IMAGE EVALUATION TEST TARGET (MT-3) k A // ^ A /. >* <• »> (meaning "CON- TINUED"), or the symbol y (meaning "END"), whichever applies. IMaps, plates, charts, etc., may be filmed at different reduction ratios. Those too large to be entirely included in one exposure are filmed beginning in the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: 1 2 3 Les images suivantes ont itA reproduites avec le plus grand soin, compte tenu de ia condition et de ia nettetA de l'exemplaire film*, et en conformity avec les conditions du contrat de fllmage. Les exemplaires originaux dont la couverture en papier est ImprimAe sont filmte en commenpant par le premier plat et en terminant soit par ia- dernlAre page qui comporte une empreinte d'Impression ou d'illustration, soit par le second plat, salon le cas. Tous les autres exemplaires originaux sont filmte en commenpant par la premiere page qui comporte une empreinte d'Impression ou d'illustration et en terminant par la dernlAre page qui comporte une telle empreinte. Un des symboies suivants apparaftra sur la derniAre image de cheque microfiche, seion le cas: ie symbols — ► signlfie "A SUIVRE", le symbols V signlfie "FIN". Les cartes, planches, tableaux, etc., peuvent Atre filmte A des taux de reduction diff^rents. Lorsque le document est trop grand pour Atre reproduit en un seul clichA, il est f iim6 A partir de Tangle supArieur gauche, de gauche d droite, et de haut en bas, en prenant le nombre d'images nAcessaire. Les diagrammes suivants illustrent ia mAthode. 1 2 3 4 5 6 w A ..>. . , - (J ^ ^ PRACTICAL MENSURATION, For the use of Students in Colleges, and Schools, and for Private Learners. F5Y C. A. FIvEMING, Principal Northern Business College. Author of "The Laws of Business"; "Expert Book- keeping"; "How to Write a Business Letter"; "Thirty Lessons in Punctuation " ; " Selt-Instruction in Penmanship " ; &c. * OWEN SOUND: Printed at the Northern Business College Steam Press. 1893. V I Entered according to Act ni P ■• / r i£>- PREFACE. ia, in the three by nculture. The design of the author of this book is to supply a work on Mensuration suitable for students in Business Colleges, Public Schools and other institutions of learning where the study of Euclid and Algebra are not taught, but where so useful and practical a subject as Mensuration should not be neglected. There is no subject in more common use by all classes of persons, nor is there one in the curriculums of our colleges and schools that has received so little attention at the hands of the Educators of our country. There is scarcely a transaction in everyday life to which it does not apply from the purchase and sale of a yard of calico to the ,;.Teat railway contracts that require the time of their execu- tion to be counted in years. In this volume the author has endeavored to combine the theoretical and practical, and to present the work in as reasonable a way as possible. It is not desirable that the student's memory be charged with a great variety of decimals, fractions, formulas, &c., which in after years he is liable to forget or mis-apply. Where such are used in this work one decimal or fraction is made to do service throughout an entire class of opera- tions instead of using a dozen or two modifications of the fraction or decimal. Chapters on square and cube root are placed at the beginning of the work as these rules are used to a considerable extent in mensuration and are placed neay the end of arithmetics when in fact they might be readily taught immediately after division. The definitions that appear immediately after the exercises in square and cube root are principally a study of the technical words pertain- ing to the subject and as such should receive considerable attention from the student so that he may readily under- stand the language of the examples and exercises throughout the book. The theoretical part is placed in the first part of the work, and common measurements in the second part, these however, may be taught simultaneously or a great many examples in the practical part (part II), may be taught before the theory in the first part with very good results if desirable. ■ The exercises in the theoretical part are not long so that the student may be advanced rapidly to the more practical work in part II. Owen Sound, October 20, 1893. It be readily initions that re and cube )rds pertain- considerabJe idily under- throughout first part of econd part, or a great y be taught d results if re not long ' the more CONTENTS. Introductory. .... • •••• t««« 9 S(iuare Ro )t and Cube Root by factoring. 10 Square Root . .... II Cube Root . . .... 13 Mensuration — Definitions 16 Tables of Measures ... 25 Measuremervt of Surfaces 27 Squares and Rectangles 27 {a) To find Area . 27 (/j) To find the side 28 Triangles . . .... 30 ( Similar Figures ... Areas of Similar Figures . . Similar Solids .... . . . . Irregular Solids .... . . . . Part II. 'OMMON Measurements Wood Measure .... Lumber Measure . . . Land Measure .... Measurement of Lathing Plastering . .... Painting, Kahomining, &c Papering . . .... Furnishing .... Stone Work .... Brick Work Shingling. . .... Carpenter Work . . . Fencing ... .... Timber Measure . . . Cisterns ... .... Measurement of Grain Measurement of Hay Gauging . . .... Shoemaker's Measure Miscellaneous Exercises Short Rules on Circles Answers . . .... Advertisements .... • • • • ■ • I • • • • • • • » • 75 77 78 79 80 80 82 86 87 90 92 92 94 95 97 99 100 101 103 104 105 106 107 108 109 120 121 127 lall 100 ictil ts If INTRODUCTORY. |A knowledge o» Square Root and Cube Root is neces- py in Mensuration. These rules, in Arithmetics, are lally placed near the end, and many pupils in our public lools who should study Mensuration do not get any in- iction in them. An introductory chapter on these sub- |ts is indispensable. Definitions. [1 The Square Root of a number is one of the two lal factors of that number. Example : 6 is the square )t of 36 — 6 X 6-36. The Cube Root of a number is one of the three lal factors of that number. 6 is the cube root of 216 6x6 = 216. The Radical sign V, when placed before a number, licates that some root ot that number is to be found. The Index is a small figure placed above the radical to denote what root is to be found. Thus v 49 means s:iuare root of 49 ; ^1728 meaiis the cube root of 1728; I means the fourth root of 81. When jw index is used understood that the square root is to be found. Thus ^4 means the square root of 64. 16 The Square of any number is the result produced by iltiplying that number by itself. Thus the square of 5 )uld be 5 X 5 or 25. >■ lO 6 The Cube of a number is the result produced by mul- tiplying twice. Thus the cube of 3 would be 3x3x3 = 27. 7 Factoring. To find the square root or cube root of a number by factoring. EXAMPLE. 1. Find the square root of 324. 2)324 Analysis. A square is the \' f- product of two equal factors. ^/ ' "^ In order to find the square root of a number wc must find the value of those factors. We find that the frime factors of 324 are 2 x 2 x 3 x 3 X 3 X 3. We can make two groups of them, viz., (2 x 3 x 3) x (2 x 3 x 3) and we find that the factors in each Square Root =2x3x3=18 group are alike, and when multipli- ed together amount in each case to 18. Thus we have two ecjual factors, which when multiplied together will produce 324, therefore the square root of 324 is 2 x 3 x 3 or 18. Rule. Resolve the number into its pi'ime factors. Then take one of every two equal factors and mul- tiply together the number taken out. The result will be the square root of the given number. EXAMPLE. 2. Find the cube root of 3375. 3)8i 3)27 3)_9 3 3)3375 3)lii5. 3)3 75_ 5)^.21 5 Analysis. A cube is the product of three equal factors. We find that ihc yime factors of 3375 niry be arranged in three equal grom s, viz , (3 >= 5) X (3 X S) X (3 X S). each of which u.ulli- plied out amounts to 15. Thus we have three equal factors which when multiplied to- gether amount to 3375, therefore the cube root of 3375 is 3x5 or 15. Rule. Resolve the number into its prime factors' Then take 07ie out of every three equal factors and multiply together the numbei's taken out. The re- sult trill be the cube root of the given number. t-tv a i> w)'i"""< ' — duced by mul- 3x3x3 = 27- ►r cube root of square is the al factors, he square root of find the value of e find that the ( are 2 X 2 X 3 X 3 lake two groups 1 X 3) X (2 X 3 X 3) : factors in each I when multipli- in each case to tiphed together 3x3ori8. 'me factor's, f and mul- The result ber. oduct of three r*" of 3375 '»'y groui s, vi/. , which uiulii- lus we have multiplied to- ore the cube te favtovH' I'tors and The re- iher. V576; ^729 ^^6561 ; ,^3375 ^9261 ; V1764 V2916; ^^35937- II EXERCISES. 8 Solve the following problems :- 1. V256;V400 5. 2. V729;'V625 6. 3- ^27; ^343 7- 4. ^125; ^216 8. 9. The foregoing method only applies to a limited number of cases. It may be advantageously used in a number that is easily factored. We now proceed to give rules for extracting the square and cube roots of numbers, that apply to all cases. 10 Gdneral Rule to find the square root of any number. EXAMPLE. Find the square root of 4225. 42,25(65 Analysis. Beginning at the right hand side, separate the 42,25 into periods of ijvo figures each. Next find the greatest square contained in the left hand period of figures. The great- est square contained in 42 is 36. The square root of 36 is 6. This will be the first figure of Place it to the right of the given numlier.s as you do in long Draw a perpendicular line to the left of the given number. 125 36 625 625 the root, division. square the first figure of the root, 6, and subtract the result 36 from the first period of figures in the given number. To the remainder, 6, afHx the next period^ 25, in your given number. Multiply that part of the root already found by 2, and place the result, 12, to the left of the perpendicubr line, using it as a trial divisor. 12 is C( itained 5 times in 62, therefore 5 will be the next figure in your root. Annex 5 to the root already found and also to the trial divisor, thus making it 125. Multijly the complete divisor by the last figure of the root, 5, and place the result, 625, under your dividend. Theie is no remainder, and 65 is the square root. Rule. Beginning at the right hand side, sepa- rate the giren number info periods of tivo figures each. i ill 12 ■ Find the greatest num:,^. j ^ O'ren number. It b,;„„ ,, J *''" '•'»'" «/ the xmare it and subtract IT •^'■"'•^^"'^ »f the root wrincl ,.f 4i " "^" the result from the n . ''^^ the remainder annerth. . nsin,, the result as a trirZ " ^''^'^' "^^t tient to that part of '"'^'^y^'- °»»«'- your.quo- . "t-o to the Mai diLor ''"m u^T'^'' f"^'"^' ««rf. ■ % t^^^mrt of the root alr^dif^ ^ "'"' "'""'"^ the result from the dividend ''""'''^^^^^^^^^^^^ I'o the remainder anne. tJ,, „,,/ .■ - Mures, and proceed as before. ^''''""' "^ ^'0"(.). Always annex 10 ih. (rial ,•• ' ' • ' you annex ,o (he root. ' ''"''^" "^^ ^"'C figure ,hat ^■"luiivij). In extracting the s -o. of b„,h „„,„e„,o, an<, de'non ITor 'Z/' . "T"""' """« ""^ ' ^•0TB(4). To exirac, ,he H,„are ,oo. of ''! '' period, of ,wo figures each, beginning .ll I '^ ''™'" °"^ « 'o <»e righ,, ,he„ proceed as in a 1,: 1 ''' """' """ ^^^ o^H "„„,„.. of di„-u co„.p,e.e 1 , :;t • ■'^^""'^ '^"= ^^ - ■ '""• "^ ■"'' ''>• """"ing a cipher 13 11 EXERCISES. Find the square root of: — I.' 1089. -w .,,9» 32041. ' 17- 123562. 2. 8796; ' 10. 14641. 18. 492804. 3-: 625. :, 11. 11881. 19. 994009. 4. 2025. 12". 13225. 20. 2050624* 5- 9025. 13. 94249- 21. 29855296. 6. 5625. 14. 86436. 22. 387420489. 7.. 7225. 15- 97344- 23- 7463824. 8. 15625. 16. 46656. 24. 3976036. Find the square ropt.of the follow ing :■ — - T - •'• - 1 :s y r\ a fi '_» s ^'' ^5iy 5- iTToOr- 9- 127T^T- , 1 . 7. 17^ II. 1 4 « 4 1 1 I TThh !•- . ••,*■ \ 7 (f T-' 18 i! 2 8. 69^. 12. 9092U. Find the square 1 root of the following num bers that in- ►Ive decimals : — . I. .061^5. 7. 62.8. • 13- 125 2. *',\f\89- ■^• 2. . 14. •1406.25. 3. .001225. 9. 3- 15- IT to 4 places. 4. .9604. 10. 196.1369. 16. 7 H to 4 II 5. .512656. II. Q. 0000994009. 17- ^ H to 4 M 6. .04,' 656. 12. 131.8084. 18. 247i\to 4 M Cube Root. 12 Find the cube root of 79507. EXAMPLE. 4^ X 300 = 4800 3x4x30= 360 3'= 9 79,507(43 64 Analysis Beginning at the units place separate the num- ber into periods oi three figuies each. The greatest number the cube of which is less than the left hand period (79) is 4, the cube of which is 64. Place this 4 to the right of your number as the *5rst figrre* of the root. Cube 4 and subtract the result (64) from the left hand period of figures (79). To the remiinder (15) annex the next period of the number ^507), forming a new dividend. Multiply the square of\he root already 5169J15507 I15507 14 foand (4) by 300, and use the result as a trial divisor. The trial di- visor (4' x 30 j= 4800) is contained inf 15507 3 times. Put 3 as the next figure of the root. Add together the trial divisor (4800) + 30 times the last figure of the root (3) multiplied by the other figure (4) + the square of the last hgure of the root (3) to form Sour complete divisor 5169. Multiply this by the last figure of the root and subtract the result from the dividend. There is no remainder, so 43 will be the cube root of 79507. Rule. i. Beginning at the units place, separate the given number into periods of three figvres each* 2. Find the greatest cube of the first period to the left, and place its cube root to the right of the given number as the first figure of the root. Subtract the cube of the first figure of the root from the left hand period of figures, and to the remainder annex the next period of figures to form your dividend. 3. To find your trial divisor multiply the square of the root already found by 300. Divide the div- idend by this trial divisor, placing the result as the second figure of the root, 4- To your complete divisor add to the trial divisor thirty times the last figure of the root mul- tiplied by the other figures of the root, and the square of the last figure of the root. Multiply this by the la^t figure of the root, and subtract the result from the dividend, 5. To the remainder annex the next period, and proceed as before. Note ( i ). If a cipher occur in the root annex two cipher to the divisor and another period to the dividend. Note (2). If there be a remainder, place a decimal point and annex three ciphers ; then proceed as before. i!'- sor. The trial di- nes. Put 3 as the divisor (4800) + 30 '■ other figure (4) + •m your complete root and subtract r, so 43 will be *^p, separate figvres each, period to the ' 0/ the given Subtract the the left hand ' annex the dend. ' the Square 3?e the div- esultasthe 15 EXERCISES. 13 Extract the cube root of: — 1. 64. 2. 216. 3- 15625- 4. 166375. 5- 474552- 6. 7529536- 7. 912673. 8. 128024064. 9. 3048625. 10. 48228544. 11. .091125. 12. .000097336. 13. 9663597. 14. 279.726264. 15 12.812904. 16. .015625. 17. 435519512. 18. 1000 T^TTT- ¥ r\ •J 1 « 19. T2 8TS- ^^\ 1 » 8 ti i zo. TJlf^J' 21. 39t/5. 22. H' 23- ''ft- 24. 18629. 25- 13788292. 26. 3336271962. 27. 2 to 5 places of decimals. 28. 3 to 5 M 1. It 29. 629.8862972. 30. 8818629.823372980. > the trial root mul- » and the Itiply this ' the result *^od, and ipher to the 1 point and i6 ^ ;> m it ii i MENSURATION. \\ Mensuration is the measuring or estimating of the areas of surfaces, the volume of solids, and their linear dimen- sions. - Definitions. 