AN MENSURATION RACTICAL GEOMETRY. I:E WITH iOUS PROBLEMS OF PRACTICAL IMPORTANCE MECHANICS. WILLIAM VOGDES, LL.IX )F MATHEMATICS W THB CENTRAL HJOH BCBOOL OF F AT'THOR OF TH? UNITED 8TATK8 AEITHM1IIO. [PAET FIES1.] PHILADELPniA: E. C\ & J. BIDDLE fe CO., No. 508 MINOR ST. (Betwee* Market, and Chtstttul, and fifth and Sixth Sts.) 1861. -7 y^-fc?^ vp/z*+ ) JfafivQ \ AN ELEMENTARY TREATISE ON MENSURATION PRACTICAL GEOMETKY, TOGETHER WITH NUMEROUS PROBLEMS OF PRACTICAL IMPORTANCE MECHANICS. WILLIAM VOGDES, LL.D. LATE PROFESSOR OF MATHEMATICS IN THE CENTRAL HIGH SCHOOL OF PHILA- DELPHIA, AUTHOR OP THE UNITED STATES ARITHMETIC. [PAET FIEST.] PHILADELPHIA: E. C. & J. BIDDLE, No. 508 MINOK ST. (Between Market and Chestnut, and Fifth and Sixth Sis.) 1862. Chamber of the Controllers of Pttblic Schools, \ First School District of Pennsylvania. / PHILADELPHIA, November 15th, 1849. AT a meeting of the Controllers of Public Schools, First School District of Pennsylvania, held at the Controllers' Chamber, on Tues- day, November 13th, 1849, the following Resolution was adopted: Resolved, That Vogdes' Mensuration be introduced as a Class Book into the Grammar Schools of the District. From the Minutes. ROBERT J HEMPHILL, Secretary. ESTEBED according to act of Congress, in the year 1846, by E. C. & J. DIDDLE, m the Clerk's Office of the District Court of the Eastern District of Pennsyl- BTEREOTYPED BY >.. JOHNSOK & CO. PHILADELPHIA. COLLINS, PRINT** PREFACE. IT has been the design of the author, in the following pages, to compile a work adapted, by its practical cha- racter, to the wants of those of the rising generation, who, not being able to command a collegiate education, are fitting themselves to fill useful stations in society as mechanics, merchants, &c. With tiiis end in view, it was deemed inexpedient to devote any portion of the book to theoretical demonstrations of the principles in- volved in the rules given, when those demonstrations are based upon principles illustrated by more advanced branches of mathematical science, with which the pupil is supposed to be unacquainted. By pursuing this course, room was afforded for the introduction of more numerous examples illustrating the respective rules, than could otherwise have been given without the enlarge- ment of the work beyond expedient limits. The intro- duction of these examples, it is believed, will enhance the value of the work in the estimation of teachers gene- rally, inasmuch as the operation of the rules is more likely to be permanently impressed on the mind of the pupil, by long continued practice, than by the solution of one or two problems only. In a treatise on Mensuration, little that is new can be looked for, other than the collection and judicious arrangement of matters not heretofore presented to the public in a form adapted to the purposes for which this 3 4 PREFACE. work is designed. The greater portion of this volume has been derived from the works of Bonnycastle, Has- well, Hutton, Gregory, and Grier, To some of these, special acknowledgment of the obligation has been made in the subsequent pages. The application of science to the arts of industry has oeen one of the most potent operative causes of the rapid increase of our country in wealth and power. In the hope that this compilation may be found a useful assistant to the teacher who is engaged in preparing pupils for an active participation in these industrial pur- suits, the author submits it to the inspection of his co- labourers in this field. Philadelphia, June 27, 1846. A KEY to this work has been published. CONTENTS. PRACTICAL GEOMETRY. Definitions 1& Instruments 23 Geometrical Problems 27 MENSURATION OF SUPERFICIES. To find the area of a square 57 To find the side of a square 58 The diagonal of a square being given, to find the area 58 The area of a square being given, to find the diagonal 59 The diagonal of a square being given, to find the side 60 To cut off a given area from a square parallel to either side. 60 To find the area of a rectangle 61 To find the side of a rectangle 62 The area and proportion of the two sides of a rectangle being given, to find the sides 63 The sides of a rectangle being given, to cut off a given area parallel to either side 64 To find the area of a rhombus 65 To find the area of a rhomboid 66 To find the side of a rhombus or rhomboid 66 To find the area of a triangle, when the base and perpendi- cular are given 67 The three sides of a triangle being given, to find the area. ... 68 Any two sides of a right-angled triangle being given, to find the third side 69 The sum of the hypothenuse and perpendicular, and the base of a right-angled triangle being given, to find the hypothe- nuse and perpendicular 70 To determine the area of an equilateral triangle 71 The area and the base of any triangle being given, to find the perpendicular height 72 The proportion of the three sides of a triangle being given, to find the sides of a triangle corresponding with a given area 73 The base and perpendicular of any triangle being given, to find the side of an inscribed square 74 The three sides of any triangle being given, to find the length of the perpendicular which will divide it into two right- - angled triangles fK 1* 5 D CONTENTS. Page The area and base of a triangle being given, to cut off a given part of the area by a line running from the angle opposite the base 76 The area and base of a triangle being given, to cut off a tri- angle containing a given area by a line running parallel to one of its sides 77 The area and two sides of a triangle being given, to cut off a triangle containing a given area, by a line running from a given point in one of the given sides, and falling on the other 78 To find the area of a trapezium 79 To find the area of a trapezoid 80 To find the area of a regular polygon 81 To find the area of a regular polygon when one of its equal sides only is given. . . . 82 When the area of any regular polygon is given to find the side 83 To find the area of an irregular right-lined figure of any num- ber of sides 83 To find the area of a mixtilineal figure, or one formed by right lines and curves 85 To find the circumference of a circle when the diameter is given, or the diameter when the circumference is given. . . 87 To find the area of a circle 88 The area of a circle being given, to find the diameter or cir- cumference 89 To find the area of a circular ring 89 The diameter or circumference of a circle being given, to find the side of an equivalent square 91 The diameter or circumference of a circle being given, to find the side of the inscribed square 9i To find the diameter of a circle equal in area to any given superficies 92 The diameter of a circle being given, to find another contain- ing a proportionate quantity 93 To find the length of any arc of a circle 93 To find the area of a sector of a circle 97 To find the area of a segment of a circle 100 To find the area of a circular zone 105 To find the area of a lune 107 To find the area of a part of a ring, or of the segment of a sector 109 CONIC SECTIONS. Definitions 110 To describe an ellipse 113 The ellipse and its parts 114 To describe a parabola 120 The parabola and its parts 120 To construct the hyperbola 124 The hyperbola and its parts 124 CONTENTS. 7 MENSURATION OF SOLIDS. Page Definitions I'M To find the area of the surface of a cube. ... 133 The area of the surface of a cube being given, to find the side 133 To find the solidity of a cube 134 To find the side of a cube the solidity being given 135 To find the solidity of a parallelopipedon 135 To find the solidity of a prism 136 To find the convex surface of a cylinder 137 To find the solidity of a cylinder 138 To find the curve surface of a cylindric ungula 139 To find the solidity of a cylindric ungula 139 To find the convex surface of a cylindric ring 140 To find the solidity of a cylindric ring ]41 The solidity and thickness of a cylindric ring being given, to find the inner diameter 142 To find the surface of a right cone or pyramid 142 To find the surface of the frustum of a right cone or pyramid 144 To find the solidity of a cone or pyramid 145 To find the solidity of the frustum of a cone or pyramid 146 The solidity and altitude of a cone being given, to find the diameter of the base 148 The solidity and diameter of the base of a cone being given, to find the altitude 149 The altitude of a cone or pyramid being given, to divide it into two or more equal parts, by sections parallel to the base, to find the perpendicular height of each part 149 To find the solidity of an ungula, when the section passes through the opposite extremities of the ends of the frustum 150 To find the solidity of a cuneus or wedge 151 To find the solidity of a prismoid 152 To find the convex surface of a sphere 153 To find the solidity of a sphere or globe 153 The convex surface of a globe being given, to find its diameter 154 The solidity of a globe being given to find the diameter 155 To find the solidity of the segment of a sphere 155 To find the solidity of a frustum or zone of a sphere 156 To find the solidity of a circular spindle, its length and middle diameter being given 157 To find the solidity of the frustum of a circular spindle, its length, the middle diameter, and that of either of the ends. being given 158 To find the solidity of a spheroid, its two axes being given . . 159 To find the solidity of the middle frustum of a spheroid, its length, the middle diameter, and that of either of the ends being given 1 CO To find the solidity of the segment of a spheroid 162 To find the solidity of a paraboloid '63 8 CONTENTS. Page To find the solidity of the frustum of a parabolo d, when its ends are perpendicular to the axis of the solid 164 To find the solidity of a parabolic spindle 165 To find the solidity of the middle frustum of a parabolic spindle 165 To find the solidity of a hyperboloid 166 To find the solidity of the frustum of a hyperboloid 167 REGULAR BODIES. To find the solidity of a tetraedron 16 To find the side of a tetraedron 1 70 To find the solidity of an octaedron 170 To find the side of an octaedron 171 To find the solidity of a dodecaedron 171 To find the side of a dodecaedron 172 To find the solidity of an icosaedron 173 To find the side of an icosaedron 173 To find the surface of any regular body 174 To find the solidity of any regular body ; . 1 75 BALLS AND SHELLS. Weights and dimensions of balls and shells 176 Piling of balls and shells 182 CARPENTERS' RULE 185 The use of the Sliding Rule 186 TIMBER MEASURE. To find the content of a board or piece of lumber one inch in thickness 188 To find the content of any piece of lumber, whose thickness is more than an inch 189 HEWN TIMBER. To find the solid content of squared or four-sided timber 190 The breadth of a board being given, to find how much in length will make a square foot, or any other required quantity 192 When the board is wider at one end than at the other, and the quantity is taken from the less end 192 To find how much in length will make a solid foot or any other required quantity 193 ROUND TIMBER. To find the solidity of round or unsquared timber 194 To find the number of cord feet in a log 196 To find the number of solid or square feet that a piece of round timber will hew to when squared 197 CONTENTS. SAW-LOGS. Page To find the Dumber o f feet, board measure, that a log contains 197 MASONS' WORK 198 BRICKLAYERS' WORK 200 CARPENTERS' AND JOINERS' WORK 202 SLATERS' AND TILERS' WORK 203 PLASTERERS' WORK 204 PAINTERS' WORK 206 GLAZIERS' WORK 207 PLUMBERS' WORK 208 PAVERS' WORK 209 VAULTED AND ARCHED ROOFS. To find the solid content of a circular, elliptical, or Gothic vaulted roof 211 To find the concave or convex surface of a circular, elliptical, or Gothic vaulted roof 213 To find the solidity of a dome 214 To find the superficial content of a dome 214 To find the solid content of a saloon 215 To find the superficies of a saloon 216 GAUGING. The gauging rule 21V The use of the gauging rule 219 The gauging or diagonal rod 221 To find the content of a cask of the first variety 222 To find the content of a cask of the second variety 223 To find the content of a cask of the third variety 224 To find the content of a cask of the fourth variety 225 To find the content of a cask by four dimensions 226 To find the content of any cask from three dimensions 227 To ullage a standing cask 228 To ullage a lying cask 228 To find the ullage by the gauging rule 229 To find the content of a cask by the mean diameter 229 TONNAGE. Carpenters' tonnage 2.'U Government tonnage 232 10 CONTENTS. CENTRE OF GRAVITY 236 FALLING BODIES 240 GRAVITIES OF BODIES 245 SPECIFIC GRAVITY 246 TABLE OF SPECIFIC GRAVITIES 249 THE PENDULUM 253 In any latitude to find the length of the pendulum which beats seconds 256 CENTRES OF PERCUSSION AND OSCILLATION 257 CENTRE OF GYRATION 259 PRESSURE OF THE AIR 260 DISTANCES, AS DETERMINED BY THE VELOCITY OF SOUND 261 MECHANICAL POWERS. The Lever 263 Wheel and Axle 265 The Pulley 267 The Inclined Plane 270 The Wedge 271 The Screw 272 WHEELS 274 HYDRAULICS 277 PUMPS. 279 WATER-WHEELS 280 MISCELLANEOUS QUESTIONS 284 TABLE CfF AREAS OF CIRCULAR SEGMENTS . . .295 TABLES. s. D 12. An acute angle is that which is less than a right angle ; as A B C. 13. An obtuse angle is that which is greater than a right angle, as ABC. 14. A figure is a space boxinded by one or more lines. 15. All plane figures bounded by three right lines are called triangles, and receive different denominations according to the nature of their sides and angles. 16. An equilateral triangle is that which has all its sides equal ; as A B C. PRACTICAL GEOMETRY. 17 17. An isosceles triangle is that which has only two of its sides equal ; as ABC. 18. A scalene triangle is that which has all its three sides unequal ; as ABC. 10. A right-angled triangle is that which has one right angle ; the side opposite to the right angle is called the hypothenuse, and the other- two sides the legs ; as A B C, where A B is the hypothenuse, and B C, A C the two legs, or base and perpendi- cular. Any triangle differing from a right-angled one is frequently called an oblique-angled triangle. 20. An obtuse-angled triangle is that which has one obtuse angle ; as A C B, where C is the obtuse angle. 21. An acute-angled triangle is that which has all its angles acute ; as A B C. 22. All plane figures bounded by four right lines, are ailed quadrangles, or quadrilaterals; and receive different ;ames according to the nature of their sides and angles. 23. A square is a quadrilateral, whose *ides are all equal, and its angles all right angles ; as A B C D. 2* 18 PRACTICAL GEOMETRY. A square is also an instrument used by artificers for what is called squaring their work ; being of various forms, as the T square, normal square, &c. A.I 24. A rhombus is a quadrilateral whose sides are all equal, but its angles not right angles ; as A B C D. Bl This figure, by mechanics, is generally called a lozenge; and both it and the square belong to the ckss of parallelo- grams. 25. A parallelogram is a quadri- lateral whose opposite sides are parallel ; as A B C D. 1 ,/ -In 26. A rectangle is a parallelogram A whose angles are all right angles ; as A B C D. 27. A rhomboid is a parallelogram whose angles are not right angles ; as A B C D. B 28. A trapezium is a quadrilateral which hath not its opposite sides parallel ; as A B C D. B 29. A trapezoid is a quadrilateral, having two of its opposite sides parallel ; as A B CD. 30. The right line joining any two opposite angles of a quadrangle, or quad- rilateral, is called its diagonal ; as B D in the figure A B C D. B c A/ 7D PRACTICAL GEOMETRY. 19 31. All plane figures contained under more -than four sides are called polygons ; and receive different names, ac- cording to the number of their sides, or angles. 32. Thus, polygons having five sides are called pentagons; those of six sides, hexagons, those of seven, heptagons ; and so on. 33. A regular polygon is that which has all its sides as well as its angles equal to each other, and if the sides or angles are unequal, it is called an irregular polygon. An equilateral triangle is also a regular polygon of three sides, and a square is one of four sides. 34. Parallel right lines are such as are everywhere at an equal distance from each A u other ; or which, if infinitely produced, would never meet to CD. thus A B is parallel 35. The base of any figure is that side on which it is sup- posed to stand ; and its altitude is the perpendicular falling upon the base from the opposite angle. 36. An angle is usually denoted by three letters, the one which stands at the angular point being always to be read in the middle, as A B C, C A_ B D, D B E, &c. 37. A circle is a plane figure bounded by a curve line called the circumference or periphery, which is everywhere equidis- tant from a point within, called its centre ; c and is formed by the revolution of a right line (O A) about one of its extremities (O), which remains fixed. 