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MENSURATION 
 
 WITH SPECIAL APPLICATION OF THE 
 PRISMOIDAL FORiMULA 
 
 Designed for Supplementary Work in 
 
 Normal Schools, High Schools, Academies, and Advanced 
 
 Classes in Grammar and Common Schools 
 
 BY 
 
 S. W. FURST 
 
 PHILADELPHIA 
 
 CHRISTOPHER SOWER COMPANY 
 1911 
 
MENSURATION 
 
 WITH SPECIAL APPLICATION OF THE 
 PEIS3I0IDAL F0E3IULA 
 
 Designed for Supplementary Work in 
 
 Classes in Grammar and Common Schools 
 
 BY 
 
 S. W. FURST 
 
 phtladelphta 
 
 CHRISTOPHER SOWER COMPANY 
 
 1911 
 
F 
 
 Copyright, 1898, 1900 
 By S. W. FUPuST 
 
 
PREFACE 
 
 A NEW text-book at the close of the luiieteeuth cen- 
 tury is no novelty'. The author, therefore, without apol- 
 ogy sends forth this little volume, hopin<^ that his co- 
 workers nia}' find at least some of the pleasure and 
 profit in using it that he has found in its preparation 
 and practical application in the school -room. 
 
 The method of treating Mensuration, as here presented, 
 has had a thorough class-room test — a safe criterion for 
 any method. Both those pupils with a knowledge of 
 Geometry and those without such knowledge have been 
 able to solve problems with facility. Without this test 
 with pupils of various ages and grades, the author would 
 hesitate long before advising the setting aside of some 
 of the standard rules used in text -books for years. 
 Pupils who have become familiar wdth this method have 
 invariably become enthused witli the work. They enter 
 examinations in this department confident of success, 
 because they have become masters of one broad prin- 
 ciple covering a great field of measurement. What was 
 vague becomes clear; the concrete supplants the abstract. 
 Concentration and simplicity are everywhere the central 
 ideas. 
 
 A few words concerning the scope of this principle 
 may not be amiss. As the title-page indicates, the 
 special feature of this w^ork is the application of the 
 prismoidal formula to finding the volume of solids. This 
 is used by civil engineers in making calculations for 
 railway embankments, excavations, etc.; but, so far as 
 we are aware, no common s<'bool text -book treats the 
 
 260bb8 
 
vi PREFACE 
 
 subject at leugth. It applies to many other solids than 
 those which fall strictly within the definition of a pris- 
 nioid. Ellwood Mivrris, a civil engineer, made known 
 this extended application about 1840: 
 
 "It applies to all solids having two parallel sides or 
 faces, where these two faces, curved or plane, are united 
 by surfaces upon which and through every point of which 
 a straight line may be made to connect the two parallel 
 faces; to prisms, wedges, pyramids, cones, etc., regular 
 or irregular, right or oblique; to the sphere, hemisphere, 
 and other spherical segments; to either section of a 
 cylinder, where the cutting plane is passed through bott 
 parallel sides ; to any section of a cone, where the cut- 
 ting plane is passed through both apex and base ; to 
 frustums of prisms, pyramids, cones, etc." Volumes of 
 frustums of pj-ramids and cones can -be found by its 
 use without the extraction of a root. It is thus seen 
 that its scope is very great. For the derivation of 
 this formula from the principles of Geometry, see 
 Supplement. 
 
 Care has been taken to use only such rules and defi- 
 nitions as have been sanctioned by the best authorities. 
 In order that comparisons may be readily made without 
 reference to other books, and that the teacher may con- 
 veniently make selections of those which best suit his 
 purpose, a number of these are given under each 
 chapter. 
 
 The work abounds in interesting problems, many of 
 which have been suggested by objects familiar to every 
 boy and girl. Actual measurements were made for many 
 of them. Both the grading of the problems and their 
 practical utility have been kept in mind. It is believed 
 that enough neiv ones have been introduced to awaken 
 interest in the dullest boy. 
 
nuci'AcE vii 
 
 It is iuteiult'd that this work be riupplemeutal to both 
 text -books on Arithmetic and Geometry. Many text- 
 books on Geometry are so abstract in treatment as to 
 leave the pupil, even after completing his course, unable 
 to apply what he has learned. Few persons can easily 
 acquire the power of abstraction necessar}^ for correct 
 geometrical conceptions. Making the subject concrete, 
 therefore, will be of great benefit to the majority of 
 pupils. As soon as the pupil discovers the practical 
 value of the ..propositions he demonstrates, he will be 
 incited to study with new interest and zeal. 
 
 Fundamentals of Measurement, the frontispiece, can be 
 made of much use to both teacher and pupil. It will be 
 noticed that the fundamental quantities of measurement — 
 points, lines, surfaces, and volumes — are so grouped 
 as to picture with great vividness the relations of these 
 fundamentals. Each block is independent of the others, 
 and yet it requires all the blocks — and in the order 
 given — to show systematically the essentials of measure- 
 ment. By starting at the top of the monument and 
 noticing in order the point, lines, plane surfaces, and 
 solids, the synthetic process of reasoning becomes ap- 
 parent. The simplicity of the arrangement is surpassed 
 only by the comprehensiveness of the whole. By it the 
 beginner can learn the elements, and the advanced 
 student, at a glance, verify his conclusions. 
 
 The author tenders his thanks to Charles Lose, A.M., 
 Superintendent of Schools of the city of Williamsport, 
 Pa., the Hon. Emerson Collins, of tlie same place, and 
 others, who have kindly given him valuable suggestions 
 and criticisms during the progress of the work. He 
 asks from his fellow teachers not only a hearty welcome, 
 but criticisms as well. 
 
 S. W. FURST. 
 
SUGGESTIONS TO TEACHERS 
 
 It is deemed best to show the application of the 
 formula to those solids for which the rules usually given 
 are short and simple, as well as to those for which the 
 rules given are more complex. This is done by means 
 of a chapter of illustrative problems, w^hich can be used 
 •for reference for all following chapters on solids. It is 
 believed that a plan for the solution of every problem in 
 the book can be found by a study of this chapter. After 
 the pupil thoroughly understands how to apply the for- 
 mula to the solution of problems involving all the com- 
 mon solids, it is well to have him use in practice the 
 shorter rules, especially for finding the volume of the 
 prism, cylinder, etc. 
 
 The pupil should be thoroughly drilled on finding the 
 area of the middle section. This can best be done by 
 liaving him draw figures of different shapes and repre- 
 senting this section by dotted lines. In all cases he 
 should write on each line the dimension given in the 
 solid. Until he is able to do this preliminary work, it 
 can not be expected that he will make rapid progress. 
 The plan is made clear in the chapter of illustrative 
 problems. 
 
 Constant reference should be made to the frontispiece. 
 Many points often vague to the pupil can be made quite 
 clear by a proper use of this. It is suggested that the 
 pupil be taught to draw and letter this from memory, 
 a block at a time, and fiuallv the whole monument. 
 
 (viii) 
 
CONTENTS 
 
 PAGE 
 
 Preface v 
 
 Suggestions to Teachers viii 
 
 Introduction 1 
 
 Mensuration Defined 2 
 
 Lines 3 
 
 Angles 3 
 
 Plane Figures Defined 4 
 
 Mensuration of Surfaces — 
 
 The Triangle 5 
 
 The Right-Angled Triangle 8 
 
 Quadrilaterals 11 
 
 The Circle 14 
 
 The Ellipse 18 
 
 Similar Plane Figures 19 
 
 Mensuration of Solids — 
 
 Prisms and Cylinders 22, 24 
 
 Surfaces of Prisms and Cylinders 25 
 
 Illustrative Problems Solved by the Prismoidal Formula . 27 
 
 Volumes of Prisms and Cylinders 37 
 
 Pyramids and Cones 39 
 
 Surfaces of Pyramids and Cones and Their Frustums . . 42 
 
 Volumes of Pyramids and Cones 44 
 
 Volumes of Frustums of Pyramids and Cones 46 
 
 The Sphere 48 
 
 The Spheroid 52 
 
 Circular Rings 55 
 
 The Wedge 56 
 
 Similar Solids 58 
 
 Miscellaneous Problems 60 
 
 Supplement 66 
 
 (ix) 
 
MENSURATION" 
 
 Mensuration was formerly a part of the course of study 
 of everv well-ordered school of higher grade. The addition 
 of a variety of subjects to the course of study gradually 
 crowded it out, so that it often became merely an insignifi- 
 cant part of the arithmetic work. 
 
 Now a reaction has set in. Tlie demand for thorough- 
 ness instead of variety is growing, and the great increase in 
 ^^ Industrial Education " has awakened an interest in men- 
 suration as a practical subject and has called for increased 
 attention to it. 
 
 A knowledge of mensuration is valuable to : 
 
 The housekeeper who buys wall-paper, curtains, carpets, 
 etc. 
 
 The farmer who measures his fields and bins to count his 
 crop values. 
 
 The carpenter who builds houses, barns, or shops. 
 
 The mason who builds walls and paves sidewalks. 
 
 The plumber who calculates pipe extension, water flow, 
 and drainage. 
 
 The paiuter who paints buildings and fences and deco- 
 rates interiors. 
 
 The contractor who paves streets and builds roads. 
 
 The coo})er who makes kegs, casks, and barrels. 
 
 The landscape gardener who lays out grounds and flower 
 beds. 
 
 The average man who employs artisans. 
 
MENSURATION 
 
 INTRODUCTION 
 
 1. A Brick (Fig. 1) is a material bodj", since it 
 occupies bounded space. The bound- 
 ary of the bricK: is its surface. Brick 
 We can conceive the brick to be 
 removed, and there would remain 
 space formerly occupied by the brick; ng.i 
 this is a geometrical solid. The sur- 
 face bears the same relation as in the material body. 
 
 2. A Plane is a surface such that a straight line con- 
 necting any two of its points 
 
 lies wholly in its surface. 
 Thus, MN (Fig. 2) is a plane, 
 since all points of any straight 
 line, as RS , lie wholly in its 
 surface. Each face of the 
 brick (Fig. 1) is a plane. 
 
 Fig. 2 
 
 3. When two planes intersect each other, their com- 
 mon intersection is a straight line. 
 Each edge of the brick (Fig. 1) is a 
 straight line, since in each case it is 
 the intersection of two planes. RS 
 (Fig. 3) is a straight line, it being the 
 common^ intersection of the two planes 
 AB and CD. 
 
 A CI) 
 
2 ML'ysrh'JTiox 
 
 4. A Straight Line is one which has the same direc- 
 tion at every point. AB (Fig. 4) 
 
 is a straight line. When two "^ Ji^ ^ 
 
 straight lines intersect, their com- 
 mon intersection is a point. In 
 Fig. 5 R is a point, it being the 
 
 intersection of the two straight ^ — -7 \ 
 
 lines AB and CD. ^ ^'^' ^ 
 
 5. A Point has no extension, but simply position. 
 A Line has only one dimension, — length. 
 
 A Surface has two dimensions, — length and breadth. 
 
 A Solid, or volume, has three dimensions, — lengthy 
 'breadth,, and thickness. 
 
 The brick (Fig. 1) is a solid. 
 
 Any face of the brick is a surface. 
 
 Any edge of the brick is a line. 
 
 Any corner of the brick is a point. 
 
 A line is limited by a point ; a surface is limited by 
 a line ; a volume is limited by a surface ; and yet we 
 can conceive of each of these as being independent of 
 the other. 
 
 Points, lines, surfaces, and solids of geometry are 
 purely imaginary. 
 
 The fundamental quantities of measurement are 
 points, lines, surfaces, and volumes. 
 
 6. Mensuration is that branch of applied geometry 
 which gives the rules and processes for finding the 
 lengtlis of lines, the areas of surfaces, or the volumes 
 of solids. 
 
DEFINITIONS 
 
 LINES 
 
 7. A Straight Line is one which has the same direc- 
 tion at all points : as MN. It is the 
 
 , M N 
 
 shortest distance between two points. 
 
 8. A Curved Line is one which 
 changes its direction at every point : 
 as RS. 
 
 9. A Broken Line is one which 
 is composed of different successive 
 straight lines : as ABODE. 
 
 10. Parallel Lines are those in the same plane which 
 have the same direction : as MN and 
 
 ST. They can never meet, however p ^ 
 
 far produced. 
 
 11. A Perpendicular Line is a 
 
 straight line joined to another straight 
 line, so as to incline equally to both 
 sides. AB is perpendicular to CD. 
 
 ANGLES 
 
 12. An Angle is the difference in direction between 
 two lines which meet at a common point, ^a 
 
 called the vertex. ABC is an angle whose 
 vertex is the point B, and w^iose sides ^' 
 are AB and BC. 
 
 (3) 
 
MKXsriiATK/X 
 
 13. A Rig*ht Ang-le is an angle 
 formed by two lines which are perpen- 
 dicular to each other. ABC is a right 
 angle. 
 
 14. An Acute Ang-le is one 
 
 which is less than a right angle. 
 MNO is an acute angle. 
 
 15. An Obtuse Ang'le is one which is greater than a 
 right angle. ABC is an obtuse angle. a 
 
 Acute and obtuse angles are N. 
 
 oblique angles. b ^ 
 
 PLANE FIGURES 
 
 16. A Plane is a surface such that a straight line 
 connecting any two of its points 
 lies wholly in its surface. MN / vR 
 is a plane, since all points of /R- 
 the straight line RS lie wholly 
 in its surface. 
 
 17. A Plane Fig'ure is a portion of a plane surface 
 bounded by straight or curved lines. 
 
 18. A Polygfon is a plane figure bounded by straight 
 lines; the lines are called the sides of the polygon. 
 
 19. The Perimeter of a polygon is the sum of its 
 sides. 
 
 20. The Area of a plane figure is the surface in- 
 cluded within the perimeter. 
 
 21. The Diag'Onal of a polygon is a line joining the 
 vertices of any two of its angles not adjacent. 
 
 22. The Altitude of a polygon is the perpendicular 
 distance between its base and a side or angle opposite. 
 
TRIAMiLES 5 
 
 23. An Equilateral Polygron has all its sides equal. 
 An Equiang-ular Polyg-on has all its angles equal. 
 
 A Reg^ular Polyg-on is both equilateral and equi- 
 angular. 
 
 24. Polygons are named according to the number of 
 sides. A polygon of three sides is a triangle ; of four 
 
 Triangle 
 
 Quadrilateral 
 
 Pentagon 
 
 Hexagou 
 
 Heptagou 
 
 Octagon 
 
 Isonagon 
 
 Decagon 
 
 sides, a quadrilateral ; of five sides, a pentagon ; of 
 
 six sides, a hexagon ; of seven sides, a heptagon ; of 
 
 eight sides, an octagon ; of nine sides, a nonagon ; of 
 ten sides, a decagon; etc. 
 
 TRIANGLES 
 
 25. A Triang'le is a plane figure having three sides 
 and three angles. ABC is a triangle. ^ 
 
 26. The Base of a triangle is the side 
 on which it stands. AB is the base. 
 
 27. The Altitude of a triangle is a line 
 drawn from the angle opposite perpendicular to the base, 
 
6 
 
 MEysuitATioy 
 
 or the base produced. CD is the altitude. The dotted 
 liues in the following figures represent the altitude. 
 
 Fig. I 
 
 Fir/. 2 
 
 ., y Equilateral Isosceles 
 
 by S> 
 
 Named \ Equiangular 
 by Angles J Acute-angled 
 
 Fig. 3 Fig. i 
 
 Scalene 
 Acute-angled Right-angled Obtuse-angled 
 
 28. A Rig'ht-ang'led Triangle is a triangle having 
 one right angle. 
 
 29. The Hypothenuse of a right-angled triangle is 
 the side opposite the right angle. 
 
 30. An Equilateral Triangle has its three sides 
 equal. 
 
 31. An Isosceles Triangle has two of its sides 
 equal. 
 
 32. A Scalene Triangle has no two sides equal. 
 
 33. An Acute -angled Triangle has three acute 
 angles. 
 
 34. An Obtuse-angled Triangle has one obtuse 
 angle. 
 
 35. An Equiangular Triangle has its three angles 
 equal. 
 
 Rule I. To find the area of a triangle when the base 
 and altitude are given: Multiply the ba.se by one -half 
 the altitude. 
 