1 A Point is that which has no parts or magnitude, but simply dlsnotes position or place. , Note. In working Mensuration a small clot (.) is generally used to denote a point, but such a dot cannot properly be called a geometrical point, for, no matter how small it i?, it must necessarily occupy a certain amount of space, and hence it has magnitude. 2 A Line ( ) is length without breadth, and is used to denote distance. Note. Here, again, it must be noticed that a line, such as is gen- erally drawn in working Mensuration, is not a geometrical line. Draw a line on paper and you will find that, besides measuring a cer- tain distance in length, it also has breadth — possibly very little, but still it has breadth, wheieas^a geometrical line denotes length without breadth. 3 The Extremities of a line are points. Lines are of two kinds — straight and curved. 4 A Straig^ht Line ( ) is one which lies evenly between its extreme points. In other words, it is the short est distance between two points. It is one that does not change its direction. 5 A Curved Line (— — ) is one that changes its direction at every point. vm g of the areas linear dimen- ignitude, but nerally used to a geometrical rily occupy a and »s used >ch as is gen- netrical hne. isuring a cer- ry little, but ngth without ines are of lies evenly the short does not anges its ^ X7 . 6 A Surface is that which has length and breadth, but no thickness. 7 A Plane Surface is a flat surface. 8 The Extremities or boundaries of a surface are hnes. 9 An Ang^le is the inclination of two lines to one an- other which meet, but are not in the same straight line. In speaking of an angle it is sometimes designated by one letter, but when two or more angles are together each of them is designated by three letters, one of which is placed at the vertex of the angle, ^. e. the place where the lines that form the angle meet, and each of the other two somewhere On one of the two lines containing the angle, the letter at Mie vertex being placed in the middle in naming the angle, 10 When a straight lin® standing on another straight line makes the adjacent an- fiorizont^ 1. «gles equal, each of these an* gles is said to be a Rig^ht Angfle, and the straight line which stands on the other is said to be perpendicular to it. A right angle contains 90 degrees. 11 An Obtuse Angle greater than a right angle. is one that is SWf-jt's HI ill m 18 1 22 An Acute Ans^le is one that \s less thafl a right angle. V. 13 A Circle is a figure contained by one h'ne, called the Circumference, and is so drawn that all lines drawn from a point with- in, called the centre, to the circumference, are equal. 14 T)ie Circumference of a circle is the distance around the circle. 1 5 The Centre of a circle is a certain point within the circle, and all lines drawn from this point to the circumfer- ence, are equal. 16 The Diameter of a circle is a straight line drawn through the centre o^ a circle and terminating at both ends at the circumference. Note.— It is the longest straight line that can be drawn within a circle. 17 The Radius of circle is a straight line drawn from the centre to the circumference. IS A Semicircle is half a circle. It is contained by a diameter and that part of the circumference cut,off by the diameter. 19 A Quadrant (a quarter of a circle,) is that part of a circle contained by two radii at right angles to each other and that part of the circumference cut off by them. »9 I*" >s one that is 20 A Chord is a straight line in a circle terminating at both ends at the cir- cumference, but not passing through the centre. (See figure, the line H K). 21 An Arc is that part of the circum- ference cut off by a chord. (See figure). 22 A Sesrment is that part of a circle enclosed by a :hord and an arc. (See figure). 28 A Sector of a circle is that part contained by two radii and the arc between them. (Any one of the parts on [the lower side of the figure). 24 Rectilineal figures are those which are contained {by straight lines. 25 A Triangfle is a figure contained or enclosed by ' three straight sides. 26 A Right Angled Triangle is one which has one of its angles a right angle. The side opposite the right angle is called the Hypothenuse, and "^ the sides containing the right angle, the Base and Perpendicular respectively, the base being the line on which it is constructed. 27 An Obtuse Angled Triangle is one which has one obtuse angle. 28 An Acute Angled Triangle is one which has acute angles only. 29 An Equilateral Triangle is one which has all its sides and all its angles equal. 20 fill i liii :r\\ Isoj:;'iS». 30 An Isosceles Triangle is one whic has two of its sides, and two of its angle (namely the angles at the base), equal. 31 A Scalene Triangle is one having I three unecjual sides, and three unequal^ ecoiuuu. 32 A Quadrilateral Figure is one that has four straight sides and four angles. 33 A Parallelogram is a figure having its opposite sides equal and parallel. I SectRoglo. 34 A Rectangle is any four-sided figure having its oppositt^ sides **equal and all its angles right angles? - 35 A Square is a four-sided figure having all its sides equal, and all its angles right angles. Sciaare. 36 An Oblong is a four-sided figure having only its opposite sides' equal, and all its angles right angles. (The figure given under 34 is an example. , :: Hi iXOB^WI. ft 37 A Rhombus is a four-sided figure ; / { having all its sides equal, but its angles y \ are not right angles. that has foi 38 A Rhomboid is a four-sided fig- l /i ure having its opposite sides etiual, but its i^<.j '\ angles are not right angles. |e. — In both the foregoing 37 and 3S the opposite angles are equal. ig its opposite jr-sided figure and all its TrapMKiW. TiapezLtun. 39 A Trapezoid is a four-sided figr ure having only two of its sides parallel. 40 All four-sided figures, besides these, are called Trapeziums. Any figure having more than four sides may be ^d a Polygon. " •>. 42 A ^entelgon is a five-sided figure. at I 11 o o o o o 48 A Hexagon is a six-sided figure. 44 A Heptagon is a seven-sided figure. 45 An Octagon is an eight-sided figure. 46 A Nonagon is a nine-sided figure. 47 A Decagon is a ten-sided figure. 48 An Ellipse is like a circle IB compressed in one direction and elongated in the other. !.' il! t r 49 ParalM lines are lines which have the same direction, and being in the same plane, equally distant from each other, they can never meet. 50 A Solid is that which has three dimensions, viz,, length, breadth, and thickness. 51 A Cube is a solid having all its sides equal, and all its angles right angles. like a circle direction and 52 A Parallelopipedon is a solid bounded by six parallelograms, of which every, opposite two are equal and parfiiHel. 58 A Cylinder is a solid having both its ends equal and parallel circles, and boun- ded by a surface curved similarly as the base. 54 A Prism is a solid, the ends of which are two sim- ilar rectangular figures, and its sides parallelograms. Prisms are called Triangular, Quadran- gular, Pentagonal, &c. according to the num- ber of their sides. 24 PTramid. 55 A Pyramid is a solid having for its base a rectilinear figure, and for its sides as many triangles as there are sides to the base; these triangles terminating in one point called the apex or vertex. 56 A Cone is a figure having a circular base and terminating in a point called the apex. i!; 57 A Frustum or a pyramid or cone is^ that part which remains after the top is cut off parallel to the base. snnoBk 58 T A Sphere or Globe is a figure bounded by a uniformly curved surface, all the points of which are equally distant from a point within called the centre. ; Hill ii 1 m ving for its fts sides as o the base; »oint called 25 ; a circular called the or cone is^ op is cut a figure surface, y dif^tant tre. TABLES OF MEASURES. 40 Rods 8 Furlongs 3 miles A thorough knowledge of the tables of measures is indispensable to every student of Mensuration. The prin- cipal tables are placed here to be thoroughly leained before beginning the study of this subject. TABLE OF LINEAR MEASUREMENT. 12 Inches (in) = i Foot (/"/.) 3 Feet == I Yard (yd.) ^}4 Yards = i Rod, Pole, or Perch {rd.^p) — I Furlong (Jur.) = I Mile {mi.) = I League {lea.) I vii. = %fur. = 320 rds. = 1^60 yds. = 5280/?. = 63360 /«. I II = 40 II = 220 II = 660 II = 7920 II I 'I = 5^ II = 16^ II = 198 M I tl = 3 M = 36 II I II = 12 II For very small measurements the line (yV inch) is some times used. Eighths, sixteenths and thirty-seconds are used more frequently. The decimal parts of an inch are also used. SURVEYORS' LINEAR MEASURE. 100 Links (/.) = I Chain (ck.) 80 Chains = i Mile (;///.) 1 mi. = 80 c//. = 320 rJs. = llQOyJs. = 5280 /A =8000/. \ f/i. = 4 rds. = 22 yds. = 66//. = 100/. 1/. = 7.92 m, The ordinary roads and streets are i Chain wide. This will help to fix on the mind an idea of the length of a chain. '11 26 .^ .« .?i .« .?i i < < 55m O o ■ l-> a o 0) 0) 3 C/2 II II li II J: O ^ p::; rt _ _ 3 = - - c/2 ■ t/3 ^ < o o o ^ = = = = O O "^ vo rj- O "^ O ON "?!■ as 00 N On M «:>H = r = = O O \ji. On M O vo M^ 00 ^ -^ -5 ^^ *^ >^ ^ ?:^ ^ = = = o o \o CO ON o 2- NT O fO j^ ►i? i^« O O w O MD M O NO u 0) g- C/3 > C P^ •g TS nd 2 »-i ^ O rt O ^ > U W -S ^ r C/3 c *rt U, (/) w l-H HH fe Ll, u 3 -• = u CO f^ 00 M N N t^ M ■v: ^* -"s: o o o o o -§ O ID O N O NO o *- v; ~ 0. o NO VO M _ -^ ill Hi' I! '( f 27 TJ 'O ,0 n ^ > U u M JO 3 ; u U '0 M H II II ^ VI U-, (1) c (U each part representing one foot. By drawing lines to the opposite dividing points you will find that you will have twenty-five equal squares, each one being a square foot. A square foot need not necessarily be a foot square. A surface two feet long aud six inches wide contains exactly one square foot, although not a loot square. Be careful therefore to distinguish between such expres- sions as five feet square and five square feet. The former means five feet long by five feet wide, containing 25 square feet, while the latter contains only five square feet. SQUARES AND RECTANGLES. To Find the Area : -^ = Rule. — To find the area of a rectangle, multiply the length by tlie breadth. 0DWM* XMtni^ i '""! I |i'i: ' i 28 Note. — If length and breadth be given in different terms, (e. g length in feet and breadth in inches,) reduce both to the same terms and multiply them together. Example. To find the area of a plot of ground 60 ft. x 88 feet. Multiply 60 by 88, and your product, 5280, (88 x 60 = 5280,) will be the area of the plot. EXERCISES. 1. Find the area of a rectangular field 30 x 70 rods. 2. Find the area of a rectangular plot 16 ft. x 10 yards. 3. How many square yards are there in a rectangular lawn 4 rods x 50 feet ? 4. How many acres in a rectangular field 30 rods x 80 rods ? 5. How many acres in a rectangular field 100 yds. x 90 rods. 6. How many acres in a rectangular piece of land i mile X 300 rods ? 7. How many yards of carpet will it take to cover a room 13 ft. X 20 ft. ; carpet i yard wide? 8. How many yards of paper will paper a room lo x 15 ft. ; ceiling 9} feet high ? Paper 18 inches wide. 9. Find in acres, k.Q.., the area of a piece of land | mile long and 800 yards wide. Rectangle and Square ™To Find a Side. The area and one of the sides of a rectangle being given, to find the other. Rule. Divide the area by the given side, and the quotient icill be the other side. 29 fierent terms, (e. ^ to the same terms 'ground 60 ft. x iuct, 5280, (88 X 30 X 70 rods. 6 ft. X 10 yards, in a rectangular d 30 rods X 80 leld 100 yds. X piece of land i ake to cover a ' a room 10 x lies wide. : of land I mile nd a Side. le being given, f'de, and the Example. The area of a rectangle is 156, one of the ides is 1 2, find the other. 12)156 Since 156 is the product of 12 multiplied by the other side 156 -j- 12 will give that side = 13. EXERCISES. 1. A piece of cloth is 2\ yards wide, and contains no quare yards. How long is the piece ? 2. A piece of road is 22 yards wide, and contains ten acres. How many rods is it in length ? 3. A field containing i6t/j acres has one of its sides ^ mile long. Find the length of the other side in rods. 4. It takes 43^ yards of carpet one yard wide to carpet a room, the length of which is 2 1 feet. Find the width of the room. f 5. What is the width of a sidewalk a quarter of a mile Jlong, containing 1173^ square yards? 6. One side of a field measures 8 chains. It contains 20 acres. Find the other side. 7. A square field contains 10 aciCs. Find the length of its sides. 8. The floor of a square hall contains 267 if square feet What is the length of its side ? Note. Extract the square root of the area. 9. A field contains 32 acres and is twice as long as it is wide. Find the length of its sides. 10. A field containing 30 acres is three times as long as it is wide. Find its length and breadth. 30 TRIANGLES. ^ Triangles are divided, (see defini- tions,) according to their angles, into three classes, viz., right angled triangles, acute angled triangles, and obtuse angled triangles. Triangles are divided, (see definitions,) according to their sides, into equilateral triangles, isosceles triangles, and scalene triangles. Right Angled Triangles. To And the Hypothenuse of a right angled triangle, the perpendicular height and the base being given. Rule. Add together the square of the two sides enclosing the right angle, and extract the square root. EXAMPLE. Find the hypothenuse of a triangle having a base 12, and perpendicular height of 16 feet. 12X12" 144 16 X 16 = 256 400 V 400=^20= hypothenuse. 31 2d, (see defini- their angles, ., right angled triangles, and ording to their triangles, and JO, thenuse of a perpendicular 'g given. r the square angle, and j base 12, and The rule for finding the hypoth enuse of a right angled triangle is susceptible of very easy experimen- tal proof. In the annexed figure it will be noticed that we have a m fight angled triangle having a base 4 feet long, and a perpendicular of 3 feet. The square constructed on the base contains 4x4 = 16 square feet ; on the perpendicular of 3 feet in length is a square containing 3 x 3 = 9 square feet, 9+ 16 = 25 square teet, and this is exactly the area of the square constructed on the hypotenuse, (5x5 = 25), hence the sum of the squares of the sides containing the right angle is equal to the square of the side opposite the right angle, and the converse is also true that the square of the side opposite the right angle minus the square on one of the sides, is equal to the square on the remaining side. EXERCISES. Find the hypothenuse ot the following triangles. I. Base 15; perpendicular height 20. 36 ; II M 40. 137 ; It If 9. 6 ft. 3 in. ; m 12 ft. 5 in. ; II 16 ft. 8 in. ; n ise. M II 3 yds. •I M 6 ft. 3 in. M II 103 ft. 6 in. Find the diagonal of a rectangle whose sides are 64 feet and 22^ feet. 8. Find the diagonal of a square whose sides are 15.75 feet long. 9. A smoke stack on a sawmill is 60 feet hi^h. Wha length of wire will I need to fasten to it 8 feet from the top and to reach a stump 70 feet from the base, allow- ing 9 feet for fastening to stump. 111! Ill illil!i 32 Right Angled Triangles. To find the Base or Perpendicular in a right angled triangle, the hypothenuse and one of the sides containing the right angle being given, to find the remaining side : Rule. From the square of the hxjpotlK'iiua 2 sub- tract the square of the given side and eatract the squa7'e root of the remairider. According to the foregoing rules, the following proposi- tions are true in a right angled triangle. 