38. The centre of a circle is the point (O) about which it is described ; and the circumference or periphery is the line or boundary A B C A, by which it is contained. The circumference itself, as well as the space which is bounded by it, is also, for the sake of conciseness, sometimes called a circle. 20 PRACTICAL GEOMETRY. 39. The radius of a circle is a right line drawn from the centre to the circumference ; asO A. A tangent is a line touching a circle, and which produced, does not cut it, as B A C. 40. The diameter of a circle is a right line passing through the centre, and A terminated both ways by the circumfe- rence ; as A B. 41. An arc of a circle is any part of its circumference, or periphery ; as A B. 42. A chord is a right line which joins the extremities of an arc ; as A B. 43. A segment of a circle is the space contained between an arc and its chord ; as A B C, 44. A sector is the space contained be- tween an arc and the two radii drawn to its extremities ; as A B C. PRACTICAL GEOMETRY. 21 45. A zone is a part of a circle included A between two parallel chords and their inter- cepted arcs ; as A B C D. 46. The versed sine or height of an arc, is that part of .the diameter contained be- A tween the middle of the chord and the arc ; as D B, or D E. 47. A lune is the space included between the intersecting arcs of two eccentric circles ; as A B C. 48. A semicircle is a half of a circle ; a quadrant is a quarter of a circle ; a sextant the sixth part of it, and an octant the eighth part, where it may be observed that these names are often applied to instruments used for taking angles. 49. The circumference of every circle is supposed to be divided into 360 equal parts called degrees ; each degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds. 50. The measure of any right-lined angle is an arc of a circle contained be- / tween the two lines which form that ' angle, the angular point being the ' x centre ; thus the angle A O B is mea- \ / sured by the arc m n. ** ..*'' The angle is estimated by the number of degrees, minutes, &c. contained in the arc ; whence a right angle is an angle of 90 degrees or | of the circumference. AXIOMS. 1. An axiom is an established principle, or self-evident truth, requiring no other conviction than that which arises from a proper understanding of the terms in which if is pro- posed. 22 PRACTICAL GEOMETRY. 2. Things which are equal to the same thing are equal to other. 3. If equals be added to equals the wholes will be equal. 4. If equals be taken from equals the remainders will be equal. 5. If equals be added to unequals the wholes will be un- equal. 6. If equals be taken from unequals the remainders will be unequal. 7. Things which are half, double, or any number of times the same thing, are equal. 8. The whole is greater than its part 9. Every whole is equal to all its parts taken together. 10. All right angles are equal to each other. 11. Angles that have equal measures, or arcs, are equal. 12. Things which coincide, or fill the same space, are identical, or mutually equal in all their parts. 13. Two straight lines cannot enclose a space. REMARKS. 1 A problem is something proposed to be done. 2. The perimeter of a figure is the sum of all its sides taken together. 3. The sum of any two sides of a triangle is greater than the third side. 4. In any triangle the sum of the three angles is equal to two right angles. f>. Every triangle is half the parallelogram which has the same base and the same altitude. 6. An angle inscribed in a semicircle is a right angle. 7. All angles in the same segment of a circle are equal to each other. PRACTICAL GEOMWPRY. 23 8. Triangles that have all the three angles of the one respectively equal to all the three angles of the other, arc called equiangular triangles, or similar triangles. 9. In similar triangles the like sides, or sides opposite to the equal angles, are proportional. 10. The areas or spaces of similar triangles are to each other as the squares of their like sides. 11. The areas of circles are to each other as the squares of their diameters, radii, or circumference. 12. Similar figures are such as have the same number of sides, and the angles contained by their sides respectively equal. 13. The areas of similar figures are to each as the squares of their like sides. 14. If three quantities are proportional, the middle one is re- peated, and the first is to the second as the second is to the third. In such a case the middle quantity is a mean proportional between the other two, and the last is a third proportional to the first and second; but, if there are four proportional quan- tities, the last is called a fourth proportional to the other three INSTRUMENTS. 2. The principal instruments used in describing or con- structing geometrical figures are as follows : THE DIVIDERS OR COMPASSES. The plain compasses consist of two inflexible rods of brass, revolving upon an axis at the vertex, and furnished with steel points. GEOMETRY. THE PARALLEL RULER. The parallel ruler consists of two flat pieces of ebony, con- nected together with brass bars, having their extremities equidistant, by which contrivance, when the ruler is opened, the sides necessarily move in parallel lines. THE SCALE OF EQUAL PARTS. The scale of equal parts consists of a certain number of equal portions of any convenient length, the extreme one or the left hand being subdivided into ten equal parts, and is called the unit of the scale, and the rest being numbered 1, 2, 3, &c. In most scales an inch is taken for a common measure, and what an inch is divided into is generally set at the end of the scale. This scale is used in laying down any distance, as inches, feet, chains, miles, &c. The several divisions may be con- sidered as feet, for example, the decimal subdivision would be tenths of a foot. So also each of the principal divisions may be regarded as ten inches, ten feet, &c., and in this case the decimal subdivision will represent inches, feet, &c., respectively. This scale is limited to two figures, or any number less than 100 may be readily taken ; but if the number should consist of three places of figures, the value oi the third figure cannot be exactly ascertained, and in this case it is better to use a diagonal scale, by which any num- ber consisting of three places of figures may be exactly found. Let it be required to take from the scale a line equal to fire inches and eight-tenths. PRACTICAL GEOMETRY. Place one foot of the dividers at 5 on the right, and ex- tend the other to. 8, which makes the eighth of the small divisions. The dividers will then embrace the required distance. THE DIAGONAL SCALE OF EQUAL PARTS. The construction of this scale is as follows : Having prepared a ruler of convenient breadth for your scale, draw near the edges thereof two right lines, a e, c f, parallel to each other ; divide one of these lines, as a e, into equal parts, according to the size of your scale, and through each of these divisions draw right lines perpendicular to a e, to meet cf; then divide the breadth into ten equal parts, and through each of these divisions draw right lines paralle. to a e and cf', divide the lines a b, c d, into ten equal parts, and from- the point a, to the first division in the line c d, draw a diagonal line ; then parallel to that line, draw diago- nal lines through all the other divisions, and the scale is com- plete. Then, if any number consisting of three places of figures, as 468, fy required from the larger sca\e,fd, you must place one foot of the compasses on the figure 4, on the line/rf, then the extent from 4 to the point d will represent 400. The second figure being 6, count six of the smaller divisions from d towards c, and the extent from 4 to that point will be 460. Mo T 'e both points of the compasses down- wards till they are on the eighth parallel line below/ d, and open them a little till the one point rests on the vertical line drawn through 4, and the other on the diagonal line drawn through 6 ; the extent, then, in the compasses will represent 468. " In the same manner the quantities 46.8, 4.68, 0.468. &c., are measured. There are generally two diagonal scales laid down on the same face of the instrument, the unit of the one being double that of the other, and commencing on opposite ends of the scale. PRACTICAL GEOMETRY. THE SCALE OF CHORDS. uTlC Draw the lines B A, B C, at right angles to each other, and with any convenient distance, B A, describe the arc A s C ; divide it into 90 equal parts, and join A C. From A as a centre, with the distances A 10, A 20, &c., describe the arcs 10, 10, 20, 20, &c., meeting the line A C. Fill up the separate degrees, which are not marked in the diagram to prevent confusion, and the scale is complete. It is evi- dent, by inspection, that the chord of 60 is equal to the radius, as shown by the letter r upon the rule ; which distance is therefore always to be taken in laying down angles. THE SEMICIRCULAR PROTRACTOR. The protractor is a semicircular piece of brass divided into 180 degrees, and numbered each way from end to end ; that is, from A to B, and from B to A. There is a small notch PRACTICAL GEOMETRY. 27 in the middle of the diameter A B, denoting the centre of the protractor. In some boxes of mathematical instruments this is omitted, and the degrees are transferred to the border of the plain scale. GUNTER S SCALE. Gunter's scale, commonly of two feet in length, contains on one side the lines of the plain scale, already described, and on the other corresponding logarithmic lines. PROBLEM I. 3. To describe from a given centre the circumference of a circle having a given radius. Let A be the given centre, and A B the given radius. Place one foot of the dividers at A, and extend the other leg until it shall reach B[ to B. Then turn the dividers around the leg at A, and the other leg will describe the required circumference. PROBLEM II. Through a given point A, to draw a line parallel to a given line, D E. Lay the edge of the parallel ruler B A c upon D E, and move it upwards till it reaches the point A, through which draw B C , and it will be parallel to D E. D ~ 28 PRACTICAL GEOMETRY. PROBLEM III. To lay off on a given line, as AB, a distance equal to C D. EXAMPLE. Let C D be the distance to be laid off, c D and A B the given line. A f T, Place one foot of the dividers at C, and extend the other leg until the foot reaches D. Then, raising the dividers, place one foot at A, and mark with the other the distance AE, this will evidently be equal to C D. PROBLEM IV. To lay down a line of given length, on a scale of a given number to the inch, to determine how many parts of it are to be represented on the paper by a distance equal to the unit of the scale. EXAMPLE. If a line 320 feet in length is to be laid down on paper, on a scale of 40 feet to the inch, what length must be taken from the scale ? Divide the length of the line by the number of parts which is represented by the unit of the scale ; the quotient wiJ. give the number of parts which is to be taken from the scale. Here 320-i-40=8 the number of parts to be taken from the scale. PROBLEM V. The length of the line being given on the paper, to deter mine the true length of the line which it represents. EXAMPLE. The length of the line on the paper is 4.75 inches, aicJ the scale is one of 20 feet to the inch ; what is the true lenyl of the line ? PRACTICAL GEOMETRY. 29 Take the line in your dividers and apply it to the scale, and note the number of units and parts of a unit to which it is equal ; then multiply this number by the number of parts which the unit of the scale represents, and the product will be the length of the line. Here 4.75x20=95 feet, the length of the line. PROBLEM VI. To make an angle of any proposed number of degrees. 1. Draw any line A B, and having taken first 60 degrees from the scale of chords, describe with this radius the arc n m. 2. Take in like manner the chord of the proposed num- ber of degrees from the same scale, and apply it from n to m, 3. Then if the line A C be drawn from the point A through m, the angle BAG will be that required. Angles greater than 90 degrees are usually made by first laying ofF90 degrees upon the arc n m, and then the remain- ing part. This problem may be performed by the protractor. Place the central notch of the instrument upon A, and the edge along A B ; make a point m against the proposed num- ber of degrees, and through it draw the line A C. 3* 30 PRACTICAL GEOMETRY. PROBLEM VH. Any angle BAG being given, to find the number of de- grees it contains. 1. From the angular point A, with the chord of 60 degrees, describe the arc n m, cutting the lines A B, A C in the points n and m. 2. Then take the distance n m, and apply it to the scale of chords, and it will show the degrees required. And if the distance n m be greater than 90, it must be taken at twice, and each part applied separately to the scale. This problem may be performed by the protractor. Place the central notch of the instrument upon A, and the edge along A C, and observe the number of degrees cut by the line A B, which will show the degrees required. PROBLEM VIII. To divide a given line AB into two equal parts. 1. From the points A and B, as centres, with any distance greater than half A B, describe arcs cutting each other in n and m. 2. Through these points, draw tne line n c ra, and the point c, where it cuts A B, will be the middle of the 'ine required. PRACTICAL GEOMETRY. 3J PROBLEM IX. To divide a given angle ABC into two equal parts. 1. From the point B, with any radius, describe the arc A C ; and from A and C, with the same, or any other radius, describe arcs cutting each other in n. 2. Then, through the point n draw the line B n, and it will bisect the angle ABC, as was required. PROBLEM X. From a given point C, in a given right line A B, to erect a perpendicular. CASE I. When the point is near the middle of the line. 1. On each side of the point C take any two equal distances Cn, Cm. 2. From n and m, with any ra- dius greater than nC or m C, de- scribe arcs cutting each other in s. 3. Then through the point s, A draw the line sC, and it will be n " c m the perpendicular required. CASE II. When the point is at, or near, the end of the line. Supposing C to be the given point, as before. 1. Take any point 0, and with the radius or distance, o C, describe the arc mCn, cutting A B in m and C. 2. Through the centre o, and the point m, draw the line m o n, cutting the arc mCn in n. 3. Then from the point n, draw the line nC, and it will be the per- A i pendicular required. 32 PRACTICAL GEOMETRY. The same otherwise. 1. Set one leg of the compasses on B, and with any extent B m describe an arc mp ; then set off the same extent from m to n. 2. Then join m n, and from n as a centre with the extent m n as radius, describe an arc q ; pro- / duce m n to q, and the line join- / ing q B will be perpendicular to /' AD A - m Another method. 1. From any scale of equal parts take a distance equal to 3 divisions, and set it from B to m. 2. And from the points B and m, with the distances 4 and 5, taken from the same scale, describe arcs cutting each other in n. 3. Through the points n, B, draw the line BC, and it will be the A perpendicular required. The same thing may also be readily done, by an instru- ment in the form of a square, or by the plain scale. PROBLEM XL From a given point C, without a given line A B, to let fall a perpendicular. CASE I. When the point is nearly opposite to the middle of the line. 1. From the point C, with any radius, describe the arc n m, cut- ting A B in n and m. 2. From the points n, m, with the same or any other radius, de- A. scribe two arcs cutting each other in 8. 3. Through the points C, draw the line C G s, and C G will \|X be the peroendicular required ** ' ** PRACTICAL GEOMETRY. A. This problem, also, may be performed like the last by means of a square. CASE II. When the point is opposite or nearly opposite to the end of the line. 1. Take any point m, in the line A B, and from C draw the Xc Jine C m. 2. Bisect the line C m, or di- vide it into two equal parts in the point n. 3. From n, with the radius nm, or nC, describe the arc CDm, A- cutting A B in D. N ^ *'' l 4. Then through the point C, draw the line C D, and it will be the perpendicular required. This method may also be used in the first case, if the line A B, when necessary, be produced. The same otherwise. 1. From A, or any other point in ^\ ^ A B, with the radius A C, describe the arcs C, D. 2. And from any other point n, in AB, with the radius nC, describe another arc cutting the former in C, D. 3. Then through the points C, D, A draw the line C E D, and C E will be the perpendicular required. Perpendiculars may be more easily raised, and let fall, in practice, by means of a square, or other instrument proper for this purpose. PROBLEM XII. To draw a perpendicular, from any angle of a triangle A B C, to its opposite side. 1. Bisect either of the sides con- taining the angle from which the per- pendicular is to be drawn, as B C in the point n 34 PRACTICAL GEOMETRY. 