) A, a be similar; their corresponding angles a equal. Because the solids A, a are similar, the surfaces BAF, baf are composed of an equ number of points disposed in the same manner. These surfaces are therefore simil figures, and consequently (Prop. 88.) equiangular. The angles B, A, F are therefore equ to the angles b, a, f. In the same manner it may be demonstrated that the other co respondent angles are equal. 983. PROP. CII. similar. Solids which have their angles equal and their sides proportional a If the solids A, a (fig. 344.) have their angles equal and their sides proportional, the are similar. But this is im- For if the solids A, a were not similar, another solid might be formed upon the line BF similar to the solid a. possible; for, in order to form this other solid, 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. 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 as many times as the cube formed upon the side BF contains the cube formed upon th side bf. Because the solid A is similar to the solid a, every point in the solid A has its cor responding 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 conse quently the cubes formed upon the sides BF, bf will be composed of an equal numbe of points. Let it then be supposed that the solid A is composed of 4000 points, and the cube o the side BF of 5000 points; the solid A must be composed of 4000 points, and the cub 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 one another as the cubes of their radii AB, ab. 985. PROP. CIV. The solid content of a perpendicular prism is equal to the product of its base and height. The solid content of the perpendicular prism ABCD (fig. 346.) is equal to the product of its base AD, and height AB. A B B eel Fig. 345 d Fig. 346. C 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 mul tiplied 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 bas and height. Let the inclined prism be CP (fig. 347.), it is equal to the product of its base RE and its height CD. M A 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 over again (Prop. 104.) as many times as there are physical points in the height CD. The base RP will therefore have been taken ver again as many times as there are physical points in CD. onsequently the solid content of the inclined prism CP is equal to the product of it e RP and the height CD. D Fig. 347. 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, bc. 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 Abc, ABC, having the angles b, B right angles, and the angle A common, are equiangular. Therefore (Prop. 61.) Ab is to AB as be to BC, and as Ac to AC. In like manner it may be proved that Ab is to AB as be to BE, and as Ae to AE. Consequently if Ab be 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 AE. Fig. 348. 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 proportional to 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 perpendicula CD as the square of the height Ab is to the square of the height AB. Draw BC, bc. 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. Fig. 349. D The planes cd, CD, being (Prop. 106.) similar figures, are to one C another (Prop. 82.) as the squares of the homologous lines cb, CB; 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. PROP. 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. m D M G E L Fig. 350. 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 Im is to the base LM as the square of the 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 basɛ CD as the section Im to the base LM; and, alternately, the section cd is to the section Im 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 Im. 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 mother. 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. Prop. 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. U Let the cube AM and the pyramid C (fig. 351.) have the same base AD, and let the ver tex of the pyramid be at the centre of the cube C; this pyramid is equal to a third part of the product of its height and base. Conceive right lines drawn from the centre of the cube to its Pight 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 example, the pyramid CABDF. F N M G C B Fig. 351. 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. PROP. 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 RS. Form a cube the height of which BL is double of the height 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, will be equal to a third part of the product of its base and height. The pyramids C and P have the same height; they are therefore (Corol. to Prop. 108.) to one another as their bases. If the base AFDB is double of the base RS, the pyramid C will therefore be double of the pyramid P. N M F Fig. 352. The But the pyramid C is equal to a third part of the product of its height and base. 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. PROP. CXI. The solid content of a cone 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. PROP. 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. PROP. CXIII. The solid content of a sphere is equal to a third part of the product of its radius and surface. D B Fig. 353. C 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 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. 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. PROP. CXIV. The surface of a sphere is equal to four of its great circles. B 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; d.aw the diagonal AC, through G, the right line FM, B parallel to AD, 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 angle, the angle G is also half a right angle; therefore (Prop. 26.) AF is equal to FG. The radius AL is equa! to the radius AD: but AD is equal to FM; therefore AL is equal to FM. In the rectangular triangle AFL the square of the hypothenuse AL is equal (Prop. 32.) to the two squares of AF and FL taken together. Instead of AL put its equal FM, and instead of AF put its equal FG; and the square of FM will be equal to the two squares of FG and FL taken together. Conceive the square ABCD to revolve about the line AB. In the revolution the square will describe a cylinder, the quadrant a hemisphere, and the triangle ABC an inverted cone the vertex whereof will be in A. Also the