1 (base)'"' + (perpendicular)- = (hypothenuse)- ; 2 (hypothenuse)- - (perpendicular)- = (base)'^ ; 3 (hypothenuse)'* - (base)- = (perpendicular)'* ; EXAMPLE. The hypothenuse of a right angled triangle is 35 and the base is 28, find the perpendicular height. 35x35 = 1225 28x28= 784 441 V441 =«: 2 1 = perpendicular height. EXERCISES. Find the bases of the following triangles : — I. Perpendicular 46 feet; hypothenuse 530 feet. ?.. II 135 II ', II 225 " 3. II 2 ft. 8 in. ; n 2 yards. Find the perpendicular of the following triangles : — 4. Base 9 ft. 9 in. ; hypothenuse 10 ft. 5 in. 5. II 12 yds. ; "3 rods. 6. It I mile ; ti i^ miles. 7. A rectangular field is 60 by 90 rods. Fird the dis- tance between its opposite corners. 33 fcs. a right angled ies containing iing side : h('nua3 sub' (.xtract the wing proposi- thenuse)- ; = (base)=^ ; idicular)^ ; is 35 and the cular height. feet. II 'ards. (Ies : — Find the cost of fencing a field in the form of a right [led triangle; base 25 rods, perpendicular 45 rods; at per yard. Find the diagonal of a plot of ground 1 J miles square. [o. A boy flying a kite had it caught in a tree. The [gth of the cord attached to the kite was 250 yards, and was 240 yards away from the tree. What was the height [the tree ? In. What must be the length uf a ladder which, when iced 12 feet from a house 50 feet high, will just reach to |e top ? 13. A ladder 60 feet long is placed against a building feet high, so as just to reach to the top. How far is the (ot of the ladder removed from the building ? 13 The bottom of a ladder is placed in a street 12 feet in. from a building, and the top of the ladder reaches the [uilding 44 feet from the ground, and on turning the ladder ^er it leaches a building at the other side of the street f6 ft. 8 in. from the ground. Find the width of the street. 14. The span of a roof is 30 feet. The length from the idge to the eaves is 26 feet. Find the height of the ridge libove the eaves, Area of Triangles. ^d the dis- To find the area of any triangle the base and the perpendicular being given. Rule. Multiply the base by the perpendicular height and di' vide by 2, ,i:il 34 EXAMPLE. Find the area of a triangle, the base of which is 20 and the perpendicular 25. -^ .^ ^ ^ '^ =250 sq. ft. area. ;'!'i| li in ! JVaHolograia. Note. It is worthy of special notice that a right angled triangle is simply half of a rec tangular parallelogram or half a s(|uare. Hence the rule to multiply the length by the breadth (base by the perpendicular,) vould give the area of the whole figure ; divide by 2 and you have the area of half of it, or the right angled triangle. 1 he same applies to the triangle in the second illustration ; and further, every tri- angle it the half of a parallelogram. EXERCISES. Eind the areas of triangles having the following dimen- sions. 1. Base 6 feet; perpendicular height 12 feet. 2. II 3 ft. 7 in. M II 10 ft. 2 in. 3. It 7 yds. I ft. II II I rod. 4. II I mile II "371 yrds. 2 ft. 8 in. 5. How many acres are there in a triangle, the base of which is 40 rods and the perpendicular 36 rods ? 6. Find the area of a right angled triangle having a base of 60 feet, and a hypothenase of 100 feet 7. The perpendicular of a right angled triangle is 16 feet and the hypothenuse is 20 feet. Find the area. 8. The base of a triangle is 32 feet and the hypothenuse 40 feet. Find the area. .15 ^hich is 20 and sq. ft. area. ecial notice that a »ly half of a rec Fa square. Hence h by the breadth f v'ould give the ide by 2 and you the right angled to the triangle in urther, every tri- 3gram. 3 wing dimen- et. 2 in. d. •ds. 2 ft. 8 in, the base of s? laving a base gle is 16 feet hypothenuse APPLICATION OF RULE TO ANY TRIANGLE Note. i. Special attention i< necessary to the wide application of lis rule. It will be noiicfd thit not only will it apply to right angled iianglcs, but to all triangles— equilateral, scalene, is-osceles, etc. m ' XsoicbIm. Ecalcno. "" 'Ivf^ral, iimp\y nmh\\)\y o/ie stWi' by the perpend'cula. drawn from it to the ittgle opposite it. In the ecjuilateraland isosceles triangles this is easily &een. In the scalene traingle the base is produced as shown by the Idotted line an.1 the perpendicular distance measured from this base [produced, to the opposite angle. Then apply the rule — the ba^'e mul- jtipliLd by the perpendicular height divided by two ecjuals area. 9. Find the area of an equilateral triangle if the base is [200 feet and the perpendicular height 171 feet. 10. Find the area in acres, &c., of an isosceles triangle, the base of which is 60 rods and the perpendicular distance to the angle opposite, 40 rods. 11. The base of a scalene triangle is 40 feet, and the perpendicular from the base produced, to the opposite angle in 52 feet. Find the area. 12. The base of a triangle is I of a niile, and the per- pendicular distance to the angle opposite is ^ mile. Find the area in acres, &c. 13. The area of a triangle is 2480 square feet, the per- pendicular distance from the base to the angle opposite is 62 feet. Find the base. 36 APPLICATION OF RULE TO ANY FIGURE EN- CLOSED BY STRAIGHT LINES. Note 1. It will be noticed that this rule is also apf licahle to the measuring of any figure enclosed by straight lines, as any ^uch figure is divitiilile ii t ) triangles. EXAMPLE. I*'ind the area of this trapezium. The hne joining opposite corners is 42 feet : the perpendicular from this to one angle is 18 feet and to the other angle 16 feet. 18 X 42 ' The area of upper triangle= =37^ 2 16 X 42 The area of lower triangle = = 336 2 The sum is the area of trapezium 714 EXERCISES. 14. Find the area in acres of a trapezium, the diagonal of which is 60 rods, and the perpendicular distance to the two opposite angles 30 and 45 rods respectively. 15. Find the area of a trapezium, the distance between two opposite angles being 45 feet, and the perpendicular distance to the remaining angles from this line being 32 and 40 feet respectively. /0* t6. Find the area of this rhom- boid in acres, the dimensions given being as foMows: diagonal 125 rods, the perpendiculai distance to the angle on each side being 40 rods. wi^ » • • f 37 find the Area of Triangles, tlie Sides being given. To find the area of any triangle, the three sides being nven. RuLi:. — From half the sum of the xides subtract hach side scparateli/.^Multlpli/ Uujether the three Wemitinders and half the sum of the sides and ex- tract tlie square root. EXAMPLE. Find the area of a triangle the sides of which are 30, 40, [and 50 feet respectively. 30 + 40 + 50=120-5-2 = 60 [60-30 = 30 ' 60 - 40 = 20 1 60 - 50 = 10 1; ^30 X 20 X 10 X 60 = 360000- 360000 = 600 square feet = area. EXERCISES. Find to three decimal places, the areas of the triangles having the following sides. 1. 15, 20, 25. 2, 18, 20, 24. '3» ^3, M- 2. 12, 13. 6, 5» 4, 6. 2 rods, I rod, 12 yards. 7. I mile, 1700 yards, 6 furlongs. 43 yards, 875 yards, 888 yards. 3501 feet, 3604 feet, 3605 feet. 8, 9 liii! Hi I I lif I, j I i! ill III! ll i j i 38 10. The sides 01 a triangle are 136, 154 and 150. ine is drawn across the triangle and parallel to the longest side and dividing each of the other sides into two equa! parts. What is the area of each of the two parts into which the triangle has been divided? 11. The sides of a triangle are 9 ft. 3 in., 14 ft. 8 in., and 14 ft. 7 in. Two lines are drawn across the triangle | parallel to the longest side and dividing the other two sides | into three equ il parts. P'ind the area of the parts into which the triangle has been divided. Note 1. — The student should give special attention to this rule for ^ maasaring tria vgles. Ic ii a mOltus^ful rale for all classes of measure- ments of land, etc. The most irregular figures bounded by any num- ber of sides which aestraight I'nes, can bf^ readily measured, as they can be divided into triangles by simply joining the angles. Find prea| of each triangle separately and add them together. In measuring landp broken by lake?, livers, &c. , the most difficult piece can be measured' without any instrument tor striking angles— a simple tape line or | measuring pole being sufficient appatatus. 12. Find the area in acres of the annexed irregular field, the dimensions being in rods. 13. Find the area of the field in the an nexed diagram. Thef ditncnsions given arc| rods. ■MM 39 154 and 150. Ilel to the longest les into two equal wo parts into which ; in., 14 ft. 8 in., Lcross the triangle he other two sides of the parts into intion to this rule for all classes of measure- ounded by any nuni- y measured, as tliey | : angles. Find pre: In measuring land :ce can be measured simple tape line or hma^mm^ltt Note 2. The use of this rule might be further illustrated by applying it to irregular figures not having its boundaries straight lines, this is best illustrated by the accomp- anying drawing. The measurement obtain- ed in this v/dy is simply an approximate, ^ the correctness depending on the judg- ment of the person making the estimate, raw the lines in such a way as to leave as much of the surface of the ace to be measured outside of the straight lines as there is taken in the adjoining surface within the lines. The greater number of lines e greater the accuracy. This is the method employed by surveyors lind others in measuring lakes, and land bordering on lakes, &c. In the above figure measure distance from A to C, then calculate rea of trianglss and add together. 14. If the side A B is 40 rods, B C 30 nods, C D 50 rods A 30 rods, and the diagonal from A to C 60 rods, find the area of the figure. THE HEXAGON. a in acres of the d, the dimension^ Find the area field in the an diagram. The; [Jions given arc o Tta»i^TL The hexagon is a figure that can easily be drawn and measured. A hexagon may be constructed bytaking a compass and drawing a circle with it ; place one point of the compass on the circum- [ference and ineasure off the length of the radius six times ^ on the circumference. It will be found that the six radii can be placed inside the circle. Join these points in the circumfrence with each other and the hexa^^on is complete. From this it will be noticed that the hexagon is composed of six Equilateral Triangles. /, n 111 11: hi I i I I In liililil 40 Rule. Find the area of one of the six triangles ii which a hexagon may l)e divided by the rule on page which reads "from half the sum of the three sides, &c and multiply by six. EXERCISES. I Find the area of a hexagonal lawn having the side t( rods long. , 2. Find the area of a hexagon, the side of which is nil, feet in length. 3. A garden plot is hexagonal in form, sides 30 feet, e^ cept that two of its opposite sides are 120 feet, making oblong in shape. Find the area, the perpendicular distaiu between the long sides 51.960. 4. Find the area of a hexagonal oblong having, four « its sides 60 rods and the two remaining opposite sides ic; rods, the pf^rpendicular distance between the opposite sidi being rodi. J 03.92 Rhombus and Rhomboid. In these figures the student may find some difiiculh sometimes by proceeding to measure them as rectangub parallelograms. An experimental proof is very easily mad as follows : Take four pieces of wood of equal lengtl: Arrange them in the form of a square and put one nail ii each corner, so as to form a flexible frame. It will Ik noticed that when the angles are right angles, that the am enclosed by the frame is greatest, and that it gradually be comes less in area as the frame is elongated by pushing t\\( opposite angles toward each other. This may be contin.uet| 41 six triangles ii ule on page hree sides, &c wing the side U le of which is nil, sides 30 feet, e^ ) feet, making endicular distanc g having, four ui •pposite sides ic;' the opposite sidi iboid. [ some difficultv m as rectangub very easily madt.| of equal lengtl: I put one nail ii ame. It will 1)^ [les, that the am lat it gradually bej 3d by pushing twc may be continueii itil the adjacent sides of the frame will coincide, andthere- >re not enclose any space. From this experiment it may be gathered : - 1. That in a four-sided figure when the angles are right igles, it has the greatest area that can be enclosed by those )ur straight lines. 2. That the area of the four-sided parallelogram becomes ;ss as two of its angles become greater, and the other two ;ss than right angles. 3. That the rule for rectangular figures will not apply. 4. That they must be measured by some such rule as a triangle is measured. It will be noticed by the accompanying figures that if a piece were cut off one end and placed on the )e formed from the rhombus and a rectangle from the [rhomboid, and that the figures thus formed may be meas- ured by the length multiplied by the breadth. The breadth in this case being the peypendicular distance hettceen the sides. Hence we have the rule for finding the area of a rhombus and rhomboid. Rur.E. MuUiphj the base hij the jui'iu ndicular distcDice to the opposide side. EXAMPLE. Find the area of a rhombus, the side of which is 24 feet, and the perpendicular distance to the opposite side 20 feet. 20 X 24:^480 s(|uave feet. !!;! il!l! m m ! Mill III i! ii lii i !i i 42 EXERCISES. 1. The sides of a rhombus are 60 feet and the perpen- dicular between the sides 36 feet. Find the area. 2. The sides of a rhombus are 40 rods and the perpen- dicular between them 24 rods. Find the area in acres. 3. The sides of a rhomboid are 3025 feet and the per- pendicular 1080 feet. Find the area in acres. 4. The area of a rhombus is 10 acres, one of the sides is 80 rods. Find the perpendicular. 5. The area of a rhomboid is 6 acres and the perpen- dicular distance between the sides 24 rods. Find the side THE TRAPEZOID. TkaptuM* The trapezoid is a four-sided figure having two sides parallel. If a line were drawn through the middle of the sides that are not parallel, at right angles to one of the parallel sides, a piece would be cut off each end, which, if placed on the opposite side, would make the figure into a rectangular parr lie logram. The length of the parallel sides would be half the sum of the two parallel sides. Hence tlie rule : — Rule. Add the parallel sides : divide by i^\ and multiply by the perpendicular distance between them. EXAMPLE. Find the area in acres of a trapezoid, the parallel sides of which are 20 rods and 30 rods respectively, and the pcrpeii- »-,'»i .fl id the perpen- area. id the perpen- :a in acres. t and the pcr- le of the sides id the perpen- Find the side ur-sided figure If a Hne were e of the sides ight angles to piece would be opposite side, )arplle lograni. [ilf the sum of le by /.^ and nee beticccn arallel sides of nd the perpcii- 43 dicular distance between them 24 rods. 20 -I- 30 = 50 -r 2 = 2 5 average length. 24x25 = 600 square rods. 600 -j- 160 = 3J acres. EXERCISES. 1. Find the area of a trapezoid 60 feet and 50 feet long on the parallel sides and 30 feet perpendicular distance be- tween them. 2. A tapering field is 80 rods long, and the ends which are parallel, are 30 and 50 rods respectively. F'ind the area in acres. 3. A board is 12 feet long, 14 inches wide at one end and 6 inches at t^he other. Find its area m square feet. 4. Find the area of a board 16 feet long, 18 inches wide at one end, and 12 at the other. 