2. Then with the radius nC, and from the centre n describe an arc cutting AB, (or AB produced if necessary, as in the second figure,) in the point D ; the line joining C D will be perpendicular to A B or A B produced. PROBLEM Xin. To trisect, or divide a right angle ABC, into three equal parts. 1. From the point B, with any radius B A, describe the arc A C, cut- ting the legs B A, BC, in A, C. 2. From the point A, with the radius A B, or B C, cross the arc A C in n ; and with the same radius, from the point C, cross it in m. 3. Then through the points m, n, draw the lines Bra, Bn, and they will trisect the angle as was required. By this means the circumference of any given circle may be divided into 12 equal parts ; and thence by bisection into 24, 48, &c. PROBLEM XIV. At a given point D, to make an angle equal to a given angle ABC. B m AD T E 1. From the point B, with any radius, describe the arc n , cutting the lines B A, B C, in the points m, n. PRACTICAL GEOME1RY. 35 2. Draw the line D E, and from the point D, with the same radius as before, describe the arc r s. 3. Take the distance m n, on the former arc, and apply it to the arc r s, from r to s. 4. Then through the points D, s t draw the line D F, and the angle E D F will be equal to the angle ABC, as was required. PROBLEM XV. To draw a line parallel to a given line A B. I. When the parallel line is to pass through a given point C. 1. Take any point m in D _ ^ _ f E the line A B, and from the > ^^^^ \ point C draw the line C m. ' ^^ /' 2. From the point m, ; ^ x / with the radius m C, de- ^ ^ m i scribe the arc C n, cutting A B in n ; and with the same radius, from the point C, de- scribe the arc m r. 3. Take the distance C n, and apply it to the arc m r, from m to r; then through the points C, r, draw the line D C r E, and it will be parallel to A B, as was required. CASE II. When the parallel line is to be at a given dis tance from A B. 1. From any two points r, s, D n c in the line A B, with a radius equal to the given distance, describe the arcs, n, m. 2. Then draw the line C D to touch these arcs without A r cutting them, and it will be parallel to A B, as was required 36 PRACTICAL GEOMETRY. s^.--**"" \ PROBLEM XVI. To divide a given line A B into any proposed number of equal parts. 1. From one end of the line A, . ** draw A m, making any angle with A B ; and from the other end B, draw B n, making an equal angle A<( A En. \ 2. In each of the lines A m, B n, \ beginning at A and B, set off as many equal parts, of any conve- nient length, as A B is to be divided into. , 3. Then join the points A 5, 1, 4, 2, 3, &c., and A B will be divided as was required. B n may be drawn parallel to A m, by means of the parallel ruler, or the same may be done by taking the aic A n equal to B m. Another method. 1. Through the point B draw the indefinite line C E, making an angle with A B. 2. Take any point E in that line, through which draw E D parallel to B A, and set off as many equal parts E H, H G, &c., from E toward D, as , from D to C, &c., and join B C, CD, D E, E F, &c., and A B C D E F G will be the polygon required. PROBLEM XXXVIII. About a given triangle A B C, to circumscribe a circle. 1. Bisect the two sides A B, BC, with the perpendiculars m O, and n O. 2. From the point of intersection O, with the distance OA, OB, or OC, de- scribe the circle A C B, and it will be that required. If any two of the angles be bisected, instead of the sides, the intersection of these lines will give, as before shown, the centre of the inscribed circle. 48 PRACTICAL GEOMETRY. PROBLEM XXXIX. About a given square, A B C D, to circumscribe a 1. Draw the two diagonals AC and B D intersecting each other in O. 2. Then from the point O, with the distance O A, O B, O C or O D, describe the circle A B C D, and it will be that required. PROBLEM XL. To circumscribe a square about a given circle. 1. Draw any two diameters n o and * r m at right angles to each other. 2. Then through the points m, o, r, , draw the lines AB, BC, CD, and D A, perpendicular to r m, and n o, and A. B C D will be the square required. PROBLEM XLI. About a given circle to circumscribe a pentagon 1 . Inscribe a pentagon in the circle ; or which is the same thing, find the points m,n, v, r,s, as in Prob. XXXIV. 2. From the centre o, to each of these points, draw the radii o n, o m, o u, o r, and o s. 3. Through the points n, m, draw the lines A B, B C, perpendicular to on, o m ; producing them till they meet each other at B. Draw in like manner the lines C D, D E, E A, perpen- dicular to o v, o r, and o s, and A B C D E will be the pentagon required. PRACTICAL GEOMETRY. 49 Any other polygon may be made to circumscribe a circle, by first inscribing a similar one, and then drawing tangents to the circle at the angular points. PROBLEM XLH. On a given line, A B, to make a regular pentagon. 1. Make Bm perpendicular to AB, and equal to one half of it. 2. Draw A m, and produce it till the part m n is equal to B m. 3. From A and B as centres, with the radius B n, describe arcs cutting each other in o. 4. Then from the point o, with the same radius, or with o A, or o B, de- scribe the circle, ABODE; and the line A B, applied five times round the circumference of this circle, will form the pentagon required. If tangents be drawn through the angular points A, B, C, D, E, a pentagon circumscribing the circle will be formed ; and if the arcs be bisected, a circumscribing decagon may be formed ; and so on. Vc Another method. 1. Produce A B towards n, and at the point B make the perpendi- cular B m, equal to A B. 2. Bisect A B in r, and from r as a centre, with the radius r m, de- scribe the arc m n, cutting the pro- duced line A B in n. 3. From the points A and B, with the radius A n, describe arcs cutting each other in D ; and from the points A, D, and B, with the radius A B, de- scribe arcs cutting each other in C and E. 4. Then if the lines B C, C D, D E, and E A, be drawn, A B C D E will be the pentagon required. 50 PRACTICAL GEOMETRY. PROBLEM XLHI. On a given line, A B, to make a regular hexagon. 1. From the points A, and B, as cen- tres, with the radius A B, describe arcs cutting each other in ; and from the point O, with the distance O A, or O B, describe the circle A B C D E F. 2. Then if the line A B be applied six times round the circumference, it will form the hexagon required. PROBLEM XLIV. On a given line, A B, to form a regular octagon. 1. On the extremities of the given line, A B, erect the indefinite perpen- diculars A F, and B E. 2. Produce A B both ways to m and n, and bisect the angles m A F and n B E with the lines A H and B C. 3. Make A H and B C each equal to A B, and draw H G,C D, parallel to A F, m ' or B E, and also each equal to A B. 4. From G, and D, as centres, with a radius equal to A B, describe arcs crossing A F, B E, in F and E ; then if G F, FE, and E D be drawn, ABCDEFGH will be the octa- gon required. PROBLEM XLV. To make a figure similar to a given figure, A B 1 . Take A b, equal to one of the sides of the figure required, and from the angle A draw the diagonals A c, A d. e 2. From the points b, c, d, draw b c, c d, E d e, parallel to B C, C D, D E, and f A b c d e will be similar to A BC D E. The same thing may also be done by A' making the angles b, c, d, e, respectively equal to the angles B, C, D, E. PRACTICAL GEOMETRY. PROBLEM XL VI. 51 To make a triangle equal to a given trapezium, A B C D. 1. Draw the diagonal D B, and make C E parallel to it, meeting the side A B produced in E. 2. Join the points D, E, and A D E will be the triangle re- quired. PROBLEM XLVII. To make a triangle equal to any right-lined figure, ABCDEFGA. 1. Produce the side A B both ways at pleasure. 2. Draw the diagonals E A, E B; and by the last problem, make the triangles A E I, B E K, equal to the figures A E F G, and BE DC. 3. Draw I L, K M, parallel to E A, E B. 4. Then if the points E, L, and E, M, be joined, ELM will be the triangle required. And in the same manner may any right-lined figure what- ever be reduced to a triangle. PROBLEM XL VIII. To describe a square that shall be any multiple of a given square, A B C D. 1. Draw the diagonal B D, and on L A B, A D, produced, take A G, A E, fT"~ each equal to B.D, then A F, the square G on A E, will be double the square AC. D 2. Draw in like manner the diagonal B G, and make A L, A H, each equal to B G ; then A K, the square on A H, will be triple the square A C. 52 PRACTICAL GEOMETRY. 3. Proceed in the same manner with the diagonal B L ; and a square will be formed which is quadruple the 'square A C ; and so on. PROBLEM XLIX. To inscribe a square in a given triangle, ABC. 1 . Through the ver- tex A, parallel to B C, draw the straight line A E, and from C raise a perpendicular to meet it in D. B 2. Make D E equal to D C ; join E B, cutting D C in F. 3. Through F draw F G, parallel to B C, and through H and G, draw H L, G K, parallel to D C. Then L H G K will be the square required. PROBLEM L. In a given circle to inscribe a polygon of any proposed number of sides. 1. Divide 360 by the numberof sides of the figure and make an angle A O B, at the centre, whose measure shall be equal to the degrees in the quotient. 2. Then join the points A, B, and apply the chord A B to the circumfe- rence the given number of times, and it will form the polygon required. PROBLEM LI. On a given line A B to form a regular polygon of any proposed number of sides. 1. Divide 360 by the number of sides of the figure, and subtract the quotient from 180 degrees. PRACTICAL GEOMETRY. 53 2. Make the angles ABO and BAG each equal to half the difference last found ; and from the point of intersec- tion O, with the distance O A or OB, describe a circle. 3. Then apply the chord A B to the circumference the proposed number of times, and it will form the polygon required. PROBLEM LIT. To describe a circle within or without a regular polygon. 1. Bisect any two angles, A ED, EDO, by the lines EG, DG, and from G let fall GF, perpendicular to the side E D. 2. Then with the radius GE de- scribe the outer circle, and with the radius G F describe the inner circle. PROBLEM LIII. To make a square that shall be nearly equal to a given circle. 1. In the given circle A C B D, draw the two diameters A B, CD, cutting each other perpendicularly in the centre O. 2. Bisect the radius O C, in E, and A /hrough the points A, E, draw the chord A E F, which will be the side of a square that is nearly equal in area to ihe circle. If F G be drawn parallel to C D, it will be nearly equal to $ of the circumference of the circle. 5* 64 PRACTICAL GEOMETRY. PROBLEM LIV. To find a right line that shall be nearly equal to any given arc A D B of a circle. 1. Divide the chord AB into four equal parts, and set one of the parts A C, on the arc from B to D. 2. Draw C D, and the double of this line will be nearly equal to the arc A D B. If a right line be made equal to 3| times the diameter of a circle, it will be nearly equal to the circumference. PROBLEM LV. To divide a given circle into any proposed number of parts, that shall be mutually equal to each other, both in area and perimeter. 1. Divide the diameter A B into the proposed number of equal parts, at the points a, b, c, d. 2. On Aa, Aft, Ac, Ad, &c., as diameters, describe semicircles on one side of the diameter A B ; and on Br/, Be, B6, Ba, &c., describe semicircles on the other side of the diameter. 3. Then the corresponding semicircles, joining each other as above, will divide the circle in the manner proposed. PROBLEM LVI. In a given circle, to describe three equal circles which shall touch one another, and also the circumference of the given circle. From the centre O, let the right- lines O H, O D. and O E be drawn, dividing the circumference into three equal parts in the points H, D, and E ; join D E, and in O E produced take E F. equal to on** PRACTICAL GF.OMETHY. 55 half of DE; draw DF, and parallel thereto draw E A meeting O D in A ; make H C and E 1? each equal D A, and upon the centres A, B and C, through the points D, H, and E, let the circles A r D, C m H and B n E be described. PROBLEM LVII. To divide a given circle into any number of equal parts by means of concentric circles. Let it be required to divide the circle A K L into three equal parts. Divide the radius A B into three equal parts ; and from the points of section c, d, draw the perpendiculars a b, d c, meeting the circumference of A| a semicircle described on A B, in b and c ; and join B c, B b. Then, if circles be described from B, as a cen- tre, with the radii B c, B b, the circle L A K L will be divided into three equal parts, as required. PROBLEM LVIII. To describe the circumference of a circle through two given points, A, B, which shall touch a right line, C D, whose position is given. 1. Draw the right line A B, joining the two given points, which bisect in E, by the perpen- dicular E F, meeting the given line C D in F. 2. Join B F, and from any point, H, in F E, let fall the per- pendicular H G ; and having made H I equal to H G, draw B K, parallel to I H, meeting F E in K. 3. Then, if a circle be described from the point K, as a centre, with the radius K B, it will pass through the points A and B. and touch the line C D, as was required. 56 PRACTICAL GEOMETRY. PROBLEM LIX. Upon a given line, A B, to describe an oval, or a figure resembling an ellipse. 1. Divide A B into three equal parts, A C, C D, D B ; and from the points C, D, with the radii C A, D B, describe A che circles A G D E, and C H B F. 2. Through the intersections m, n, and centres C, D, draw the lines m H, n E ; and from the points n, m, with the radii n E, m H, de- scribe the arcs E F, H G, and A G H B F E will be the oval required. Another method. c 1. On A B, as a common base, describe the two equal isosceles triangles A C B, A D B, produc- ing their sides C A, C B, D A, and DB. 2. From D and C, as centres, with any convenient radius, de- scribe the arcs E G, F H ; and from A and B, with the radius A E, B G, describe the arcs G H, E F, and the figure E F H G will be the oval required. PROBLEM LX. To divide a given straight line A B in extreme and mean ratio ; that is so that the whole line shall be to the greater part as the greater part is to the less. 1. From A, draw AC at right angles to A B, and make A C = A B ; produce B A to D, making A D = half A B ; and about D, at the distance D C, describe the arc C E. Then A B is divided in extreme and mean ratio at the point E; and AB AE::AE:EB. MENSURATION OF SUPERFICIES. 4- THE area of any figure is the measure of its surface, or of the space contained within the bounds of that surface, without any regard to the thick- ness. A square whose side is 1 inch, 1 foot, or 1 yard, &c., is called the measuring unit ; and the area, or content of any figure, is estimated by the number of squares of this kind that are contained in it, A as in the rectangle, ABCD. THE SQUARE. PROBLEM I. 5- To find the area of a square. RULE. Multiply the side by itself, and the product will be the area. EXAMPLES. 1. What is the area of the square ABCD, whose side is 21 inches ? Here, 21 x 21 = 441 inches : the area required. 2. What is the area of a square whose side is 4 ft. 2 in. ? Jlns. 17 ft. 4 in. 4". 67 58 MENSURATION OF SUPERFICIES. 3. What is the area of a square field whose side is 50 perches ? Ans. 15 a. 2 r. 20 p. 4- What te the area of a square meadow whose side is 35.25 chains? Ans. 124 a. 1 r. 1 p. 5. How many yards are contained in a square whose side is 30 feet? Ana. 144yds. (3. How many men can stand on acres of land, each occupying a space of 3 feet square ? Ans. 29040 men. PROBLEM II. The area of a square being given, to find the length of the side. RULE. Extract the square root of the area. EXAMPLES. 1 . The area of a square is 2025 feet ; what is the side ? Here w many pieces, each 4 inches square, may be cut from {^fegular decagon whose side is 12 inches ? fins. 69.2178 pieces. MENSURATION OF SUPERFICIES. 83 PROBLEM HI. When the area of any regular polygon is given, to find the side. RULE. Divide the area by the number in the table cor- responding with the figure, and the square root of the quotient will be the length of the side. EXAMPLES. 1. The area of a regular pentagon is 4 a res ; how many perches are contained in the side ? Here 160 x 4 = 640 perches, Then V (640 -j- 1.720477) = ,/ 371.9898 = 19.2870 perches, the length of the side. 2. Required the length of the side of a regular hexagon containing one acre. rfns. 7.8475 perches. 