TUIANGLES 7 
 
 Rule II. To find the area of a triangle when the 
 three sides are given : ISubtnict each side sejmrately from 
 half the sum of ike three sides; multij)ly the continued 
 jyroduct of these three remainders bij the half -sum ; 
 the square root of the product is the area. 
 
 Rule III. To find one dimension of a triangle when 
 the area and the other dimension are given: Divide the 
 area by one -half the given dimension. 
 
 PROBLEMS 
 
 1. A triangle has a base of 60 feet, and an altitude 
 of 20 feet ; how many square feet does it contain ? 
 
 Ans. 600 sq. ft. 
 
 2. Find the area of a triangle, the length of whose 
 base is 50 feet, and whose altitude is 12 feet 4 inches. 
 
 Ans. 308i sq. ft. 
 
 3. How many acres in a triangular lot whose base is 
 28 rods and altitude 18 rods? Ans. 1 A. 92 sq. rds. 
 
 4. What is the cost of a triangular piece of land 
 whose base is 30.96 chains and altitude 4.835 chains, at 
 $120 an acre? Ans. $898.15—. 
 
 5. Find the area of a triangle whose sides are respec- 
 tively 60, 80, and 100 feet. Ans. 2,400 sq. ft. 
 
 6. Find the base of a triangle whose area is 40 acres 
 and altitude 80 rods. Ans. 160 rds. 
 
 7. A triangle contains 48 square feet 63 square 
 inches ; the base is 12 feet 6 inches ; required the 
 altitude. Ans. 7 ft. 9 in. 
 
 8. Find the cost of sodding a triangular plot of 
 ground whose sides are respectively 60, 75, and 80 feet, 
 at 15 cents per square yard. Ans. $35.60+. 
 
8 M EX DURATION 
 
 9. Mr. A built a barn in the form of the letter L ; 
 the width of the barn at each end is 60 feet, and the 
 ridge of the roof is 24 feet higher than the foot of the 
 rafters ; how many sqnare feet of boards were required 
 to cover the three gables? Ans. 2,160 sq. ft. 
 
 10. Owing to the difference in the soil of a farmer's 
 field, which is square, he decided to plant part in corn 
 and part in potatoes. He first made a straight furrow 
 from the northeast corner of the field to the southwest 
 corner, and found it to be 88 rods long ; from the 
 middle of this furrow to the southeast corner of the 
 field is 44 rods. He planted all to the eastern side of 
 the first furrow drawn in potatoes and the balance in 
 corn. How many acres in the field"? How many acres 
 did he plant in potatoes ? In corn ? 
 
 Ans. field, 24i A.; potatoes, 12to A.; corn, 12to A. 
 
 11. Find the area of the penta- 
 gon as shown in the accompany- 
 ing diagram. The diagonal AC is 
 48 inches, and the diagonal AD, 36 
 inches ; the altitude of the triangle 
 ABC is 16 inches, of the triangle 
 ACD, 20 inches, and of the triangle 
 AED, 12 inches. Ans. 1,080 sq. in. 
 
 THE RIGHT-ANGLED TRIANGLE 
 
 36. The following principles relating to right-angled 
 triangles have been proved by Geometry. An examination 
 of Figures 5, 6, and 7 will be helpful to the student in 
 fixing them in mind. 
 
 Principle I. The square of the hijpothenuse is equal 
 to the sum of the squares of the other two sides. 
 
EIOHT-AXGLKD TRIAXGLES 
 
 9 
 
 
 
 
 
 \X^-'=:^ 
 
 
 
 
 A ^' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 io 
 
 \| 
 
 24 
 
 10 
 
 ciS 
 
 Fig. 5 
 
 Fig. a 
 
 Fig. 7 
 
 Principle II. The square of the base, or of the per- 
 pendicular, is equal to the square of the hypothenuse 
 diminished by the square of the other side. 
 
 Rule I. To find the hypothenuse : Extract the square 
 root of the sum of the squares of the base and perpendicular. 
 
 Rule II. To find the base or perpendicular : Extract 
 the square root of the difference between the square of the 
 hypothenuse and tli square of the given side. 
 
 PROBLEMS 
 
 1. What is the length of the hypothenuse of a right- 
 angled triangle (Fig. 5) whose base is 4 feet, and whose 
 perpendicular is 3 feet ! 
 
 Operation. — "i/4"^ -f- 32 = 5 feet. Ans. 
 
 In a right-angled triangle, given : .... 
 
 2. The base 5 feet, perpendicular 12 feet, to find the 
 hypothenuse. Ans. 13 ft. 
 
 3. The base 8 inches, perpendicular 15 inches, to find 
 the hypothenuse. Ans. 17 in. 
 
 4. The perpendicular 20 rods, the base 21 rods, to find 
 the hypotheuusc. Ans. 29 rds. 
 
10 MENSURATION 
 
 0. The perpendicular 3 yards, the base 4 yards, to find 
 the hypothenuse. Ans. 5 yds. 
 
 6. The hypothenuse 10 feet, the base 8 feet, to find 
 the perpendicular. Ans. 6 ft. 
 
 7. The hj-pothenuse 78 feet, the perpendicular 30 feet, 
 to find the base. Ans. 72 ft. 
 
 8. Find the diagonal of a square whose side is 20 feet. 
 
 Ans. 28.28+ ft. 
 
 9. Find the diagonal of a cube whose edge is 20 feet. 
 
 Ans. 34.64+ ft. 
 
 10. The gable end of a house 50 feet wide is 14 feet 
 high; what is the length of the rafters ? Ans. 28.65+ ft. 
 
 11. A flag -pole 120 feet high casts a shadow 110 feet 
 in length ; required the distance from the end of the 
 shadow to the top of the pole. Ans. 162.78+ ft. 
 
 12. A hunter stood 40 feet from the foot of a tree 
 and shot a squirrel on the top of a tree 55 feet high ; 
 how far did the bullet go before hitting the squirrel, 
 provided the starting point of the bullet was 5 feet 
 from the ground? Ans. 64.03+ ft. 
 
 13. Thomas Lighty, a carpenter, directed his appren- 
 tice to cut a brace of the proper length to secure a per- 
 pendicular in a sill, bj^ measuring 6 feet each way from 
 the corner where the two sticks joined; how long was 
 the braced Ans. 8.48+ ft. 
 
 14. A deiTick at a granite quarry had an upright 
 48 feet long, and a horizontal ar^ji 2Q feet Jong ; a rope 
 ran on pulleys from the foot of the perpendicular to the 
 extremity of the arm and thence to the ground ; what 
 length of rope was required, no allowance being made 
 fi/r fastening the rope? Aus 100 ft. 
 
Q UA DRILA TERALS 1 1 
 
 15. A's horse travels a mile in 3 minutes, and B*s a 
 mile in 4 minutes ; A drives at this rate due north for 
 24 minutes, and B, starting at the same place, due east 
 for the same time ; how far are they apart after driving 
 24 minutes f After driving at the same rate and same 
 direction, respectively, 1 hour? Ans. 10 mi.; 25 mi. 
 
 16. Williamson Porter in hunting bees placed his bait 
 on a stump 5 feet high; one bee flew in a "bee line" 
 from the bait 75 feet to a tree due west from the stump, 
 alighting 45 feet from the foot of the tree ; another flew 
 80 feet east to a tree, alighting 65 feet from the foot of 
 the tree ; if the ground is level, and both trees per- 
 pendicular, what is the shortest distance between the 
 trees? Ans. 116.35-f ft. 
 
 17. Mr. A, a politician, desired his three neighbors, 
 B, C, and D, to vote at a certain election. A lived 100 
 rods south; B, 120 rods east; C, 140 rods north; and D, 
 150 rods west of the polling place. A, with his carriage, 
 drove from home in a straight line for B, thence in a 
 straight line for C, and thence in a straight line for D ; 
 from D's residence he drove in a straight line to the 
 polling place, and thence in a straight line home : how 
 far did he drive, provided he allowed his neighbors 
 to walk home? Ans. 2 mi. 155.77+ rds. 
 
 QUADRILATERALS 
 
 37. A QuadrHateral is a portion of a plane bounded 
 by four straight Hues. 
 
 38. A Parallelogram is any plane figure of four 
 straight sides, the opposite ones of which are parallel. 
 
 39. A Rectang'le is any parallelogram having all its 
 angles right angles. 
 
12 MENSURATION 
 
 40. A Square is a rectangle all ot* whose sides are 
 equal. 
 
 41. A Rhombus is a parallelogram having all its 
 sides equal and all its angles oblique. 
 
 42. A Rhomboid is a parallelogram whose opposite 
 sides only are equal and whose angles are oblique. 
 
 Square Eectangle Rhoaibus Rhomboid Trapezoid Trapezium 
 
 r 1. Parallelograms ■{ 
 
 r 1. Square 
 * 2. Rectangle 
 
 (3. Rhombiis 
 4. Rhomboid 
 (_ 3 Ti-apezium 
 
 43. A Trapezoid is a quadrilateral which has only 
 two of its sides parallel. 
 
 44. A Trapezium is a quadrilateral having no sides 
 parallel. 
 
 45. The Lower Base of a parallelogram is the side 
 upon which it stands. The Upper Base is the side oppo- 
 site the lower base. 
 
 46. The Altitude of a parallelogram or of a trapezoid 
 is the perpendicular distance between two bases. The 
 dotted lines in the figures represent the altitude. 
 
 47. The Diag'Onal of a quadrilateral is a straight line 
 joining two opposite vertices. 
 
 • Rule I. To find the- area of any parallelogram : Muh 
 fiphj the base hy the altitude. . . 
 
 Rule II. To find the area of a trapezoid : MuUij)^ 
 half the sum of the pfirallel sides hf/ Ute altitude. 
 
or A I>li I LA TJi'JU L S 1 3 
 
 Rule III. To find the area of a trapezium : Multiply 
 the diagonal hij half the sum of the perpend iculai^s drawn 
 to it from the vertices of the op2)Osite angles. 
 
 Note. — The area of any polygon may be found by dividing it 
 into triangles and finding the sum of their areas. 
 
 PROBLEMS 
 
 1 . The side of a square is 20 feet ; what is its area ? 
 
 Ans. 400 sq. ft. 
 
 2. A has a rectanguhir field 80 rods long and GO rods 
 wide; how many acres does it contain? Ans. 30 A. 
 
 3. Find the area of a parallelogram whose base is 150 
 feet and altitude 80.5 feet. Ans. 12,075 sq. ft. 
 
 4. Find the area of a rhombus whose base is 20 feet 
 and altitude 6 feet 3 inches. Ans. 125 sq. ft. 
 
 5. A field in the form of a rhomboid has a base 136 
 rods long and an altitude of 550 yards ; hoAv many acres 
 does it contain? Ans. 85 A. 
 
 6. The parallel sides of a trapezoid are 70 rods and 
 90 rods and the altitude 20 rods ; find the area in 
 acres. Ans. 10 A. 
 
 7. Find the area of a trapezium, a diagonal of which 
 is 50 feet, and the perpendiculars of which to this 
 diagonal are 10 feet and 35 feet. Ans. 1,125 sq. ft. 
 
 8. Required the area of a trapezium, the lengths of 
 whose sides are respectively 20, 30, 25, nnd 35 chains, 
 and the diagonal 40 chains, making two triangles whose 
 sides are respectively 35, 20, and 40; and 30, 25, and 40. 
 
 Ans. 72 A. 71 sq. rds. 17 sq. yds. 5 sq. ft. 143.136 sq. in, 
 
 9. Required the area of a regular hexagon whose side 
 is 6 inches. Ans. 93.528+ sq. in. 
 
 Suggestion. — Divide into 6 equilateral triangles, or 3 rhom- 
 buses, or 2 trapezoids. 
 
14 
 
 MENS UH ATI ON 
 
 THE CIRCLE 
 
 48. A Circle is a plane figure bounded by a curved 
 line, every point of which is equally distant from a point 
 within called the center. 
 
 C\tc«ni£e^^^ 
 
 Circle 
 {Fig. 1) 
 
 The Circle 
 
 Clrcumscribecl Square 
 
 
 ^ ~^ 
 
 
 
 Inscribed 
 
 \ 
 J 
 
 Scfiiare 
 
 
 ^ ^ 
 
 
 {Fig. 2) 
 
 Lines 
 
 Curved 
 
 Straight . 
 
 Circumference 
 Arc 
 
 1. Diameter 
 
 2. Radius 
 
 3. Secant 
 
 4. Tangent 
 
 5. Claord 
 
 f 1. Semi-circle 
 
 Surfaces of portions of Circle J "' ^^^ ^ ^ 
 I 3. Sector 
 
 4. Segment 
 
 49. The Circumference of a circle is the line which 
 bounds the circle. 
 
 50. An Arc is any portion of the circumference. 
 
 51. The Diameter of a circle is a straight line passing 
 through the center, and having its extremities in the cir- 
 cumference. 
 
 52. The Radius of a circle is a straight line drawn 
 from the center to any part of the circumference. It is 
 one -half the diameter. 
 
HIE CIliCLK 15 
 
 63. A Secant is a straight line which cuts the cir- 
 cumfereuce at two points. 
 
 54. A Tangfent is a straight line which touches the 
 circumference, but does not cut it. 
 
 55. A Chord is a straight line joining the extremities 
 of an arc. 
 
 56. A Semi-circle is half a circle. 
 
 57. A Quadrant is a quarter of a circle. 
 
 58. A Sector of a circle is a portion of a circle 
 included by two radii and the arc intercepted by them. 
 
 59. A Seg'ment of a circle is a portion of a circle 
 included between an arc and its chord. 
 
 60. The ratio of the circumference to the diameter is 
 the same for all circles. For convenience tiiis ratio is 
 represented by the Greek letter tt (pi). The approximate 
 numerical value of ^ is 3.1416, which is the value used 
 in this work. 
 
 Rule I. To find the circumference of a circle when the 
 diameter is given : Multiply the diameter by 3.1416. 
 
 Rule II. To find the diameter of a circle when the cir- 
 cumference is given: Divide the circumference hy 3.1416. 
 
 Rule III. To find the area of a circle when its diameter, 
 radius, and circumference are given, or when either is 
 given : 
 
 (a) Multiply the circumference hy one- fourth of its 
 diameter ; or, 
 
 (b) Multipthj the square of the diameter hy .7854; or, 
 
 (c) Multiply the square of the radius hy 3.1416; or, 
 
 (d) Multiply half the distance roimd hy half the dis 
 tance through. 
 
Rule IV. To find either dimension, when the area is 
 given : 
 
 Let A represent the area of a circle, C the circumference, D the 
 diameter, and K the radius ; then. 
 
 I>-Vw-:-.-.', 0=(T/:y^)X 3.1416; R = i/^-, 
 Rules I, II, and III may be stated as follows : 
 
 C = 7rD or 27rR ; D=-^ ; A = tt R2 or ^. 
 
 Rule V. To find the side of an inscribed square when 
 the diameter is given : 
 
 (a) Extract the square root of one -half the square of 
 the diameter; or, 
 
 (b) Multiply the diameter hy .7071. 
 
 Rule VI. To find the area of a circular ring formed bj'- 
 two concentric cii'cles: Find the areas of the circles 
 separately and tal'e their difference. 
 