5. Find the area in acres of a trapezoid, the parallel sides of which are 180 and 220 yards and the perpendicular distance between them 1 2 1 yards. The Trapezium. AND OTHER FIGURES. The trapezium and other irregular figures contained by straight lines are treated under the heading of triangles — pages 36 and 38, and as all such figures are easily divisible into triangles they can be measured by either rule. The annexed diagram of a pentagon illustiates the process of dividing regular figures into tri- angles for convenient measurement. Ill I I III I m I !!; !li ill i ! 44 CIRCLES. A circle is a plane figure bounded l)y one line of such curvature that all lines drawn from the centre to it are equal. A line drawn from the centre to the cir- cumference is called a radius, (plural radii.) A line drawn from one point in the cir- cumference to another through the centre is called a diameter. Not?. The diameter is equal to two radii. CIRCULAR TABLE. 60 seconds, marked " = i minute, marked ' 60 minutes = i degree marked " 360 degrees = i circle. EQUIVALENTS. 1 circle 360' r= 21600' -= 1296000" I = 60' — 3600" l' ^^ 60" Circles — The Circumference. To find the circumference of a circle the diameter bcipu given. Rule. Multiphj tlie diameter ht/^l, {or more cor- recti ij by 3. 141 6.) Note i. There are various modifications of the number 3^. (r.-) and of the decimal 3. 1416, that are sometimes given in rules for measuring circles. We have, however, been careful to use only the one fraction or one decimal to avoid burdening the memory with A, 39 iiii 45 e line of such : to it are equal. lire to the cir- , (plural radii.) nt in the cir- h the centre is irked ' ■ked° many decimals and fractions. Some of them would surely be for- j;olten or misapplied, and further, by the use of one the mind is led to trace the relation between the various operations much better than if separate decimals were given for each operation. Note 2. The fraction v- (3I) is not exactly conect, but is near enough for all practical purposes. By reference to the fifth question following this, it will be seen that it makes a differtnee of about ten miles in the circumference of the earth, the v- (^j) being too great. Note 3. By experiment it has been found that the decimal 3. 1416 is also a little too large, and that 314159 is a little nearer correct as expressing the relation between tie diameter and circumference. This is only necessary in very fine calculations. EXAMPLE. Find the circumference of a circle having a diameter of 35 feet. 35 X 22 35 X 3t= = 1 »o feet. snce. iameter hcinu; (or more cot- lumber 3! (y) yen in rules for to use only the e memory with EXERCISES. 1. Find the circumference of a circle 42 yards indiameter. 2. A circle is one mile in diameter. Find the distance around it. .^ . . 3. Find the distance around a circular race course 126 yards in diameter. 4. A lawn in the form of a semicircle is to be fenced at a cost of 30 cents per yard. Find the total cost if the diameter of the circle is 168 feet. 5. Find the distance around the earth if the diameter is 8000 miles, using separately the fraction Y-, (3i)the decimal 3.1416 and the decimal 3.14159. Note the difference in your answers. ! I t li ' ' • II lllll I'llili iiif 46 Circles— The Diameter. To find the diameter of a circle from the circumference. Rule. Divide the circumference by V-, (3l)> (or more correctly 3. 141 6). EXAMPLE. The circumference of a circle is no miles. Find the diameter. 110x7 iio-T-3|= =35 miles. X 22 EXERCISES. 1. The circumference of a circle is 88 feet. Find the diameter. 2. The circumference of a circular lawn is 242 feet. Find its diameter. 3. A circular race course is 891 yds. i ft. in circumfer- ence. Find its diameter. 4. A circular cistern is 22 feet in circumference. Find its diameter by both fraction and decimal methods. 5. A circular field is fenced at the rate of 25 cents per rod at a cost of $66. Find the diiimeter of the field. 6. A circular lake is 220 miles in circumference. Find its diameter. Cireles— The Area To find the area of a circle the radius being given. Rule. Square the radius and multiply by ^-(3|), or more accurately 61/3.1416. SWi ■»>, 47 ter. ircumference. ?/-V-.(3t)» {or les. Find the =35 miles. ;et. Find the I is 242 feet. in circumfer- erence. Find thods. 25 cents per be field. erence. Find given. EXAMPLE. Find the area of a circle having a radius of 35 feet. 35x35x22 =3850 sq. ft. = area. 7 NoTK. It will be noticed that in hnding the area of anything, two |mensions are always multiplied together. In the square or oblong vo sides are multiplied together. In the circle two radii are multi- li^d together. EXERCISES. 1. Find the area of a circle having a radius of 42 feet. 2. Find the area of a circular fish pond in square yards [hat is 14 yards in diameter. 3 Find the cost of sodding a semi-circular plot of ground, [he radius of which is 27 feet, at 35 cents per square yard. 4. Find the cost of sodding a semi-circurlar lawn, the I'adius of which is 63 feet, at ten cents per square yard. 5. Find the area of a circle 280 feet radius, by decimal IS well as fractional method. 6. Find the area of a circular fish pond 105 feet in Iiameter. 7. Find the cost of cementing the bottom of a circular :istern 7 feet in diameter, at 4 cents per square foot. 8. Find the cost of the stained glass for a circular win- [10 ft. 6 in. in diameter, at 32 cents per square foot. No illowance for sash. 9. A ventikitor in the ceiling of a hall is 21 inches in Iiameter. One-third of the space is taken up by an orna- nental iron screen. Find the actual ventilating surface. 10. Find the area of the end of a barrel 2 feet in diam- iter, by both fractional and decimal methods. 11. Find the value of a circular island i mile 6 furlongs fn diameter, at $4 per acre. 12. Find the area of a circular race track 140 rods in Iiameter. i J'' I , I i I'!''!!! Mill 48 Circles — Find the Radius. To find the radius of a circle the area being given. Rule. Divide the area by -y {^\) or 3.1416, and extract the square root. ^ Note. This rule is simply the converse of the preceding rule, hence the operation is just reversed. EXAMPLE. Fi(id th« radius of a circle having an area of 5544 square feet. - ■ 22 5544-^ =1764 Vi764 = 42 = radius. 7 EXERCISES. 1. Find the radius of a circle having an area of 246400J square feet. 2. Find the radius of a circle having an area of 1386 square feet. 3. Find the radius of a circular field containing 31 ^ acres, The Area of Rings. In finding the area of a ring it will be noticed that it is formed by one circle within another drawn from the same centre. These are soinetimes called Con centric Circles. Rule. Find the area of each circle and subtract the lesser from the greater, the difference unll be the area of the ring. i>\ 49 ladius. eing given. or 3. 14 1 6, ami m Jreceding rule, hena .?' e^ of 5544 square' 764 = 42 = radius. 1 area of 246400] m area of 1386 taining 31^ acres. 1 >. a ring it will be i by one circle from the saiiie imes called Con- and subtract nee ivill be the NoTR I. The above rule may be shortened by finding the differ- ice of the squares of the radii of the large and sniull circles and udtiplying by '^j-- NoTtt 2. The student who has studied algebra will find the above very nice exercise. Let x= the radius of large andj' cf small circle, i^-- ~ y- = {x - ?/) (.r 4- Ij). Hence the rule may be Bgain modified to the following : MitUipy the'[prodttct of the sum and iiffeience 0/ the radii by .-,-- . EXERCISES i: Find the area of a ring, the radius of the inside being 14 rods and of the outside circle 21 rods. 2. A circular race track has a radius of 20 rods for the smaller circle and 30 rods for the outer circle. Find the area, 3. Find the area of a ring, the radius of the inside being 40 rods and the outside being 60 rods. Parts of Circles —The Sector. ^ A Sector of a circle is bounded by two radii and the part of the circumference between them. A sector riiay be greater, equal to, or less than a semicircle. To find the area of a sector, the radius and the part of the circumferenc being given. Rule i. Midtiphj the radius by the part of the circurnfei'ence enclosing the sector and divide by two- Note I. It will be noticed that this rule deitical with that fo finding the area of a triangle. If a circle were divided into an in- definite number of sectors the curvature would be so impreceptible that nfinitely small .sectors could not be called any thing but triangles- This rule ii found to be accurate by experimental pioof. 'III II illiiil' III i II I in I I i I Ji Mill 50 NoTK 2. When the niiml)er of degrees are given between the radii, also the length of tlic radius, the area of the sectoi may be readily found, there l)ting 360 degrees in a circle. I^ for example the sector contained 30 degrees, then the sector would contain AHr or J_ of tli^" area of the circle. RuLK 2. Find the area of the circle, multiplied bif the degrees in the angle of the sector and divide hij ::(iO. \ EXAMPLE. Find the area of a sector of a circle, the radii being 10 feet lon^' and the angle 35°. ^'9 =301, square feet 7 X 3^>o EXERCISES. I, Find the area of a 90' sector of a circle having a radius of 35 rods. 2 Find the .»rea of a sector of a circle, the part of the circumference being 60 feet and the radius 45 feet. 3. Find the area, in acres, of a 75° sector of a circle having a radius of 40 rods. 4. A sector of a circle having a radius of 21 feet, con- tarns 240'. Find the area in square feet. 5. The r.rea of a 45' sector is 180 square feet. Find the radius of the circle. 6. Find the area of a sector of a circle, if the radius is 32 feet, and the part of the circumference 80 feet. 7. A sector of a circle is bounded by 120 feet of cir- cumference, and two radii of 120 feet each. Find the area. 51 >l)er of degrees are o the length of tlic loi may be readily ees in a circle. jf tained 30 degrees, 1 /k . . no o. iV "^ ''•'' multiplied bij %nd divide btj Find the number of degrees in the sector cf the circle Itioned in question No. 6. ' How many degrees are there in the sector mentioned i)roblem No. 7 ? How many acres are there in a sector bounded by radii 30 rods long and 45 rods of circumference? II. The sector of a circle, 42 feet in diameter, contains square feet. How many decrees are there between the lii ? radii being ,0 1 Pafts of Clrcles ! The Segment. ;] square feet circle having a the part of the 5 feet. tor of a circle f 2 1 feet, con- e feet. Find f the radius is feet. o feet ofcii- find the area. A segment of a circle is a part cut off by a line that does not pass through the centre. It may be either less or greater than a semicircle. In the annexed dia- gram what is above the line H K is a segment less than a semicircle, and the remainder of the circle below the line H K a segment greater than a semicircle. To measure a segment less than a semicircle, take the jgment H G K, join the ends of the chord H and K with le centre O, and we have a sector. The sector contains le segment and a triangle. Find the area of the sector and fie triangle, and subtract the triangle area from the sector [rea, and the segment area remains. To find the area of le segment larger than a semicircle, notice that the large sgment is composed of the triangle we deducted from le sector and the remainder of the circle, add there. )re the triangle to the portion that remained after the 1 II '"I ! 1^ I ! ; mill I i MlitI I I I illll!!! Ii I l!l l| i ! Ili'h 52 sector was taken away, and the result is the large segmen: or simply find the area of the small segment and subtra from the area of the circle. As the foregoing are only applications of the rules alreac given, it is not deemed necessary to give exercises. A Square and Hexagon Inscribed in Circle. To find the side of a square inscribed in , circle. It will be observed from the figur annexed, that the diameter of the circle fornix the diagonal of the square, and it is the hy | pothenuse for two equal right angled triangles.! A hexagon may be inscribed in a square, and measured| as per rule on page 39. The area of the six segments ma}l be found by subtracting the area of the hexagon from the] circle. Rule. Square the hypothenuse, divide hy t(ro,\ and extract the square root. ' * EXAMPLE. Find the side of a square inscribed in a circle 50 feet in| diameter. 50 x 50 = Vi25o = 35.35 feet. Ans. 2 EXERCISES. 1. How large a stick of square timber can be hewnj from a log 20 inches in diameter at the small end ? 2. How large a stick of square timber can be sawn froni| a log 32 inches in diameter at the small end ? Solve the above exercises as indicated in the example. 53 the large segmen: ment and subtra: Df the rules alreac exercises. )TE. A short rule is sometimes given for finding the side of an ribed square. "Multiply the diameter by .7071068." The author ses the student to work by general rules, rather than to attempt lorizing such decimal quantities. The area of a circle is 98.56 rods. A regular hexagon Inscribed in it. Find the difference between the area of hexagon and that of circle. scribed in 1 A Circle Inscribed In a Square. lare inscribed in i from the figur of the circle form; ', and it is the hy tit angled triangles, .re, and measured six segments inayj hexagon from the! divide by t(vo,\ a circle 50 feet inj 55 feet. Ans. ber can be hewnj lall end ? can be sawn from | nd? n the example. It will be - observed by the accom- panying diagram that the side of the square is equal to the diameter of the circle, and conversely the diameter of the circle is always equal to the side of the square in which it is inscribed, [here is therefore, scarcely any exercise necessary on this )int. It might be noticed also that the area of the corners imaining can be found by subtracting the area of the cir- \e from the area of the square. To Find the Area of a Cylinder. This point may perhaps be best treated in .connection with circles. The area consists 'of the areas of two ends and the area of sur- face in the form of a rectangle, (if cut and Ipread out,) the length being equal to the length of the Cylinder and the breadth to its circumference. :!ii!l Ji "ill If" I { ll«i !lll{ 'III i i' -v 54 EXAMPLE. , Find the area of the outside of a cistern 7 feet in diar eter and 8 feet long. 7 7 22x2 Area of two ends= — x — x = 77 square feet. 227 7x22 Area of outside = = 2 2 ft. circumference. 7 22x8=176 square feet. 176 + 77=2153 square feet. Ans. EXERCISES. 1. Find the area of a cylindrical cistern 10 feet dianietl and 10 feet deep. 2. Find the area of the inside, (no ends,) of a condu pipe 4 feet in diameter and 2^ miles long. 3. Find the area of the inside of the City of Toront water conduit from beyond Hanlan's Island, -/(^ of a mil lOng and 6 fict in diameter. (No ends). The Ellipse— To Find the Area. An Ellipse is a regular figure, anc easily drawn and measured. The line A B we will call the lon^ diameter, and the half of it the loiu radius ; the line O E the short diam-| eter, and half of it the short radius. To draw an ellipse, draw the long and short diametersj^on J to each other the exact length required, length of the long radius from the endoftheHete r O to where it would strike the long diameter Bsm; ^5^ 55 :n 7 feet in diar n lo feet diamet^ nds,) of a condu! \i C and C on each side. Now stick pins at C, C and O. Fie a piece of string, that will not stretch, tightly around (he three pins, then remove the pin at O, and with a lead )encil inside the string, pressing outwards against it, draw [he ellipse, the string around the pins at C and C will con- Irol the direction. To find the area, the rule is practically the same as the Ine for finding the area of a circle, the difference being that re have a long and a short radius. Rule. Multiply together the long and short radii md multiply by --/ (sD- Nc I-; Tm find the area of an elliptical I'ingT, find the areas ^f the i.. >. ' > .nd smaller ellipses and find their difference, which is the irea of the ring. EXAMPLE. Find the area of an ellipse, the diameters being 30 and ^o feet respectively. Diameters being 30 and 40 the radii will be 15 and 20. 