3. The area of an octagonal floor is 560 feet ; what is the length of the side ? Ans. 10.7693 feet. 4. The area of a regular hexagon is 73.9 feet ; what is the side ? Jins. 5i feet. 5. The area of a regular heptagon is 1356.6 yards; what is the length of the side ? Jlns. 19.3214 yards. 6. The area of a regular decagon is 3233.4912 feet ; what is the length of the side ? rfns. 20$ fret. IRREGULAR FIGURES. PROBLEM I. 13. To find the area of an irregular right-lined figure of any number of sides. RULE 1. Divide the figure into triangles and trapeziums, and find the areas of each of them separately by Prob. I. p. 67, and Prob. I. p. 79 ; then add these areas together, and their sum will give the area of the whole figure. 84 MENSURATION OF SUPERFICIES. EXAMPLES. 1 . Reqi lired the area of the i rregular right-lined figure, A B C D E F, the dimensions of which are as follows : F B = 20.75,F C = 27.48. E C = 18.5, B n = 14.25, E m = 9.35, D r = 12.8, and A s = 8.6 perches respectively. Here (F B x A *) -=- 2 = (20.75 A X 8.6) -f- 2 = 178.45 -~ 2 = 89.225 perches, the area of the triangle, A B F. And (E C x D r) -r- 2 = (18.5 x 12.8) ~- 2 = 236.8 -r- 2=1 18.4 perches, the area of the triangle, DEC. Also, [(Bra -f Em) x F C] -r- 2 = [(14.25 -f 9.35) x 27.48] -7- 2 = (23.6 x 27.48) -r- 2= 648.528 -~ 2 = 324.264 perches, the area of the trapezium, F B C E. Whence 324.264 -f 89.225 -f 118.4 = 531.889 perches = 3 a. 1 r. 11.889 p., the area of the whole figure. 2. Required the area of an irregular hexagon, like that in the last example, supposing the dimensions of the different lines to be the halves of those before given. Ans. 3 r. 12.9722 p. RULE 2. The area of any irregular right-lined figure may also be determined, by drawing perpendiculars from all its angles, to one of its diagonals, considered as a base ; and then adding the areas of all the triangles and trapezoids to- gether for the content. EXAMPLES. 1. Required the area of the irregular right-lined figure, A B C D E F, the dimensions of which are as follows : E n <= 4.54, E m = 8.26, E r =20.01, E s = 26.22, E B = 30.15, D m = 10.56, C r = 12.24, F n = 8.56, and A s = 9.26 chains respectively. Here m r = E r E m = 20.01 8.26 =11.75, n* = Es En== 26.22 4.54 = 21 .68, B r = E B E r -= 30.15 20.01 = 10.14, and B*=EB E*= 30.15 26.22 <= 3.93. MENSURATION OF SUPERFICIES. 85 Whence (F n x E n) -4-2= (8.56 x 4.54) -s- 2 = 38.8624 -i- 2 = 19.4312 chains, area of the triangle E F n. And ^D m x E m) -r- 2 = (10.56 x 8.26) -4- 2 = 87.2256 .4- 2 =43.6 128 chains, area of the triangle, E D m. Also (C r x B r) -4- 2 = (12.24 x 10. 14) -- 2 = 124.1 136 -i. 2 = 62.0568 chains, area of the triangle C B r. And (A x B s] -4- 2 = (9.26 x 3.93) -5- 2 = 36.3918 -r- 2 = 18.1959 chains, area of the triangle A B s. Then [(D ra + C r) x m r] -f- 2 = [(10.56 + 12.24) x 11.75] -4- 2= (22.8 x 11.75) -r- 2 = 267.9 -4-2= 133.95 chains, area of the trapezoid D C r m. And [(F n + A s) x n s] -=- 2 = [(8.56 -f 9.26) x 21.68] ~ 2 = (17.82 x 21.68) -=- 2 = 386.3376 -4- 2 = 193.1688 chains, area of the trapezoid F A s n. Hence 19.4312 + 43.6128 + 62.0568 + 18.1959 + 133.95 + 193.1688= 470.4155 chains = 47 a. r. 6.648 p., the area of the whole figure. 2. Required the area of an irregular figure, like that in the last example, only doubling the dimensions of the diagonal, and the several perpendiculars, rfns. 188 a. Or. 26.592 p. PROBLEM II. To find the area of a mixtilineal figure, or one formed by right lines and curves. RULE 1. Take the perpendicular breadths of the figure in several places, at equal distances from each other, and divide their sum by their number, for the mean breadth ; and this quotient, multiplied by the length, will give nearly the true area of the figure. 2d. Or, if greater accuracy be required, take half the sum of the two extreme breadths, for one of the said breadths, and add it to the others, as before ; then divide this sum by the number of parts in the base (instead of by the number of breadths), and the result multiplied by the length will give the area, with a sufficient degree of correctness to answer most questions of this kind that can occur. When the curved, or mixti'ineal, boundary meets the base, as is 8 86 MENSURATION OF SUPERFICIES. frequently the case in surveying, the area is found by dividing the sum of all the breadths by the number of parts in the base, and then multiplying the result by the length, as before ; observing, in each of these cases, that the greater the number of parts into which the base is divided, the nearer will the approximation be to the exact area. It may likewise be further remarked, that if the perpen- diculars, or breadths, be not at equal distances from each other, the parts should be computed separately, as so nutny trapezoids, and then added together, for the area. EXAMPLES. 1. The perpendicular breadths of the irregular mixtilineal figure, A B C D, at 5 equidistant places, AEGIB, being 9.2, 10.5, 8.3, 9.4, and 10.7 yards, and its length, A I 1 ' IT A B, 20 yards ; what is its area ? Here (9.2 -f 10.5 + 8.3 + 9.4 + 10.7) -i- 5 = 48. i -,- 5 = 9.62, And 9.62 X 20 = 192.4 yards, area by the first part of the rule. Or, (AD + BC) -v- 2 = (9.2 + 10.7) -+ 2 = 19.9 -=- 2 = 9.95. Then (9.95+10.5+8.3 + 9.4)-=- 4=38.15 -r- 4 =9.5375, and 9.5375 x 20 = 190.75 yards, the area by the second part of the rule. 2. Required the area of the figure, A B C D, of which the part A C is D a rectangle, whose sides are 20.} and 10^ chains respectively; and the perpendicular breadths of the curvilineal spaces, reckoning from A DC at 4 equidistant places, are 10.2, 8.7, 10.9, and 8. 5 chains respectively. Here (10.2 + 8.7 + 10.9 + 8.5) -r- 5 =38.3 -4- 5 = 7.06, And 7.66 x 20.5 = 157.03 chains, the area of the curved space. Then 20.5 x 10.5 = 215.25 chains, the area of the rect' angle. Whence 215.25 + 157.03 = 372.28 chains = 37 a Or 36.48 p., the area of the whole figure. MENSURATION OF SUPERFICIES. 87 3. The length of an irregular mixtilineal figure is 47 chains, and Us breadth, at 6 equidistant places, beginning at the left hand extremity of the base, 5.7, 4.8, 7.5, 5.1, 8.4, and 6.5 chains respectively ; what is its area? Am. 29 a. 3 r. 2 p.. by the first rule. 4. The length of an irregular mixtilineal figure, of which the curvilinear boundary meets the base, is 37i chains, and its breadth, at 7 equidistant places, is 4.9, 5.6, 4.5, 8.2, 7.3, 5.9, and 8.5 chains respectively ; what is its area ? Jlns. 24 a. r. 8.571 p. by the first rule. And 23 a. 3 r. 20 p. by the second rule. 5. The length of an irregular field is 39 rods, and its breadths, at five equidistant places, are 2.4, 2.6, 2.05, 3.05 3.6 rods respectively ; what is the area ? Jlns. 111.54 rods by the first rule. 6. The length of an irregular piece of land being 42 chains, and the breadths, at six equidistant points, being 8.7, 10.3, 7.1, 8.34, 10.04, 12.2 chains respectively ; what is the area ? Jlns. 38 a. 2 r. 39.872 p., by the second rule. THE CIRCLE. PROBLEM I. 14. To find the circumference of a circle, when the diameter is given, or the diameter, when the circumference is given. RULE. Multiply the diameter by 3.1416, and the product will be the circumference ; or divide the circumference by 3.1416, or multiply the circumference by .31831, and the result will be the diameter. EXAMPLES. 1. What is the circumference of the circle, A C B D, whose diameter, A B, is 7 fi'Pt ? A Here 3.1410 x 7 = 21.9912 feet, the cir- cumference. 88 MENSURATION OF SUPERFICIES. 2. What is the diameter of a circle whose circumference is 100 yards? Here 100 -j- 3.1416 = 31.831 yards, the diameter. Or, 100 X .31831 = 31.831 yards, the diameter as before. 3. If the diameter of a circle be 17 chains ; what is the circumference ? Jlns. 53.4072 chains. 4. If the circumference of a circle be 354 perches ; what is the diameter ? Jlns. 112.0817 perches. 5. What is the circumference of the earth, supposing its diameter to be 7935 miles, which it is very nearly ? Jlns. 24928.590 miles. 6. If the circumference of a carriage-wheel be 16 feet 6 inches; what is its diameter? Jlns. 5.2521 feet. PROBLEM II. To find the area of a circle. RULE. Multiply the square of the diameter by .7854 ; or the square of the circumference by .07958, and the pro- duct in either case will be the area. EXAMPLES. 1. How many square feet are there in a circle whose diameter is 5 feet 6 inches ? Here (5.5) 3 X .7854 = 30.25 x .7854 = 23.75835 sq. feet. 2. Required the area of a circle, the circumference of which is 9| yards. Here (9.2) 2 x .07958 = 84.64 X .Q7958 = 6.7356512 sq. yds. 3. How many square yards are there in a circle whose radius is 15* feet? Jlns. 81.1798 sq. yards. 4. Hov many square feet are there in a circle whose circumference is 10* yards ? Jlns. 82.7081 sq. feet. 5. The diameter of a circle is 16 chains; how many acres does it contain ? Jlns. 20 a. r. 16.U9M p. 6. What is the value of a circular garden whose diameler is 6 perches, at the rate of 75 cents per square yard ? Jlns. 9641.471* MENSURATION OF SUPERFICIES. 89 PROBLEM III. The area of a circle being given, to find the diameter or circumference. RULE. Divide the area by .7854, and the square root of the quotient will be the diameter. Or, divide the area by .07958, and the square root of the quotient will be the cir- cumference. EXAMPLES. 1. The area of a circle is 5 acres, 3 roods, and 26 perches ; what is the diameter ? Here 5 a. 3 r. 26 p. = 946 perches ; and ^7(946 -f- .7854) = v/ 1204.48179271 = 34.7056 perches, the diameter. 2. The area of a circle being 2 acres, 3 roods, and 12 perches, what is the circumference ? Here 2 a. 3 r. 12 p. = 452 perches ; and ^/ (452-*- .07958) = v/ 5(579.09 = 75.3637 perches, the circumference. 3. The area of a circle is 5028f square yards ; what is its diameter ? Ans, 80.0160 yds. 4. The area of a circular garden being 1 acre, what is the length of a stone wall which will enclose it ? Ans. 44.8392 p. 5. It is required to find the radius of a circle whose area is an acre. Jlns. 39.2507 yds. 6. The value of a circular piece of ground, at $4 per square perch, is $46.50. How many dollars will encircle it if the diameter of a dollar be 1 inches? Ans. $1595.3916. PROBLEM IV. To find the area of a circular ring, or the space included between two concentric circles. RULE. Find the areas of the two circles separately. Then the difference of these areas will be the area of the rug. Or, multiply the sum of the diameters by their difference. 90 MENSURATION OF SUPERFICIES. and this product again by .7854, and it will give the area required. EXAMPLES. 1. The diameters of the two circles are, A B 20, and D C 12 yards ; re- quired the area of the ring. Here 400 x .7854 = 314.16, area of A( the outer circle. 144 x .7854= 113.0976, area of the inner circle. And 314.16113.0976 = 201.0624 yards, area of the ring. Or, 20 + 12 = 32, sum of diameters. 20 12 = 8, difference of diameters. And 32 x 8 X .7854 = 201.0024 yards, area as before. 2. What is the area of a circular ring, the diameters of the concentric circles being 20 and 30 feet 1 Jlns. 392.7 ft. 3. The diameters of two concentric circles are 8 and 12 yards ; what is the area of the ring contained between their circumferences ? Jlns. 62.832 yds. 4. The diameters of two circles are 21| and 9 feet; re- quired the area of the ring. Jlns. 300.6009 ft. 5. The diameter of the inner circle is 6, and the outer 10 chains ; what is the area of the ring 1 Ans. 50.2656 chs. 6. The area of the outer circle contains 100 acres, and the diameter of the inner is equal to f of the diameter of the greater ; what is the area of the ring ? Jins. 555.55 chs. PROBLEM V. The diameter or circumference of a circle being given, to find the side of an equivalent square. RULE. Multiply the diameter by .8862, "or the circum- ference by .2821, the product in either case will be the side of an equivalent square. MENSURATION OF SUPERFICIES. 91 EXAMPLES. 1. The diameter of a circle is 200 yards ; what is the side of a square of equal area ? Here 200 x .8862 = 177.24 yards. 2. The circumference of a circle is 316 yards ; what is he side of a square of equal area ? Here 310. x .2821 = 89.1436 yards. 3. The diameter of a circle is 1 142 feet ; what is the side of a square of equal area ? Ans. 1012.0-104 ft. 4. The circumference of a circle is 18.8 chains ; what is the side of a square of equal area ? rfns. 5.3034 chs. 5. The diameter of a circular fish-pond is 22 perches ; what would be the side of a square fish-pond of an equal area ? dns. 19.4964 p. 6. The circumference of a circular walk is 64 rods ; what !s the side of a square containing the same area ? fins. 18.0544 r. PROBLEM VI. The diameter or circumference of a circle being given, 19 find the side of the inscribed square. RULE. Multiply the diameter by .7071, or the circum- ference by .2251, and the product in either case will be the side of the inscribed square. EXAMPLES. 1. The diameter, A B, of a circle is 614 feet ; what is the side, A C, of the inscribed square ? Here 614 x .7071 = 434.1594 feet =AC. 2. The circumference of a circle is 1804 feet; what is the side of the in- scribed square? Ans. 406.0804 ft. 3. The diameter of a circle is 239 feet ; what is the side of the inscribed square ? tfns. 168.9909 fl. 92 MENSURATION OF SUPERFICIES. 4. The circumference of a circle is 98 chains ; what is the side of the inscribed square ? Jlns. 22.0598 chs. 5. The diameter of a circle is 65 rods ; what is the side of the inscribed square ? Ans. 45.9015 r. 6. The circumference of a circular walk is 721 perches ; what is the side of an inscribed square ? tins. 162.2971 p. PROBLEM VII. To find the diameter of a circle equal in area to any given superficies. RULE. Divide the area by .7854, and the square root of the quotient will be the diameter. EXAMPLES. 1. The length and breadth of a rectangle are 24 and 10 chains ; what is the diameter of a circle which contains the same area? Here 24 x 16 = 384, the area of the rectangle. Then v/(384 -=- .7854) = v/488.9228 = 22. 1 1 16 chains, the diameter. 2. The side of a square is 16 perches ; what is the diame- ter of a circle containing the same area ? Jlns. 18.054 p. 3. The base and perpendicular of a right-angled triangle are 16 and 20 feet ; what will be the diameter of a circle which contains the same area ? rfns. 14.2/29 ft. 4. The three sides of a scalene triangle are 14, 18, and 24 yards ; what is the diameter of a circle containing the sam area? -Ans. 12.6267 yds. 5. The three sides of a triangle are 18, 20, and 26 feet ; what is the diameter of a circle containing three times as much? dns. 26.1919 ft. 6. The length and breadth of a parallelogram are 32 and 18 yards : what is the diameter of a circle that contains the same area ? -fins. 27.0810 \ds. MENSURATION OF SUPERFICIES. 93 PROBLEM VIII. The diameter of a circle being given, to find another con- taining a proportionate quantity. RULE. Multiply the square of the given diameter by the given proportion, and the square root of the product will be the diameter required. EXAMPLES. 1. The diameter of a circle is 24 chains ; what is the diameter of one containing one-fourth of the area ? Here x /(24 2 x *) = v/(576 x |) = v/144 = 12 chains. 2. The diameter of a circle is 16 perches ; what is the diameter of one containing nine times as much ? rfns. 48 p. 3. The diameter of a circle is 36 yards; what is the diameter of one containing four limes as much ? rfns. 72 yds. 4. The diameter of a circle is 81 feet ; what is the diame- ter of one containing five times as much ? Ans. 181.1215 ft. 5. A gentleman has a circular grass-plat in his yard, the diameter of which is 25 yards; required the length of the string that would describe a circle to contain sixteen times as much. Jlns. 50 yds. 6. The diameter of a circle is 9 rods ; what is the diame- ter of one containing six times as much? Jlns. 22.0454 r. PROBLEM IX. To find the length of any arc of a circle. RULE. 1 . From eight times the chord of half the arc, s tract the chord of the whole arc, and one-third of the mainder will be the length of the arc nearly. 