 PROBLEMS 
 
 1. Find the circumference of a circle whose diameter 
 is 20 inches. Ans. 62.832 in. 
 
 2. Find the circumference of a circle whose diameter 
 is 39 feet. Ans. 122.522 ft. 
 
 3. Find the diameter of a circle whose circumference 
 is 128.8056 yards. Ans. 41 yds. 
 
 4. Find the diameter of a circle whose circumference 
 is 37.6992 inches. Ans. 12 in. 
 
 5. Find the radius of a circle whose circumference is 
 119.3808 inches. Ans. 19 in. 
 
 6. Find the radius of a circle whose circumference 
 is 100.5312 rods. Ans. 16 rds. 
 
 7. How far is it round a tree which measures 4 feet 
 over the stump? Ans. 12.5664 ft. 
 
THE ClKCLi: 17 
 
 8. The spokes of tlie fore wheel of a buggy are 
 2 feet long, and of the hind wheel 2 feet 2 inches ; no 
 allowance being made for the thickness of the hub, how 
 much longer is the tire of the hind wheel than of the 
 fore wheel? Ans. 1.0472 ft. 
 
 9. Four flower -pots each with a lower diameter of 
 
 1 foot were placed on a bench 10 feet long and 16 
 inches wide ; how much of the bench remained un- 
 covered? Ans. 1467.6096 sq. in. 
 
 10. John threw a base -ball 2 inches in diameter at a 
 paste -board card 2 feet square, and made in the card 
 15 holes of the same circumference as the ball ; how 
 much of the surface of the card remained unbroken ? 
 
 Ans. 528.876 sq. in. 
 
 11. Certain trustees asked for bids to fresco the ceil- 
 ing of their church, which is 60 feet long and 40 feet 
 wide. A contractor offered to place on the ceiling 8 
 circular designs 12 feet in circumference at $3 each, and 
 paint the remaining surface at 9 cents per square j-ard, 
 or to place on the ceiling 8 square designs, the perimeter 
 of each to be 12 feet, at $3 each, and paint the remain- 
 ing surface at 1 cent per square foot. The trustees 
 chose the second offer. How much did it cost them ! 
 How much did they save or lose by taking the second 
 offer instead of the first? Ans. cost, $47.28; lost, 20c. 
 
 12. The minute hand of a town clock is 2 feet long ; 
 over how much surface does it pass in 20 minutes ? 
 In 15 minutes? Ans. 4.1888 sq. ft.; 3.1416 sq. ft. 
 
 13. How many times will a wheel whose radius is 
 
 2 feet revolve in going 1 mile? Ans. 420.168+ times. 
 
 14. Find the width of the ring between two concen- 
 tric circles whose circumferences are respectively 15.708 
 and 31.416 feet. Ans. 2^ ft. 
 
 B 
 
IS MENSV RATION 
 
 15. Find the area of a circle inscribed in a square 
 containing 576 square feet. Ans. 452.39+ sq. it. 
 
 16. Find the area of a circle circumscribed about a 
 square containing 256 square feet. Ans. 402.109+ sq.ft. 
 
 17. The diameter of a cii'cle is 100 feet ; find the 
 side of the inscribed square. Ans. 70.71 ft. 
 
 18. The circumference of a circle is 62.832 feet; find 
 the side of an inscribed square. Ans, 14.142 ft. 
 
 19. The area of a circle is 1,963.44 square inches ; 
 find the side of an inscribed square. Ans. 35.355 in. 
 
 THE ELLIPSE 
 
 61, An Ellipse is a curved plane formed by an 
 oblique section of either a cone or a cylinder, passing 
 through its curved surface, but not 
 touching its base; the sum of the dis- 
 tances from every point of the bound- 
 ing line to two fixed points, called 
 foci, is equal to a line drawn through 
 these points and terminated by the Ellipse 
 
 curve. The line drawn through the 
 foci is a transverse axis ; a perpendicular to this axis 
 at its middle point is the conjugate axis. 
 
 Rule. To find the area of an ellipse : Find the product 
 of the semi -axes, a7id multiply this iwoduct by 3.1416. 
 
 PROBLEMS 
 
 1. What is the area of an ellipse whose transverse 
 axis is 40 inches and conjugate axis 32 inches! 
 
 Ans. 1,005.312 sq. in.. 
 
SIMILAH rLAXK FIG V RES 19 
 
 2. The axes of an elliptical flower bed are 20 and 
 16 feet; what is its area? Ans. 251.328 sq. ft. 
 
 3. The transverse axis of an elliptical fish pond is 
 80 feet and conjugate axis 60 feet ; find the area. 
 
 Ans. 3,769.92 sq. ft. 
 
 4. How much larger is a circle whose diameter is 
 10 inches than an ellipse whose transverse axis is equal 
 to the diameter of the circle, and conjugate axis f of 
 the length of the transverse axis? Ans. 31.416 sq. in. 
 
 SIMILAR PLANE FIGURES 
 
 62. Similar Plane Fig-upes are those which have the 
 same form. Their angles are equal and their like dimen- 
 sions proportional. Squares, equiangular triangles, regu- 
 lar polygons of the same number of sides, and circles 
 are similar. The surfaces of similar solids are also 
 proportional. The like dimensions of circles are their 
 radii, diameters, and circumferences, and of other plane 
 figures, those which are like placed. The following 
 principles are derived from Geometry : 
 
 Principle I. The areas of similar surfaces are to 
 each other as the squares of their like dimensions. 
 
 Principle II. The like dimensions of similar sur- 
 faces are to each other as the square roots of their areas. 
 
 PROBLEMS 
 
 1. How many circles eacli 8 inches in diameter will 
 equal the area of a circle 4 feet in diameter? Ans. 36. 
 
 2. John's slate is 14 by 10 inches, and William's 7 by 
 5 inches ; how do they compare in size ? 
 
 Ans. John's = 4 times William's. 
 
20 MENSURATloyf 
 
 3. Two farms of exactly the same shape contain 
 respectively 144 and 169 acres. One side of the former 
 is 480 rods ; required the corresponding side of the 
 latter. Ans. 520 rds. 
 
 4. A has a square field, the side of which is 40 rods, 
 worth $50 an acre, and B a square field, i as long, worth 
 $100 an acre ; which is the more valuable, and how 
 much? Ans. A's, $250 more. 
 
 5. If it cost $40 to (-arpet a rectangular room 20 feet 
 long, how much will it cost to carpet a room of similar 
 shape 25 feet long? Ans. $62.50. 
 
 6. A lady has two circular flower beds ; the one 62 feet 
 in diameter, the other Si feet ; the first is how many times 
 the size of the second? Ans. 4 times. 
 
 7. The area of a rectangular field is 2,880 square rods ; 
 its sides are as 4 to 5 ; what is the length of each side ? 
 
 Ans. 48 rds.; 60 rds. 
 
 8. The hypothenuse of a right-angled triangle is 25 
 inches ; what is the hypothenuse of a similar triangle 
 which contains twice the area? Ans. 35.35+ in. 
 
 9. The area of a trapezoid is 216 square rods, and its 
 altitude 12 rods ; find the altitude of a similar trapezoid 
 whose area is 384 square rods. Ans. 16 rds. 
 
 10. A battle ship whose anchor weighs 4,000 pounds 
 requires a cable 3 inches in diameter ; what should be the 
 diameter of the cable when the anchor weighs 4 tons? 
 
 Ans. 4.242+ in. 
 
 11. If a cistern can be filled hy a pipe 3 inches in 
 diameter in 24 minutes, in what time can it be filled by 
 a pipe 9 inches in diameter ? Ans. 2 min. 40 sec. 
 
 12. How many saucers each 5 inches in diameter will 
 equal in area the surface of a round table 5 feet in 
 diameter ? Ans. 144. 
 
SIMILAR PLANU FIGURES 21 
 
 13. If the circumference of a circle is 2 feet, what is 
 the circumference of a circle Sro times as great ? 
 
 Ans. 3^ ft. 
 
 14. The altitudes of two similar triangles are 21 feet 
 and 5i feet ; what is the relation of their areas ? Ans. 16. 
 
 15. The area of a triangle is 5,400 square feet, and its 
 sides are proportional to the ' numbers 9, 12, and 15; 
 required the length of the sides. 
 
 Ans. 90 ft., 120 ft., 150 ft. 
 
 16. Mr. Bowers wished to sow clover seed on two 
 fields, one 40 rods square and the other 60 rods square; 
 he estimated that one bushel would be sufficient for the 
 smaller field ; what should have been his estimate for 
 the larger field? Ans. 2i bu. 
 
 17. Sarah baked for John 2 buckwheat cakes each 6 
 inches in diameter, and for Rufus 8 cakes of the same 
 thickness each 3 inches in diameter; John accused Sarah 
 of unfairness ; was she unfair ? 
 
 18. If it requires 96 square inches of tissue paper to 
 cover a cube whose edge is 4 inches, how much would be 
 required to cover a cube whose edge is 8 inches ? 
 
 Ans. 384 sq. in. 
 
 19. It cost Mr. McDowell $80 to pave his walk, which 
 is 6 feet wide and 100 feet long ; Mr. Blair paid $320 
 for paving his walk similar in shape to Mr. McDowell's, 
 with the same material; what are the dimensions of Mr. 
 Blairls walk? Ans. 12 ft. wide; 200 ft. long. 
 
MENSURATION OF SOLIDS 
 
 PRISMS AND CYLINDERS 
 
 63. A Solid or Body has three dimensions, — length, 
 breadth, and thickness. The bounding planes are its 
 faces, and their intersecting lines its edges. 
 
 64. A Prism is a solid whose two ends are parallel, 
 similar, and equal, and whose sides are parallelograms. 
 
 They are named according to the form of their bases; 
 as triangular, quadrangular, pentagonal, hexagonal, etc. 
 
 65. The Altitude of a prism is the perpendicular dis- 
 tance between its bases. In Figs. 4, 5, and 6 the per- 
 pendicular dotted lines represent the altitude. 
 
 66. A Rig'ht Prism is a prism whose lateral edges 
 are perpendicular to the bases. 
 
 67. An Oblique Prism is a prism whose lateral edges 
 are not perpendicular to the bases. 
 
 Right Prisms 
 
 Oblique Prisms 
 
 Frustums of Prisms 
 
 V 
 
 w 
 
 a 
 
 t^ 
 
 Q 
 
 Fig. 1 Fig. 2 Fig. 3 Fig. i Fig. 5 Fig. 6 J'lg. 
 
 
 ■ U 
 
 Fig. 8 Fig. 
 
 68. A Frustum of a prism is that part which remains 
 after cutting olf a part of it by passing a plane through 
 it not parallel to either base. This is also called a 
 trunmUd prism. 
 
 (22) 
 
MENSURATION OF SOLIDS 23 
 
 69. A Parallelepiped is a prism bounded by six 
 parallelograms, the opposite faces being parallel. 
 
 70. A Rig'ht Parallelepiped is one whose lateral 
 edges are perpendicular to its bases. 
 
 Rectangular 
 
 
 Parallelopipeds 
 Cuhc Rhomb 
 
 Rhombic Prism 
 
 
 \ 
 
 \ 
 
 
 ^ 
 
 N. -^ .^i 
 
 
 
 
 a 
 
 
 1 
 
 ■/ /^\ 
 
 a 
 
 \ 
 
 
 k 
 
 
 
 Right 
 Fig. 10 
 
 
 Right 
 Fig. 11 
 
 Ol.lique 
 Fig. 12 
 
 Oblique 
 Fig. Vi 
 
 
 71. A Rectang-ular Parallelepiped is a right paral- 
 lelopiped whose bases are rectangles. All of its faces 
 are rectangles. 
 
 72. A Cube is a parallelopined whose faces are all 
 equal squares. 
 
 73. A Rhomb is a parallelopiped whose faces are all 
 equal rhombuses. 
 
 74. A Rhombic Prism is a parallelopiped, whose 
 faces are rhombuses or rhomboids, having each pair of 
 opposite faces equal, but not all its faces equal. In each 
 figure above, a represents the altitude. 
 
 75. A Cylinder is a solid bounded hy a uniformly 
 curved surface, and having for its ends two equal par- 
 allel circles. Any section of a cylinder parallel to the 
 ends is a circle equal to either end. 
 
 76. A Right Cylinder is one whose ends are per- 
 pendicular to its sides. 
 
 77. An Oblique Cylinder is one whose ends are not 
 perpendicular to its ^ides. 
 
24 
 
 MENSURATION 
 
 78. The Bases of a cylinder are the two equal and 
 parallel circles. 
 
 79. The Altitude of any cylinder is the perpendicu- 
 lar distance between' the planes of its bases. 
 
 80. The Axis of a cylinder is a line joining the 
 centers of its bases. 
 
 81. A Cylindroid is a solid resembling a cylinder, 
 but having its bases elliptical. 
 
 Fia. L4 
 
 Oblique Cyliader 
 
 Mff. 15 
 
 Cylindroid 
 
 Fig 
 
 The right cylinder (Fig. 14) may be conceived to be 
 generated by the revolution of the rectangle ABCD 
 about its side BD, as an axis. 
 
 A rigJit section of a prism or a cylinder is a sec- 
 tion made by a plane perpendicular to the axis ; as X 
 (Fig. 15). 
 
 0iM 
 
 
 FigAI 
 
 m 
 
 
 Fig.\^ 
 
 Fig.l^ 
 
 82. By examining Figs. 17, 18, and 19, it will be 
 observed that the convex surface of each maj' be represented 
 by a rectangle, the perimeter of the prism or cylinder being 
 
SURFACES OF miSMS AND CYLINDERS 25 
 
 equal to the base of the rectangle, and the altitude of the 
 prism or cylinder equal to the altitude of the rectangle ; 
 hence the following rule : 
 
 Rule. To find the convex surface of a right prism or 
 right cylinder : Multiply the j^erimeter of the base by the 
 altitude. 
 
 For the entire surface, add to the convex surface the 
 areas of the two bases. 
 
 The convex surface of any prism may be found by 
 adding together the areas of the lateral faces. 
 
 The convex surface of any cylinder may be found by 
 finding the product of the perimeter of a right section 
 of the cvlinder and the axis. 
 
 SURFACES OF PRISMS AND CYLINDERS 
 
 PROBLEMS 
 
 1. What is the convex surface of a prism whose alti- 
 tude is 10 feet and perimeter of the base 20 feet ? 
 
 Ans. 200 sq. ft. 
 
 2. What is the convex surface of a prism whose alti- 
 tude is 10 yards and perimeter of the base 10 yards ? 
 
 Ans. 100 sq. yds. 
 
 3. What is the convex surface of a hexagonal prism 
 whose altitude is 20 feet and each side of the base 12 
 inches? Ans. 120 sq. ft. 
 
 4. What is the convex surface of a cylinder, the circum- 
 ference of whose base is 32.66 feet and altitude 10 feet ? 
 
 Ans. 326.6 sq. ft. 
 
 5. What is the convex surface of a cylinder whose alti- 
 tude is 20 feet and the radius of the base 5 feet ? 
 
 Ans. 628.32 sq. ft. 
 
26 MEXSURATION 
 
 6. The diameter of the base of a ej'linder is 20 feet 
 and the altitude 30 feet ; required the convex surface. 
 
 Aus. 1,884.96 sq. ft. 
 
 7. Find the entire surface of a quadrangular prism 
 whose altitude is 20 feet and each side of the base 3 feet. 
 
 Ans. 258 sq. ft. 
 
 8. What is the convex surface of a pentangular prism 
 whose altitude is 30 feet and each side of the base 4 feet? 
 
 Ans. 600 sq. ft. 
 
 9. What is the convex surface of a cylinder whose 
 altitude is 4 yards and diameter of the base 3 feet ? 
 
 Ans. 113.0976 sq. ft. 
 
 10. The altitude of a triangular prism is 5 inches and 
 each side of the base 6 inches ; what is the entire sur- 
 face f Ans. 121.1768 sq. in. 
 