line FM will form a circular section of a cylinder, the line FL will form a circular section of a hemisphere, and the line FG a circular section of a cone. These circular sections, or circles, are to each other (Prop. 83.) as the squares of their radii ; therefore, since the square of the radius FM is equal to the squares of the radii FL and FG, the circular section of the cylinder will be equal to the circular sections of the [hemisphere and cone. B, A L In the same manner it may be demonstrated that all the other sections or circular surfaces whereof the cylinder is composed are equal to the corresponding sections or surfaces of the hemisphere and cone. Therefore the cylinder is equal to the hemisphere and cone taken together: but the cone (Prop. 112.) is equal to a third part of the cylinder; the hemisphere is therefore equal to the remaining two thirds of the cylinder; and consequently the hemisphere is double of the cone. The cone BSC (fig. 356.) is (Prop. 111.) equal to a third part of the product of the radius and base BC, which is a great circle of the sphere: the hemisphere ALD is therefore equal to a third part of the product of the radius and two of its great circles; and consequently the whole sphere is equal to a third part of the product of the radius and four of its great circles. Lastly, since the sphere is equal (Prop. 113.) to a third part of the product of the radius and surface of the sphere, and also to a third part of the product of the radius and four of its great circles, the surface of the sphere is equal to four of its great circles. M SECT. II. PRACTICAL GEOMETRY. 996. Practical Geometry is the art of accurately delineating on a plane surface any plane figure. It is the most simple species of geometrical drawing, and the most generally useful; for the surfaces of buildings and other objects are more frequently plane than curved, and they must be drawn with truth, and of the required proportions, before they can be properly executed, unless in cases where the extreme simplicity of the form renders t improbable that mistakes should arise. It has been defined as the art which directs the mechanical processes for finding the position of points, lines, surfaces, and planes, with the description of such figures on diagrams as can be intelligibly understood by definition, according to given dimensions and positions of lines, points, &c. No part of a building or drawing can be laid down or understood without the assistance of practical geometry, nor can any mechanical employment in the building department be conducted without some assistance from this branch of the science. Cases frequently occur requiring a knowledge of very complex problems, as in masonry, carpentry, and joinery; but these will be given in other parts of this work. The demonstration of most of the following problems will be found in the preceding section; we therefore refer the reader back to it for definitions, and for the proof of those enunciations which will follow. PROBLEMS. 997. PROBLEM I. To bisect a line AB; that is, to divide it into two equal parts. From the two centres A and B (fig. 357.) with any equal radii describe arcs of circles intersecting each other in C and D, and draw the line CD. This will bisect the given line in the point E. 998. PROB. II. To bisect an angle BAC. From the centre A (fig. 358.) with any radius describe an arc cutting off the equal lines AD, AE; and from the two centres D, E, with the same radius describe arcs intersecting in F, then draw AF, and it will bisect the angle A, as required. 999. PROB. III. At a given point C in a line AB to erect a perpendicular. C From the given point C (fig. 359.) with any radius cut off any equa parts CD, C of the given line; and from the two centres D and E with any one radius describe arcs intersecting in F. Then join CF, and it will be the perpendicular required. Otherwise - When the given point C is near the end of the line. From any point D (fig. 360.) assumed above the line as a centre, through the given point C describe a circle cutting the given line at E, and through E and the centre D draw the diameter EDF; then join CF, and it will be the perpendicular required. Fig. 337. Fig. 358. P Fig. 339. 1000. PROB. IV. From a given point A to let fall a perpendicular on a line BC. From the given point A (fig. 361.) as a centre with any convenient radius describe an arc cutting the given line at two points D and E; and from the two centres D and E with any radius describe two arcs intersecting at F; then draw AF, and it will be the perpendicular to BC required. Otherwise When the given point is nearly opposite the end of the line. From any point D in the given line BC (fig. 362.) as a centre, describe the arc of a circle through the given point A cutting BC in E; and from the centre E with the D B radius EA describe another are cutting the former in F; then draw AGF, which will be the perpendicular to BC required. 1001. PROB. V. At a given point A, in a line AB, to make an angle equal to a given angle C. From the centres A and C (fig. 363.) with any radius describe the arcs DE, FG; then with F as a centre, and radius DE, describe an arc cutting FG in G; through G draw the line AG, which will form the angle required. 1002. PROB. VI. Through a given point C to draw a line parallel to a given line AB. E When the parallel is to be drawn at a given distance from AB. BA B C Fig. 364. From any two points c and d in the line AB, with a radius equal to the given distance describe the arcs e and f; draw the line CB to touch those arcs without cutting them, and it will be parallel to AB, as required. C F E 1003. PROB. VII. To divide a line AB into any proposed number of equal parts. Draw any other line AC (fig. 365.), forming any angle with the given line AB; on the latter set off as many of any equal parts AD, DE, EF, FC as those into which the line AB is to be divided; join BC, and parallel thereto draw the other lines FG, EH, DI; then these will divide AB, as required. 1004. PROB. VIII. To find a third proportional to two other A lines AB, AC. Let the two given lines be placed to form any angle at A (fig. 366.), and in AB take AD equal to AC; join BC, and draw DE parallel to it; then AE will be the third proportional sought. |