94 MENSURATION OF SUPERFICIES. The chord of the whole arc, or sim- ply the chord, is a right line which joins the extremities of an arc. Thus in the figure, D C is the chord of the whole arc, D F C ; D F the chord of half the A[ arc, D F C ; and D E, or E C, half the chord of the arc D F C. The height of an arc, called its versed sine, is that part of the diameter contained between the middle of the chord and the arc. Thus, in the case of the arc D F C, the versed sine, or the height of the arc, is the line E F. The chord of half the arc may be found by adding together the square of half the chord and the square of the versed sine, and extracting the square root of the sum. Or, by taking the square root of the product of the diame- ter and the versed sine. Half the chord of the whole arc may be found by sub- tracting the versed sine from the diameter, multiplying the remainder by the versed sine, and taking the square root of the product. By doubling this we get the chord of the whole arc. To find the versed ine, or height of the arc, subtract the square of the chord from the square of the diameter, and ex- tract the square root of the remainder ; subtract this root from the diameter, and one-half of the remainder will be the versed sine. Or, from the square of the chord of half the arc, subtract the square of half the chord of the arc, and the square root of the remainder will give the versed sine. Again, to obtain the versed sine, divide the square of the chord of half the arc by the diameter. The diameter may be known by adding together the square of the versed sine and the square of half the chord of the arc, and dividing the sum by the versed sine. The diameter may likewise be obtained by dividing the square of the chord of half the arc by the versed sine. It may here likewise be observed, as another rule for the parne purpose, that if the number of degrees in the arc be multiplied by radius, and that product again by .01745, the result will give the length of the arc nearly. MENSURATION OF SUPERFICIES. 95 EXAMPLES. 1. The chord A B of the whole arc A C B, is 48.74 feet, and the chord A C of half the arc 30.25 feet ; what is the length of the arc ? Here [ (30.25 x 8) 48.74) ] -r- 3 = (242 48.74) -j- 3 = 193.26 -=- 3 = 64.42 ft. the length of the arc A C B nearly. 2. If the chord A B, of the arc A C B, be 30 yards, and the versed sine, or height C D, 8 yards, what is the length of the arc ? Here */ (15 9 + 8*} = A C, the chord of half the arc. And ^[(30 4) x 4] x2 = ^/(26x4) x2 = v/(104) X 2= 10.19803 x 2 =20.39606 = A B, the chord of the whole arc. Then [(10.95445 x 8) 20.39606] +- 3 = (87.6356 20.39606) -4-3 = 67.23954 -7- 3 = 22.41318 feet, the length of the arc nearly. 4. Required the length of an arc of 30 degrees, the radius of the circle being 14 feet. (30 x 14 X .01745) = 7.329 feet, the length of the arc. 5. The chord of the whole arc is 50f yards, and the chord of half the arc is 33| yards ; required the. length of the arc. Ans, 71.6 yards. 6. The length of the chord of the whole arc is 36| feet, and the length of the chord of half the arc is 231 feet ; what is the length of the arc ? Jlns. 49.6166 feet. 7. The chord. of the whole arc is 48k feet, and its versed sine, or height of the arc, 18| feet ; what is the length of the arc ? ns. 64.7667 fee-.. 96 MENSURATION OF SUPERFICIES. 8. If the versed sine or height of the arc be 2 yards, and the diameter of the circle 36 yards ; what is the length of the arc? Ans. 17.1299 yards. 9. If the chord of the whole arc be 16 chains, and the radius of the circle 10 chains ; what is the length of the arc ? Jlns. 18.518 chains. 10. The chord of half the arc is 10 perches, and the diameter of the circle 16 perches ; what is the length of the arc? rfns. 21 5 perches. 11. Required the length of an arc of 57 17' 44*", the diameter of the circle being 50 feet. Jlns. 25 feet, which equals the radius. 12. Required the length of a degree of a great circle of the earth, supposing its circumference to be 25000 miles. Ans. 69j miles nearly. RULE 2. Let d = the diameter C E of the circle, and t; = the versed sine or height C D of half the arc, then will the length of the arc be expressed by the following series : 6_, , v." I r^ f\ i "-'7 vwv ' X 7rf 8 x9e? Where A, B, C, &c. represent the terms immediately pre- ceding those in which they first occur. Which rule is to be used, by substituting in the above series the numerical values of the given parts, in the place of the letters by which they are denoted, and then finding the sum of such a number of its terms as may be thought suffi- cient for determining the length of the arc, to a degree of accuracy required. EXAMPLES. 1. Required the length of the arc, A C B, whose versed sine or height, DC, is 5 feet, and the diameter, CE, of >he circle 25 feet. Here d . 25, and v = 5. MENSURATION OF SUPERFICIES. 97 Then 2 v/ dv = 2 v/(25 x 5) = 2 v/ 125 = 22.3606798 == A , " , X A = ^-^ X 22.3606798 = .7453559= B 2x3 d 6 x 25 *' v x B = J! X JL x .7453559 = .0670820 = C x 5 d 20 x 25 f^i 71 6x7<* xc =4ra x 0670820= OOT9859= 7 3 u 49 x 5 8 9~d xD= 72 ox x -0079859 = .0010869= E The sum = 23.1821905" fret, = the length of the arc, A C B. 2. Required the length of the arc, A C B, whose versed sine or height, DC, is 2 feet, and the diameter, C E, of the circle 52 feet. Ans. 20.5291 feet. 3. Required the length of the arc whose versed sine is 9 yards, and the diameter of the circle 100 yards. Ans. 60.9385 yards. 4. It is required to find the length of the arc whose chord is 16 perches, and its height 4 perches. Ans. 18.5459 perches. 5. It is required to find the length of the arc whose height is 6 inches, and the chord of half the arc 1 foot. Ans. 2.0943 feet. 6. It is required to find the length of the arc whose chord is 48 chains, and the radius of the circle 25 chains. Ans. 64.3501 chains. PROBLEM X. To find the area of a sector of a circle. ROLE 1. Multiply the radius or half the diameter of the circle by half the length of the arc of the sector, as found by the last Problem, and the product will be the area. To this we may add, that as 360 is to the number of de- grees in the arc of the sector, so is the area of the circle, to the area of the sector. 9 MENSURATION OF SUPERFICIES. EXAMPLES. 1 The chord, A B, of the whole arc, A C B, is 24 feet, and the chord, A C, of half the arc 13 feet ; what is the area of the sector, OBC AO? Here v/ (13 3 12 3 ) = v/ (169 - 144) = ^/ 25 = 5 =C D, the versed sine. And (13 a -=- 5) =169 -T- 5 = 33.8 = C E, the diameter. E [713 x 8) 24] -4- 3 = (104 24) -=- 3 = 80 -v-3 =20 = the length of arc A C B. Then 13 x 16.9 = 225! feet, the area of the sector. 2. Required the area of the sector, the arc of which is 30, and the diameter 3 yards. Here .7854 x 9 = 7.0686 = the area of the circle. As 360 : 30 : : 7.0686 : .58905 yards, the area of the sector. 3. The chord of the whole arc is 8 yards, and its height 3 yards ; what is the area of the sector ? Ans. 22.2222 yards. 4. What is the area of a sector whose chord is 18| chains, and the diameter of the circle 20 chains ? Ans. 118.954 chains. 5. Required the area of the sector whose height is 4 perches, and the radius of the circle 8 perches. Ans. 66.8581 perches. 6. Required the area of the sector whose arc is 17 15', and the diameter of the circle 19 feet. Ans. 13.5857 feet. RULE 2. Let d = the diameter C E of the circle, and v = the versed sine or height, C D, of half the arc, then will the area of the sector be expressed by the following series. D, &c. MENSURATION OF SUPERFICIES. 99 When A, B, C, &c., represent the terms immediately preceding those where they first occur. This Rule is to bo used in the same manner as directed in regard to Rule 2, Problem IX. p. 96. EXAMPLES. 1. Required the area of the sector, A C B O, whose versed sine, or height, C D, is 7 feet, and the diameter, C E, of the circle 28 feet. Here d = 28, and v = 7. Then J v=14 v /(28x7)= 7 XA = xB = C = XD = 6x28 = 196.000000 = A = 8.1 60666 = B 9X ~X8. 166666 = 20 x 28 25x7 42x28 49x7 72x28 .918750 = X .918750= . 136718 = D X .136718 = = E 9 a u 81 x 7 10 X lid X " 110x23 X IPu 121x7 v T^ v .023261= .004282 =F .004282= .000830 = G The sum = 205.250507 feet = the area of the sector O A C B O. 2. Required the area of the sector O A C B O, whose versed sine, C D, is 2 yards, and the diameter, C E, of th circle 52 yards. Ans. 266.8787 yds. 3. Required the area of the sector, the radius of the cir- cle being 10 feet, and the chord of the arc 12 feet. tins. 64.3501 ft. 4. Required the area of the sector of a circle whose diame- ter is 20 perches, and the chcrd of its arc 16 perches An* 92.7295 p 100 MENSURATION OF SUPERFICIES. 5. Required the area of the sector of a circle whose versed sine is 9 feet, and the chord of half the arc 30 feet. tins. 1523.4632ft. 6. Required the area of the sector of a circle whose chord is 48 feet, and versed sine 18 feet. rfns. 804.370 ft. PROBLEM XT. To find the area of a segment of a circle. RULE 1. Find the area of the sector, having the same arc as the segment, by the last problem. Also find the area of the triangle formed by the chord of the segment, and the two radii of the sector, by Problem I. page 67. The sum, or difference of these areas, according as the segment is greater or less than a semicircle, will be the area of the segment required. EXAMPLES. 1 . The chord A B is 24 feet, and the versed sine or height C D, of half the arc A C B, is 5 feet ; what is the area of the segment A B C A ? Here ^ (A D a + C D a ) = v/ (12 a -f 5 3 ) = v' (144 + 25) = ^ 169 = 13 = A C, the chord of half the arc. Again, (A C 3 -7- C D) = (169 -4- 5) = 33.8 = C E, the diameter; therefore 16.9 = C O, the radius. Then [ (13 x 8) 24] -7- 3 = 26f , the length of the arc AC B. And (13 x 16.9) =225.3333 feet, the area of the sec- tor O A C B O. But (C O C D) = (10.9 5) = 11.9 = perpendicular OD. Whence (11.9 x 12) = 142.8, area of the A A O B. And, consequently, (225.3333142.8) =82.5333 feet area of the segment A B C A. 2. What is the area of a segment of a circle whose arc is 00, and the diameter of the circle 10 feet ? Here .7854 x 100 = 78.54, area of the whole circle. MENSURATION OF SUPERFICIES. 101 Then, as 360 : 60 : : 78.54 : 13.09, area of the sector <> A C B O. And since the chord A B (to an arc of 60) is = radius O A or O C, which = 5, then by Problem II., page 82, the area of the A A O B = 10.8253. Whence (13.09 10.8253) = 2.2647 feet, the area of the segment required. 3. Required the area of the segment of a circle, whose versed sine, or height, of half the arc is 5 yards, and the diameter of the circle 20 yards. Am. 61.1(345 yds. 4. Required the area of the segment of a circle, whose chord is 16 chains, and diameter of the circle 16| chains. Am. 70.2222 chs. 5. What is the area of the segment of a circle, whose arc is a quadrant, the diameter being 24 perches ? Am. 41. 0976 p. 6. Required the area of a segment of a circle, whose chord is 18.9 feet, and height 2.4 feet. Am. 30.601 ft. RULE 2. Let d equal the diameter C E of the circle, and v equal the versed sine C D, of half the arc, or height of tne segment ; then will the area of the segment be expressed by the following series. Area of the segment A C B A = x9cT 8xll^ And, 2. When the two parallel chords are equal to each other, the diameter d or E F will be = ^/ (6 a + c 2 ) ; and the versed sine mn = 5 (d c). And when one of the parallel chords is the diameter, the versed sine mn - d ^ / (18 x 32) : E F. Or, E F = nn o yo v/576 = - X 24 = = 14.4, the ordinate required. 50 5 & 2. If the transverse diameter be 40, the conjugate 30, and the abscissa 24, what is the ordinate ? rfns. 14.6969. 3. If the transverse diameter be 120, the conjugate 40, and the abscissa 24, what is the ordinate ? Jlns. 16. 4. If the transverse diameter be 35, the conjugate 25, and he abscissa 28, what is the ordinate ? jins. 10. CASE 2. When the transverse, conjugate, and ordinate are given, to find the abscissa. RULE. As the conjugate diameter is to the transverse, so is the square root of the difference of the squares of the ordi- nate and semi-conjugate to the distance between the ordinate and centre. CONIC SECTIONS. 115 And this distance being added to, and subtracted from, the semi-transverse, will give the two abscissas required. EXAMPLES. 1. The transverse diameter A B is 60, the conjugate diameter C D 40, and the ordinate F E 12; what is the length of each of the two abscissas B E and A E ? Here A B = 60, C D = 40, and F E = 12. Whence 40 : 60 :: v /(20 2 12 2 ) : O E. Or O E = v/(400- 144)=| ^ 256 = | x 16 = I 8 = 24. And 30 24 = 6 = B E, and 30 + 24 = 54 =A E. 2. What are the two abscissas to the ordinate 10, the diameters being 35 and 25 ? Jlns. 7 and 28. 3. What are the t\vo abscissas to the ordinate 16, the diameters being 120 and 40 ? Jlns. 24 and 96. 4. What are the two abscissas to the ordinate 14.4, the diameters being 50 and 30 1 Jlns. 18 and 32. CASE 3. When the conjugate, ordinate, and abscissa are given, to find the transverse diameter. RULE. To, or from the semi-conjugate, according as the less or greater abscissa is used, add or subtract the squaio root of the difference of the squares of the ordinate and semi- conjugate. Then, as the square of the ordinate is to the product of the conjugate and abscissa, so is the sum or difference above found to the transverse diameter required. EXAMPLES. 1. The conjugate diameter C D is 60, the ordinate E F 24, and the less abscissa B E 36 ; required the transverse diameter A B. Hete O C = 30, E F = 24, and B E = 36. Whence 30 + ^/ (30 2 24 2 ) = 30 +^(900 576) =^0 -f v/ 324 = 30 + 18 = 48. And 24 2 : (00 x 36) : : 48: (60 x 36 x 48) -4- 24 2 = 103080 -i- 576 = 180, the transverse diameter reauired. 116 CONIC SECTIONS. 2. The conjugate diameter is 60, the ordinate 18, and the less abscissa 24 ; required the transverse diameter. . 240. 3. The conjugate diameter is 40, the ordinate 16, and the frreater abscissa 96 ; required the transverse diameter. AM. 120. 4. The conjugate diameter is 25, the ordinate 10, and the greater abscissa 28 ; required the transverse diameter. Am. 35. CASE 4. The transverse, ordinate, and abscissa being given, to find the conjugate diameter. RULE. As the square root of the product of the two abscissas is to the ordinate, so is the transverse diameter to the conjugate. EXAMPLES. 1. The transverse AB is 82, the ordinate EF 8, and the abscissa BE 32; required the conjugate diameter C D. Here A B = 82, EF = 8, BE = 32, and AE = 82 32 = 50. Hence ^/ (32 x 50) : 8 : : 82 : (8 x 82) -j- v/ (32 x 50) = 656 -r- ^/1600 == 656 -r- 40 = 16.4, the conjugate diame- ter required. 2. The transverse diameter is 35, the ordinate 10, and the abscissa 28 ; what is the conjugate ? Am. 25. 3. The transverse diameter is 120, the ordinate 16, and its abscissa 24 ; what is the conjugate ? Jlns. 40. 4. The transverse diameter is 50, the ordinate 14.4, and the abscissa 32 ; what is the conjugate ? Ans. 30. PROBLEM III. The transverse and conjugate diameters being given, to find the circumference. RULE. Multiply the square root of half the sum of the squares of the two diameters by 3.1416, and the product will be the circumference nearly. CONIC SECTIONS. 117 EXAMPLES. 1. The transverse is 40, and the conjugate 80;' required the circum- ference of the ellipse. Here ^ [(40 3 + 30 3 ) -4- 2] = / v' [(1600 + UOO) -4- 2] = v' (2500 -4- 2) = v/ 1250 = 35.3553. Whence 35.3553 x 3.1416 = 111.0722 = the circum ference. 2. The transverse diameter is 24, and the conjugate 20; required the circumference of the ellipse. Jim. 69.4001. 3. The transverse diameter is 60, and the conjugate 40 ; what is the circumference? Jlns. 160.1907. 4. The transverse diameter is 24, and the conjugate 18 ; what is the circumference 1 Jlns. 66.6433. PROBLEM IV. The transverse and conjugate diameters being given, to find the area. RULE. Multiply the transverse diameter by the conjugate, and this product again by .7854, and the resiilt will be the area. EXAMPLES. 1. Required the area of an ellipse whose transverse di- ameter is 24 perches, and the conjugate 18 perches. Here 24 x 18 X .7854 = 339.2928 perches = 2 a. r. 19.2928 p., the area required. 2. Required the area of an ellipse whose two diameters are 60 and 40 rods. Jlns. 11 a. 3 r. 4.96 p. 3. Required the area of an ellipse whose two diameters are 40 and 36 chains. Jlns. 113 a. r. 15.616 p. 4. The transverse and conjugate diameters are 66 and 22 yards ; what is the area 1 Jlns. 1 140.4 yds. 118 CONIC SECTIONS, PROBLEM V. The transverse and conjugate diameters of an ellipse being given, to find the diameter of a circle containing the same area. RULE. Multiply the transverse and conjugate diameters together, and extract the square root of their product. EXAMPLES. 1. The transverse and conjugate diameters are 70 and 50 chains ; what is the diameter of a circle containing the same area? Here ^/ (70 x 50) = -v/3500 = 59.16079 chains, the diameter. 2. The transverse and conjugate diameters are 36 and 24 chains ; what is the diameter of a circle containing the same area ? rfns. 29.3938 chs. 3. The transverse and conjugate diameters are 49 and 16 rods ; what is the diameter of a circle containing the same area ? jlns. 28 rods. 4. The transverse and conjugate diameters are 87 and 52 feet ; what is the diameter of a circle containing the same area ? rfns. 67.2606 ft. PROBLEM VI. To find the area of an elliptic segment whose base is parallel to either of the axes of the ellipse. RULE. Divide the height of the segment by that axis of the ellipse of which it is a part, and find in the table, p. 295, a circular segment whose versed sine is equal to the quotient. Then, multiply the segment thus found and the two axes of the ellipse together, and the product will be the area required. iG CONIC SECTIONS. 119 EXAMPLES. 