 11. Required the entire surface of a parallelopiped 8 
 feet long, 5 feet wide, and 3 feet high. Ans. 158 sq. ft. 
 
 12. What is the entire surface of a cube whose edge is 
 7 feet "? Ans. 294 sq. ft. 
 
 13. What is the entire surface of a rhomb, each face of 
 which is a parallelogram, whose base is 2 feet 2 inches and 
 altitude 2 feet? Ans. 26 sq. ft. 
 
 14. A door 8 feet high and 4 feet wide revolves around 
 one of its sides as an axis ; what is the convex surface of 
 the cylinder generated by it f Ans. 201.0624 sq. ft. 
 
 15. A door 8 feet long and 4 feet wide revolves upon a 
 point in its center ; what is the convex surface of the 
 cylinder generated by it ; what is the entire surface ? 
 
 Ans. 100 5312 sq. ft.; 125.664 sq. ft. 
 
 16. The diameter of the base of a cjdinder is 16 
 inches, and the altitude 2o- feet ; how many square feet 
 ai-e there in the lateral surface? Aus. 10.472 sq. ft. 
 
ILLUSTRATIVE PROBLEMS 27 
 
 ILLUSTRATIVE PROBLEMS 
 
 Solved by the Prismoidal Formula 
 
 83. The special feature of this work is the applica- 
 tion of the Prismoidal Formula to finding the volume of 
 solids. Since this one rule covers so broad a field of 
 measurement, it is thought that a thorough knowledge 
 of the method of using it in finding the volume of the 
 cube, cylinder, etc., as well as of those solids for which the 
 rules usually given are more complex, will be beneficial. 
 The author does not advise the teacher to discard the 
 well known rules for finding the volume of a prism, 
 cylinder, etc.; but he does recommend that this rule be 
 substituted for those usually quoted for finding the 
 volume of many other solids, as the frustum of a pyra- 
 mid or of a cone. Both rules are given under each 
 subject; this will enable the teacher in each case to take 
 his choice. In order that the pupil may be made quite 
 familiar with the rule, a number of illustrative problems 
 are solved, and such suggestions made as will enable the 
 pupil to use it intelligently in following chapters. These 
 are arranged in a separate section at this place so that 
 comparisons may be more advantageously made. Future 
 reference is made to this chapter. For a derivation of 
 the formula, etc., see Supplement. 
 
 Rule. — To find the volume of a prismoid, etc.: 
 
 84. Multiply the sum of both end areas and four times 
 the area of a section half way between them^ by one-si^h 
 the altitude. 
 
 Letting a = the altitude or length ; B, the area of either base, 
 and, when unequal, the larger base; b, the area of the opposite 
 
28 
 
 MEXSURATWX 
 
 base, aud, when imequa], the smaller ba.se; and 4m, fonr times the 
 area of a section half way between the bases, the rule may be 
 stated as follows: |(B + b + 4m). 
 
 This formula will be used throughout this work. 
 
 In all figures in the following chapter, the middle sec- 
 tion is represented by dotted lines and the altitude is 
 marked with the lettei* a. 
 
 1. Required the volume of a cube 
 whose edge is 6 feet. 
 
 Operation.— B = 6 X 6 = 36 sq. ft. 
 b=6 X 6=36 " 
 4m =4 (6 X 6) = 144 sq. ft. 
 a = 6 
 Substituting in formula, -|(B -|- b + 4m; 
 we have, f (36 + 36 + 144)= 216 cubic feet, volume 
 
 / 
 
 b 
 
 z 
 
 
 6 
 
 ni 
 
 4 
 
 / 
 
 B 
 
 A 
 
 Remark,— It will be observed that in all uniform solids not 
 tapering or having an apex, such as prisms and cylinders, the area 
 of both bases and of once the middle section is the same ; this neces- 
 sitates only one calculation for these areas. 
 
 2. Required the number of cubic feet of air in a room 
 20 feet long, 8 feet wide, and 
 12 feet high. 
 
 Operation. — 
 
 B= 20X8= 160 sq. ft. 
 
 b = 20X8 = 160 
 
 4m = 4 (20X8) =640 sq. ft. 
 a=12 
 
 
 / 
 
 b 
 
 'a 
 
 
 y 
 
 20 
 
 
 a 
 
 m 
 
 '8 
 
 
 
 - 20 
 
 \i 
 
 / 
 
 B 
 
 A 
 
 20 
 
 Substituting in formula, | (B + b + 4m) 
 
 we have, V" i 160 -f 160 -\~ 640)= 1,920 cubic feet. 
 
IL L I 'SIR A TJ I K I'JxUBLKMS 
 
 29 
 
 8. What is the vohniie of a triaiigiihu' prism whose 
 altitude is 10 feet, and each side of the base 
 4 feetf 
 
 Operation. — B = 6.928 sq. ft. (Art. 35, R. II) 
 b= G.928 " " *' 
 
 4m = 27.712 " 
 a = 10 
 Substituting in formula, | (B + b + 4m) 
 we have, \^ (6. 928 + 6. 928 4-27. 712) = 69.28 cubic 
 feet, volume. 
 
 
 \,. VI 
 
 
 K /41 
 
 a 
 
 
 
 
 ■-. m 
 
 4/ 
 
 10 
 
 4^N 
 
 /'4 
 
 
 \ B 
 
 4/ 
 
 
 N 
 
 /\ 
 
 4. What is the cost of a piece of timber 18 inches 
 square and 20 feet long, at $.30 a cubic foot? 
 
 18^ 
 
 iPl 
 
 18"'; 
 
 18 B 
 
 :i 20 
 
 Operation. — 18 inches = f ft. 
 
 B=tXf = f sq. ft. 
 b=tXf=f " 
 4m = 4 (IX f) =9 sq- ft. 
 a = 20 
 Substituting in formula, |(B + 1^ -f 4m) 
 we have, -^ (1 + t 4- 9) = 45 cubic feet, volume. 
 
 5. What is the volume of a cylinder whose radius is 
 1 foot 3 inches, and altitude 30 feet? 
 
 Operation. — 1 foot 3 inches = 1 ft. 
 
 B= (I) 2 X 3.1416= 4.90875 sq. ft. (Art. 60, R. Ill c) 
 
 b= (1)2X3.1416= 4.90875 
 4m = 4(f) 2X3.1416 = 19.635 " " 
 
 a = 30 
 Substituting in formula, | (B + b + 4m) 
 we have, ^8^(4.90875 + 4.90875 + 19.635)=147.2625 cubic feet, volume. 
 
30 
 
 MEX.<L'liAnu\ 
 
 G. What is the volume of a pyramid whose base is 6 
 feet square, and altitude 24 feet ? 
 
 Operation. — In the triangle X Y Z, 
 
 1 = 
 
 (5 _|_ -J- 2 or 3 is the length of the line RS, 
 or the side of the middle section. 
 B = 6X6 = 36 sq. ft. 
 
 4m = 4 (3X3) =36 sq. ft. 
 a = 24 
 Substituting in formula, | (B + b + 4m) 
 we have, V- (36 + + 36) = 288 cubic 
 
 feet, volume. 
 
 Remarks.— (a) It is apparent that in the 
 pyramid and cone the area of the smaller 
 
 base is Ze:'o. (b) Since similar surfaces are to each other as the 
 squares of their like dimensions, and since in the pyramid and cone 
 the side or diameter of the lower base is twice the length of the side 
 or diameter of the middle section, it follows that the area of the larger 
 base is equal to four times the area of the middle section. There- 
 fore, only one calculation is necessary for both areas. 
 
 7. Find the volume of the frustum of a square pyramid 
 whose altitude is 40 feet, each side of the lower base 12 
 feet, and of the upper base 10 feet. 
 
 Operation.— The face W X Y Z, is a 
 trapezoid. Therefore, i the sum of the 
 parallel sides, or i (12 + 10) or 11 feet is 
 the length of the line RS, the side of the 
 middle section. 
 
 B = 12X12 = 144 sq. ft. 
 
 b = ioxio = ioo *' 
 
 4m = 4 (11 X 11) = 484 sq. ft. 
 
 a = 40 
 Substituting in formula, f (B + b + 4m) 
 we have, \^ (144 + 100 + 484)= 4,853^ 
 
 jubic feet, volume. 
 N. B. — By this rule no root is extracted. 
 
ILL i'STRA Tl ] E moliLKMS 
 
 31 
 
 8. Find the volume of a eoiie, the diameter of whose 
 base is 20 feet, and altitude 36 feet. 
 
 Operation. — 20 -(-0-^2 = 10, diameter of middle 
 section. 
 B == lO*'' X 3.141G = 314. IG sq. ft, (Art. 60, R. Ill e) 
 o-=0 
 4m = 4(52X3.U16)=314.16 " 
 
 u = 3G 
 Substituting in formula, f (B + b + 4m) 
 we have, ^^ (314.16 + + 314.16) = 3,769.92 cubic 
 feet, volume. 
 
 b=o 
 
 
 \ 
 
 
 lUA 
 
 /^.,_ 
 
 ']>\ 
 
 / ^ 
 
 _\ 
 
 
 20 ^ 
 
 9. The altitude of the frustum of 
 a cone is 60 feet, the diameter of the 
 lower base 12 feet, and of the upper 
 base 8 feet; required the volume. 
 
 Operation.- 12"+^ h- 2 = 10, diameter of 
 middle section. 
 B = 62X3. 1416 = 113. 0976 sq.ft. (Art. 60, R.IIIc) 
 b = 4^X3.1416= 50.26.56 " *' 
 
 4m = 4(5-X3.1416) = 314.1G '' '' " 
 
 a = 60 
 Substituting in formula, |(B -f- b -f- 4m) 
 we have, ^i- (113.0976 + 50.2656 + 314.16)' = 
 4,775.232 cubic feet, volume. 
 
 From the principles of Geometrj' we have the follow- 
 ing : The volume of the frustum of a cone is equal to 
 the sum of the volumes of three cones, having for a 
 common altitude the altitude of the frustum, and for 
 bases the two bases of the frustum and a mean propor- 
 tional between them. This involves the extraction of 
 a root. In the above solution no root is extracted. 
 
32 
 
 Mi!:ysuiiATJoy 
 
 10. Find the volume of a sphere whose diameter is 40 
 inches. 
 
 Operation. — In the sphere any 
 diameter may be re-garded as the alti- 
 tude, and the area of any great circle, the 
 area of a middle section. The area of 
 each base is Zero. 
 
 b = 
 4m = 4 (20^ X 3.1416) = 5,026.56 
 sq.ft. (Art. 60, R. Ill c) 
 a = 40 
 
 trO 
 
 BiO 
 
 Substituting in formula, | (B -(- b -f- 4m) 
 
 we have, ^(0 + + 5,026.56)= 33,510.4 cubic inches, volume. 
 
 11. What is the volume of a prolate spheroid whose 
 shorter axis is 10 feet 
 and longer axis 15 
 feet '? 
 
 Operation.— Regard 
 the longer axis as the alti- 
 tude, and the shorter axis 
 as the diameter of the mid- 
 dle section. The area of 
 each base is Zero. 
 B = 
 b = 
 4in = 4 (5^X3.1416)= 314.16 sq. ft. (Art. 60, R. Ill c) 
 a = 15 
 ••Substituting in formula, -|(B + b + 4m) 
 we have, -^ (0 + + 314.16) = 785.4 cubic feet, volume. 
 
 Remark.— The problem can also be solved by using the shorter 
 axis as the altitude, and finding the area of the elliptical middle 
 section. 
 
ILL isTHA TI f L LROBLKMS 
 
 33 
 
 12. Required the volume of a wedge, the dat end of 
 which is a rectaiio:le 4 by 8 inches, 
 and the height 24 inches. 
 
 Operation,— B = 4 X S = 32 sq. in. 
 b = 
 4m = 4 (2X8) = 64 sq. in. 
 a = 24 
 Siibstitutincr in formula, -„- (B + b -f 4m 
 we have, -^o* (32 + + 64) = 384 cubic 
 inches, volume. 
 
 13. Required the contents 
 of a cylindrical ring, the 
 thickness of which is 2 
 inches, and the inner diame- 
 ter 12 inches. 
 
 Operation. — 12 + 2 + 2 = 16 
 inches, diam. of outer circle. 
 
 16X3.1416 = 50.2656 in. eir. outer circle. 
 12 X 3.1416 = 37_^992 " inner " - 
 
 2; 87.9648 [other two. 
 
 43.9824 "■ of a circle half the sum of the 
 
 Conceive the ring to be cut and then made straight. 
 
 It is now an equal solid in the form of a cylinder, any 
 
 cross section of the ring being equal to either base or 
 
 middle section of the cylinder, and the altitude of the equal 
 
 solid in the cylindrical form being equal to one -half the sum 
 
 of the inner and outer circumferences of the ring. Hence, 
 
 B= 12X3.1416 = 3.1416 sq. in. (Art. 60, R. Ill c) 
 
 b= 1-X3.1416 = 3.1416 *' " " 
 
 4111 = 4(1^X3.1416) =12.5664 '' " '' 
 
 a = 43.9824 
 Substituting in formula, f (B -|- b -|~ 4m) 
 we have, ^^-^--'-i (3.1416 + 3.1416 +12. 5664)=138. 1751+ 
 
 cubic inches, volume of the cylinder. 
 The volume of the cylinder is equal to the volume of 
 the cylindrical ring ; hence, the volume of the cylinirieal 
 ring equals 138.1751+ cubic inches. 
 
 C 
 
34 
 
 ME^' a U RATION 
 
 P\] 
 
 14. A certain monument has the 
 form of the frustum of a square 
 pyramid ; how many cubic feet of 
 material does it contain if it is 240 
 feet high, the lower base 3G feet 
 square, and the upper base 24 feet 
 square, provided that through the 
 center, from top to bottom, there is 
 an opening in the form of a frustum 
 of a cone whose lower diameter is 
 20 feet, and upper diameter 16 feet ? 
 
 .30N 
 
 -18-t 
 
 Operation. — Frustum of pyramid : 
 
 B= 36X3G =1,296 sq. ft. 
 b= 24X 24 = 576 " 
 4m = 4(30X30)=3,600 " 
 
 Substituting in formula, | (B -f- b + 4m) 
 
 we have, ^ (1,296 + 576 + 3,600)= 218,880 
 
 cubic feet volume if there were no opening. 
 
 -20- 
 
 Frustum of cone : 
 
 20-f-2 = 10 R of B. 
 16-4-2= 8 R of b. 
 18 -f- 2= 9 R of M. 
 B = 10^ X 3.1416 = 314.16 sq. ft. 
 b= 8- X 3.1416 = 201.0624 " 
 4m = 4 (9- X 3.1416) = 1,017.8784 sq. ft. 
 
 Substituting in formula, ^ (B-|-b + 4m) 
 we have, ^ (314.16 + 201.0624 + 
 1,017.8784) = 61,324.032 cubic feet in 
 frustum of cone to be deducted. Hence, 
 218,880 — 61,324.032 = 157,554.968 
 cubic feet of material. 
 
ILL UHTltATl \ K rUOBLKMS 
 
 15. Mr. Brown contracted 
 to build an embankment for 
 a railroad up a sliglitly 
 inclined plane, the bed for 
 the track to be level. The 
 embankment was to be 1 mile 
 long, and the width of the 
 bed for the ties 12 feet wide ; 
 the perpendicular measure- 
 ment of the higher end was 
 to be 20 feet, and at the lower 
 end 16 feet ; the width of 
 the embankment at the higher 
 end at the bottom was to be 
 3G feet, and at the lower 
 end 24 feet ; what was the 
 cost at 20 cents per cubic 
 yard ? 
 
 Operation.— ADEG, IJKL, and 
 STUV, which are B, b, and M, are 
 all trapezoids. Hence, 
 
 B= (36+12^-2)20 = 480 sq.ft. 
 b= (24+12 -J- 2)10 =288 " 
 4m = 45 (30+12 -^2)18i=l, 512 " 
 a = 5,280 ft. 
 