15 X 20 X 22 =942 J square ft. 7 half of it the Ions E the short diam-l EXERCISES. 1. Find the area of a fish pond in the form of an ellipse, 1 1 54 feet long and 60 feet wide. 2. Find the area, in acres, of an elliptical field 84 rods |longand 40 rods wide. 3. Find the area of an elliptical race course, the diam- leters of the larger ellipse being 280 and 140 yards, and of the smaller ellipse 266 and 126 yards. !■■:•: -ill i:ii! i lii!:i III ''i'ii-V' I! ' '■ II iliiilli illlt il!! Mill 11! . iiii iillllll!li!llll|il| i ! < 'il' <:iin> mill II ijlllHill 56 The Circumferenee of an Ellipse. The figure is akin to the circle, and so is the rule, except rhat two diameters must be used, and half the sum taken. Rule. Add together the long and short diameters:]^ divide by 2 and Multiply by -y (3:)- EXAxMPLE. Find the circumference of an ellipse, the diameters ofj which are 43 and 27 feet respectively. 43 + 27 ■ 35x22 • _- ^ c =110 leet. 2 • • ■ . 7 EXERCISES. 1. Find the circumference of an ellipse having its di- ameters 18 and 24 feet respectively. 2. Find the circumference of an ellipse having its di- ameters 48 and 71 feet respectively. The Area of a Sphere. Rule.- Square the diameter and EXAMPLE. Find the area of a sphere, the diameter of which is 28 in. 28 X 28 X 22 1=2464 square inches. 7 EXERCISES. I. A football is 10 inches in di.imeter. Find the area of the surface. 57 2. The earth is 8000 miles in diameter. Find the sur- face in square miles. 3. yind the surface of a sphere 84 feet in circumference. 4. The circumference of a sphere is 88 inches. Find the area of the surface. s diameters of Find the aica The Area of the Curved Surface of Cones. COMu Suppose we take a circular tent and open up one of the seams and spread it out on the ground. We will find that the tent is a sec- tor of a circle. The higher the pitch of the tent the smaller the angle between the radii of the sector. Draw a circle on paper, and cut it out. Cut out a sector from it, and try the experiment of setting up a tent. The slant height of the tent corresponds wnth the radius of the circle. The rule, therefore, for finding the area of the cur- ved surface of a cone is the same as for finding the area of a sector, page 49, viz., — Rule. — Multiply the slant height of the cone hy the circumferejice mid divide hy tivo. EXERCISES. 1. Find the area of a cone, the slant height of which is 36 feet and the circumference 80 feet. 2. A tent is 42 feet m circumference and 15 feet slant height. How many yards of canvas are required to make it, (no allowance for waste). 3. How many yards of canvas in a tent 42 feet in di- ameter and 30 feet slant height ? i'lliilHH \\m III i! I ill ! l> ■ .ill liiii 58 Simpson's Rule. y r"irL~l5~i:i'~'i' The following rule, generally known as Simpson's rule, sometimes as the rule of Equi-distantOrdinates, is useful in finding the areas of irregular fig- ures approximately. The greater | the number of ordinates the nearer the approximation. Take for an ex- ample an areahaving one side curved, the others straight lines at right angles to the base line. Divide the line A G into an even number of equal parts at B, C, D, E, F. Draw straight lines through the points at right angles to A G, meeting the curved line at b, c, d, e, f. These lines are called ordinates. Rule. Add together the first and last ordinates twice the sum of all the other odd ordinates and four ti^nes the sum of the even ordinates. Multiply the result hy one-third of the distance between the ordinates. Let the following be the ordinates in their order, begin- ning with A a. 4, 5, 6, 7, 8, 9, 10 feet, and the distance between the ordinates i^^ feet : — 4-1-10= 14 6 + 8=14x2= 28 5 + 7 + 9=21 X4= 84 126 126 X i^ =63 sq. feet. ■MMHMM -, generally known Jometimes as the )rdinates, is useful of irregular fig- The greater nates the nearer Take for an ex- one side curved, s to the base lumber of equal t right angles to ^' These lines afif ordinateH ^dinates and f?.5. Multiply ■ hetiveen the r order, begin- I^H^H • H 1 H EXERCISES. j^H I. The ordinates in a figure a''e 4, 14, 36, 76, 140 feet, ^■id the common distance one foot. Find the area. i th e distance 3 sq. ft^t. C M Simpson's rule may be applied to almost any irregular figure. Take the annexed nregular oval for an example. The first and last ordinates will of course be o. 2. Suppose the ordinates are o, 20, 32, 36, 32, 20, o, |nd the common distance 2 feet. Find the area. 3. The ordinates of a certain irregular field are o, 12, 16, :8, 20, 24, 22, [8, 16, 15, o rods, and the common distance )et\veen them 2] rods. Find the area in acres, ^:c. Mensuration of Solids. A solid has length, breadth, and thickness or height. Volume, soli iity, or solid contents of any body is the [number of cubic inches, feet, yards, &c., it contains, It will be remembered that to find area tivo dimensions were multiplied together. To find vol- ume ^A^Tf^ dimensions must be mul- tiplied together — length, breadth» and thickness. In the accompany- ing engraving of a cube it will be noticed that three dimensions are placed on three oi the sides— 25 X 25 = 625. Suppose these dimeneions are feet. 625 square feet is the area of the base 625 x 25=* 15625 cubic feet, the volume of the cube. 6o liiiiii!" 'Hill iiliiil llUlli 'I ,„ il „ The connection between the tables of long, square, and cubic measure, should also be noticed : — 12 in. = I foot. 12 X 12 = 144 sq. inches=i sq. foot. 12 X 12 X 12 = 1728 cubic inches= i cubic foot. 3 feet=i yard. 3 x 3=^9 sq. feet:= i sq. yd. ^ 3 X 3 X 3=27 cubic feel=i cubic yard. The following are a few equivalents in weights and meas- ures : — I cubic foot of water weighs 62^ lbs, I imperial bushel contains 2218.19 cubic inches. I imperial gallon contains 277.274 cubic inches. Where absolute accuracy is not required 25 quarts maybe counted a cubic foot. A short rule for reducing cubic feet to bushels is — Mul- tiply the cubic feet by 100 and divide by 128. This rule is not a half a bushel astray per thou'^and bushels. The Cube. CMbt. To find the volume of a cube. Rule. Multiply together the lengthy breadth, and height. (In a cube these are equal.) EXAMx^LE. Find the volume of a cubical block of wood, each side being 15 inches. 15 x 15 x 15=3375 cubic inches. EXERCISES. 1. Find the volume of a cube, the side of which is 8 ft. 2. Find the volume of a cube 5 feet square. 6i g, square, and his and meas- 3. How many cubic yards in a cube 15 feet square? 4. Find the length of the side of a cube containing 13824 cubic feet. Note. Extract the cube root of the volume. 5. A farmer borrows a piece of hay i yard long, i yard wide, and i yard thick from his neighbor, and pays him back in pieces half a yard long, half a yard wide, and half a yard thick. How many such pieces should he r<;turn him for what he borrowed ? The Parallelopipedon which is 8 ft. To find the volume of a parallelo- pipedon. llteSgiSiii^^^ Rule. MuUlply torjether the ra.iiikiBi?U«»5^ lenxjth, breadth and thickness {or Jteight.) EXAMPLE. Find the volume of a piece of marble 6 feet long, 3 feet wide, and 2 feet thick. 6 x 3 x 2 = 36 cubic feet. If the volume and two of the dimensions of a parallelo- pipedon are given, the third dimension may be found by dividing the volume by the product of the given dimensions* EXAMPLE. The volume of .1 parallelopipedon is 180 cubic feet, the breadth is 5 feet, and the thickness 4 feet. Find the length. t8o = 9 feet long. 5x4 62 IHltil lull I i I' immF ! ^1! i!i lliiii EXERCISES. 1. Find the cubic yards of earth excavated from a cel- lar 30 feet long, 24 feet wide, and 6 feet deep. 2. Find the cost of excavating a cellar 45 feet long, 36 feet wide, and 5 feet deep, at 20 cents per cubic yard. 3. Fnd the volume, in cubic feet, of a stick of timber 60 inches long, 21 inches wide, and 15 inches thick. 4. Find the cubic feet in a parallelopidedon, 7 feet 9 in. long, 4 feet 6 inches wide, and 2 feet 3 inches high. 5. How many bricks, each 8 inches long, 5 J- inches wide, and 2] inches thick, are there in a pile 3 feet 6 inches long, 3 feet wide, and 2 feet high ? 6. Prove that a 6 inch cube is equal in volume to the sum of the volume of a 3 in«:h, a 4 inch, and a 5 inch cube. 7. A swimming bath is 24 feet 8 inches long by 12 feet 9 inches wide. How many gallons of water are required to fill it 4 feet deep? (25 ([uarts to cubic foot.) 8. 144 cubic yards of earth were excavated from a cel- lar 24 > 36 feet. Find its dei)th. 9. A cubical block of stone weighing 23040 lbs., weighs 15 lbs., to 72 cubic inches. Find the size of the block. Diagonal of Cube or Parallelopipedon. To find the diagonal of a cube or a parallelopipedon. Rule. Find first t]n> diagonal {Jij/pofhenusc) of the base by squaring the sides, adding togetliei' and extracting the squa re roof. Next use this diagonal as a base of a triangle and the height as the other side and repeat the operation of finding the diagon- al (hypothenuse). The last found is the required diagonal of the cubical figure. 63 EXAMPLi:. A cul)e is 40 feet long, 30 feet wide, and 20 feet thick. Find its diagonal. First find diagonal of base ^(30- +40'^) = 50 feet the di- agonal of the base V (50- 4- 20-) -53.85 feet, the diagonal required. 1. The north and south sides of a room are 30 ft. long ; the east and west sides arc 24 ft. The height is 16 ft. Find the distance from the north-east top corner to the south, east bottom corner. The Prism. Triansrnlar fzlBin, Uuadraugulaf .. Frism. Pentagonal ftlua. Hexagonal Prisms. 12 3 4 5 A prism is a solid whose ends are plane figures, equal, parallel and similar. Its sides are always parallelograms. The shape of the base determines the name. If the base is triangular we have a triangular prism. Rule. Ma/fiply the area of the base by the per- 2)endicular heujht. Not 7. That the above rule is true for four of the above prisms will be evident to any student. It is likewise true of the prism in an oblique position (see No. 5). The student who has studied Euclid will find suggestions towirds the proof of the correctness of this in proposi- tions XXX', xxxvi Eiclid, Book I. il i||lii!*lli|i ,!i!l '|'*'M!' 1 iillllii ill HI i'lliil Wii„ .,„„„„, ipii 1 I! li .iiiiiiijiiiii 64 EXAMPLE. Find the volume of a triangular prism 70 feet high, the sides of the base being 30, 40, and 50 feet. Area of base 600 feet. 600 x 70 = 42000 cubic feet. EXERCISES. I Find the cubical contents of a scjuare prism 8 feet high, the area of the base being 6 square feet. 2. Find the volume of a hexagonal prism 15 inches high, if the are.i of the base is 12J sfjuare inches. 3 Find the volunie, in cubic feet and inches, of hexag- onal prism having a base of 35 square feet 123 square inches area, and a height 5 feet 5 inches. 4. The volume of a pentagonal prism is 28 cubic feet and 500 cubic inches. The area of the base is 7 scjuare feet and 103 scjuare inches. Find the height. 5 A triangular prism measures 12 inches on every edge. Find the volume. The Cylinder. 'i'iie cylinder is a round solid having c.rcular csuls. As the figure is like a prism, except that the base Cylinder. q^ ^^^ j^ circular, it might t)e called a circular prism. To find the volume of a cylinder. Rule. Multiply the area of the end by the length or height. OjUadBi r EXAMPLE. A cylindrical cistern is 7 feet in diameter and 10 feet |dccp. Find its volume. •i X A X -V = y- = 38'j square feet area of base. 38 J x 10 = ;85 cubic feet = volume. EXERCISES. 1. A length of 12 inch water pipe is 14 feet long. How pany cul)ic feet of water will it contain ? 2. Find the volume of water in a cylindrical well 4 feet 111 diameter, depth of water 35 feet. 3. The length of a round log is 18 feet, and its circum- ference 96 inches. Find its volume in cubic feet. 4. Suppose a 25 cent piece is ^ of an inch in diameter uid -jV of an inch thick. If $25000 in quarter dollars were licked into a cube, find its size. 5. How many cubic feet in a log 12 feet long and 18 Inches in diameter? C). A water pipe is one inch thick and the inner diam- jter is )4 inch. How many cubic inches of metal are there |n 20 ft. of pipe. 7. What length of wire could be made from i cubic foot >f brass if the wire is .rV inch in diameter. 8 A piece of brass contains 6000 cubic inches. Suppose were made into wire ^jj inch thick, find how long the 'ire would be. 9. How many cubic feet are contained in the wall of a [ound tower, the height of which is 60 feet, the thickness of rail 2j4 feet. The tower covers a piece of ground 20 feet in diameter. 10. How many cubic inches of metal are contained in a V inch water pipe, the thickness ot the metal being ^ inch Ind the length of the pipe ten feet. e-1 I .llilill I ! i'llJliiiiil illl ! i 'I ! 11 III 66 The Ring. It will be evident to the student that the same rule that applies to the cylin- der will apply to a ring if made from square material, viz., multiply the area of the base by the thickness. For area of the base see page 48. A ring made from round material, however, may present some little difficulty. Rule. Multiply the area of a erofis section of the ring by the length of the ring. The cross section would be the end of the metal, or other material, if the ring were cut through. The length of the ring would be the distance around at the middle of tie metal substance of which it is composed. Thus if the di.i.neter of the outside were 20 inches and of the inside 16 inches, the diameter for finding the length of the ring would be found thus : — 20 + 16 _ <. 2 EXAMPLE. If the inside and outside diameters of a ring made from round iron are 16 and 20, find the volume. It is evident that the diameter of the iron is tvvo inches, and the radius one inch 1x1x22 7 3 f square inches area of cross section. 18 X 22 ^06 ^1-1 1 .1 r • ___ = ^^/_ =561 mches = length of nng. 77' •*5^f X3r=i77jlr- Ans. 67 . may present section of the metal, or other of cross section. EXERCISES. 1. A cylindrical ring is 20 inches long and i^ inches diameter cross section of metal. Find the volume. 2. A cylindrical ring is 7 inches thick. The inner di- ameter is 20 inches. Find the volume. Portions of Cylinders. Any portion of a cylinder having a base, a segment, or sector, may be measured by the same rule, viz. : — Multiply the area of the base by the perpendicular height. The Wedge. The wedge is simply a triangular prism turn- ed on its side, and may be measured as such. It, however, the sides of the base and the perpendicular height be given, the contents are easily found. XoFE. By inspection it will be observed that the wedge is just ha I ofa pirallelipijja 1)11 or a ca')2. H-'n:; thj following rule : — Rule. Malt f ply the area of the base by the per- pendieulav helcfht and divide by two EXAM P I.E. The base of a wedge is 4x6 inches, and the perpendic- ular height 10 inches. Find the volume. 