1. Required the area of the elliptic segment whose height A G is 20 chains, and the axes of the ellipse, 2 A B and C D, 70 and 50 chains respectively. Here 20 -r- 70 = .285f , the tabular versed sine. And the tabular segment ; belonging to this, as found byProb.XI., c l - : D Rule 5, p. 104, is .185166. Whence .185166 x 70 (2AB) x 50 (CD) == 648.081 chains = 64 a. 3 r. 9.296 p., the area of the segment E A F. 2. What is the area of an elliptic segment cut off by an ordinate parallel to the transverse diameter whose height is 20 feet, the axes being 80 and 50 feet ? fins. 1173.476 ft. 3. What is the area of an elliptic segment cut off by an ordinate parallel to the transverse diameter whose height is 10 chains, the axes being 35 and 25 chains ? Ana. 25 a. 2 r. 27.166 p. 4. What is the area of an elliptical segment cut off by an ordinate parallel to the conjugate diameter whose height is 10 feet, the axes of the ellipse being 35 and 25 feet ? Ans. 162.0202 ft. 5. What is the area of an elliptic segment cut off by an ordinate parallel to the transverse diameter whose height is 5 yards, the axes being 35 and 25 yards ? dns. 97.8451 yds. 6. What is the area of an elliptic segment cut off by an ordinate parallel to the conjugate diameter, the axes of the ellipse being 60 and 40 feet, and the height of the segment 15 feet? dns. 368.51 ft. 7. What is the area of an elliptic segment cut off by a double ordinate parallel to the conjugate axis, at the distance of 36 yards from the centre, the axes being 120 and 40 yards ? tins. 536.7504 yds. 120 CONIC SECTIONS. THE PARABOLA. PROBLEM L 17. To describe a parabola, any ordinate to the axis and its abscissa being given. Construction. 1 . Let V R and R S be the given abscissa and ordinate ; bi- sect the latter in m, join Vra, and draw m n perpendicular to it, meeting the axis in n. 2. Make V C and V F each equal to Rn, and F will be the focus of the curve. 3. Take any number of points r, r, &c. in the axis, through which draw the double ordinates-S r S, &c., of an indefi- nite length. 4. With the radii C F, Cr, &c., and centre F, describe arcs cutting the corresponding ordinates in the points s, s, &c., and the curve S V S drawn through all the points of in- tersection will be the parabola required. The line s F s passing through the focus F is called the parameter. PROBLEM H. In a parabola, any three of the four following terms being given, viz., any two ordinates and their two abscissas, to find the fourth. RULE. As any abscissa is to the square of its ordinate, so is any other abscissa to the square of its ordinate. Or as the square root of any abscissa is to its ordinate, so is the square root of any other abscissa to its ordinate, and conversely. CONIC SECTIONS. 121 EXAMPLES. The abscissa V F is 18, and its or- dinate E F 12 ; required the ordinate G H, the abscissa of which, V H, is 32. Here VF = 18, EF = 12, and V H = 32. Whence 18: 12 2 :: 32: 12 * * 32^ 18 144x32 OK _ , , , A = 256 = the square oi the 18 ordinate GH, and ^/ 256 = 16 = the ordinate required. 2. The abscissa VF is 25 and its ordinate EF 16; required the ordinate G H, the abscissa of which, V H. is 49. ._ 16^/49 16x7 112 Here 25 solid feet. 2. The length of a parallelopipedon is 8 feet, Us breadth 3 feet, and thickness 2 feet ; how many solid feet does it contain 1 J) ns . 48 ft. 136 MENSURATION OF SOLIDS. 3. The length of a parallelopipedon is 36 inches, the width 20 inches, and depth 18 inches ; how many solid feet does it contain ? Ana. 7.5 feet. 4. The length of a parallelopipedon is 15 feet, and each side of its square base 21 inches ; what is the solidity ? Ana. 45.9375 ft. 5. What is the solidity of a block of marble, whose length is 12 feet, breadth 5? feet, and depth 2 feet ? Am. 172 ft. PROBLEM VI. To find the solidity of a prism. RULE. Multiply the area of the base by the perpen- dicular height of a prism, and the product will be the solidity. EXAMPLES. 1. What is the solidity of the triangular prism A B C D E F, whose length A B is 20 feet, and either of the equal sides B C, C F, or F B, of one of its equilateral ends B C F, 5 feet ? Here, by Problem II., page 82, the area of the base B C F is = 5 a x .433013 = 25 X .433013 = 10.825325. And consequently, 10.825325 x 20 = 216.5065 feet, the solidity required. 2. What is the solidity of a triangular prism whose length is 18 feet, and one side of the equilateral end H feet? Am. 17.5370 ft. 3. What is the solidity of a triangular prism whose length is 40 feet, and one side of the equilateral end 18 inches ? Ana. 38.9711 ft. 4. What is the solidity of a triangular prism whose lenglh is 24 feet, and one side of the equilateral end 16 inches ? Ana. 18.4752 ft. 5. What is the value of a prism whose height is 32 fe. % and each side of the equilateral end 14 inches, at 20 CGI i per solid foot ? Ana. $3.772. MENSURATION OF SOLIDS. 137 6. ^Required the solidity of a prism whose base is a hexa- gon, supposing each of the equal sides to be 1 foot 6 inches, and the length of the prism 10 feet. rfns. 93.5307 ft. PROBLEM VII. To find the convex surface of a cylinder. RULE. Multiply the circumference or periphery of the base, by the height of the cylinder, and the product will be the convex surface required ; to which add the area of each end, and the sum will be the whole surface of the cylinder EXAMPLES. 1. What is the convex surface of the right cylinder A BC D, whose length, BC, is 24 feet, and the diameter of its base, A B, 16 feet ? Here 3.1416 x 16 = 50.2650, the circum- ference of the base. And 50.2656 x 24 = 1200.3744 square feet, the convex surface required. 2. What is the whole surface of a right cylinder, the diameter of whose base is 21 feet, and the height 5 feet ? Here 3.1410 x 2.5= 7.854, the circumference of the base. And 7.854 x 5 == 39.27 square feet, the convex surface. Then (2.5 2 x .7854) x2 = (6.25 x .7854) x 2 = 4.90875 x.2 = 9.8175 square feet, the area of the ends. Whence 39.27+9.8175 = 49.0875 square feet, the whole surface. 3. Required the convex surface of a right cylinder, whose circumference is 8 feet 4 inches, and its length 18 feet. Ana. 150 ft. 4. What is the convex surface of a right cylinder, the diameter of whose base is 2 feet, and its length 30 feet? Am. 188.490ft. 12* 138 MENSURATION OF SOLIDS. 5. What is the whole surface of a right cylinder, the diameter of whose base is lo inches and its height 20 feet ? Jlns. 80.5685 ft. 6. How many square yards are contained in the whole surface of a cylinder, the diameter of whose base is 4 feet and its length 10 feet ? ,. Jlns. 16.7552yds. PROBLEM Vin. To find the solidity of a cylinder. RULE. Multiply the area of the base by the perpendicu lar height of the cylinder, and the product will be the solidity. EXAMPLES. 1 . What is the solidity of the cylinder A B C D, the diameter of whose base, A B, is 30 inches, and the height, B C, 55 inches? Here (30 2 X .7854) x 55 = (900 X-785 1) X 55=706.86x55 =38877.3 cubic inches. 2. What is the solidity of a cylinder, whose height is 30 feet, and the circumference of its base 20 feet ? Here (20 3 x .07958) x 30 = (400 x .07958) x 30 = 31.832 x 30 = 954.90 cubic feet. 3. What is the solidity of a cylinder whose height is> 4 feet, and the diameter of its base 10 inches ? Ans. 2.1816 ft. 4 The diameter of a cylinder is 16 inches, and the lengtl 20 feet ; what is the solidity ? rfns. 27.9253ft. 5. The circumference of a cylinder is 2 feet, and the length 5 feet ; what is the solidity ? dns. 1.5910 ft. MENSURATION OF SOLIDS. 139 C. The circumference of a cylinder is 20 feet, and the height 19.318 feet ; what is the solidity ? fins. 614.9305ft. PROBLEM IX. i To find the curve surface of a cylindric ungula, when the section passes obliquely through the sides of the cylinder. RULE. Multiply the circumference of the base by half the sum of the greatest and the least heights of the ungula, and the product will be the curve surface. EXAMPLES. 1. What is the curve surface of a cylin- dric ungula, the diameter of whose base is A B, 16 feet, and the greatest and least heights are B F, 17, and A E, 14 feet ? Here 3.1416 x 16 = 50.2856, the cir- cumference of the base. Then 50.2656 X (17 + 14) -4- 2 = 50.2656 x (31 -s- 2) =50.2656 x 15.5 = 779.1168 feet, the curve surface required. 2. What is the curve surface of a cylindric ungula, the circumference of whose base is 21 feet, and the greatest and least heights are 13 and 8 feet? rfns. 220.5 ft. 3. What is the curve surface of a cylindric ungula, the diameter of whose base is 19 feet, and the greatest and least heights are 13| and 11| feet? rfns. 746.13 ft. PROBLEM X. To find the solidity of a cylindric ungula, when the sec- tion passes obliquely through the opposite sides of the cylin der. RULE. Multiply the area of the base of the cylinder by half the sum of the greatest and least heights of the unguia, and the product will be the solidity. 140 MENSURATION OF SOLIDS. EXAMPLES. 1. What is the solidity of a cylindric ungula, the diame- ter of whose base, A B, is 12 feet, and the greatest and least heights are BF, 6, and AE, 4 feet? Here (12 2 x.7854) = 144 x. 7854 = 113.0976 square feet, the area of the base. Then 113.0976 x (6 + 4) -*- 3 = 113.0976 x (10 -4- 2) = 113.0976 x 5 = 565.488 feet, the solidity required. 2. What is the solidity of a cylindric ungula, the diameter of the base of which is 10 feet, and the greatest and least heights are 4 and 3 feet ? Am. 274-89 ft. 3. What is the solidity of a cylindric ungula, the circum- ference of the base of which is 24 feet, and the greatest and least heights 18 and 12 feet? Am. 687.5712 feet. PROBLEM XI. To find the convex superficies of a cylindric ring. RULE. To the thickness of the ring add the inner dia- meter, and this sum being multiplied by the thickness, and the product again by 9.8696 (or the square of 3.1416) will- give the superficies required. EXAMPLES. 1. The thickness, A c, of a cylindric ring is 3 inches, and the inner diameter, cd, 12 inches ; what is the convex su- perficies ? Here [(12 + 3) x 3] x 9.8696 = (15 x 3) x 9.8696 = 45 x 9.8696 = 444.132 square inches. 2. The thickness of a cylindric ring is 4 inches, and the inner diameter 18 inches ; what is the convex superficies ? fins. 868.5248 inches. 3. The thickness of a cylindric ring is 2 inches, and the inner diameter 1 loot 6 inches; what is the convex super- ficies? Ans. 394.784 inches MENSURATION OF SOLIDS. 141 4. The thickness of a cylindric ring is 3 inches, and its inner diameter 9 inches ; what is the convex superficies ? JIns. 355.3056 inches. 5. The thickness of a cylindric ring is 2 inches, and the inner diameter 12 inches ; what is the convex superficies? rfns. 276.3488 inches. 6. The thickness of a cylindric ring is 3.5 inches, and its inner diameter 18.765 feet; what is its convex superficies? Ana. 7899.4304 inches. PROBLEM XII. To find the solidity of a cyiindric ring. RULE. To the thickness of the ring add the inner diame- ter, and this sum being multiplied by the square of half the thickness, and the product again by 9.8096, will give the solidity. EXAMPLES. 1. What is the solidity of an anchor ring, whose inner diameter is 8 inches, and thickness in metal 3 inches ? Here [(8 + 3) x 1.5 s ] X 9.8696 =(11 X 2.25) x 9.8696 = 24.75 x 9.8690 = 244.2726 cubic inches. 2. What is the solidity of an anchor ring whose inner dia- meter is 9 inches, and the thickness of metal 3 inches ? dns. 266.4792 inches. 3. The inner diameter of a cylindric ring is 12 inches, and its thickness 4 inches ; what is its solidity ? dns. 631.6544 inchts. 4. Required the solidity of a cylindric ring whose thick- ness is 2 inches, and its inner diameter 16 inches. rfns. 177.6528 inches. 5. Required the solidity of a cylindric ring whose inner diameter is 12 inches, and thickness 5 inches. Jlns. 1048.645 inches. 6. What is the solidity of a cylindric ring whose thick- ness is 4 inches, and inner diameter 16 inches ? Jlns. 789.568 inches. 142 MENSURATION OF SOLIDS. PROBLEM XIH. Thi solidity and thickness of a cylindric ring being given, to find the inner diameter. RULE. Divide the solidity by 9.8696, and that quotient by the square of half the thickness, from which subtract the thickness, and the remainder will be the inner diameter of the ring. EXAMPLES. 1. The thickness of a cylindric ring is 4 inches, and its solidity 789.568 solid inches. What is its inner diameter? Here 789.568 -r-9.8696 = 80, and SO-s-2 2 = 80-i-4=20, then 20 4 = 16 inches the diameter. 2. Required the inner diameter of a cylindric ring whose solidity is 138. 1744 inches, and thickness 2 inches. JLns. 12 inches. 3. What is the inner diameter of a cylindric ring whose solidity is 1 solid foot, and thickness 4 inches ? fins. 39.77 inches. 4. If the solidity of a cylindric ring be 4 solid feet, and the thickness 3.5 inches, what is the inner diameter ? rfns. 18.765 feet. 5. What must be the inner diameter of a cylindric ring whose solidity is 1 solid inch, and thickness J- of an inch ? rfns. 25.8132 inches. 6. What is the inner diameter of a cylindric ring whose wlidity is 244.2726 inches, and the thickness 3 inches ? Ans. 8 inches. PROBLEM XIV. To find the surface of a right cone or pyramid. RULE. Multiply the circumference or perimeter of the i ase by the slant height or length of the side of the cone or pyramid, and half the product will be the surface required. And if this be added to the area of the base, it will give the whole surface. MENSURATION OF SOLIDS. EXAMPLES. 1. The diameter of the base A B, of a right cone C A B, is 6 feet, and the slant height A C or B C, 21 feet. Required the convex surface of the cone. Here 3.1416 x 6 = 18.8496 = the cir- cumference of the base. And (18.8496x21)-*- 2=395.8416-=-2 =197.9208 square feet, the convex sur- face required. 2. The circumference of a right cone is 10 feet, and the slant height 12 feet. What is the whole surface of the cone ? Here (10 x 12) -r- 2 = 120 -j- 2 = 60 square feet, the con- vex surface. And (10 2 x .07958) = 100 X .07958 = 7.958 square feet, the area of base. Whence 60 + 7.958 = 67.958 square feet, the whole sur- face required. 3. Required the whole surface of a triangular pyramid, each side of its base being 5 feet, and its slant height 171 feet. Here 5.5 X 3 = 16.5 = the perimeter of the base. And (16.5 x 17.5) -h 2 = 288.75 -=- 2 = 144.375 square feet, the outward surface of the pyramid. Also (5.5 2 x. 433013) = 30.25 x. 433013 = 13.0986 square feet, the area of 'he base. Whence 144.375 -f 13.0986 = 157.4736 square feet, the whole surface required. 4. The slant height of a right cone is 20 feet, and the dia meter of the base 8 feet ; required the convex surface. Ans. 251.328 ft. 5. The circumference of a right cone is 27.5 feet, and the slant height 11 feet; required the convex surface. rfns. 151.25 ft. 6. The slant height of a right cone is 20 feet, and the diameter 3 feet ; what is the whole surface ? Am. 101 3166 ft J44 MENSURATION OF SOLIDS. 7. Required the outward surface of a triangular pyramid, each side of its base being 85 feet, and its slant height 14 feet. tins. 73.5 ft. 8. The circumference of a right cone is 10 feet, and the perpendicular height 12 feet ; required the convex surface. Ans. 00.525 ft. PROBLEM XV. To find the surface of the frustum of a right cone or pyramid. RULE. Multiply the sum of the perimeters of the two ends by the slant height of the frustum, and half the pro- duct will be the surface required. EXAMPLES. 1. In the frustum of the cone A B D E, the circumferences of the two ends A B and ED, are 22.75 and 15.5 feet respect- ively, and the slant height, A E, is 26 feet ; required the convex surface. Here [(22.75 + 15.5) x 26] -f- 2 = (38.25 x 26) -^ 2 = 994.5 -^ 2 = 497.25 square feet, the convex surface required. 2. Required the surface of the frustum of a square pyramid, one side of the base being 12 feet, and of the upper end 5| feet, and its slant height 40* feet. Here 12 X 4 = 50 = the perimeter of the base, and 5? x 4 = 23 = the perimeter of the upper end. Then [(50 + 23) x 40.25] -f- 2 = (73 x 40.25) ~ 2 = 2938.25 -T- 2 = 1469.125 square feet, the surface required. 3. What is the convex surface of the frustum of a right cone, the circumference of the greater end being 23| feet, and that of the less end 16| feet, and the length of the slant side 12 feet ? Ans. 240 ft. 4. What is the convex surface of the frustum of a right cone, the diameters of the ends being 8 and 4 feet, and the length of the slant side 20 feet ? Ans. 376.992 ft. MENSURATION OF SOLIDS. 145 5. Required the surface of a hexagonal pyramid, one side of the base being 85 feet, and of the upper end 3^ feet, and the slant height 20 feet. Ans. 733.5 ft. 6. What is the convex surface of the frustum of a right cone, the diameters of the ends being 5 and 4 feet, and the slant height 6 feet ? tins. 84.8232 ft. PROBLEM XVI. To find the solidity of a cone or pyramid. RULE. Multiply the area of the base by the perpendicu- lar height of the cone or pyramid, and one-third of the pro- duct will be the solidity. EXAMPLES. 