 Substituting in formula, 4(B+ b +4m) 
 we have, H" (4S0 + 2S8 + 1,512) = 
 2,000,400 cubic feet. 2,000,400 
 -^- 27 = 74,311^ cubic yards, 
 74,3lU X 20 cents = $14,802.22^ 
 cost. 
 
 ^^\ 
 
 m 
 
 ^*"4':-^ 
 
 "^h^i^ 
 
 
 
 en: 
 
36 
 
 Mt\\\ii:}L-irwy 
 
 16. A bridge -pier of solid masonry has a rectangular 
 base 10 by 40 feet ; the top of the pier is a rectangle 6 
 by 28 feet, and the perpendicular distance between the 
 two bases is 18 feet ; how many perches of material in 
 the pier ? 
 
 Operation.— 
 
 40 + 28h- 2 = 34 ft. side of middle section. 
 10+ 6-f-2= 8 ft. end " " " 
 
 B = 40 X 10 = 400 sq. ft. 
 
 b= 28 X 6= 168 
 4m = 4 (34 X 8) =1,088 " 
 a = 18 
 
 Substituting in formula, |. (B + b -(- 4m) 
 we have,^o^(400 + 168 + 1,088)= 4,908 cubic feet. 
 4,968 -f- 24.75 = 200.72+ perches, Ans. 
 
 17. A saddler uses a block for mak- 
 ing horse collars shaped like a cylin- 
 droid. Find the volume of such a 
 block whose lower base is an ellipse 
 having a transverse axis 8 inches long, 
 conjugate axis 4 inches long, and whose 
 upper base is an ellipse, the trans- 
 verse axis being 4 inches long, the con- 
 jugate axis 2 inches long, if the altitude 
 of the block is 30 inches. 
 
rOLUMES OF PRISMS AND CYLINDERS 37 
 
 Operation. — 
 
 B= 4X2 X 3.1416 =25.1328 sq. in. (Art. 61. R) 
 
 b= 2X1 X 3.1416 = 6.2832 " 
 4m = 4 (3 X HX 3.1416) =56.5488 " 
 
 a = 30 
 Substituting in formula, |(B + b + 4ra) 
 we have, ^(25. 1328 + 6.2832 + 56.5488)= 439.824 cubic inches, Ans. 
 
 {8 + 4-i-2 = 6 transverse axis of M. 
 4 + 2-^2 = 3 conjugate " " 
 
 VOLUMES OF PRISMS AND CYLINDERS 
 
 85. Rule I. To find the volume of a prism or cylin- 
 der : MuUlphj ilie area of the base by the altitude. 
 
 Rule II. Multiply the sum of both end areas and four 
 times the area of a section half-tuay between them, by one- 
 sixth the altitude. KB + b + 4m) . 
 
 PROBLEMS 
 
 In a prism or cylinder, given : 
 
 1. Side of square base 4 feet, altitude 30 feet, to find 
 the volume. Ans. 480 cu. ft. 
 
 2. Side of square base 3i feet, altitude 8 feet, to find 
 the volume. Ans. 98 cu. ft. 
 
 3. Side of regular hexagonal base 8 inches, altitude 10 
 feet, to find the volume. Ans. 11.546+ cu. ft. 
 
 4. Sides of a triangular base 12, 15, and 24 feet, alti- 
 tude 20 feet, to find the volume. Ans. 1,472.676+ cu. ft. 
 
 5. Diameter of the base 4 feet, altitude 30 feet, to find 
 the volume. Ans. 376.992 cu. ft. 
 
 6. Diameter of the base 6 inches, altitude 5 feet, to 
 find the volume. Ans. 1,696.464 cu. in. 
 
38 MENSURATION 
 
 7. Radins of the base 6 feet, altitude 20 feet, tu find 
 the vohime. Ans. 2,261.952 cu. ft. 
 
 8. Radius of the base 10 feet, altitude 30 feet, to find 
 the volume. Ans. 9,424.8 cu. ft. 
 
 9. Circumference of the base 31.416 feet, altitude 40 
 feet, to find the volume. Ans. 3,141.6 cu. ft. 
 
 10. Circumference of the base 80 feet, altitude 50 feet, 
 to find the volume. Ans. 25,464+ cu. ft. 
 
 11. Find the cost of a stick of timber 20 inches square 
 and 30 feet long, at 25 cents a cubic foot. Ans. $20.83i. 
 
 12. Find the number of cubic feet in a log 30 feet 
 long and 24 inches in diameter at each end. 
 
 Ans. 94.248 cu. ft. 
 
 13. Find the capacity in gallons of a cylindrical cis- 
 tern measuring 8 feet across and 20 feet deep. 
 
 Ans. 7,520.256 gals. 
 
 14. Find the value of a piece of moulding, the end of 
 which is a triangle whose sides are respectively 3, 4, and 
 5 inches, and length 24 feet, at 50 cents per cubic foot. 
 
 Ans. 50 cts. 
 
 15. What is the comparative weight of two poles of 
 the same material, one being 3 inches in diameter and 10 
 feet long, and the other 6 inches in diameter and 20 feet 
 long? Ans. latter =8 times former. 
 
 16. Find the volume of a prism whose base contains 
 8i square feet, and the square of whose height equals six 
 times the number of square feet in the base. 
 
 Ans. 57i cu. ft. 
 
 17. A lime kiln in the form of a prism is 12 feet 
 square aiid 24 feet high ; through the center of this, from 
 top to bottom, there is a cylindrical opening 6 feet in 
 diameter; required the number of perches of masonry in 
 the kilu. Ans. 112 218+P. 
 
PIBAMIDS AND CONES 39 
 
 PYRAMIDS AND CONES 
 
 DEFIWITIOWS 
 
 86. A Pyramid is a solid having for its base a poly- 
 gon and for its sides plane triangles, all terminating at 
 one point, called the vertex. 
 
 When the base is regular the pj-ramid is regular ; 
 otherwise irregular. 
 
 87. A Cone is a solid whose base is a circle, and 
 whose lateral curved surface tapers uniformly to a point, 
 called the vertex. 
 
 88. The Altitude of a pyramid or of a cone is the 
 perpendicular distance from the vertex to the plane of the 
 base. 
 
 89. The Axis of a pyramid or of a cone is the straight 
 line that joins the vertex with the center of the base. 
 
 90. A Rig'ht Pyramid or a Rig-ht Cone is one whose 
 axis is perpendicular to the base of the pyramid or cone. 
 The base of the pyramid is a regular polygon. 
 
 91. An Oblique Pyramid or an Oblique Cone is one 
 whose axis is oblique to the base of the pyramid or cone. 
 
 92. The Slant Heig-ht of a pyramid is the perpen- 
 dicular distance from its vertex to either of the sides of 
 the base. 
 
 93. The Slant Height of a cone is a straight line 
 drawn from the vertex to the circumference of the base. 
 
 94. A Rig-ht Cross Section of a pyramid or of a 
 
 cone is a section made by a plane perpendicular to the 
 axis. 
 
40 
 
 MENSURATION 
 
 95. A Frustum of a pyramid or of a cone is that part 
 which remains after cutting off the top by a plane paral- 
 lel to the base. 
 
 96. The Altitude of a Frustum of a pyramid or of 
 a cone is the perpendicular distance between the bases. 
 
 97. The Slant Heigrht of the Frustum of a right 
 pyramid or of a right cone is the shortest distance 
 between the perimeters of the bases. 
 
 98. A pyramid is triangular, quadrangular, pentago- 
 nal, hexagonal, etc., according as its base is a triangle, 
 quadrilateral, pentagon, hexagon, and so on. 
 
 Riglit Pyramid 
 
 (Triangular) 
 
 (.FUJ- 1^ 
 
 Eight Pyramid 
 
 (Hexagonal) 
 
 {Fig. 2) 
 
 Oblique Pyramid 
 (Quadrangular) 
 {Fig. 3) 
 
 Eight Cone 
 {Fig. 4) 
 
 In Fig. 4, if the right-angled triangle MON be 
 revolved around MO as an axis, it will generate the cone 
 NQP— M. 
 
 In Fig. 7, if the trapezoid MNOP be revolved around 
 MO as an axis, it Avill generate the frustum of the cone 
 QRP— N. 
 
 In Figs. 1, 2, 3, 4, 5, 6, and 7, the line representing 
 the altitude is marked a. 
 
 In Figs. 1, 2, 4, 6, and 7, the line representing the 
 slant hei'^ht is marked b. 
 
rVUAMlDS AM) (UM'IS 
 
 41 
 
 All the lateral faces of a ri^lit pyramid are equal tri- 
 angles ; the sum of tlie bases of these triangles is equal 
 to the perimeter of the base of the right pyramid. 
 
 The altitude of each lateral triangle is equal to the 
 slant height of the pyramid ; moreover, a cone may be 
 regarded as a pyramid of an infinite number of sides. 
 The reason for the following rule is, therefore, obvious : 
 
 Rule I. To find the conrex surface of a right pyra- 
 mid or of a right cone : Multiply the perimeter ov circum- 
 ference of the base hy one-half the slant heUjht. 
 
 To find the entire surface, add to the convex surface 
 the area of the base. The lateral surface of any })yra- 
 mid may be found by adding together the areas of the 
 lateral faces. 
 
 Oblique Coue 
 (Fig. 5) 
 
 rrubtiiiu of Right 
 Pyramid 
 
 r 
 
 
 
 
 / 
 
 
 
 y-^ 
 
 
 1 
 
 ru; 
 
 -llMll 
 
 of 
 
 i\ 
 
 :•?;» 
 
 ll Co 
 
 lie 
 
 
 c^ 
 
 lij. < 
 
 .1 
 
 99. Each lateral face of the frustum of a right pyra- 
 mid is a trapezoid. The slant height of the frustum is 
 equal to the perpendicular distance between the parallel 
 sides of the trapezoid. The perimeter of the lower base 
 of the frustum is equal to the sum of the lower bases 
 of the trapezoids forming the convex surface, and the 
 perimeter of the upper base of the frustum is equal 
 to the sum of the upper bases of the same tra]>ezoids. 
 Now, the frustum of a cone may be regarded as the 
 
42 MENS URA Tl ON 
 
 frustum of a pyramid of an iiitiuite iium])er of sides. 
 The rule for finding the area of a trapezoid is, therefore, 
 applicable. 
 
 Rule II. To find the convex surface of the frustum 
 of a rig-ht pj'ramid or cone : Multiply the sum of the perim- 
 eters or circumferences of the two bases hij one -half the 
 slant height. 
 
 For the entire surface, add to the convex surface the 
 area of both ends or bases. 
 
 Rule III. To find the volume of a pj'ramid, cone, 
 frustum of a pyramid, or frustum of a cone : Multiplf/ 
 the sum of hoth end areas and four times the area 
 of a section half ivay l)etween them, by one- sixth the 
 altitude. -^ (B + b + 4m) , or. 
 
 To find the volume of a pyramid or cone : Multiply 
 the area of the base by one-third, the altitude. 
 
 To find the volume of the frustum of a pyramid or of a 
 cone : To the sum of the areas of both bases add the 
 square root of their product, and multiply this sum by 
 one -third the altitude. 
 
 SURFACES OF PYRAMIDS AND CONES 
 AND THEIR FRUSTUMS 
 
 PROBLEMS 
 
 1. What is the lateral surface of a quadrangular pyra- 
 mid, the sides of whose base are each 5 feet and slant 
 height 30 feet? Ans. 300 sq. ft. 
 
 2. What is the entire surface of the pyramid mentioned 
 in Problem 1 ? Ans. 325 sq. ft. 
 
PYRAMIDS AXl) COXES 43 
 
 3. Wliat is the lateral surface of a hexagonal pyramid, 
 the sides of whose base are each 8 feet and slant height 
 40 feet? Ans. 960 sq. ft. 
 
 4. Find the cost of painting an octagonal church -spire 
 at 22 cents per square yard, the sides of whose base are 
 each 4 feet and slant height 60 feet. Ans. $23.46i. 
 
 5. Find the convex surface of a cone, the diameter of 
 whose base is 10 feet and slant height 20 feet. 
 
 Ans 314.16 sq. ft. 
 
 6. Find the entire surface of the cone mentioned in 
 Problem 5. Ans. 392.7 sq. ft. 
 
 7. How many square yards of canvas will it take to 
 make a conical tent whose center -pole is 8 feet high and 
 the diameter of the base 12 feet! Ans. 20.944 sq. yd. 
 
 8. Required the lateral surface of the frustum of a 
 square pyramid whose slant height is 40 feet, the side of 
 the lower base 20 feet, and upper base 16 feet. 
 
 Ans. 2,880 sq. ft. 
 
 9. AVhat is the entire surface of the frustum men- 
 tioned in Problem 8? Ans. 3,536 sq. ft. 
 
 10. What is the lateral surface of the frustum of a 
 hexagonal pyramid whose ^lant height is 20 feet, the sides 
 of the lower base each 6 feet, and upper base 4 feet ? 
 
 Ans. 600 sq. ft. 
 
 11. Find the convex surface of the frustum of a cone 
 whose slant height is 30 feet, the circumference of the 
 lower base 40 feet, and of the upper base 30 feet. 
 
 Ans. 1,050 sq. ft. 
 
 12. Find the entire surface of the frustum mentioned 
 in Problem 11. Ans. 1,248.943+ sq. ft. 
 
 13. What is the convex surface of the frustum of a 
 square pyramid whose slant height is 10 inches, the side 
 of the upper base being 18 inches and of the lower base 
 20 inches? Ans. 760 sq. in. 
 
44 MENSURATION 
 
 VOLUMES OF PYRAMIDS AND CONES 
 
 PROBLEMS 
 
 lu a pyramid, given : 
 
 1. The side of a square base 4 feet, altitude 30 feet, 
 to find the volume. Ans. 160 cu. ft. 
 
 2. The side of a square base 8 inches, altitude 21 feet, 
 to find the volume. Ans. 3 cu. ft., 192 cu. in. 
 
 3. The area of the square base 16 square j^ards, alti- 
 tude 30 yards, to find the volume. Ans. 160 cu. yd. 
 
 4. The sides of a rectangular base 60 X 40 feet, alti- 
 tude 90 feet, to find the volume. Ans. 72,000 cu. ft. 
 
 5. Find the volume of a square pyramid whose alti- 
 tude is 60 feet, and the side of whose base is 9 feet. 
 
 Ans. 1,620 cu. ft. 
 
 6. The base of a right triangular pyramid is an 
 equilateral triangle, each side of which is 12 feet and the 
 altitude 18 feet ; find the cubic contents. 
 
 Ans. 374.1228+ cu. ft. 
 
 7. The altitude of a pyramid is 8 feet, and its base is 
 a rectangle 3 feet by 2 ; find the volume. Ans. 16 cu. ft. 
 
 8. A certain pyramid has an altitude equal to twice 
 the diagonal of its base, which is a rectangle 60 by 80 
 feet; find its volume. Ans. 320,000 cu. ft. 
 
 9. Find the volume of a pyramid whose base is a rec- 
 tangle 40 feet by 30, and the length of the edges which 
 meet at the vertex 65 feet. Ans. 24,000 cu. ft. 
 
 10. A granite monument consists of a pedestal 24 
 inches square and 4 feet high, on which stands a pyramid 
 18 inches squai-e and 9 feet high; what did it cost at 
 $15 per cubic foot ? Ans. $341.25. 
 
In a cone, given : 
 
 11. The diameter of the base 6 feet, altitude 18 feet, 
 to find the volume. Ans. 1G9.6464 cu. ft. 
 
 12. The diameter of the base 2 feet, altitude 12 feet, 
 to find the volume. Ans. 12.5G64 cu. ft. 
 
 13. The radius of the base 10 inches, altitude 15 
 inches, to find the volume. Ans. 1,570.8 cu. in. 
 
 14. The circumference of the base 62.832 feet, altitude 
 30 feet, to find the volume. Ans. 3,141.6 cu. ft. 
 
 15. Find the solid contents of a cone whose altitude 
 is 48 feet and the diameter of whose base is 5 feet. 
 
 Ans. 314.16 cu. ft. 
 