4x6x10 1 • • 1 1 ^ =120 cubic mches= volume. 2 EXERCISES. 1. The base of a wedge is a foot square, and the per- pendicular height three feet. Fnd the volume. 2. The base of a wedge is 8 x 12 inches, and the volume 210 cubic inches. Find the perpendicular. m mi .■ill iiiii lii m Mm ill lii !i; Mi 'i 'IM I f , III Ml II I! lii" III I I '^ ! ' I i 68 Pyramids and Cones. i\,;-iiiiL OOMl Pyramid. A pyramid is a solid having a plane figure for its l)ase, and triangular sides that meet in a point called the vertex. Pyramids are named accord- ing to the shape of their bases. A triangular pyramid has a triangular base. A quadrilateral pyramid has a quadrilateral l)ase, &c. A cone is a pyramid with a circular base. A pile of sand or grain, or any such substance, is a cone. The wedge, iji the preceding chapter, sloped from two sides, the volume was found by mult;plvin,«; the area of ihe base by the height and dividing by two. The pyramid slopes to a point from all its sides, therefore the divisor must be uTeater than two. To fiiid the volume of a pyramid or cone. Rule. — MultlyJij the area of the base by the per- pcruticnhtr iteifjJit (cxJ diride by th ree XoiK. It will be observed the words "perpendicular height "are used. From the accompaning diagrams it '.vill be seen how this is four.d. The proof will be evident to any student \\ho has studied the first book of Euclid. 69 EXAMPLES. Find the volume of a pyramid having a base 6 feet square and 1 5 feet high. 6 x6_x_25 _ , g^ ^^^,^ f^^^^ 3 Find the volume of a cone 14 feet in diameter and 24 feet high. Radius^ 7 feet 7 X 7 X 22 X 24 = 1232 cubic feet. 7x3 EXERCISES. 1. A cone is 16 feet in diameter and 15 feet high. Find the vol'ime in '-ubic feet. 2. The base of a pyramid is 12 feet squire, height 20 feet. Find the volume. 3. The sides of the base of a triangular prism are 4, 5, and 7 feet, and the height 6 feet. Find the volume. 4. The base of a rectangular pyramid is 15x20 feet^ and the height is 18 feet. Find the volume. 5. The base of a pyramid has an area of 20 square feet 120 square inches, and the height is 5 feet 8 inches. Find the volume. 6. A ship's mast 100 feet high, is 3 feet in diameter at the base, and pointed at the top. Find the volume in cubic feet. 7. A side of the base of a hexagonal pyramid is 6 feet, the oerpendicular height is 15 feet. Find the volume 8 A pile of grain is t3 feet in diameter, aid 8 fiet high. How many bushels does it contain? (2218.19 cubic inches to the bushels). 9. A circular reservoir is 30 feet in diameter, 3 feet deep at the edge, and 8 feet in the centre. How many gallons vTill it contain? (277.274 cubic inches to the gallon). 1 1 I' ! mm 11 II lit '■iiiiiiii I'll 'II I. ■'ii';iil;;i I IlliPllllllllllljlliJIPj :i II! Wli iili Hi! i! '!f!!|ij il lii'n 1 1 1 .,,„Jiiir ill li iliil! ii''' '"'■' iliiil I'iiiliiiii '111 lir ' I 1 1 10. Find the volume of a pyramid 300 feet square at the base, the slant height from the middle of a side to the vertex being 200 feet. 11. The great pyramid of E^ypt was 481 feet higa when complete 1, and its base] was a square, 764 feet in length. Find its volume in cubic yards. 12. The spire of a church is a pyramid on a regular hex- agonal base. Kach side of the base is 10 feet, and the height 60 feet. The spire is built of brick, hollow, the same shape inside as outside. The base on each side is 9 ft., aid the hei-^ht inside 54 feet. Find the volume of the brick work in cubic feet. The following diagram will illustrate a very com- mon figure, and will illus- trate a very nice appli- ^ ^ c cation of the rules for measuring the volume of a wedge and pyramid. If this fig- ure were sawn down through perpendicularly at A and B, the middle portion would be a wedge, and the two ends brought together would form a pyramid — B, I), G, H, C. EXERCISE. 13. Suppose the base of the figure above is 20 feet long and 8 feet wide, and the perpendicular height 18 feet. Find the volume. 71 The Frustum of the Pyramid and Cone. d. If this fig- A frustum of a pyramid or cone is simply that part of a pyramid or cone that remains after the top has been cut off parallel with ,^ the base. To find the volume of a frustum of a pyramid or cone. Rule. Add toifether the areas of the ends and the square root of the product of the areas of the nds. Multiply hij the perpendicular height and \Hcide by three. EXAMPLE. Find the volume of the frustum of a pyramid 12 feet |quare at the base, 3 feet square at the top and 15 feet high. Area of the small end 3x3= 9 II II large end 12 x 12=144 Square root of their product=V'9 x 144= 36 189x15 =^945 cubic feet =volume. 189 EXERCISES. I. Find the volume of a frustum of a pyramid 10 x 10 ft. base, 4x4 feet at top, and 15 feet high. 1 2. Find the volume of a frustum of a pyramid 10 x 10 ft- base, 3x3 feet at top, and 12 feet high. 'i'«f!^!'!i, 'm. , 'Hi P :iil| (III: mi'il;' III': ||||j||||!!!|iii|l| „.illiiiii ijliiiii'lil I lii piiiiiiiiil .:il lilililill 72 »\ 3 Find the volume of a frustum of a cone 14 feet radius | at ba*-;^- 7 feet radius at top, and 30 feet high. 4. Find in cubic feet the volume of a frustum of a cone 12 feet high, the area of the large end being 36 square feet,| and the small end 16 square feet. 5. The mast of a ship is 66 inches in circumference at the base, 44 inches at top and 45 feet high. Find the| volume in cubic feet. 6. How many gallons will a circular tank contain that isj 10 feet diameter at bottom, 8 feet at top and 6 feet deep. 7. Find the volume of the frustum of a pryamid, the! base of which is 18 x 24 feet, the top 6x8 feet, and thej height 15 icet. 8. A cistern 5x12 feet at the bottom, 8 x 15 feet at tojj and 10 feet deep will contain how many gallons? 9. A frustum (»f a triangular pyramid is 20 feet high, thj sides of the base are 6, 8, and 10 feet, and the sides of thj top 3, 4, and 5 feet i;espectively. Find the volume. 10. A circular stack of new hay is 44 feet circumferencj at the base, 66 feet circumference at the largest part, 15 fj from the ground. The top part is 30 feet high above thif Find the number of tons in it, if 24 cubic yards make a toil 73 The Sphere. A sphere is a round solid body. To find the volume of a sphere. Rule. Cube the diaineter. Multi- ply by V", (or 3.1416 more accurately), and divide by six. Note i. The rule for finding the volume of a sphere may be stated in many ways. One very common way is, volume = (radius)"* x 4 x -iy^. Note 2. The volume of a spherical shell will be found by subtract- ing the volume of the inside from the volume of the shell, or by finding the difference of the cubes of the radii, multiplying by -%^- and divid. ing by six. EXAMPLE Find the volume of a sphere 4 feet in diameter. 4 X 4 X 4 X 22 =33^T cubic feet = volume. 7x6 EXERCISES. 1. Find the volume of a football i foot in diameter. 2. If the diameter of the earth is 8000 miles, find its volume in cubic yards. 3. Find the volume of a globe 3 feet in diameter. 4. Find the volume of a sphere 15 inches in diameter. 5. Find the volume of a sphere 16 inches in diameter. 6. Find the weight of a cannon ball 5 niches in diam- eter, made from metal weighing 700 lbs. per cubic foot. 7. A cannon ball, 9 inches in diameter, is made from I metal weighing 840 lbs. to the cubic foot. Find its weight. 8. A spherical balloon is filled with a light gas that has la lifting power of 21 ounces to the cubic foot. What size 74 ilill should it be so as to carry two men and fittings, weighing in all 600 lbs? 9. Find the lifting power of a spherical balloon 50 feet in diameter, filled with gas that has a lifting power of 22 oz. per cubic foot. 10. A spherical balloon is filled with gas that has a lift- ing power of 1 7I oz. per cubic foot, and carries a load of 1000 lbs. Find its diameter. NOTE TO THE STUDENT. The following four rules are not often used. We will simply give the rules for finding their volume without giv- ing any examples or exercises on them. The Zone of a Sphere. To find the volume of a zone of a sphere. Rule. Add together three times the .Sinn of the squares of the radii of the two ends and the square of the height. Multiply hy the height and by -y-, «^i<^ divide by six. MM ill m {I ! I'll! ill iili I, "I! m The Segment of a Sphere. To find the volume of the segment of a sphere. Rule. Add together three times the square of the radius of the base and the square of the height Multiply by the height and by V-, «^f^ divide by six. 75 The Oblate Spheroid. A sphere partly flattened, or a curling stone, is a common example of an oblate spheroid. The curling stone, however, is flattened at the top and bottom, and a spheroid is not. To. find the volume of an oblate spheroid. Rule. Multiply the square of the short diame- ter by the long diameter. Multiply by V", ^^^ divide by six. Prolate Spheroid. A prolate spheroid is like a ball elongated, or like a bird's egg, except that the egg is larger at one end than at the other, and the prolate spheroid is the same at both ends. To find the volume of a prolate spheroid. Rule. Multiply the square of the short diameter by the long diameter. Multiply by -j- and divide by six. Similar Figures. In order that figures may be similar they must be propor- tionate in their dimensions, and those containing angles must have their corresponding angles equal. 76 iiiJiliiii i i ' '' ! I] ; i 1 1 if ; ! i 1 '';!;•! i i M All circles are similar. All squares are similar ; So are pentagons, hexagons, octagons. Triangles, and parallelo- grams, are similar if their corresponding angles are equal. Similar triangles can be used in determining distances by the use of simple articles, such as a square and measuring pole or tape line, and a few pegs or pickets when the object to be measured is out of reach. Similar triangles may be formed by drawing a line through any tri- angle parallel to a side. Take the triangle A E D, draw through it the line B C. The small triangle is similai to the whole triangle because the angle at C is equal to the angle at E. The angle at B to the one at D, and the angle at A common to both. Rule. The corresponding or homologous sides of similar triangles are proportionate. To find the side D E of the large triangle, the side A D being given, and the two corresponding sides of the smaller one given. EXAMPLE. If the side A B is 8 feet and B C 6 feet and the side A D 40 feet. Let X = the side D E, then 8: 40:: 6: x .-. X = 30 feet = D E. The annexed figure illustrates a method of finding the breadth of a river by laying out similar triangles with a few pegs and a carpenter's square. It will always be found convenient to lay * ■- out right angled triangles in such work as this. :.77 The subjcn'ned illustrations show a method of measuring the distance of an inaccessible object — ? tree beyond a river, and the height of a standing tree. The latter may also be conveniently measured sometimes by setting up a stake or post and measuring the shadow cast by each* Then knowing the height of the post, the height of the tree may be found by an application of the triangle and pro- portion. Areas of Similar Figures. Rule. The areas of similar figures are in pro- portion as the squaj'ea of corresponding or homolo- gous dim^ensions. EXAMPLE. The area of a circle, 14 feet in diameter, is 154 square feet. What would be the area of a circle 25 feet in diam- eter ? The corresponding dimensions given, are diameters — 14'' : 154 :: 25^ : x «'. 196 : 154 :: 625 : x, £P = 49ij*j = area of circle. m I t % i.M i:r.- :! 7» EXERCISES. 1. A tract of land in the form of a right angled triangle is owned jointly and equally by A B and C. 'Ihe sides con- taining the right angle are 15 and 20 miles respectively. Two lines are surveyed at right angles to the 20 mile side, and parallel to the 1 5 mile side. A has the small triangle, B the middle portion, and C the portion lying along the 15 mile side. Find the frontage each man has on the 20 mile boundary. 2. A carriage wheel, 3 feet in d'ameter, is set in motion on the ground at an inclination of one inch from the p**r- pendicular. Find the diameter of the circle it will traverse' Similar Solids. • One solid body is similar to another if they are alike in form, though they differ in size. In other words, when one is the model of the other. All cubes are similar solids All spheres are similar. Other solids may be similar ; but for present purposes the above will serve as examples. The following principle will hold good in all similar solids. Rule. The volumes of similar solids arc in pro- portion to the cubes of their similar dim,ensionSt EXAMPLE. If a sphere 3 J inches in diameter weighs 6 lbs., find the weight of a sphere 5! inches in diameter. (3^)'^ : 6 :: (5I)'* : x x =• 20]- The model of a machine weighs 100 lbs. Find the weight of a machine made from' the same material five times its linear dimensions. r* : 100 :: $^ X aj= 12500 lbs. the weight of the machine* 79 Irregular Solids, A simple plan to find the volume of an irregular body, is to fill a vessel, sufficiently large, with water. Immerse the solid in the water and measure the water that was run over the vessel. An application of Simpson's rule is sometimes used in es- timating earth excavations, &c. Rule. Divide the aolul into any even number of portions, {by length). Ascertain the areas of the several cross sections at the points of division per- pendicular to the length of the solid. Add togetber the first area and the last ai'ea, and tivice the sum of the other odd areas, and four times the sum of all the even areas, and multiply the sum by one- third of the common distance betiveen them. M: m 8o COMMON MEASUREMENTS. The following pages contain many every day problems that are so simple that it has not been deemed necessary to add much explanation nor to work out many problems, almost all the work being applications of the rules already learned, if the student has worked from the beginning. A great many of the examples may, however, be worked by a person who has simply learned the "four simple rules in arithmetic." Wood Measure. A cord of wood contains 128 cubic feet. A standard cord pile is 8 feet long, 4 feet wide and 4 feet high — 8 X 4 X 4= 128 cubic feet. Rule. Find the voluine of the pile in cubic feet by multiplying together the length) breadth and height, and divide by 128. EXAMPLE. How many cords of wood are there in a pile of 4 foot wood, 40 feet long, 6 feet high. 40 X 4 X 6 I ^ J V— = 7^ Cords. 128 EXERCISES. 1. How many cords of wood are there in a pile 20 feet long, 4 feet wide and 4 feet high ? 2. How many cords in a pile 60 feet long, 4 feet wide and 4 feet high? 8i roblems issary to roblems, already ling. A ced by a rules in md 4 feet ihic feet [ith and )f 4 foot b 20 feet feet wide 3. How many cords in a pile 64 feet long, 4 feet vride, 7 feet high ? 4. How many cords in a pile 1 19 feet long, yf feet high, 1 8 feet wide ? 5. How many cords in a pile 77 feet long, 4 feet wide, 9 feet high ? 6. Find the cost of the wood in exercise No. 4, at $2.75 per cord. 7. How high must 4-foot wood be piled on a sleigh rack, 1 1 feet long, to make i ^ cords ? 8. My wood-shed is 20 feet long by 16 feet wide. How high must I pile wood in it in order to put 18 cords in it ? 9. How high should wood be piled between two stakes* 6 feet 4 inches apart so as to make a cord ? 