1. Required the solidity of a cone CAB, whose diameter, A B, is 30 feet, and its per- pendicular height, G C, 36 feet. Here (.7854 X 30 2 ) = .7854 X 900 = 706.86 = the area of the base. And (706.86 x 36) -=- 3 = 25446.96 -^ 3 = 8482.32 feet, the solidity. A 2. Required the solidity of the hex- agonal pyramid H A D H, each of the equal sides of its base being 20 feet, and the perpendicular height, H G, 50 feet. Here, by the table, page 82, for poly- gons we have 2.598076 (multiplier when the side is 1) x 20 3 = 2.598076 x 400 = 1039.2304 = the area of the base. And (1039.2304 x 50) -j- 3 = 51961.52 -*- 3 = 17320.5066 feet, the solidity. 3. What is the solidity of a cone, the diameter of whose base is 18 inches, and its altitude 24 feet ? Ans. 14.1372ft. 13 1-iO MENSURATION OF SOLIDS. 4. If the circumference of the base of a cone be 60 feet and its height 72 feet ; what is the solidity ? rfns. 6875.712 ft. 5. What is the solidity of a triangular pyramid, whose height is 30 feet, and each side of the base 3 ft. ? dns. 38.9711 ft. 6. What is the solidity of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet ? Ans. 27.5276 ft. 7. What is the solidity of a cone, whose diameter of the base is 14 feet, and the slant side being 25 feet ? Ans. 1231.5072ft. PROBLEM XVII. To find the solidity of the frustum of a cone or pyramid. 1. For the frustum of a cone, the diameters of the two ends and the height being given. RULE. Divide the difference of the cubes of the diameters of the two ends by the difference of the diameters, and this quotient being multiplied by .7854, and again by one-third of the height, will give the solidity. Or divide the difference of the cubes of the circumferences of the two ends, by the difference of the circumferences, and the quotient being multiplied by .07958, and again by one- third of the height, will give the solidity. Or to the product of the diameters add one-third of the square of their difference, and that sum being multiplied by .7854, and again by the height, will give the solidity. Or to the product of the circumferences add one-third of the square of their difference, and this sum being multiplied by .07958, and again by the height, will give the solidity. 2. For the frustum of a pyramid, the sides of the base ond the height being given. RULE. To the areas of the two ends of the frustum add the square root of their product, and this sum being multi- plied by one-third of the height, will give the solidity. MENSURATION OF SOLIDS. 147 EXAMPLES. 1. What is the solidity of the frustum of the cone A B D E, the diameter of whose greater end, A B, is 6 feet, that of the less end, E D, 4 feet, and the perpendicular height, F G, 9 feet ? Here (6 1 4 s ) -f- (6 4) = (216 64) -4-2 = 152 -4-2 = 76. And (76 x .7854) x (9 -j- 3) = 59.6904 X3 = 179 .0712 feet, the solidity required. Or [(6 x 4) + (6 - 4) a + 3] = 24 + (2'-4- 3) = (24+-*) =25i. And (25 x.7854) x 9 = 19.8968 x 9=179.0712 feet, the solidity, as hefore. 2. What is the solidity of the frustum of a cone, the cir- cumference of the greater end being 40 feet, and that of the less 20 feet, and the length or height 51 feet ? Here (40 3 20 5 ) -f- (40 20) = (64000 8000) ~ 20 = 56000 -4- 20 = 2800. And (2800 x .07958) x (51 -4- 3) = 222.824 x 17 = 3783.008 feet, the solidity required. Or [(40 x 20) + (40 20) 2 -4-3] = 800 + (20 a -^3)=800 + 133^ = 933|, and (9331 x .07958) x 51 = 74.274666 X 51 =3788.008 feet, the solidity, as before. 3. What is the solidity of the frustum aADd of an hexagonal pyramid, the side A B of whose greater end is 4 feet, that ab of the less end 3, and the height G g 9 feet .' Here 2.598076 (the tabular multiplier) x 3 2 = 2.598076 x 9 = 23.382684 the area of the less end. And 2.598076 x 4 C = 2.598076 x 16 = 41.5(39216 the area of the greater end. Whence ^(23.382684x41.569216) = v/971.999341 = 31.176912. And (23.3S2G84 + 41.569216 +31. 176912) x (9 -4- 3) = 96.128812 x 3 = 288.3864556 feet. 4. What is the solidity of the frustum of a cone, the dia- meter of the greater end being 4 feet, that of the less end 2, und the altitude 9 feet ? Jlns. 65.9736 feet 148 MENSURATION OF SOLIDS. 5. What is the solidity of the frustum of a cone, the dia- meter of the greater end of which is 5 feet, that of the less end 3 feet, and the altitude 4 feet ? Ans. 51.3128 feet. 6. What is the solidity of the frustum of a cone, the cir- cumference of the greater end being 20 feet, that of the less end 10 feet, and the length or height 21 feet ? tins. 389.942 feet. 7. What is the solidity of the frustum of a cone, the cir- cumference of the greater end being 12 feet, that of the less end 8 feet, and the height 5 feet ? Am. 40.3205 feet. 8. What is the solidity of the frustum of a square pyra- mid, one side of the greater end being 18 inches, that of the less end 15 inches, and the altitude 60 inches ? Ans. 9.4791 feet. 9. What is the solidity of the frustum of an equilateral triangular pyramid, one side of the greater end being 14 inches, that of the less end 8 inches, and the height 10 feet? Ans. 3.7287 feet. 10. What is the solidity of the frustum of a pentagonal pyramid, the side of whose greater end is 18 inches, that of the less end 12 inches, and the height 4 feet 6 inches ? Ann. 12.2583 feet PROBLEM XVm. The solidity and altitude of a cone being given, to I nd the diameter. RULE. Divide the solidity by the product of .7854, nd one-third of the altitude, and the square root of the quol snt will be the diameter. EXAMPLES. 1. The solidity of a cone is 16 feet, and the altitui 9 feet ; what is the diameter ? Here x/{16 -- [.7854 x (9 + 3)]} = v /[16-f-(.7854x-<)] feet, the diame>r. 2. The, altitude of a cone is 15 feet, and the solidity 30 feet ; what is the diameter ? Ans. 2.7639 ft. *. MENSURATION OF SOLIDS. 149 3. The solidity of a cone is 18 feet, and the altitude 9 feet; what is the diameter? Ans. 2.9316 feet. PROBLEM XDC. The solidity and diameter of a cone being given, to find the altitude. RULE. Divide the solidity hy the product of .7854 and the square of the diameter, and the quotient, being multiplied by 3, will give the altitude. EXAMPLES. 1. The solidity of a cone is 30 feet, and the diameter 2 feet ; what is the altitude ? Here [30 -4- (.7854 x 2*)] X 3 = [30 - (.7854 x 4)] x 3 = (30 -=- 3.1410) x 3 = 9.5492 x 3 = 28.6476 feet, the al- titude. 2. The diameter of a cone is 20 inches ; what must be the altitude, to make 20 solid feet ? Ans. 27.5019 ft. 3. The solidity of a cone is 2513.28 feet, and the diame- ter 20 feet ; what is the altitude 1 Ans. 24 ft. PROBLEM XX. The altitude of a cone or pyramid being given, to divide it into two or more equal parts, by sections parallel to the base, to find the perpendicular height of each part. RULE. Multiply the cube of the altitude by the numera- tor of the proportion left at the vertex, and divide the pro duct by the denominator; the cube root of the quotient will be the altitude of the cone or pyramid left at the vertex. EXAMPLES. 1 . The altitude of the cone A B C is 10 feet, to be divided into three equal parts by sections parallel to the base ; required the perpendicular height of each part. Here ^[(10x 1) -5-3] = N V[(1000xl) -5-3]= v 3/( 1000 -r- 3) = x 3/333 = 6.9336 feet, the altitude of the first section = C H. 13* 150 MENSURATION OF SOLIDS. And # [(10 3 x2)-=-8] = ^[(1000 x 2) -r-3] = ^(2000 -f. 3) = # 666f = 8.7358 feet, the altitude C I. Now C I C H == 8.7358 6.9336 = 1.8022 feet, the altitude of the second section I H. Then C K - C I = 10 - 8.7358 = 1.2642 feet, the alti- tude of the third section I K. 2. The altitude of a pyramid is 12 feet, to be divided into three equal parts by sections parallel to the base ; required the perpendicular height of each part. Ans. 8.3203, 2.1626, and 1.5171 feet, the altitudes. 3. The altitude of a cone is 20 feet, to be divided into four equal parts by sections parallel to the base ; required the perpendicular height of each part. Ann. 12.5992, 3.2748, 2.2972, and 1.8288 feet, the altitudes. PROBLEM XXI. To find the solidity of an ungula when the section passes through the opposite extremities of the ends of the frustum. RULE. From the square of the greater diameter subtract the square root of the product of the two diameters, multi- plied by the less diameter. This difference being divided by the difference of the diameters, and the quotient, multiplied by the greater diame- ter, that product by the height, and the last product by .2618 will give the solidity. EXAMPLES. 1. Required the solidity of a conical ungula, the diameter of the greater end being 5 feet, that of the less end 1 .8 feet, and the height 12 feet. Here [5 1.8 */ (5 X 1.8)] -~ (5 1.8) = [(25 1.8^9) -T- 3.2] = [25 (1.8 x 3) H- 3.2] = [(25 -5.4) -5-3.2]= (19 6-1-3.2) = 6.125. MENSURATION OF SOLIDS. 151 And (6.125 x 5 x 12 x .2618) = 96.2115 feet, the so- lidity. 2. Required the solidity of a conical ungula, the diametei of the greater end being 10 feet, that of the less end 2| feet, and the height 15 feet. tfns. 458.15 ft. 3. Required the solidity of a conical ungula the diameter of the greater end being 4.23 inches, that of the less end 3.7 inches, and the height 5.7 inches. rfns. 38.7692 in. PROBLEM XXII. To find the solidity of a cuneus or wedge. RULE. Add twice the length of the base to the length of the edge, then multiply this sum by the height of the wedge, and again by the breadth of the base, and one sixth of the last product, will be the solidity. EXAMPLES. 1. How many solid feet are there in a wedge whose base is 5 feet 4 inches long, and 9 inches broad, the length of the edge being 3 feet 6 inches, and the perpendicular height*2 feet 4 inches ? Here 5 ft. 4 in. = 64 in., 3 feet 6 in. = 42 in. and 2 ft. 4 in. = 28 in. Then (64 x2+42)x28 =(128+42) A X 28 = 170 x 28 = 4760. And (47(50 x 9) -j- 6 = 42840 -- 6 = 7140 solid inches. Whence 7140 -i- 1728 = 4.1319 solid feet. 2. The length and breadth of the base of a wedge are 35 and 15 inches, the length of the edge 55 inches, and the perpendicular height 18 inches; what is the solidity ? rfns. 3.2552 ft. 3. The length and breadth of the base of a wedge are 27 and 8 inches, the length of the edge 36 inches, and the perpendicular height 3 feet 6 inches ; what is the solidity? rfns. 2.9166 ft. 162 MENSURATION OF SOLIDS. PROBLEM XXIII. To find the solidity of a prismoid. RULE. Multiply the sum of the lengths of the two ends by the sum of the breadths, and add this product to the sum of the areas of the two ends ; multiply the result by one sixth of the height, and the product will be the solidity. EXAMPLES. 1. What is the solidity of a rectan- gular prismoid, the length and breadth of one end being 14 and 12 inches, and the corresponding sides of the other 6 and 4 inches ; and the perpendicular 30-a feet? Here (by Prob. I, p. 61) E F x E H = 6 x 4 = 24 square inches, the area of the less end, E F G H. And, by the same problem ABxAD = 14xl2 = 168 square inches, the area of the greater end, A B C D. Also 30<|ft. = 366 inches, calling the height h, [(A B + E F) x (A D + E H) + (E F G H + A B.C D)] x = [(14 + 6) x (12 + 4) + (168 -f 24)] '^. = [(20 X 16) -f 192] X 61 = (320+192) X 61 =512 x 61 = 31232 cubic inches, the solidity required. 2. What is the solidity of a rectangular prismoid, the length and breadth of one end being 12 and 8 inches, and the corresponding sides of the other 8 and 6 inches, and the perpendicular height 5 feet? fins. 2.4537 ft. 3. What is the solidity of a stick of hewn timber, whose ends are respectively 30 by 27 inches, and 24 by 18 inches, and whose length is 48 feet ? Ans. 204 ft. 4. What is the capacity of a coal wagon, whose inside dimensions are as follows : at the top the length is 7 feet, and breadth 6 feet, at the bottom the length is 5 feet, and breadth 3 feet, and the perpendicular depth is 4 feet? tins. 110ft MENSURATION OF SOLIDS. 153 PROBLEM XXIV. To find the convex surface of a sphere. RULE. Multiply the diameter of the sphere by its cir- cumference, and the product will be the convex superficies required. The curve surface of any zone or segment will also be found by multiplying its height by the whole circumference of the sphere. EXAMPLES. 1. What is the convex surface of a globe A D B C, whose diameter, A B, is 16 inches ? Here (3.1416x16) x 16 =50.2650 X 16 = 804.2496 square inches, the surface required. 2. What is the convex surface of a sphere whose diameter is 10 feet ? Ans. 314.16 ft. 3. What is the convex surface of a sphere whose diameter is 4 feet ? Ana. 50.2656 ft. 4. The diameter of a globe is 21 inches ; what is the con- vex surface of that segment of it whose height is 4% inches? Jlns. 296.8812 inches. 5. What is the convex surface of a sphere whose diame- ter is 6 feet ? Ana. 1 13.0976 ft. 6. If the diameter of the globe we inhabit be 7935 rniles; what is the convex surface ? Jlns. 197808409.26 miles. PROBLEM XXV. To find the solidity of a sphere or globe. RULE. Multiply the cube of the diameter by .5236, and the product will be the solidity. 154 MENSURATION OF SOLIDS. EXAMPLES. 1. What is the solidity of a globe whose diameter is 4.5 feet? Here (4.5 s x .5236) = 4.5 x 4.5x4.5x.5236=91.125x .5236 =47.71305 solid feet. 2. What is the solidity of a globe whose diameter is 3$ feet? Ans. 22.4493ft. 3. What is the solidity of a. globe whose diameter is 17 inches? Ans. 1.4886ft. 4. What is the solidity of a globe whose diameter is 3 feet 4 inches ? Ans. 19.3925 ft. 5. How many cubic miles are contained in the solidity of the earth, if its diameter be 7935 miles ? Ans. 261601621246.35 miles. PROBLEM XXVI. The convex surface of a globe being given, to find its dia- meter. RULE. Multiply the given area by .31831, and the square root of the product will be the diameter. EXAMPLES. 1. What is the diameter of that globe, the area of whose convex surface is 14 square feet ? Here ^(14 x .31831) = ^4.45634 = 2.1110 feet, the diameter required. 2. The convex surface of a sphere is one square rood ; what is its diameter ? Ans. 3.5682 rods. 3. The expense of gilding a ball at $1.80 per square foot is thirty-four dollars ; what is its diameter ? Ans. 2.452ft. MENSURATION OF SOLIDS. 155 PROBLEM XXVII. The solidity of a globe being given, to find the diameter. RULE. Divide the solidity by .5236, and extract the cube toot of the quotient. EXAMPLES. 1. The solidity of a globe is 2000 solid inches ; what is its diameter ? Here & (2000-=-. 5236) == ^/3819.7097 = 15.631 inches the diameter. 2. The solidity of a globe is 10 solid feet ; what is its dia- meter 1 j3ns. 2.67 ft. 3. What is the circumference of a globe whose solidity is 8 solid feet ? rfns. 7.7952 ft. ' PROBLEM XXVIII. To find the solidity of the segment of a sphere. RULE. To three times the square of the radius of its base add the square of its height ; and this sum multiplied by the height, and the product again by .5236, will give the solidity. Or, from three times the diameter of the sphere subtract twice the height of the segment, multiply by the square of the height, and that product by .5230 ; the last product will be the solidity. EXAMPLES. 1. The radius An of the base of the segment CAB is 7 inches, and the height Cn 4 inches ; what is the soli- dity? Here [(7 a x 3) + 4 2 ] x4= [(49 x 3) + 16] x 4 = (147 + 16) x 4 = 163 X 4 == 652. Then 652 x .5236 = 341.3872 solid inches 156 MENSURATION OF SOLIDS. 2. The diameter of a sphere is 6 inches. What is the solidity of the segment whose height is 2 inches ? Here Q6 x 3) - (2 x 2)] x 2 9 = (18 4) x 4 = 14 x 4 = 56. Then 56 x .5236 = 29.3216 solid inches. 3. What is the solidity of a spherical segment, the dia- meter of its base being 40 inches, and the height 10 inches I 3.9391 feet. 4. The diameter of a sphere is 18 inches ; what is the soli- dity of a segment cut from it, the height being 3 inches ? Ans. 226. 1952 inches. 5. The diameter of a spherical segment is 20 inches, and the height 6 inches ; how many gallons of water will it hold, each gallon containing 282 cubic inches ? Ans. 3.743 gallons. PROBLEM XXIX. To find the solidity of a frustum or zone of a sphere. RULE. To the sum of the squares of the radii of the two ends, add one-third of the square of their distance, or of the breadth of the zone, and this sum multiplied by the said breadth, and the product again by 1.5708, will give the so- lidity. EXAMPLES. 1. What is the solidity of the zone A BC D, whose greater diameter, A B, is 1 foot 8 inches, the less diameter, D C, 1 foot 3 inches, and the distance nm of the two ends 10 inches ? Here [(Am a + Dn 9 ) + $ (nm) 8 ] X nm x 1.5708 = [(10 3 + 7 9 ) + (10 8 -^-3)]xlO x 1.5708=(100 + 561+331) x 15.7(58=189 T \ X 15.708 = 2977.975 cubic inches the solidity of the zone required. 2. What is the solidity of a zone whose greater diameter is 9 feet 3 inches, less diameter 6 feet 9 inches, and height 5 feet 6 inches. Ans. 370.3242 feet. 3 What is the solidity of a zone, whose greater diameter MENSURATION OF SOLIDS. 