 16. How many tons of hay in a conical stack 30 feet 
 high and 40 feet across the base, there being 500 cubic 
 feet to the ton? Ans. 25.1328 tons. 
 
 17. Mr. Johnston's crop of wheat piled on his barn 
 floor formed a heap in the form of a cone 10 feet high 
 and 62.832 feet in circumference ; what was it worth at 
 $1.25 per bushel? Ans. $1,051,865. 
 
 18. Mr. Brown's granary sprang a leak 2 feet above 
 the level of his barn floor ; the clover seed that ran out 
 of this opening formed a heap in the shape of a half -cone 
 2 feet high, and extended 2 feet at the bottom from the 
 perpendicular partition between the barn floor and the 
 granary; how many bushels leaked out? Ans. 3.3659+ bu. 
 
 19. The circumference of the base of a cone is 15.708 
 feet, and the altitude is equal to the diameter of the base ; 
 find its volume. Ans. 32.725 cu. ft. 
 
 20. Compare the volumes of a cone and a cylinder, the 
 altitude of each being 12 feet and the radius of the base 
 of each 4 feet. Ans. Cone = i vol. eyl. 
 
4G MKXSURATIOX 
 
 VOLUMES OF FRUSTUMS OF PYRAMIDS AND CONES 
 
 PROBLEMS 
 
 Given, in the frustum of a pyramid : 
 
 1. Upper side of square base, 10 feet; lower side, 
 12 feet ; altitude, 20 feet ; to find the volume. 
 
 Ans. 2,426f cu. ft. 
 
 2. Upper sides of rectangular base, 12 and IG feet ; 
 lower sides, 18 and 24 feet ; altitude, 30 feet ; to find 
 the volume. Ans. 9,120 cu. ft. 
 
 3. Upper sides of triangular base, each 6 feet ; lower 
 sides of triangular base, each 8 feet; altitude, 18 feet; 
 to find the volume. Ans. 384.50+ cu. ft. 
 
 4. Find the volume of the frustum of a square pja^a- 
 mid whose altitude is 30 feet, each side of the lower 
 base 24 feet, and of the upper base 18 feet. 
 
 Ans. 13,320 cu. ft. 
 
 5. Find the volume in cubic feet of a stick of timber 
 36 feet long, the larger end being 18 inches square and 
 the smaller end 15 inches square. Ans. 68t cu. ft. 
 
 6. How many cubic feet in the upright of a crane 
 30 feet high, the lower end being 2 feet square and the 
 upper end 18 inches square ? Ans. 922 cu. ft. 
 
 7. The lower base of the frustum of a pyramid is a 
 rectangle 20 by 40 inches ; the upper base, a rectangle 
 15 by 30 inches, and the altitude 12 feet ; what is the 
 volume I Ans. 51 cu. ft. 672 cu. in. 
 
 8. The smaller base of the frustum of a pyramid is 
 a rectangle 3 by 4 feet ; the longer side of the lower 
 base is equal to the diagonal of the upper base, and the 
 shorter side of the lower base bears the same ratio to 
 the longer side as the shorter side of the upper "base 
 
} DLrvKs or rinsTiwfs or ryu.t.yins jxn ro.v/;\ 47 
 
 hears to its longer side ; the altitude is equal to the 
 sum of the perimeters of both bases. Required the 
 volume. Ans. 480.375 cu. ft. 
 
 9. What are the solid contents of the frustum of a 
 cone Avhose upper l)ase is 10 feet in diameter, lower base 
 16 feet, and altitude 18 feet? Ans. 2,431.5984 cu. ft. 
 
 10. Find the number of eubie feet in a log 42 feet 
 long, the diameter of the lai-ger end being 3 feet and 
 of the smaller end 2 feet. Ans. 208.9164 eu. ft. 
 
 11. A telephone pole 48 feet long is 5 feet in cir- 
 cumference at the bottom and 3 feet at the top; required 
 the contents. Ans. 62.387+ cu. ft. 
 
 12. Find the capacity in gallons of a pail 18 inches 
 cieep, measuring 14 inches across the top and 12 inches 
 across the bottom. Ans. 10.363+ gal. 
 
 13. A tub 2 feet deep is 26 inches in diameter at the 
 bottom and 30 inches at the top ; how many pails of 
 water the size of the one described in Problem 12 will 
 be required to fill the tub! Ans. 6.1+ pails. 
 
 14. Mr. Long's cistern is 7 feet deep, the bottom 
 being 6 feet in diameter and the top 4 feet ; how many 
 barrels of water will it hold f Ans. 33.07+ bbl. 
 
 15. How many quarts of wine will be required to fill 
 2 dozen tumblers, each 2 inches across the bottom, 3 inches 
 across the top, and 4 inches deep? Ans. 8.26+ qt. 
 
 16. A butcher's pickle -stand is 5 feet deep, 6 feet in 
 diameter at tlie bottom, and 4 feet at the top ; how^ many 
 gallons will it contain '? Ans. 744.192 gal. 
 
 17. A farmer had a conical stack of hay 314.16 feet 
 in circumference at the bottom and 36 feet high. A 
 storm blew away the top so that the remainder was fo\md 
 
4S 
 
 Mi:XS(JiATI()X 
 
 to be only 18 feet high ; how many cubic feet remained 
 in the stack "? How many cubic feet blew away ? 
 
 Ans. Remained, 82,467 cu. ft.; blew away, 11,781 cu. ft. 
 
 18. At $15 per cubic foot, find the difference in the 
 cost of two granite monuments, one of which is the frus- 
 tum of a square pyramid 3 feet square at the top, 4 feet 
 square at the bottom, and 30 feet high ; the other is the 
 frustum of a cone 30 feet high, the diameter of the upper 
 base being 3 feet and of the lower base 4 feet. 
 
 Ans. $1,191.03. 
 
 THE SPHERE 
 
 DEFimXIOWS 
 
 100. A Sphere is a solid bounded by a uniformly 
 curved surface, evei-y point of which is equally distant 
 from a point within, called the 
 
 center. It may be generated by 
 the revolution of a semi- circle 
 about its diameter as an axis. 
 
 101. A Diameter of a sphere 
 is a straight line passing through 
 the center of the sphere, and termi- 
 nated at V)oth ends bv the surface. 
 
 ^.Ktcua 
 
 If^^e 
 
 /m 
 
 '■-HX\ 
 
 ?^rilJn^ 
 
 e t el V ^"^1 
 
 \\V" 
 
 ^W7 
 
 \\\ \ 
 
 1 f/y 
 
 \^ 
 
 ^>^ 
 
 Sphere 
 FigX 
 
 102. A Radius of a sphere is 
 a straight line drawn from the 
 center to any point in the surface. It is half the diameter. 
 
 103. A Zone is the portion of the surface of a sphere 
 included between two parallel planes. 
 
 104. A Spherical Segment is the solid portion of a 
 sphere included between two parallel planes. 
 
THE s rut: HE 49 
 
 105. A Spherical Sector is a volume generated by 
 the revolution of a circular sector al)out the diameter. 
 
 106. A Great Circle is a section made by any plane 
 passing through the center of a sphere. 
 
 Rule I. To find the surface of a sphere : Multiply the 
 circumferenci of the sphere hy the diameter. Or, S = 7rD^. 
 
 Rule II. Multiply the square of the radius by four times 
 3.1416. Or, 8 = 47^R^ 
 
 Rule III. To find the volume of a sphere : Multiply 
 the sum of both end areas and four times the area 
 of a section half way between them, by one- sixth the alti- 
 tude. |(B-f-b+4m) 
 
 Rule IV. Multiply the surface by one- sixth the diam- 
 eter, or one -third the radius. 
 
 Rule V. Multiply the cube of the diameter by one-sixth 
 
 of 3.1416, or by .5236. Or, V = 'L^. 
 
 107. The spliere may be regarded as a polyedron 
 consisting of an infinite number of pyramids whose com- 
 bined bases form the surface of the sphere and whose 
 equal altitudes are equal to the radius of the sphere. 
 Since the volume of each pyramid is equal to its base 
 multiplied by one -third of its altitude, and since the 
 whole is equal to the sum of all its parts, Rule IV is 
 readily derived. 
 
 The cylinder, the cone, and the sphere are the three 
 round bodies of Geometry. There is a fixed relation 
 existing between these three solids, both as to their sur- 
 faces and their volumes. See Problems 22 and 23, Art. 
 il4. 
 
.')( ) MEXS URA TI ON 
 
 SURFACES OF THE SPHERE 
 
 PROBLEMS 
 
 111 a sphere, given : 
 
 1. The diameter 8 inches, to find the surface. 
 
 Ans. 201.06+ sq. in. 
 
 2. The diameter 18 inches, to find the surface. 
 
 Alls. 1,017.9 — sq. in. 
 
 3. The diameter 10 inches, to find the surface. 
 
 Ans. 314.16 sq. in. 
 
 4. The radius 3 inches, to find the surface. 
 
 Ans. 113.1 — sq. in. 
 
 5. The radius 8 inclies, to find the surface. 
 
 Ans. 804.25 — sq. in. 
 
 6. The radius 10 inches, to find the surface. 
 
 Ans. 1,256.6+ sq. in. 
 
 7. The circumference 62.832 inches, to find the 
 surface. Ans. 1,256.6+ sq. in. 
 
 8. The circumference 15.708 inches, to find the 
 surface. Ans. 78.54 sq. in. 
 
 9. The circumference 251.328 inches, to find the 
 surface. Ans. 20,106.24 sq. in. 
 
 10. How many square inches of tissue paper will be 
 
 required to wrap 20 oranges each 2 inches in diameter, 
 
 provided no allowance is made for waste ? 
 
 Ans. 251.328 sq. in. 
 
 11 . How many square inches of pig skin in a spherical 
 football 9 inches in diameter? Ans. 254.47 — sq. in. 
 
 12. How many square inches of bronze will it take 
 to cover the three balls used as a sign of a pawn- 
 broker, if each ball is 4 inches in diameter ? 
 
 Ans. 150.796+ sq. in. 
 
voiA.MEs OF THE sriiu:!-: 51 
 
 VOLUMES OF THE SPHERE 
 
 PROBLEMS 
 
 In a sphere, given : 
 
 1. The diameter 2 feet, to find the volume. 
 
 Ans. 4.1888 en. ft. 
 
 2. The diameter 5 inches, to find the volume. 
 
 Ans. 65.45 cu. in. 
 
 3. The radius 23 inches, to find the volume. 
 
 Ans. 50,965.1296 cu. in. 
 
 4. The circumference 37.6992 feet, to find the volume. 
 
 Ans. 904.7808 cu. ft. 
 
 5. How much will 16 croquet balls each 4 inches in 
 diameter weigh, if a cubic inch of the wood of which 
 they are made weighs i ounce! Ans. 16 lbs. 12.08+ oz. 
 
 6. If the croquet balls mentioned in Problem 5 were 
 placed in a rectangular box 16 inches long, 8 inches deep, 
 and 8 inches wide, how many quarts of sand (dry measure) 
 would be required to fill all the remaining space and make 
 the box level full f Ans. 7.259+ qt. 
 
 7. Admiral Dewey at the battle of Manila used a num- 
 ber of 8 -inch guns ; what would be the weight of a solid 
 spherical ball made to fit these guns, if a cubic foot of 
 iron weighs 450 pounds? Ans. 69.81251b. 
 
 8. A sphere is 20 inches in diameter ; what is the 
 volume of an inscribed cube f Ans. 1,539.598+ cu. in. 
 
 9. The edge of a cube is 20 inches ; what is the volume 
 of an inscribed sphere ? Ans. 4,188.8 cu. in. 
 
 10. The outside diameter of a hollow globe of uniform 
 thickness is 16 inches and the inside diameter 14 inches ; 
 how roany cubic inches of metal in the shell ? 
 
 Ans. 707.9072 cu. in. 
 
52 MEy.suiuriox 
 
 11. The main battery of the battleship Oregon has 
 four 13 -inch, eight 8 -inch, and four 6 -inch guns ; if solid 
 splierical balls were used instead of projectiles, what 
 weight of metal would be required to load these once 
 round, there being 450 pounds of the metal used to the 
 cubic foot? Ans. 1,874.59+lb. 
 
 12. A hollow sphere having an inside diameter of 12 
 inches was filled with water ; if a solid ball 3 inches in 
 diameter, a solid cylinder having a base 2 inches in 
 diameter and 4 inches high, and a solid square pyramid 
 whose base is 3 inches on a side and altitude 6 inches, 
 were placed inside the sphere so as to displace part of 
 the water, how many cubic inches of water would 
 remain! Ans. 860.0772 cu. in. 
 
 THE SPHEROID 
 
 DEFINITIONS 
 
 108. A Spheroid is a volume formed by the revolu- 
 tion of an ellipse about one of its axes. A revolution 
 about the longer axis forms a 
 prolate spheroid ; and about the 
 shorter axis, an oblate spheroid. 
 
 Rule I. To find the volume of 
 a spheroid : Multiply the sum 
 of both end areas and four times 
 the area of a section half way be- 
 tween them, by one -sixth the alti- 
 tude. i(B + b + 4m). 
 
 Note. — For the spheroid each end area is Zero, and either axis 
 may be used as the altitude, according as the area of the middle 
 section is found for the prolate or oblate spheroid. 
 
 Rule II. Multiply the square of the revolving axis hu 
 the fixed, axis and this product by .5236. 
 
 Spheroid 
 
 or 
 
 Ellipsoid 
 
 Fig.l 
 
VOLUMES OF THE SPHEROID 53 
 
 VOLUMES OF THE SPHEROID 
 PROBLEMS 
 
 1. What is the vohime of a spheroid whose longer 
 diameter is 24 iuches and shorter diameter 20 inches ? 
 
 Ans. 5026.56 cu. in. 
 
 2. Find the volume of a stone in the shape of a 
 spheroid whose longer diameter is 4 feet and shorter 
 diameter 2 feet. Ans. 8.'J776 cu. ft. 
 
 o. A watermelon is shaped like a spheroid ; what 
 are its contents, if its longer diameter is 18 inches and 
 shorter diameter 12 inches"? Ans. 1,357.1712 cu. in. 
 
 4. How many cubic inches of air in a spheroidal 
 football whose longer inside diameter is 14 inches and 
 shorter diameter 9 inches? Ans. 593.7624 cu. in. 
 
 5. Which has the greater volume and how much, a 
 sphere 12 inches in diameter, or a spheroid whose longer 
 axis is 8 inches and shorter axis 4 inches f 
 
 Ans. The sphere, 837.76 cu. in. greater. 
 
 6. Which has the greater volume and how much, a 
 spheroid whose longer diameter is 30 inches and shorter 
 diameter 20 inches, or the frustum of a square pyramid 
 whose lower base is 10 inches square, upper base 8 
 inches square, and whose altitude is 30 inches ? 
 
 Ans. The spheroid, 3843.2 cu. in. greater. 
 
 7. A sphere whose radius is 8 inches and a sphei'oid 
 having a longer diameter of 18 inches and shorter diam- 
 eter of 12 inches, were placed in a tub in the form of a 
 frustum of a cone ; the lower diameter of the tub was 
 2 feet, the upper diameter 26 inches, and depth 18 
 inclies ; how many gallons of water would have been 
 required to fill the remaining space? Ans. 23.11-f gai. 
 
54 
 
 MENSURATION 
 
 CIRCULAR RINGS 
 
 DEFINITIONS 
 
 109. A Circular Ring" is formed by bending a cylin- 
 der or a bar nut 11 the 
 
 Circular Rings 
 
 two ends meet. 
 
