The radius AL is equal 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. 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 A L C hemisphere and cone taken together: but the cone (Prop. 112.) is B, 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. III. 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 it 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. C From the given point C (fig. 359.) with any radius cut off any equal parts CD, CE 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. E B D F с Fig. 358. Fig. 359. A 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 AGF, and it will be the perpendicular to BC required. Otherwise When the given point is nearly opposite the end of the line. Fig. 360. B F Fig 361. 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 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 ares DE, FG; then with F as a centre, and radius DE, de- B D scribe 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. When the parallel is to be drawn at a given distance from AB. 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 eutting them, and it will be parallel to AB, as required. 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. F E D H Fig. 365. 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. B A A B A 1005. PROB. IX. To find a fourth proportional to three lines AB, AC, AD. Let two of the lines AB, AC (fig. 367.), be so placed as to form any angle at A, and set out AD or AB; join BC, and parallel to it draw DE; then AE will be the fourth proportional required. 1006. PROB. X. To find a mean proportional between two lines AB, BC. Place AB, BC (fig. 368.) E Fig. 366. B B C E Fig. 367. B joined together in one straight line AC, which bisect in the point 0; then with the centre O and radius OA or OC describe the semicircle ADC, to meet which erect the perpendicular BD, which will be the mean proportional between AB and BC sought. 1007. PROB. XI. To (fig. 369.), and bisect perpendicularly with the line CD, which bisected in O will be the centre required. 1008. PROB. XII. To describe the circumference of a circle through three points A, B, C. From the middle point B (fig. 370.) draw the chords BA, BC to the two other points, and bisect these chords perpendicularly by lines meeting in O, which will be the centre; from the centre O, with the distance of any one of the points, as OA, describe a circle, and it will pass through the two other points B which, as a diameter, describe a semicircle cutting the given circumference in D, through which draw BADC, which will be the tangent required. 1010. PROB. XIV. To draw an equilateral 374.) describe an arc; with the centre B and distance BC describe another arc cutting the former in C; draw AC, BC, and ABC will be the triangle required. Bisect the angles at A and B with the two lines AD, BD (fig. 376.); from the intersection D, which will be the centre of the circle, draw the perpendiculars DE, DF, DG, and they will be the radii of the circle required. Bisect any two sides with two of the perpendiculars DE, DF, DG (fig. 377.), and D will be the centre of the circle. 1015. PROB. XIX. To inscribe an equilateral triangle in a given circle. Through the centre C draw any diameter AB (fig. 378.); from the point B as a centre, with the radius BC of the given circle, describe an are DCE; join AD, AE, DE, and ADE is the equilateral triangle sought. 1016. PROB. XX. given circle. To inscribe a square in a E Fig. 377. E B Fig. 378. Draw two diameters AC, BD (fig. 379.) crossing at right angles in the centre E; then join the four extremities A, B, C, D with right lines, and these will form the inscribed square ABCD. 1017. PROB. XXI. To describe a square about a given circle. Draw two diameters AC, BD crossing at right angles in the centre E (fig. 380.); then through the four extremities of these draw FG, IH parallel to AC, and FI, GH parallel to BD, and they will form the square FGHI. F B G C E E Fig. 379. D G Fig. 380. 1018. PROB. XXII. To inscribe a circle in a given square. Bisect the two sides FG, FI in the points A and B (see fig. 380.); then through these two points draw AC parallel to FG or IH, and BD parallel to FI or GH. Then the point of intersection E will be the centre, and the four lines EA, EB, EC, ED radii of the inscribed circle. 1019. PROB. XXIII. To cut a given line in extreme and mean ratio. Let AB be the given line to be divided in extreme and mean ratio (fig. 381.); that is, so that the whole line may be to the greater part as the greater part is to the less part. Draw BC perpendicular to AB, and equal to half AB; join AC, and with the centre C and distance CB describe the circle BD; then with the centre A and distance AD describe the arc DE. Then AB will be divided in E in extreme and mean ratio, or so that AB is to AE as AE is to EB. 1020. PROB. XXIV. To inscribe an isosceles triangle in a given circle that shall have each of the angles at the base double the angle at the vertex. Draw any diameter AB of the given circle (fig. 382.), and divide the radius CB in the point D in extreme and mean ratio (by the last problem); from the point B apply the ADBCE will be the inscribed equilateral triangle required. 1022. PROB. XXVI. To inscribe a regular hexagon in a circle. Apply the radius of the given circle AO as a chord (fig. 384.) quite round the circum. ference, and it will form the points thereon 1024. PROB. XXVIII. To inscribe a circle in a regular polygon. Bisect any two sides of the polygon by the perpendiculars GO, FO (fig. 386.), and their intersection O will be the centre of the inscribed circle, and OG or OF will be the radius. 1025. PROB. XXIX. To describe a circle about a regular polygon. Bisect any two of the angles C and D with the lines CO, DO (fig. 387.), then their intersection O will be the centre of the circumscribing circle; and OC or OD will be the radius. 1026. PROB. XXX. To make a triangle equal to a given quadrilateral ABCD. Draw the diagonal AC (fig. 388.), and parallel to it DE, meeting BA produced at E, and join CE; then will the triangle CEB B E be equal to the given quadrilateral ABCD. 1027. PROB. XXXI. To make a triangle egual to a given pentagon ABCDE. Draw DA and DB, and also EF, CG parallel to them (fig. 389.), meeting AB produced at F and G; then draw DF and To make a square equal to a given rectangle ABCD. A c d B Produce one side AB till BE be equal to the other side BC (fig. 391.). On AE as a diameter describe a circle meeting BC produced at F, then will BF be the side of the square BFGH equal to the given rectangle BD, as required. A catenary being a curve into which a perfectly flexible cord or chain will arrange D itself when suspended by its two extremities, it may thus be described. Let AB (fig. 392.) be a given line from two points c, d, whereof the curve is to fall, and let C be the lowest point in the curve. From two pins inserted at the points c and d suspend a fine cord or chain, lengthening it till the lowest point of the crown touches C, then a pencil tracing its line on the paper; the curve thus formed will be the catenary required. 1031. PROB. XXXV. To draw a cycloid. C If the circumference of a circle be rolled along a right line AB (fig. 393.) until any point b, b in the circumference which was in contact with the line come again in contact with it, the point b will describe a curve called a cycloid. Let the circle be BC, and the semibase be AB, which must be equal to the semi-circumference of the circle. Draw any chords Cb, Cb, and parallel to AB drawthe horizontal lines ab, ab, making them respectively equal to the length of the arcs cut off by the chords. Then through the points a, a, so obtained, draw a curve line, and it will be the cycloid required. α Fig. 393. 1032. PROB. XXXVI. To draw a diagonal scale. many times as necessary, the number of feet by equal Fig. 394. C |