157 is 2 feet, the less diameter 1 foot 8 inches, and the distance of the ends 4 inches ? rfns. 1560.0112 inches. 4. Required the solidity of the middle zone of a sphere, whose top and bottom diameters are each 3 feet, and the breadth of the zone 4 feet? Ans. 61.7848 feet. PROBLEM XXX. To find the solidity of a circular spindle, its length and middle diameter being given. RULE. To the square of half the length of the spindle, or longest diameter, add the square of half the middle diameter, and this sum divided by the middle diameter will give the radius of the circle. 2. Take half the middle diameter from the radius thus found, and it will give the central distance ; or that part of the radius that lies between the centre of the circle, and that of the spindle. 3. Find the area of the generating circular segments, by Problem XL, rule 5, page 104. 4. From one-third of the cube of half the length of the spindle, subtract the product of the central distance, and half the area last mentioned, and the remainder multiplied by 12.5064, will give the solidity of the spindle. EXAMPLES. I. The longest diameter, AB, of the circular spindle A DBG, is 48 inches, and the middle dia- meter, C D, 36 inches ; what is the solidity of the spindle ? F' _ j ! G First (Ae 2 +Ce 2 ) -=- CD =(24 a -f IS 2 ) ~S6 = (576 + 324) -=- 36 = 900 -=- 36 = 25, the radius OC. Second, OC Ce = 25 18 = 7, O e, the central dis- tance. Third, (by Problem XL, rule 5, page 104,) Ce-r-2OC -= 18 -i- 50 = .36, the tabular versed sine, against which .stands .254550 the tabular segment. 14 MENSURATION OF SOLIDS Here .254550 x 50 a =. 254550 x 2500 = 636.375, the area of the segment A B C A. Fourth, AAe 3 - (i A EGA x Oe)x 12.5664= [(24 3 -i-3) - (636.375-r-2)x7]x 12.5664 = [(13824-5-3) (318.1875 X7)] x 12.5664= (4608-2227.3125) x 12.5664=2380.6875 X 12.5664 = 29916.6714, solidity of the spindle. 2. If the length of a circular spindle be 40 inches, and its middle diameter 30 inches, what is its solidity ? tins. 17312.8886 inches. PROBLEM XXXI. To find the solidity of the middle frustum of a circular spindle, its length, the middle diameter, and that of either of the ends, being given. RULE 1. Divide the square of half the length of the frus- tum by half the difference of the middle diameter, and that of either of the two ends ; and half this quotient, added to one- half of the said difference, will give the radius of the circle. 2. Find the central distance, and the revolving area, as in the last problem. 3. From the square of the radius take the square of the central distance, and the square root of the remainder will give half the length of the spindle. 4. From the square of half the length of the spindle take one-third of the square of half the length of the frustum, and multiply the remainder by the said half length. 5. From this product take that of the generating area and central distance, and the remainder multiplied by 6.2832 will give the solidity of the frustum. EXAMPLES. 1. What is the solidity of the middle frustum, A B C D, of a circular spindle, whose middle diameter, nm, is 36 inches, the diameter D A or C B, of the end i6 inches, and its length or 40 inches ? MENSURATION OF SOLIDS. 159 First, [Oe 9 ~ (we Do)] + | (ne Do) = % [20* -=-(18 8)]+ (18 8) = i (400 -H 10) + (10 -T- j =20+5 = 25, the radius of the circle. Second, 25 ne = 25 18 = 7, the central distance. And ne Do = 18 8 = 10, the versed sine of the arc On. Then, by Problem XL, rule 5, page 104, 10 -4- (25 X 2) = 10 -T- 50 = .2, the tabular versed sine; against which stands .111823 the tabular segment. Hence .11 1828 x 50 a =.l 118*23 X 2500 = 279.5575, the area of the revolving segment DCwD. Again, by Problem 1, page 01, or x Do = 40 x 8 = 320, the area of the rectangle DorCD. And DCD -f D or CD = 279.5575 + 320 = 599.5575, the generating area OrCnDo. Third, ,/(26 7 s ) = >/(625 49) = ^7576 = 24 = Ee, half the length of the spindle. Fourth, {[(Ee 8 | Oe 2 ) x Oe] (O/nDox7)}X 6.2832 = {[(24 2 133 j) x 20] (599.5575 X 7)} X 6.2832 = [(442f x 20) 4196.9025] x 6.2832 = (8853.3333^ 4196.9025) x 6.2832 = 4656.4308J x6.2832 = 29257.2862 cubic inches, the solidity of the middle frustum AmBCnDA required. 2. The middle diameter of the frustum of a circular spin- dle is 2 feet 8 inches, the diameter at the end is 2 feet, and the length 3 feet 4 inches ; what is the solidity ? Ans. 27285.0882 inches. PROBLEM XXXII. To find the solidity of a spheroid, its two axes being given. RULE. Multiply the square of the revolving axis by the fixed axis, and this product again by .523(5, or one-sixth of 3.141(5, and it will give the solidity required. EXAMPLES. 1. In the prolate spheroid A BCD, the transverse or fixed axis, AC, is 3 ft, and the conjugate or revolving axis, DB, is" 2 feet ; what is the solidity ? Here (2- x 3) x .523(5 = (4 X 3) X.5-.J30 = 12 X .523G = :i.2b32 feet, the solidity required. 1GO MENSURATION OF SOLIDS. 3. What is the solidity of a prolate spheroid whose trans- verse or fixed axis is 4 feet 2 inches, and conjugate or re- volving axis 3 feet 4 inches? rfns. 24.2407 ft. 3. What is the solidity of a prolate spheroid whose fixed axis is 8 feet 4 inches, and its revolving axis 5 feet ? Am. 109.0833ft. 4. What is the solidity of an oblate spheroid whose con- jugate or fixed axis is 5 feet, and its transverse or revolving axis 8 feet 4 inches ? Ans. 181.8055 ft. 5. What is the solidity of an oblate spheroid whose fixed axis is 30 inches, and its revolving axis 40 inches ? Ans. 14.5444ft. PROBLEM XXXIII. To find the solidity of the middle frustum of a spheroid, its length, the middle diameter, and that of either of the ends, being given. CASE 1. When the ends are circular, or perpendicular to the fixed axis. RULE. To twice the square of the middle diameter add the square of the diameter of either of the ends, arid this sum multiplied by the length of the frustum, and the product again by .2618 (or one-twelfth of 3.141(5), will give th solidity. EXAMPLES. 1. In the middle frustum of a prolate spheroid E F G H, the mid- dle diameter, B D, is 50 inches, and / that of either of the ends E F or G H, is 40 inches, and its length, nm, 18 inches ; what is its solidity? Here [(50 2 x2 + 40-)xl8] X .2618 = [(2500x2+1600) Xl8]x.2618= [(5000+ 1600) x 18] x. 2618= (6600 x 18) X .2618 = 11&800 x .2618 = 31101.84 cubic inches, the solidity required. 2. What is the solidity of the middle frustum of a prolate spheroid, the middle diameter being 5 feet, that of either of the two ends 3 feet, and the distance of the ends 6 feet 8 inches Ans. 102.9746 feet. MENSURATION OF SOLIDS. 161 3. What is the solidity of the middle frustum of an oblate spheroid, the middle diameter being 100 inches, that of either of the ends 80 inches, and the distance of the ends 36 inches ? Ans. 248814.72 inches. 4. What is the solidity of the middle frustum of a prolate spheroid, the middle diameter being 5 feet, that of either oi the ends 3 feet, and the distance of the ends G feet ? Ans. 92.6772 ft. CASE 2. When the ends are elliptical or perpendicular to the revolving axis. RULE 1. Multiply twice the transverse diameter of the middle section by its conjugate diameter, and to this product add the product of the transverse and conjugate diameters of either of the ends. 2. Multiply the sum thus found, by the distance of the ends or the height of the frustum, and the product again by .2018, and it will give the solidity required. EXAMPLES. 1. In the middle frustum, A BC D, of a prolate spheroid, the diameters of the middle section are 50 and 30 inches; those of the end 40 and 24 inches ; and its height, 2 on, 18 inches ; what is the solidity ? Here {[(50x2 x 30) + (40x24)]x 18} x .2618=[(3000 + 9(50) x 18] x .2618 = (3060 x 18) x .2618 = 71280 X .2618 = 18661.104 cubic inches, the solidity required. 2. In the middle frustum of a prolate spheroid, the diame- ters of the middle section are 100 and 60 inches ; those of the end 80 and 48 inches ; and the length 36 inches ; what is the solidity ? tfns. 86.3940 ft. 3. In the middle frustum of an oblate spheroid, the diameters of the middle section are 100 and 60 inches ; those of the end 60 and 36 inches ; and the length 80 inches ; what is the solidity of the frustum ? 171.6244ft. 14' A TABLE OF THE AREAS OF THE SEGMENTS OF A CIRCLE, Whose diameter is Unity, and supposed to be divided into 1000 equal Parts. Versed Sine. Seg. Area. Versed Sine. Seg. Area. Versed Sine. Seg. Area. .001 .000042 .034 .008273 .007 .022652 .002 .000119 .035 .008638 .008 .023154 .003- .000219 .036 .009008 .069 .023659 .004 .000337 .037 .009383 .070 .024168 .005 .000470 .038 .009763 .071 .024080 .006 .000618 .039 .010148 .072 .025195 .007 .000779 .040 .010537 .073 .025714 .008 .000951 .041 .010931 .074 .020236 .009 .001135 .042 .011330 .075 .026761 .010 .001329 .043 .011734 .076 .027289 .011 .001533 .044 .012142 .077 .027821 .012 .001740 .045 .012554 .078 .028356 .013 .001968 .046 .012971 .079 .028894 .014 .002199 .047 .013392 .080 .029435 .015 .002438 .048 .013818 .081 .029979 .010 .002685 .049 .014247 .082 .030526 .017 .002940 .050 .014681 .083 .031076 .018 .003202 .051 .015119 .084 .031629 .019 .003471 .052 .015561 .085 .032186 .020 .003748 .053 .016007 .086 .032745 .021 .004031 .051 .016457 .087 .033307 .022 .004322 .055 .016911 .088 .033872 .023 .004618 .056 .017369 .089 .034441 .024 .004921 .057 .017831 .090 .035011 .025 .005230 .058 .018296 .091 .035585 .026 .005546 .059 .018766 .092 .036162 .027 .005867 .060 .019239 .093 .036741 .028 .006194 .061 .019716 .094 .037323 .029 .006527 .062 .020190 .095 .037909 .030 .006865 .063 .020680 .096 .038490 .0531 .007209 .064 .021108 .097 .039087 1 .032 .007558 .065 .021659 .098 .039680 .033 .007913 .006 .022154 .099 .040276 205 96 THE AREAS OF THE SEGMENTS OF A CIRCLE. Versed Sine. Seg. Area. Versed Sine. Seg. Area. 1 Versed Sine. Seg. Aifa. .100 .040875 .141 .067528 .182 .097674 .101 .041476 .142 .068225 .183 .098447 .102 .042080 .143 .068924 .184 .099221 .10* .042687 .144 .069625 .185 .099997 .104 .043296 .145 .070328 .186 .100774 .105 .043908 .146 .071033 .187 .101553 .106 .044522 .147 .071741 .188 .102334 .107 .045139 .148 .072450 .189 .103116 .108 .045759 .149 .073161 .190 .103900 1 .109 .046381 .150 .073874 .191 .10^685 .110 .047005 .151 .074589 .192 .105472 .111 .047632 .152 .075306 .193 .106261 .112 .048262 .153 .076026 .194 .107051 .113 .048894 .154 .076747 .195 .107842 .114 .049528 .155 .077469 .196 .108636 .115 .050165 .156 .078194 .197 .109430 .116 .050804 .157 .078921 .198 .110226 .117 .051446 .158 .079649 .199 .111024 .118 .052090 .159 .080380 I .200 .111823 .119 .052736 .160 .081112 .201 .112624 ; .120 .053385 .161 .081846 .202 .113426 .121 .054036 .162 .082582 .203 .1142530 .122 .054689 .163 .083320 .204 .115035 .123 .055345 .164 .084059 .205 .115842 .124 .056003 .165 .084801 .206 .116650 .125 .056663 .166 .085544 .207 .117460 .126 .057326 .167 .086289 .208 .118271 .127 .057991 .168 .087036 .209 .119083 .128 .058658 .169 .087785 .210 .119897 .129 .059327 .170 .088535 .211 .120712 .130 .059999 .171 .089287 .212 .121529 .131 .060872 .172 .090041 .213 .122347 .132 .061348 .173 .090797 .214 .123167 .133 .06202o .174 .091554 .215 .123988 .134 .062707 .175 .092313 .216 .124810 .135 .063389 .176 .09:1074 .217 .125634 .136 .004074 .177 .093836 .218 .126459 .137 .064760 .178 .094601 .219 .127285 .138 .005449 .179 .095366 .220 .128113 .139 .066140 .180 .096134 .221 .128942 .140 .0668*3 .181 .096903 .222 .129773 THE AREAS OF THE SEGMENTS OF A CIRCLE. 297 f Versed Sine. Seg. Area. Versed Sine. Seg. Area. Versed Sine. Seg. Area. .223 .130005 .264 .165780 .305 .202761 .224 .131438 .265 .166663 .306 .203683 .225 .132272 .266 .167546 .307 .204605 .226 .133108 .267 .168430 .308 .205527 .227 .133945 .268 .169315 .309 .206451 .228 .134784 .269 .170202 .310 .207376 .229 .135624 .270 .171089 .311 .208301 .230 .136465 .271 .171978 .312 .209227 .231 .137307 .272 .172867 .313 .210154 .232 .138150 .273 .173758 .314 .211082 .233 .138995 .274 .174649 .315 .212011 .234 .139841 .275 .175542 .316 .212940' .235 .140688 .276 .176435 .317 .213871 .236 .141537 .277 .177330 .318 .214802 .237 .142387 .278 .178225 .319 .215733 .238 143238 .279 .179122 .320 .216666 .239 .144091 .280 .180019 .321 .217599 .240 .144944 .281 .180918 .322 .218533 .241 .145799 .282 .181817 .323 .219468 .242 .146655 .283 .182718 .324 .220404 .243 .147512 .284 .183619 .325 .221340 .244 .148371 .285 .184521 .326 .222277 .245 .149230 .286 .185425 .327 .223215 .246 .1500!)! .287 .186329 .328 .224154 .247 .150953 .288 .187234 .329 .225093 .248 .151816 .289 .188140 .330 .226033 .249 .152680 .290 .189047 .331 .226974 .250 .153546 .291 .189955 .332 .227915 .251 .154412 .292 .190864 .333 .228858 .252 .155280 .293 .191775 .334 .229801 .253 .156149 .294 .192684 .335 .230745 .254 .157019 .295 .193596 .336 .231689 .255 .157890 .296 .194509 .337 .232634 .256 .158762 .297 .195422 .338 .233580 .257 .159636 .298 .196337 .339 .234526 .258 .160510 .299 .197252 .340 .235473 .259 .161386 .300 .198168 .341 .236421 .260 .162263 .301 .199085 .342 .237369 .261 .103140 .302 .200003 .343 .238318 .2(52 .164019 .303 .200922 .344 .239268 .263 .164899 .304 .201841 .345 .2402 is 298 THE AREAS OF THE SEGMENTS OF A CIRCLE. Versed Sine. Seg. Area. Versed Sine. Seg. Area. Versed Sine. Seg. Area. .346 .241169 .387 .280668 .428 .320948 .347 .242121 .388 .281642 .429 .321938 .348 .243074 .389 .282617 .430 .322928 .349 .244026 .390 .283592 .431 .323918 .350 .244980 .391 .284568 .432 .324909 .351 .245934 .392 .285544 .433 .325900 .352 .246889 .393 .286521 .434 .326892 .353 .247845 394 .287498 .435 .327882 .354 .248801 .395 .288476 .436 .328874 .355 .249757 .396 .289453 .437 .329866 .356 .250715 .397 .290432 .438 .330858 .357 .251673 .398 .291411 .439 .331850 .358 .252631 .399 .292390 .440 .332843 .359 .253590 400 .293369 .441 .333836 .360 .254550 401 .294349 .442 .334829 .361 .255510 .402 .295330 .443 .335822 .362 .256471 403 .296311 .444 .336816 .363 .257433 404 .297292 .445 .337810 .364 .258395 405 .298273 .446 .338804 .365 .259357 .406 .299255 .447 .339798 .366 .260320 .407 .300238 .448 .340703 .367 .261284 408 .301220 .449 .341787 .368 .262248 409 .302203 .450 .342782 .369 .263213 410 .303187 .451 .343777 .370 .264178 .411 .304171 .452 .344772 .371 .265144 .412 .305155 .453 .345768 .372 .266111 .413 .306140 .454 .346764 .373 .267078 .414 .307125 .455 .347759 .374 .268045 .415 .308110 .456 .348755 .375 .269013 .416 .309095 .457 .349752 .376 .269982 .417 .310081 .458 .350748 .377 .270951 .418 .311068 .459 .351745 .378 .271920 .419 .312054 .460 .352741 .379 .272890 .420 .313041 .461 .353739 .380 .273861 .421 .314029 .462 .354736 .381 .274832 .422 .315016 .463 .355732 382 .275803 .423 .316004 .464 .356730 .383 .276775 .424 .316992 .465 .357727 .384 .277748 .425 .317981 .466 .358725 .385 .278721 .426 .318970 .467 .359723 ' .386 .279694 .427 .319959 .468 .360721 THE AR-EAS OF THE SEGMENTS OF A CIRCLE. 299 Versed Baa. Seg. Area. Versed Sine. Seg. Area. Versed Sine. Seg. 4 r ea. .469 .361719 .480 .372764 .491 .383699 .470 .362717 .481 .373703 .492 384699 .471 .363715 .482 .374702 .493 .385699 .472 .364713 .483 .375702 .494 .386699 .473 .365712 .484 .376702 .495 .387699 .474 .366710 .485 .377701 .496 .388699 .475 .367709 .486 .378701 .497 .389699 .476 .368708 .487 .379700 .498 .390699 .477 .369707 .488 .380700 .499 .391699 .478 .370706 .489 .381699 .500 .39269U .479 .371705 .490 .382699 THE END. UC SOUTHERN REGIONAL LIBRARY FACILITY A 000 655 659 1 ^ BOOKS FOE COMMON SCHOOLS, ACADEMIES, & COLLEGES, Published by E. C. & J. BIDDLE. ALSOP. First Lossons in. Algebra. KEY to ditto- ALSOP. A a. For High Schools, Colleges, Ac. KEY ALSOP. A 'i BERQJTIN. Th n : in French. CLEVELAND. A .Con< tcrature, chronologically ar- i, from the se of the 18th century. CLEVELAND. English Literature of the Nineteenth Century: bekv quel to lift "Compendium of English Literature." CLEVELAND. A Compondiuii' MI Literature. 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The Cbss Book of Etymology. 2S[el!URTBIE.--A Dictionary of Terms used in the Various branches of Science^: designed for the uso of schools and private students. MAP OF THE WORLD, as known to the Ancients : adapted to the Treatise on Geography in the "Manual of Claasical Literature-." Printed in ;ed,.Wue, and' brown and mounted on rollers. Size 61 "by 50 inches. MAl/RY. A Theoretical and Practical Treatise oil Navigation. (The text- book of the United States Navy.) BLONGE. An Elementary treatise on S^|MS, by Gaepard Monga. Trans- rom the French by Woods Bak^A.M. OaWALD. An Ely; ry of the English Language. OUTLINES OF SACRED B3STORY. From the Creation of the World to the | ,m. PEALE. i duced to its most simple principle*, intro- ihmetio. KHT to ditto. THOMA n the basis f "The First Book of .ynd. TREGO VOGDES & ALSOP'.-- s Arithmetic. - ANSWERS to ditto, to ditto, i. VOGDES & ALSOP. The Fir, ites Arithmetic. VOGDES. An Elementary "'> ionsuration and Practical Geometry. KKY to ditto.