 It will be observed by 
 examining Figs. 1 and 3 
 that if the ring is cut 
 at any place and made 
 straight the same volume 
 will be changed into the 
 form of a cylinder. The 
 altitude of the cylinder 
 will equal one -half the 
 sum of the inner and 
 outer circumferences, and either 
 base of the cylinder will be equal 
 to any cross section of the ring. 
 The rules for finding both the 
 surface and the volume, therefore, 
 are practically the same as those 
 for finding the convex surface 
 and volume of a cylinder. 
 
 Rule I. To find the surface of a circular ring : 
 Multiply the perimeter of a cross section of the ring, 
 or its girt, by one -half the sum of the inner and outer 
 circumferences. 
 
 Rule II. To find the volume of a circular rhig : 
 Multiply both end areas (conceiving the ring to he pi^ 
 made straight) and four times the area of a section 
 half way between them, by one -sixth the altitude. 
 
 |(B + b-j-4m). 
 
 Rule III. Multiply the area of a rrosH sortion of the ring 
 by one -half the sum of the inm r and outer circumfprrnces. 
 
 Fig.Z 
 
 X 
 
SriiFACl'JS OF ClRl'CLAR JilXGS 55 
 
 SURFACES OF CIRCULAR RINGS 
 
 PROBLEMS 
 
 1. What is the surface of a pneumatic bicycle tire 
 ■whose outside diameter is 28 inches and thickness 
 li inches? Ans. 392.3186+ sq. in. 
 
 2. A hose whose outside diameter is 2 inches was 
 placed in the form of a ring- ; the outer circumference 
 measured 62.832 feet ; required the surface of the hose. 
 
 Ans. 32.6246 + sq.ft. 
 
 3. The felloes of a heavy lumber wagon were made 
 2 inches deep and 4 inches wide ; the outside diameter 
 of each hind wheel, inside the tire, was 4 feet 4 inches ; 
 how many square f ^et of surface are there to paint 
 on each felloe of the hind wheels, provided the four sides 
 are painted before putting on the tire, and no allowance 
 is made for shaving off the corners on the inside ? 
 
 Ans. 13.09 sq. ft. 
 
 4. A semi -circular white oak arch of uniform thick- 
 ness and 3 inches in diameter, was made to span the 
 main street of a town ; the distance on the inside 
 between the ends of the poles where buried in the ground 
 was 30 feet ; required the number of square feet of bun- 
 ting needed to cover the arch. Ans. 37.3196+ sq. ft. 
 
 VOLUMES OF CIRCULAR RINGS 
 
 5. What is the volume of a solid iron ring of the 
 same dimensions as the pneumatic bicycle tire in 
 Problem 1 ? Ans. 147.1194+ cu. in. 
 
 6. Find the solid contents of a grape vine of exactly 
 the same shape and dimensions as the hose in Problem 2. 
 
 Ans. 1.3593+ cu. ft. 
 
56 
 
 MENSURATION 
 
 7. Find the number of cubic inches in each felloe 
 described in Problem 3. Ans. 1,256.64 cu. in. 
 
 8. What is the weight of the white oak arcii in Prob- 
 lem 4, if a cubic foot of wood weighs ol pounds? 
 
 Ans. 118.956+ lb. 
 
 THE WEDGE 
 
 DEFHriTIOWS 
 
 110. A Wedg'e is a volume of five sides, having a 
 rectangular base, two rectangular or trapezoidal sides 
 meeting in an edge, and two triangular ends. 
 
 Fig.\ 
 
 Fi(/.2 
 
 Wcdgt 
 
 Fig:d 
 
 FigA 
 
 Rule I. To find the surface of a wedge : Find the 
 sum of the areas of the jive faces. 
 
 Rule XL To find the volume of a wedge : Midtiphj 
 the sum of the two end areas and four times the area of 
 a section half way between them, by one-sixth the altitude. 
 i(B + b + 4m). 
 
 In the wedge, the end area where the sides meet is Zero. 
 
 SURFACES OF THE WEDGE 
 PROBLEMS 
 
 To find the surface of a wedge, given ; 
 1. Base of wedge, a rectangle 6x4 inches; sides, equal 
 •ectangles ; ends, equal isosceles triangles ; altitude of 
 
SUIiFACliS or THE WKDCE 57 
 
 wedge, 12 inches; altitude of rectaiio;le, 12.1655 inches. 
 (Fig-. 2.) Ans. 217.986 sq. in. 
 
 Note. — The altitudes of the wedgos are given in these surface 
 problems ; from these the pupil may be required to find the alti- 
 tudes of the ends and sides, when time will permit. 
 
 2. Base of wedge, a square 4x4 inches; sides, equal 
 trapezoids 4 inches along the base and 8 inches along 
 sharp edge ; ends, two equal isosceles triangles ; altitude 
 of wedge, 6 inches ; altitude of each trapezoid, 6.3246 
 inches ; altitude of each triangle, 6.3246 inches. (Fig. 3.) 
 
 Ans. 117.1936 sq. in. 
 
 3. Base of wedge, a square ^^Q> inches ; sides, 2 equal 
 trapezoids whose edges next the base are each 6 inches 
 and sharp edge 4 inches ; ends, two equal isosceles tri- 
 angles ; altitude of wedge, 8 inches ; altitude of trape- 
 zoids, 8.544 inches ; altitude of triangles, 8.0623 inches. 
 (Fig. 4.) Ans. 169.8138 sq. in. 
 
 VOLUMES OF THE WEDGE 
 
 To find the volume of a wedge, given : 
 
 4. The base of the wedge, a rectangle 16 x 12 inches ; 
 the sharp edge, 16 inches ; triangular ends both perpen- 
 dicular to the base of the wedge ; altitude of the base, 21 
 inches. (Fig. 2.) Ans. 2,016 cu. in. 
 
 5. The base of the wedge, a square 12 x 12 inches; the 
 sharp edge, 20 inches ; the altitude of the wedge, 30 
 inches. (Fig. 3.) Ans. 2,640 cu. in. 
 
 6. The base of the wedge, a square 8x8 inches ; sharp 
 edge, 6 inches ; altitude of wedge, 12 inches. (Fig. 4.) 
 
 Ans. 352 cu. in. 
 
 7. A plane was passed from the upper edge of a cube, 
 which edge is 6 inches, to the lower parallel edge on the 
 opposite side of the cube : i*e(iuired the volume of eitlier 
 wedge thus made. (Fig. 1.) An.->. 108 cu. in. 
 
58 MEX!SURAri(JX 
 
 , SIMILAR SOLIDS 
 
 DEFINITIONS 
 
 111. Similar Solids are such as have the same 
 form, and differ from each other only in size ; as, cubes, 
 
 i spheres, etc. 
 
 112. A Dimension of a solid is a radius, diameter, 
 circumference, height, length, breadth, etc. 
 
 113. The following principles of similar solids are 
 derived from geometry : 
 
 Principle I. Similar solids are to each other as the 
 cubes of their Wke dimensions. 
 
 Principle II. Lil-e dimensions of similar solids are to 
 each other as the cube roots of those solids, 
 
 PROBLEMS 
 
 1. If the volume of a cube 4 inches on each side is 
 64 cubic inches, what is the volume of one 5 inches on 
 each side ? 
 
 Operation. — 4^ : 5^ :: (34 cu. in. : X. X = 12.5 eii. in., Ans. 
 
 2. How many baseballs each 2 inches in diameter 
 are equal in volume to one large ball 1 foot in 
 diameter! Ans. 21G. 
 
 3. If a conical stack of wheat in the sheaf 20 feet 
 high contains 400 bushels, how many bushels should a 
 similar stack 30 feet high contain? Ans. 1,350. 
 
 4. A farmer can load on wagons the contents of a 
 cubical bin 10 feet on a side in half a day ; how long, 
 at the same rate, will it take him to load the contents 
 of a cubical bin 20 feet on a side ? Ans. 4 days. 
 
SIMILAR SOLIDS 59 
 
 5. If a colt 13^- hands high weighs 1,000 pounds, 
 how much should he weigh when he becomes a horse 
 15 hands high? Ans. 1,371.7+ lbs. 
 
 G. If a barrel whose bung -diameter is 38 inches, 
 when filled contains G-1 gallons, what would be the bung- 
 diameter of a similar cask that holds but 27 gallons ? 
 
 Ans. 28^ in. 
 
 7. If the volume of a pyramid whose altitude is 25 
 feet is 729 cubic feet, what is the altitude of a similar 
 pyramid whose volume is 9,261 cubic feet? 
 
 Ans. 58i ft. 
 
 8. If a cylindrical tank 2 feet in diameter contains 
 4 barrels of water, what is the capacity of a similar 
 tank 4 feet in diameter? Ans. 32 bbl. 
 
 9. John is 4 feet tall and weighs 64 pounds ; what 
 should be his height when he weighs 216 pounds, pro- 
 vided he develops proportionately? Ans. 6 ft. 
 
 10. If it requires 311 tons of ice to fill a cubical 
 ice house whose edge is 10 feet, how many tons will be 
 required to fill one whose edge is 20 feet ? 
 
 Ans. 250 tons. 
 
 11. The diameters of three globes are respectively i, «, 
 and f of an inch ; what is the diameter of a globe equal 
 to the volume of the three? Ans. f in. 
 
 12. The patriotic citizens of a certain city wished to 
 raise $8,000 for the purpose of erecting a monument to 
 the heroes Avho went down with the "Maine" in Havana 
 liarbor. The monument was to be the frustum of a 
 square pyramid 50 feet high. Patriotism, ran so high, 
 however, that the sum of $15,625 was raised; how high 
 a monument did the latter sum erect, provided the orig- 
 inal design of the monument was carried out? 
 
 Ans. 62^ ft. 
 
GO MENSURATION 
 
 MISCELLANEOUS PROBLEMS 
 
 114. Review. 
 
 1. Required the distance between an upper corner and 
 an opposite lower corner of a cliurch 96 feet long, 72 feet 
 wide, and 25 feet high. Ans. 122.576+ ft. 
 
 2. How many square yards of plastering were required 
 to cover the walls and ceiling of the church described in 
 the preceding problem, deducting 80 square yards for 
 openings? Ans. 1,6213" sq. yds. 
 
 3. John's bicycle wheels are each 28 inches in diame- 
 ter ; how many times will they revolve in going 8,796.48 
 inches ? Ans. 100 times. 
 
 4. A gas receiver in a certain city is a cjdinder 40 feet 
 in diameter ; on an average during the j'ear each morning 
 it is 6 feet deeper in the water than the evening before ; 
 how many cubic feet of gas are consumed each night? 
 
 Ans. 7,539.84 cu. ft. 
 
 5. The diameter of a cylindrical vessel filled with 
 water is 10 inches. An immersed stone displaces 2 
 inches of the depth of the water. How many cubic 
 inches are there in the stone? Ans. 157.08 cu. in. 
 
 6. A teamster hauled from a mill 40 quadrangular 
 posts, each 6 by 8 inches and 12 feet long. How many 
 triangular posts, the base of each of which is a triangle 
 measuring 6, 8, and 10 inches on a side respectively, and 
 the same length, could he have hauled, provided the 
 loads were equal in weight ? Ans. 80 posts. 
 
 7. A hall 60 by 120 feet has a roof in the form of a 
 wedge. Both ends are equal triangles and both sides 
 equal trapezoids ; tlie two sides meet at the ridge, which 
 
MiSCKLLAXEors PROBIJIMS 61 
 
 is 80 feet long and 30 feet above the top of the walls. 
 Required the length of a rafter extending from a corner 
 of the building to the ridge f Ans. 46.904+ ft. 
 
 8. Required the number of square feet in the entire 
 roof as described in the preceding problem. 
 
 Ans. 10,648.58-1- sq.ft." 
 
 9. How many cubic feet of air are enclosed by the 
 roof and attic floor of the hall described in Problem 7 ? 
 
 Ans. 96,000 cu. ft. 
 
 10. In cutting the grass on a lawn 60 feet wide and 
 containing 4,800 square feet, how many times must a 
 lawn-mower 18 inches wide be drawn over it lengthwise 
 so that the whole may be cut? Ans. 40 times. 
 
 11. Mr. Clark, wishing to give his three boys prac- 
 tical lessons in both industry and foresight, made the 
 following proposition : " You may each have sufficient 
 lumber to build 60 rods of fence, with which you may 
 enclose as much land as you can ; all that you raise on 
 the land thus enclosed ma}^ be your own ; I should like 
 to see you make as much of my offer as you possibly 
 can." The oldest enclosed a cirr^ular piece ; the second, 
 a square piece, and the youngest, a rectangular piece 
 twice as long as it was broad ; how many square rods 
 did each have in his field f 
 
 Ans. Oldest, 286.478+ sq. rds.; second, 225 sq. rds.; 
 youngest, 200 sq^ rds. 
 
 12. Find the length of a hand-rail for a flight of 
 stairs having 20 steps, each step ll inches high and 
 9 inches wide. Ans. 19.525+ ft. 
 
 13. In Brandon Park there is a regular octagonal 
 stand. Surrounding this stand there is a porch of uni- 
 form width. The perimeter of the stand is 6I3 feet; the 
 
{\'2 MJ:Xsri!ATf()X 
 
 perimeter of the outer edge of the porch is 94f feet ; the 
 perpendicular distance between the stand and the outer 
 edg-e of the porch, or the width of the porch, is 5 feet 2 
 inches. How many square feet of flooring in the porch 
 floor! Ans. 403 sq. ft. 
 
 14. The frame and handle of a corn -popper is made of 
 a continuous piece of wire. The handle is 12 inches long, 
 and the frame upon which the wire sieve is fastened is a 
 rectangle 8 by 4 inches ; how long a wire was required to 
 make this frame and handle, making no allowance for the 
 thickness of the wire ? What is the combined length of 
 the handle and frame? Ans. 36 in.; 20 in. 
 
 15. A lamp having a paper shade is placed on a stand 
 in the center of a room 18 by 20 feet ; the shade casts on 
 the ceiling a shadow in the form of a circular ring extend- 
 ing the entire width of the ceiling and uniformly 5 feet 
 wide. How many square feet of the ceiling are not 
 shaded? Ans. 155.796 sq. ft. 
 
 16. Mr. Jones gave each of his two sons half a dollar 
 to spend. The younger invested his in 5 -cent tickets for 
 a merry-go-round, each ticket allowing him to ride on 
 the merry-go-round, 80 feet in circumference, 10 times. 
 The older bought a railroad ticket at 3 cents a mile. How 
 many miles more did the older ride than the younger ? 
 
 Ans. 15.15 mi. 
 
 17. The width of Mr. F's buggy from center of tire to 
 
 center of tire is 4 feet 9 inches ; the diameter of the fore 
 wheel is 43 inches and of the hind w^heel 46 inches. This 
 buggy is drawn round a circular race -course so that the 
 outside fore wheel leaves a track exactly one mile long. 
 The buggy is so coupled that the track made by the out- 
 side hind wheel, from center of track to center of track, 
 is at all places one inch distant from the track rnade by 
 
Mis(ELLAX/:()rs I'uoni.KMs 68 
 
 the out? Idc fore wheel. How many revolutions does each 
 
 wheel of the buggy make in going once round the track? 
 
 Ans. Outside fore wheel, 469.02+ times ; outside 
 
 hind wheel, 438.39+ times ; inside front wheel, 
 
 406.37+ times ; inside hind wheel, 435.91+ times. 
 
 18. A mechanic converted a cylindrical piece of iron 2 
 inches in diameter and 2 feet long into a wire i of an inch 
 in diameter; what was the length of the Avire? Ans. 128 ft. 
 
 19. Lamar 'township purchased tiling for drainage. 
 Each section of the tiling is a hollow cylinder 28 inches 
 high, the outer diameter being 36 inches and the inner 
 diameter 31 inches ; how many cubic feet of material in 
 each section, no allo^vance being made for the flange? 
 