10. A sleigh rack is 9 feet 6 inches long. How high must wood be piled in it to make a cord. 11. If wood is piled 7 feet 4 inches high in a shed 30 feet long, 22 feet wide, how many rords does it contain ? 12. How many cords in a pile of tan bark 300 feet long, 8 feet wide, 5 feet high? 13. A wood rack is 11 feet 6 inches long. How many loads of 4-foot wood, 4 feet high on such rack will be re- quired to make a pile 46 feet long, 40 feet wide, 6 feet high? T4. Find the value of the wood in exercise No. 13, at $3.15 per cord. 15. Find the value of a load of wood on a rack 10 feet 8 inches long, 4 feet wide and 4 feet 4 inches high, at $3.25 I)er cord. 16. What length of a pile of 30 inch wood, 4 feet high, would be equivalent to a cord ? 82 \ 17. What length of a pile of 18 inch wood, 4 feet high, would be equal to a cord. 18. What length should a pile of 20 inch wood, 10 feet long, be to contain a cord ? A single pile of 20 inch wood is 5 feet high, 30 feet How many cords does it contain ? A pile of wood, 12 feet long, 6 feet high, contains a 1 ind the length of the wood. 21. A flat car, 33 feet long, 8 feet wide, is loaded 8 feet high with wood. How many cords are on it ? 22. Find the value of a load of 4-foot wood, 3 feet 9 inches high, on a rack 9 feet 8 inches long, at $2.75 per cord. 19 long. 20 cord. Lumber Measure. Lumber is usually sold by the 1000 square feet. A foot of lumber is i foot square and an inch thick. If less than an inch in thickness it is counted an inch thick, that is, surface measure is counted. If more than an inch in thick- ness, the thickness is counted. A piece of board a foot square, 2 inches thick, would contain 2 feet of lumber ; if 3 inches thick, 3 feet of lumber ; if li inches thick, i^ feet,&c. Lumber includes boards, scantlings, joists, planks, &c. Rule. Multiply the length by the breadth for all lumber not more than one inch thick to find the ar3a. If thicker than one inch, inultiply the area by the thicknes in inches. 83 et big^» 1, 10 feet ;b, 30 ^e^^ contains a aded 8 feet , 3 fe^^ 9 $2.75 P^"^ Note. Although the exact mathematical answer is given in these Jumber exercises, in practice the nearest ev .n inches are counted in breadth and the nearest even feet in length, and in fractions of teet the nearest even feet are taken in final results for finding cost. EXERCISES. 1. How many feet of lumber in a board 12 feet long, 12 inches wide, i inch thick ? 2. How many in the above board if 2 inches thick ? if 3 inches thick ? if 3^ inches thick ? 3. How many feet in the board in exercise No. i, if ^ ^n inch thick? 4. How many feet of lumber in 12 boards, 12 feet long, 5 inches wide, i inch thick ? 5. How many feet in 90 scantlings, 12 feet long, 2x4 inches? 6. How many feet of lumber is 100 scantlings, 2x4 inches, 14 feet long. 7. How iTiany feet in 12 4x4 scantlings, 16 feet long. 8. How many feet in 24 half-inch boards, 10 inches wide, 1 5 feet long ? 9. How many feet in the following bill ? also find the cost at $7.00 for all that is less than 18 feet long and $8.00 for all that is 1 8 feet long and longer. 200 boards, 12 f^et long, 10 inches wide, i inch thick. 100 tl 14 M 9 M I t, 25 M 18 II 12 II I M 40 joists 20 II 10 II 9 « 180 scantlings 16 m 4 n 2 h 10. The following are the breadths of ten planks, each |i4 feet long, 2 inches thick : 11, 12, 10, 12, 11, 9, 8, 6, 7, 10. [ow many feet do they contain ? 84 11. How many feet are there in 30 boards, i inch thick, 10 feet long ? The following are their wi^^ths in inches : 12, 13, 16, 10, 9, 13, 8, 7, 12, 6, 8, 4, 9, 12, 13, 13, 14, 15. Mr i5» i3» 9j 8, 9, 9, 9, 8, 7, 5, 3. 12. How many feet of lumber are required to lay the floor of a stable, 20 x 30 feet ; the floor to be 2 inches thick? 1 3. How many feet of matched lumber are required to lay an inch floor in a house 24 x 36 feet, if one eighth of the surface is lost in matching ? 14. How many scantlings are required for a partition 40 feet long if they are placed at 16 inch centres ? 15. How many feet of lumber are required for the sheeting of a building 22 feet long, the sheeting to project a foot over each end, the rafters being 12 feet long ? 16. Find the cost of the lumber for a slatted walk, 160 feet long, 3 feet wide, the slats being 2 inches wide, i^ inches thick, placed i inch apart, at $18.50 per 1000 feet, and three 4x4 scantlings, the entire length underneath, at $8.75 per 1000. 17. A sidewalk make of 2 inch plank is 220 rods long, 12 feet wide. There are 5 scantlings, 3x6 inches, under- neath the entire length. Plnd the cost of the lumber at| $11.00 per 1000. 18. How many feet of lumber are required to close in the sides and ends of a barn, 40 x 60 feet, 18 feet high, the] gable ends being 15 feet high ? 85 11 II 11 M II 19. Find the cost of the following bill of lumber. Under 16 feet long at $7.50, and 16 and over at $8.00 per 1000. Make out the bill neatly. 3^ joists 2 X 10 20 feet long. 32 '» 2 X 8 20 64 rafters 2 x 615 220 scantlings 2 X 4 14 125 M 2X412 200 boards 1x12 14 2 sticks 9 X 9 30 2 II 9 X 9 20 II 200 battens at 3 cents each. 20. A piece of timber in the form of a wedge is 1 2 inches square at one end and sharp at the other, 14 feet long. How many feetof lumber are there in it ? 21. A 3-inch plank, 20 feet long, is 12 inches wide at one end and 20 inches at the other. How many feet of lumber are there m it ? 22. Find the number of feet of lumber in the following bill and the cost of same at $8.00 per 1000 feet when less than xo feet long, and ,$10.00 for all that is 20 feet and over. 300 pieces 2 x 416 feet long. 200 200 100 100 400 300 210 2 2x412 2x410 2x10 28 2 X 8 28 2 X 212 I X 12 12 2x618 6x 8 28 6x 8 36 It II II 86 Land Measure. At present all government lands are laid out in Townships of 6 miles square. All the older settled parts of Canada are laid out irregularly. Under the new system of survey a Township contains 36 square miles. Each square mile is called a section and contains 640 acres. These are again sub-divided into half sections (320 acres), quarter sections (160 acres), &c. EXERCISES. 1. What are the dimensions in rods of a quarter section of land ? 2. Find the value of a Township of land in Manitoba, at 50 cents per acre. 3. If a quarter section of land is worth $240.00, what is the price per acre ? 4. How iiiiiny quarter sections of land are there in a tract 12 miles square? 5. Find the area in acres of a field 30 x 40 rods. 6. Find die area in acres of 5 miles of road, 66 feet wide. 7. How many acres are there in a field 40 x 60 rods ; Find the side of a square field equal to it in area. 8. A piece of land in the form of a sector of a circle has a radius of 42 rods and an angle between the radii of 40°- Find the area. 9. A trapezium has its sides 40, 60, 70, 80, rods, the diagonal is 100 rods, forniing triangles having their sides 60, 70, ICO, and 80, 40, 100. Find the area. 87 ownships ■ Canada survey a ; mile is are again r sections ler section [anitoba, at oo, what is there in a d, 66 feet X 60 rods ; I. a circle has •adii of 40°' 10. A farm has a frontage of 30 rods on a concession and contains 50 acres. Find its length. 11. A quarter acre lot is 210 feet in depth. Find its frontage on the street. 12. How many acres, tS:c., are there in a circular plot of ground 28 rods in diameter ? 13. A circular field contains 5^ acres. F'':.d its, diameter 14. A hexagon is drawn in a circle having a radius of 42 rods. How much land is cut off as a segment of the circle by one of the sides of the hexagon ? Measurement of Lathing. Latns are sold by the 1000. They are put up in bunches of 50. A lath is 4 feet long and 1} inches wide. They are put on an average of '^- of an inch apart. Contractors usually estimate 1000 laths as covering 60 square )ards, or 3 yards to the bunch. This mak^s suffi- cient allowance for waste. Cood lath, carefully put on, will usually cover from 65 to 68 square yards. The con- tractors rule of 60 yards to the 1000 lathes is to be used in these exercises unless otherwise specified. 7.t11)s. of lath naiis will be required to nail on 1000 laths. In estimating the cost of the work only, of lathing, deduct half the o2)enin(/.s, such as windows, doors, &c. 1-athing is measured by the square yard. Less than half a yard is not counted. A half yard or over is counted the next above. This only applies to the final number of yards when about to multiply by the price. 88 In measuring lathing and plastering, a very rapid way is to measure the length of all walls with a tape line and take together all walls of the same height in one operation, also the ceiling of the whole house in another. A whole house in this way may be calculated almost as easily as one room. QpKNiNds. In making deductions for openings it will be noticed that all outside openings appear once only, and that inside doors appear t»vice, that is, in two rooms. EXAMPLE. A two story house is 24 x 36 feet inside. The first story ceiling is 10 feet, and the second story 9 feet high. The total length of the walls on the first flat 320; second flat, 360 feet. There are 4 outside doors, 3x7 feet ; 15 windows, 3 X 6 ; 16 inside doors, 3x7 feet. How many yards of lathing are there in it? Ceilings, 24 x 36 x 2 = 1728 Walls, first flat, 320 x 10= 3200 II second fiat, 360x9= 3240 8168 4 outside doors 15 windows. Deductions. 4x3x7 _ 2 1 5 X 3 X 6 _ 42 135 16 inside doors appearing twice, 16x3x7 = ^^6 513 7655 7655 square feet -f 9 = 850;', scjuare yards. The above example applies to plasu nng, wall painting? kalsoinining, »5s:c., as well as lathing. for w 8. laths bunc 9- 420 doorj 3^7 yard. lol lathsf lalh'^f ^vasi( 89 .1 EXERCISES. 1. How much space on a wall will a lath cover including the usual space between ? 2. How much space will a bunch of lath actually cover, allowing proper space between, but nothing for waste ? 3. How many laths are required for a room 10x15, ceiling 10 feet high? No allowance for waste. 4. How many laths arc required for a room 25 x 40 feet, (ceiling 15 feet high), in which there are 4 doors, 3x8 feet, 6 windows, 3.} x 7 feet ? Contractor's rules. 5. How many bunches should be bought in exercise No. 4? 6. How many laths are required in exercise No. 3 ? (Contractor's rules). 7. Suppose a house 24x30 feet, 16 feet high, gables 6 feet high, were lathed outside, first perpendicularly and then horizontally, as it is done in the Province of Quebec for rough-casting ; the laths in both cases being placed i inch apart. How many laths would be required ? No allowance for waste. 8. Allowing one-tenth of laths to be watted, how many laths would be required in exercise No. 7 ? How many bunches should be purchased ? 9. A cottnge 30x30 feet, ceiling 10 feet high, measures 420 ft. in length of lathed walls inside. There are 2 outside doors and 6 inside doors, each 3 >^ 8 feet, and 6 windows, 3x7 feet. Find the cost of lathing at 3 J- cents per square yard. 10. A fence 4; feet long, 3 feet high, is made by putting laths I inch apart, diagonally, in two directions. How many laths are re(piired, .Jlowiiig one-tenth of laths to be wasted ? •i"ii gc Plastering. In the incasurcment of plastering the same methods nre used as in the previous chapter on lathing. When the total yards in a building are found, take the neaiesteven yards in finding the cost. 235^ yards would be counted as 236, (S:c. Less than half a yard is left out. 7 bushels of lime, i bushel of hair and 1 ;| cubic yards of sand will make enough mortar to first-coat 100 yards. 2' bushels lime and ^ barrel of plaster of Paris will be sufficient to finssh 100 yards. Half t!ie openings only are counted out. KXKRCISE. 1. How many scjuare yards of plastering are there in the ceiling of a room 20 >; 40 feet ? 2. How many sciuare yards of plastering are there in the walls of a room 28\ 18 feet, 10 feet high. 3. How many yards of plastering are there in the walls and ceii.ii^ of a room, 36^26, 10 feet high ? 4. Allowing for an eighteen inch base board, find the number of yards of i)lastering in a room 38 -^ 27 feet, 14 feet high ? 5. Find the cost of plastering the walls and ceilings of a room, 18 >: 26 feet, 12 feet high, at 25 cents per yard. 6. Find the cost of plastering a ceiling 24X 18 feet, at 10^ cents perstjuaie yard. 7. At 12.^ cents per square yard, find the cost of plas- tering the walls of a room 30 x 18 and 12 feet high. 8. Find the cost of plastering the walls and ceilings of a room, 18 ^ 12 feet, lo.l feet high, at 12 cents per square yard, making allowance for four openings, each 4x8 feet. 9> 9. Find the cost of plastering a room 24x15 feet, it feet high, with 18 inch base, and havfng openings averaging 26 square feet, at 10 cents per s(|uare yard, (even yards). 10. Find the cost of cementing the bottom and sides of a cistern S < 1 1 feet, S feet deep, at 4 cents per s(|uare foot. 1 1. Find the cost of lathing and i)lastering the ceiling of a room 25 ^ 32 feet, tiie laths costing 8 i^' cents per bunch. Putting on laths and |)las(er 15 cents per s()uare yard. (Even bunches and even yards). 12. At 13.! cents per yard, find the cost of plastering the walls only of the following rooms. No allowance for openings. 30 >: 20 feet, 10 feet high ; 50^42 feet, 14 feet high ; 18-15 feet, 9 feet high. 13. .\ two story house 27 x 37 feet, has one story kitchen attached, 20^30 feet, is to be lathed and plastered at 10 cents per square yard for doin^^ the work. The ceilings are all 10 feet high, the length of the walls 760 feet. There are three outside and 13 inside dt»ors 3 x 8 feet, and 16 windows 3^6 feet. l'"ind the cost. 14. Find the cost of cementing a circular cistern, 8 feet in diameter, 10 feet deep, at 2 2 cents per square foot. 15. Find the cost, at 2I cents per square foot of cement- ing a circular cistern, 7 feet in diameter, 10 feet deep. 16. A circular cistern is 10 feet in diameter and 4 feet deep. P'ind the cost 01 cementing at 3 cents per square foot? 17. In a two story house 27x37 feet inside, the first story is 10 feet high, the second 9 feet, the length of plastered walls on first flat 325 feet, and on second story 410 feet. There are 12 windows, 3x6 feet, 3 outside doors, 3x8feet, and 10 inside doors, 3x7 feet. Find the cost at 18 cents per yard. (Even yard^). 5^ >^ I\^ IMAGE EVALUATION TEST TARGET (MT-3) A 1.0 I.I us 1^ ■ii lU 12.2 SB, I US *^ I. «lla L2I iu 1 17 Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 87^-4503 m'^ i\ ^^n 1 ^^^ -^A. '^\ Z ? .-o^ ^ Painting, Kalsomining, &e. The cost of painting, kalsomining, tuck-pointing, etc., is computed by the square yard. EXERCISES. 1. Find the cost ot painting a floor ^^ x 50 feet, at 15 cents per square yard. 2. A close board fence 210 feet long, 7 feet high, is to be painted on one side. Find the cost at 9 cents per square yard. 3. Find the cost at 1 2^ cents per square yard, of kalso- mining the walls and ceilings of a room 15 x 20 feet, 10 feet high. There is a base board i foot high around the room, and 120 feet to be deducted for openings. 4. Estimate the cost of tuck -pointing a shop front 30 feet wide, 60 feet high, at 90 cents per square yard. 350 square feet to be deducted for openings. 5. Find the cost of painting a shingled roof 62 feet long, each side being 28 feet wide, at 22 cents per square yard. Papering. Wall paper is usually 18 inches wide and put up in single rolls of 24 feet in length, or double rolls of 48 feet in length. The number of strips that can be cut from a roll, or double roll, will vary according to the height of the ceiling and the waste in matching. It is almost impossible to get at the exact quantity of paper required for a room on account of the waste in match- ing patterns, and the openings. The following is the rule generally used. !