 Ans. 4.2633+ cu. ft. 
 Note. — The pupil should work the above problem by using the 
 area of the circular ring as a base, and also by finding the volume 
 of two cylinders and taking their difference, 
 
 20. Each side of the lower base of the frustum of a 
 square pyramid is 4 feet, of the upper base 3 feet, and 
 slant -height 20 feet. The perimeter of the lower base 
 of the frustum of a cone is 16 feet, of the upper base 
 12 feet, and slant-heigat 20 feet. How do the lateral 
 surfaces compare ? How^ do the entire surfaces compare ? 
 
 Ans. 6.83+ sq. ft.; frus. cone greater. 
 
 21. Mr. Johnson's heater has a cylindrical maga- 
 zine 2 feet deep and 10 inches in diameter ; how many 
 shovelfuls of coal will l)e required to fill it, if 1 shovelful 
 weighs 8 pounds, and coal weighs 60 pounds to the 
 cubic foot? Ans. 8.1812.3 shovelfuls. 
 
 22. Compare the volumes of a cylinder, a sphere, and 
 a cone, whose dimensions are as follows : Cylindei — 
 diameter of base and altitude, each 20 inches ; sphere— 
 
64 MKNSUKATIOX 
 
 diameter, 20 inches ; cone — diameter of base and alti- 
 tude, each 20 inches. 
 
 Ans. 2 vol. cyl.= o vol. sphere; 1 vol. cyl.= 3 vol. 
 cone ; 1 vol. sphere = 2 vol. cone. 
 
 23. Compare the entire surfaces of the cylinder and 
 sphere given in the preceding: problem. 
 
 Ans. 2 sur. cyl.= 3 siir. sphere. 
 
 24. A globular piece of wax 6 inches in diameter, 
 a conical piece whose base is 6 inches and altitude 
 12 inches, and a cylindrical piece w^hose diameter is 
 6 inches and altitude 12 inches, were melted and then 
 made into one cubical piece ; required the edge of 
 the cube. Ans. 8.26+ in. 
 
 25. A farmer placed on the level ground a heap of 
 apples in the form of a cone whose diameter was 6 feet 
 and altitude 6 feet ; this he covered with earth of uni- 
 form thickness, making when complete a cone similar 
 to the heap of apples ; the diameter of the larger cone 
 thus formed was 8 feet. How many cubic feet of earth 
 in the covering? Ans. 77.4928 cu. ft. 
 
 26. A granite monument consists of three pieces — a 
 rectangular solid, a frustum of a square pyramid, and a 
 globe ; the rectangular solid is 6 feet square and 4 feet 
 high ; the lower base of the frustum of the pyramid is 
 5 feet square, the upper base 3 feet square, and altitude 
 24 feet ; the globe is 2 feet in diameter. Required the 
 weight of the entire monument, if a cubic foot of granite 
 weighs 168 pounds? Ans. 45.3758+ tons. 
 
 27. A hollow squash shaped like an ellipsoid has an 
 outside longer diameter of 12 inches and outside shorter 
 diameter of 8 inches ; the average thickness of the sub- 
 stance is 2 inches. How many cubic inches of material 
 in the squash? Ans. 335.104 cu. in. 
 
.viscKLLAyKocs i'i:(}r>ij:Ms 
 
 (if) 
 
 28. The lower joint of a stove pipe is 2 feet long ; 
 the upper end is a circle 6 inches in diameter, and the 
 lower end an ellipse whose transverse axis is 7 inches and 
 conjugate axis 5 inches. How many cubic inches of 
 sand would be required to fill it? Ans. 672.3024 cu. in. 
 
 29. The accompanying diagram represents a regular 
 hexagon inscribed in a circle, the radius of which is 6 
 inches. Each side of a regu- 
 lar inscribed hexagon is equal 
 to the radius of the circum- 
 scribing circle. Hehce, UB = 
 6 inches, and any side of the 
 hexagon, as BC = 6 inches. 
 
 (a) Find the length of 
 GH. Ans. 5.19615+ in. 
 
 (h) Find in two ways the 
 area of the triangle GBC. 
 
 Ans. 15.58845 sq. in. 
 
 (c) Find in two ways the area of the rhombus ABCG. 
 
 Ans. 31.1769 sq. in. 
 
 (d) Find in two ways the area of the trapezoid ABCD. 
 
 Ans. 46.76535 sq. in. 
 
 (e) Find the area of the hexa- 
 gon by using the area of GBC as 
 a basis. 
 
 (/) Find the area of tlie hexa- 
 gon by using the area of ABCG as 
 a basis. 
 
 {(j) Find the area of the hexa- 
 gon by using the area of ABCD as 
 a basis. 
 
 (h) Find the area of the circumscribing circle. 
 
 Ans. 113.0976 sq. in. 
 
 Ans. 93.5307 sq. in. 
 
66 M i:\siiunox 
 
 (l) Fiud the area of the 6 segments. 
 
 Ans. 19.5669 sq. in. 
 
 30. A circle whose radius is 6 inches is circumscribed 
 b}' a square, and also has a square inscribed (Art. 48, 
 
 Fig-. 2): 
 
 (a) Compare the length of the side of the circum- 
 scribed square, the diameter of the circle, and the 
 diagonal of the inscribed square. 
 
 (b) Find the perimeter of the circumscribed square. 
 
 Ans. 48 in. 
 
 (c) Find the circumference of the circle. 
 
 Ans. 37.6992 in. 
 
 (d) Find the perimeter of the inscribed square. 
 
 Ans. 33.9408 in. 
 
 (e) Find the area of the circumscribed square. 
 
 Ans. 144 sq. in. 
 (/) Find the area of the circle. 
 
 Ans. 113.0976 sq. in. 
 
 ig) Find the area of the inscribed square. 
 
 Ans. 71.9986+ sq. in. 
 
 31. In 1899 the Williamsport Gas Company erected a 
 cylindrical vat 100 feet in diameter and 100 feet high ; 
 required its capacity. Ans. 785,400 cu. ft. 
 
 SUPPLEMENT 
 
 DERIVATION OF THE PRISMOIDAL FORMULA FROM THE 
 PRINCIPLES OF GEOMETRY 
 
 A Prismoid is composed of prisms, wedges, and pyra- 
 mids. It is desired to find a common expression for find- 
 ing the volume of these three volumes. From the prin- 
 ciples of GeoRietry we get the following : 
 
srrrfj:yi:\r 
 
 07 
 
 Volume of Prism = area of base X V)y altitude. 
 Volume of Wedgo = area of base X by -^ altitude. 
 Volume of Pyramid = area of base X ^^y ^ altitude. 
 
 area of either base of prism, 
 area of the base of wedge, 
 area of the base of pyramid, 
 length or altitude of each. 
 
 Let B: 
 
 and let a 
 
 Then, Vol. Prism = Ba 
 Vol. Wedge =-^\ 
 Vol. Pyramid = -^* 
 
 Rediicing these 
 three expres- 
 sions to u 
 common de- 
 nominator, we 
 have the fol- 
 lowing : 
 
 Vol. Prism = Ba = -(j- 
 Vol. Wedge = 4=^- = «.|^ 
 Vol. Pyramid =4=*-=-!* 
 
 Now, let b = 
 
 and let m 
 
 the area of a cross -section of the pyramid at the apex. 
 
 the area of a cross-section of the wedge at its cutting 
 edge. 
 
 the area of a cross -section of the prism at corre- 
 sponding end of prism. 
 
 the area of a cross -section of the pyramid half way 
 
 between base and apex, 
 the area of a cross -section of the wedge half way 
 
 between base and sharp edge, 
 the area of a cross -section of the prism half way 
 
 between the two bases. 
 
 In the Prism 
 
 In the Wedge 
 
 im = 
 
 = B 
 B 
 
 b = 
 
 iB 
 
 f b = 
 
 In the Pyramid -^, , _. 
 
 i m = iB 
 
 Substituting these values in the fractions having a common denomi- 
 nator, we have the following : 
 
 Vol.Prism=«-r = (f- + t- + ^-r) = |(B-fB + 4B) = ^(B-fb + 4m). 
 Vol.Wedge=T = (4^- + ^H-^-r-)=|(B-fO-f2B)=f(B+b-f4m). 
 Vol. Pyramid =^l:^=(4^-+^-f 4^-)= ^(B-fO + B ) = f (B+ b + 4m). 
 Hence, the volume of the Prism, or Wedge, or Pyramid = 
 § (B -|- b -|- 4m), and the volume of a Prismoid, therefore = 
 |(B4-b-f-4m). 
 
(l.s MJty.suiiATioy 
 
 The rule may be expressed in words as follows : M^d' 
 tiply the sum of both end areas and four times the area 
 of a section half way between them, by one-sixth the altitude. 
 
 A cylinder may be divided into an infinite number of 
 prisms ; a sphere or a cone into an infinite number of 
 pyramids ; frustums of pyramids and cones into prisms 
 and p3^ramids, etc. It is evident, therefore, that this rule 
 is very wide in its application. 
 
 If, for the purpose of association, w^e consider the 
 capital letter B as the initial letter of the term "Base" 
 (large), and the small letter b as the initial letter of 
 "base" (small), and m the initial letter of the term 
 "middle section," the rule may be quite easily remem- 
 bered. In some solids, as the pyramid, cone, and their 
 frustums, these letters will have special significance, since 
 one base is larger than the other. This is not true, how- 
 ever, of all solids. 
 
 The following abbreviated process of using the formula 
 may be used to advantage after the pupil has become 
 thoroughly grounded in the process used in the chapter 
 on Illustrative Problems. 
 
 Problem.— What are the solid contents of the frustum 
 
 of a square pyramid whose upper base is 6 feet square, 
 
 lower base 10 feet square, and altitude 30 feet f 
 
 Operation. — 
 
 B= lOX 10 =100 sq. ft. 
 b= GX 6 = 36 
 
 4m = 4 (8X 8) = 256 
 
 392 X¥ = 1,960 eu. ft., Ans. 
 
 Probleim.— What is the volume of a sphere whose 
 
 radius is 10 feet"? 
 
 Operation. — 
 
 B= 0.0 
 
 b= 0.0 
 
 4iii = 47rr-= 4(3.1416 X 100) =1,256.64 
 
 1, 256. 54X¥ = 4,188.8 eu. ft., Ans. 
 
MEASURES OF EXTENSION 
 
 69 
 
 MEASURES OF EXTENSION 
 
 LINEAR MEASURE 
 
 TABLE 
 
 12 inches (in.) = 1 foot (ft.) 
 3 feet = 1 yard (yd.) 
 
 5i yards 
 
 16i feet 
 320 rods =1 mile (mi.) 
 
 } = 
 
 1 rod (rd.) 
 
 1 mi.= 320 rds.= 1,700 yds.= 5,280 ft. = 63,360 in. 
 Scale.— 320, 5^, 3, 12. 
 
 OTHER DENOMINATIONS 
 
 3 barleycorns 
 
 4 inches 
 
 9 inches 
 
 8 spans 
 
 6 feet 
 
 18 inches 
 
 21.888 inches 
 
 20 fathoms 
 
 3 feet 
 
 8 fnrlongs 
 
 1.152f common miles 
 
 3 geographical miles 
 60 geographical, or | 
 69.16 statute miles j 
 360 d'"2rees 
 
 == 1 inch (Used by shoemakers) 
 
 = 1 hand (Used to measure the 
 height of horses) 
 
 = 1 span 
 
 = 1 fathom (Used by sailors) 
 
 = 1 fathom (Used to measure the 
 depths at sea) 
 
 = 1 common cubit 
 
 = 1 sacred cubit 
 
 = 1 cable's length 
 
 = 1 pace 
 
 = 1 mile 
 
 = 1 geographical, or nautical 
 
 mile = 1 knot 
 = 1 league 
 
 r of latitude on a me- 
 
 = 1 degree "^ ridian, or of longi- 
 
 (_ tude on the equatoi 
 
 = the circumference of the 
 
 earth 
 
70 MENS URA TION 
 
 SURVEYORS' LINEAR MEASURE 
 
 TABLE 
 
 7.92 inches =1 link (1.) 
 25 links =1 rod (rd.) 
 
 4 rods =1 chain (ch.) 
 
 80 chains =1 mile (mi.) 
 1 mi.= 80 eh.= 320 rds.= 8,000 1.= 63,360 in. 
 
 A Gunter's Chain is the unit of measure, and is 4 rds. or 66 ft. 
 long, and consists of 100 links. 
 
 SQUARE MEASURE 
 
 TABLE 
 
 144 square inches (sq. in.) =1 square foot (sq. ft.) 
 9 square feet = 1 square yard (sq. yd.) 
 
 3O4 square yards = 1 square rod or perch 
 
 160 square rods or perches= 1 acre (A.) [(sq. rd.,P.) 
 
 640 acres = 1 square mile (sq. mi.) 
 
 1 sq. mi.= 640 A.= 102,400 sq. rds.= 3,097,600 sq. yds.= - 
 27,878,400 sq. ft. = 4,014,489,600 sq. in. 
 
 Scale.— 640, 160, 30i, 9, 144. 
 
 SURVEYORS' SQUARE MEASURE 
 
 TABLE 
 
 625 square links (sq. 1.) =1 pole (P.) 
 
 1(3 poles =1 square chain (sq.ch.) 
 
 10 square chains =1 acre (A.) 
 
 640 acres = 1 square mile, or sec- 
 
 tion (sq. mi., see.) 
 36 square miles (6 miles square) = 1 township (Tp.) 
 
 1 Tp.= 36 sq. mi. = 23,040 A. = 230,400 sq. eh. = 3,686,400 P.= 
 2,304,000,000 sq. 1. 
 
 Scale.— 36, 640, 10, 16, 625. 
 
MI':ASiHES.OF fJXTWSfO^ , 71 
 
 CUBIC MEASURE 
 
 TABLE 
 
 1,728 cubic inches (cu. in.) = l cubic foot (cu. ft.) 
 
 27 cubic feet = 1 cubic j'ard (cu. yd.) 
 
 10 cubic feet = 1 cord foot (cd. ft.) 
 
 8 cord feet, or) ^ , „ i , . 
 
 y =1 cord of wood (cd.) 
 
 128 cubic feet 
 
 24i cubic feet = 1 { P^''^^ °* ^^°"«' | (Pch.) 
 
 I or or masonry ) 
 
 1 cu. yd.= 27 cu. ft.= 46,656 cu. in. 
 Scale.— 27, 1,728. 
 
 CIRCULAR OR AKGULAR MEASURE 
 
 TABLE 
 
 60 seconds (") =1 minute (') 
 
 60 minutes = 1 degree (°) 
 
 30 degrees =1 sign (S.) 
 
 12 signs, or 360 degrees = 1 circle (cir.) 
 
 1 eir.= 12 S.= 30°= 21,600= 1,296,000". 
 Scale.— 12, 30, 60, 60. 
 
 DUODECIMALS 
 
 TABLE 
 
 12 fourths ("") = 1 third ('") 
 
 12 thirds =1 second ( ') 
 
 12 seconds = 1 prime (') 
 
 12 primes, or inches = 1 foot (ft.) 
 
 1 ft.= 12 '= 144"= 1 ,728'"== 20,736"". 
 Scale.— 12, 12, 12, 12. 
 
72 MEXSUR.ITION 
 
 MISCELLANEOUS 
 2,150.42 cubic inches = 1 Winchester bushel (bu.) 
 
 COMPARATIVE TABLE OF WEIGHTS 
 
 Atwirdupois Troy Apothecaries' 
 
 1 pound = 7,000 grains = 5,760 grains = 5,760 grains 
 1 ounce = 437.5 grains = 480 grains = 480 grains 
 
 COMPARATIVE TABLE OF MEASURES OF CAPACITY 
 
 C'u. in. in Cu. in. in Cu. in. in Cu. in. in 
 
 one gallon one quart one pint one gill 
 
 Liquid Measure 231 574 288^ lit^ 
 
 Dry Measure (half -peck) . 2685 675 33 5^ St 
 
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