•/ 93 .-^i->. Rule. Measure the distance around the room in feet. Deduct 3 feet for each windoiv or door, and divide by i\feet, (the icidth of the paper) to find the number of strips. To find the I'olls requir^edj find how many strips can be cut from a roll and divide into the total number of strips required. NOT5. A pait of a r »11 is counted as a whole roll In computing the cost. EXERCISES. 1. How many strips of paper are required for a room 26 X 20 feet, if there are three windows and two doors ? 2. How many strips are required for a room io>< 15 ft. with two windows and one door ? 3. How many strips are required for the two sides of a hall 30 feet long, if there are six doors in the sides ? 4. If a ceiling of a room is 14 feet high, how many strips* for the wall, can be cut from a double roll ? ,5. How many single rolls are required to paper a ceil- ing 20 X 30 feet ? 6. How many rolls will paper a ceiling 10X14 feet ? 7. How many single rolls are required for the wall of a room 14x16 feet, 12 feet high, if there are two windows and one door ? 8. How many double rolls are required for the walls of SI room 30 X 40 feet, 5 feet high, having in it four windows and three doors. 9. How many yards of cornice paper, that has two copies of the cornice patterns to the width, are required for the room in question. No. 8 ? 10. At 40 cents per double roll for the paper, and 30 cents per yard for cornice paper, three patterns to the width* find the cost of the paper for a room 18x20 feet, 9 feet ceiling, allowing for three doors and two windows. '''inii '■^fi ... :*l!i :l.:3 II ;/ '• i' $4 Furnishing. EXERCISES. 1. How many strips of carpet, i yard wide, are required for a room 1 5 feet wide ? 2. How many strips of carpet, 27 inches wide, are re- quired for a room 18 feet wide ? 3. How much will be turned under or cut off from a carpet 27 inches wide, when laid on a room 15 feet wide? 4. If the strips of a 27 inch carpet are laid lengthwise on a room, 13 ft. 6 in. by 18 feet, and a half yard is lost each time tor matching strips, how many yards are required for the room ? N)TC. If there are six strips of carpet on a room, it will only be matched Bve tim;;, nothing being wasted on the first strip, as there is nothing to match it with. 5. If the strips of a carpet one yard wide are laid length- wise on a room 2o|^x28^ feet, how many yards will be Re- quired, allowing 10 inches waste each matching. 6. Find the cost of sewing and binding the carpet in No. 5, at 5 cents per yard for binding the ends, and 3 cents per yard for sewing together. 7. Find the cost of sewing the carpet and binding the ends in question No. 4, at 4 cents per yard for seams, and 7 cents per yard for binding. 8. Find the cost of felt carpet lining, one yard wide at 4J cents per yard for the room in question No. 5. , 9. Find the cost of a stair carpet at 90 cents per yard for a stair consisting of 16 steps, 12 inches wide, with 7 in. rises, allowing i^ feet to go on both upper and lower floors. ? ' - 95 10. Find the cost of moulding for the following rooms, at 6 cents per lineal foot : (a) 12x16 feet ; (b) 8x12 feet ; capacity in gallons. 6 i\ cistern somewhat similar in shape to the annexed engraving, is 2 feet diameter at top, 8 feet diameter at a depth of 3 feet, and 6 feet diameter at bottom a depth of 1 1 feet. Find how many gallons it will contain. I feet 2. pyraii 3. The bottoi pyran: 4. high. / 105 Measurement of Grain. The estimating of the quantity of grain in a bin or pile is a very useful and easy part of mensuration. Rule. Find the volume of tlw (jrain in cnhiv feet and rednve to bu.sJie/.s. To reduce grain in cubic feet to bushels, bring cubic feet to cubic mches and divide by 2218.19 — the number of cubic inches in a bushel. A practical and useful rule is to multiply the cubic feet by 100 and divide by 128 to find bushels. This rule is almost absolutely correct. A common rule is to multiply cubic feet by 4 and divide by 5. This, however, is not very accurate. A heaped bushel, such as a bushel of apples, potatoes, &c., contains about 2500 cubic inches. EXERCISES. I How many bushels in a bin of grain 4x8 feet, and 8 feet deep ? 2. How many bushels in a pile of grain in the form of a pyramid, 6 feet square at the base, 1 5 feet high ? 3. An elevator hopper is 20 feet square for 6 feet deep. The bottom part is 10 feet deep and 2 feet square at the bottom, the lower part being in the form of the frustum of a pyramid inverted. How many bushels will it contain ? 4. A circular pile of grain is 18 feet in diameter, 8 feet high. How many bushels does it contain ? n\ i.r 1 06 Ml PH- Measurement of Hay. The contents of mows or stacks of hay may be easily es- timated. The accuracy will depend on the good judgment of the person making the estimate. Old hay from the bottoms of stacks or mows may be es- timated from 15 to 17 cubic yards to the ton. Taking a mow or a stack entire, estimate from 16 to j8 cubic yards to the ton. New hay, freshly put in mows or stacks, 19 to 22 yards to the ton. On loads 22 to 24 cubic yards to the ton. '^ EXERCISES. 1. How many tons of hay in a mow 18 x 20 feet and 24 feet deep, estimating 16 cubic yards to the ton ? 2. How many tons of hay in a new stack 50 feet long throughout ; 20 feet wide at the base ; 30 feet wide at a height of 12 feet ; total height of stack 32 feet ; estimated at 20 cubic yards to the ton ? 3. How many tons of hay in an old circular stack 50 ft- in circumference at the base; 75 feet in circumference at a height of 15 feet; the top in the form of a cone above this ? Total height of stack 40 feet. (One ton estimated at 16 cubic yards.) 4. A circular stack 40 feet high is 42 feet in circumfer- ence at 15 feet high, and 28 feet in circumference at the base. At 18 cubic yards to the ton, how many tons does it contain ? 5. A new stack of hay is 77 feet in circumference at the largest part 21 feet from the ground, and 55 feet in circum- ference at the base — total height 63 feet. Find tons at 24 cubic yards to the ton. 107 Gauging. IL By gauging is meant the estimating the capacity of barrels, casks, &c. On account of the variation in the shapes of barrels, it is impossible to give a rule that will measure all casks exactly, a close approximate is all that is to be be expected. Note the difference in shape of the two casks illustrated. The diameter of the barrel inside at the end is known as the head diameter. The diameter at the middle is called the bungf diame- ter. Rule. To find the average diaineter, add to the head diameter two-thirds of the difference between the head and the hung diameters. Use average di- meter and m,easure as a cylinder, EXAMPLE. The head diameter of a barrel is 24 inches, the bung di- ameter 30 inches, and the cask is 40 inches long. Find its volume in cubic inches. Head diameter = 24, +§ of difierence between head and bung diameter. =| x 6 = 4. 24 + 4 = 28 average diameter. — — = 24640 cubic inches = volume. 7 EXERCISES. 1. Find the volume of a barrel 16 inches head diameter and 20 in. bung diameter and 3 feet long. 2. Find the gallons in a keg 30 inches long, 15 inches head and 18 inches bung diameter. M hi'} fi.f ;■ / i •! io8 %\ 3. A keg is 12 inches diameter at the head, 15 inches at the bung, and 18 inches long. How many gallons will it contain ? 4. Find the contents in gallons, of 100 kegs of vinegar, each 24 inches lon^., 14 inches diameter at the head and 16 Inches at the uung. Shoemaker's Measure. A size of shoemaker's measure is one-third of an inch. Number one children's is 4^^ inches. There are thirteen numbers in children's wear, that is, 12 sizes above No. i. 12 sizes = 4 inches, 4,^ + 4 = 8,! inches. No. i in adults' measure is one-third of an inch greater than No. 13 child- ren's foot wear = 8^ -H I = 8.1 inches. There is a regular in- crease of one-third of an inch for each size. Note. The length in the above refers to the "last" on which the boot is made — not to the length of the sole. EXERCISES. 1. What is the length of No. 7 childrens' size ? 2. What is the length of a No. 9 boot, adults' size? 3. What is the length of a No. 4 boot, children's size? 2 2i M m 109 f-. Miscellaneous Exercises. 1. How many acres more in a piece of land 4 miles square, than in one containing 4 square miles? 2. How many quarter inch cubes may be cut from a solid foot of marble ? (No allowance for waste in cutting.) 3. A man in plowing makes a furrough 9 niches wide. How far will he walk in plowing 2 acres ? 4. Your lot of .?j of an acre is 33 feet in width. Find its depth in yards. 5. The difference between the diameter of a circle, and the circumference, is 30 rods. How many square rods in the circle ? 6. Which will fill a cistern quicker, 6 three inch pipes or two 6 inch pipes? Explain why. 7. A cistern is 8 ft. long, 3 ft. wide, and 6 ft. deep- How many 12 qt. pailfuls will it hold? (10 lbs. to gallon.) 8. Find the value of a tree, 48 ft. long, 3 ft. diameter at small end, at 45 cents per cubic foot when squared. 9. 90 cubic yards of clay were thrown out of a cellar, 2 2^ feet long, by 15 feet wide. How deep was the cellar? 10. A cask full of water weighs 126 lbs. The cask weighs 14 il)s. How many gallons of water are there? (Cubic foot of water = 62^ fibs. 11. How many gallons will a cistern hold, the diameter being 3^} feet, and depth 6 feet ? 12. Find how many yards of cloth will be required for a circular tent 15 feet high, with a diameter of 14 feet. tfli ' iii i^ H I', . ! ' M IIO 13. How large a square stick may be hewn from a log 3 feet in diameter at the small end? 14. A bird hovering abov© a steeple which is 90 feet high, was shot by a man who was 80 feet from the foot of the steeple. He was 125 feet from the bird. How far was the bird above the steeple ? 15. A cow is fastened to a stake in the centre of a grass plot so that she may eat the grass off an acre and a half. How long is the rope with which she is fastened ? 16. Find the volume of a pyramid 8 ft. high, standing on a square base whose edge is 3 ft. 17. A and B start from the corner of a square field of ten acres to reach the opposite corner. Ii both travel at the same rate, B going across and A going round, how far is A away when B reaches the point ? 1 18. Find the distance through the opposite corners of a cube whose height is 8 ft. 19. A field which is three times as long as it is broad, contains 4^ acres. Find the length of it. 20. How much lumber i^ in. thick, is required for a lidless box, 2 ft. long, 2^ feet deep and i ft. 8 in. wide inside ? 21. If a boy 4 feet high, well proportioned weighs 60 B>s. How much will a man 6 ft. 3 in. weigh in proportion ? 22. C borrowed from D a cubic block of hay 10 ft. long. He paid back 2 cubical blocks 6 feet square, and 4 cubical blocks 4 fc. square. How large a cube will pay the balance ? 23. A penny being if inches in diameter, what part is covered by a cent one inch in diameter ? r it> III I a log lo feet foot of ar was a grass a half. anding field of ravel at how far ers of a broad, Id for a n. wide [ighs 60 )ortion ? (ft. long. cubical balance ? part is 24. Allowing 25 qts. to a cubic foot, how many gallons will pass a given point in 10 minutes, if the stream be 3 ftr deep and 20 ft. wide, and running 6 miles per hour ? 25. Two hunters, A and B, killed a deer and balanced him on a pole. A on the short arm just balanced the deer on the long arm, but B on the long arm just balanced the deer on the short arm. The weight of A and B being 192 and 147 respectively. Find the weight of the deer. 26 If 4 men were to grind off an equal portion of a grindstone 3 ft. in diameter ; how many inches would each grind off? 27. A street lamp is 24 ft. above the ground. How long a shadow will a man 6 ft. cast, if he stand 20 ft. from the lamp post ? 28. Find the cost per acre of land if $56 be paid for a plot 44 ft. by 140 ft. f 29. How many cubic teet of masonry in a hollow round tower, 48 ft. high, 1 1 ft. in diameter, if the thickness of the wall be two feet ? 30. If laid 5 inches to the weather, how many shingles will be required for the roof of a building 60 by 40 feet. The roof projecting over ihe sides and ends 1 ft., height of roof from eave to ridge being 15 feet. No allowance for waste. 31. On a pair of scales an article put on one end seems to weigh 22^ Il)s. and on the other end 18 lbs. What is its true weight ? 32. Find the cost of surveying 48 miles of railroad at $1.12} per chain. 112 . ■ ,jli 33. A field in the form of a rectangle is 15 chains long, and 40 rods wide. How many acres does it contain ? 34. How many loads of gravel will be required for a road 3.5 miles long, if it is spread 9 ft. wide and 8 in. deep? 35. Find the cost of fencing ^ acre in the form of a rectangle, at 50 cents per rod, if it is 33 ft. wide. 36. Find the number of tf-et of lumber in the following ; a stick of timber 12x12 inches, 48 ft. long ; a stick 9x9 inches, and 27 ft. long. 37. Find the number of perches of stonework in a wall under a building 20 ft. wide and 30 ft. long, 7.' ft. high, i^ ft. thick, allowing for 2 doors, 7 x3V feet, and 2 win- dows 3 X 3 ft. 38. Find the number of bushels in a bin 6x6 feet and 4}^ feet deep. 39. Find the diameter of a cistern 6 ft. deep, which will contain 1572 gallons. 40. Find the cost of carpeting a room 18x12 ft. with carpet 27 inches wide, at $1.80 per yard. 41. A contractor undertakes to dig a ditch 3^} miles long, 12 feet wide at top, and 6 ft. at botfom, and 12 feet deep, at 18 cents per cubic yard. How much did he receive for the work ? 42. Find the cost of plastering a hall 40 ft. long 24 ft. wide, 5 ft. high, allowing for 4 doors 37} ft. x 8 ft., and 12 windows 3 ft. x 7 ft., at 24 cents per square yard. 43. Find the number of gallons contained in a cistern 8 ft. in diameter and 6 ft. deep. 44. Find the number of cords of stone required to build a semi-circular wall measuring 90 feet in length, around the outside, 12 feet high, and 2 feet thick. "3 45- I'*"^ ^^^ number of bushels contained in an eleva- tor bin of the following dimensions : The top part is a regular parallclopipedon 8 ft. square at top, and 6 feet deep. It then slopes inward and at the bottom is only 2 ft. square. The bottom part being 4 feet deep. 46. Define Hexagon, Parallclopipedon, Cube, Multi- lateral figure, Frustum of Cone. 47. Give the rules for finding the volume of three different kinds of solids, and exemplify each by two prob- lems, worked out. 48. Find how many cubic yards of sand there are m a heap 35 feet in circumference, and 5.' feet high. 49. Find tlie cost of digging a ditch 500 yards long, 4 feet deep, 4I feet wide at top, and 3 feet wide at bottom, at 30 cents per cubic yard. 50. How many feet of lumber ^ inch thick, can be sawed out of a log 12 feet long, 3 feet diameter at bottom, and 2 ft. 9 in. at top. Making no allowance for waste except the slabs. 51. Find the estimated cost of building a side walk 300 feet long, 1 2 feet wide, there being 36 days' work of men at $1*25 per day, 10 days of men at $1.50, 6 days of foreman at $2.00 per day, 3 days of man and team at $3.00 per day, 300 lbs. nails at 3 J cents per ft, the plank to be two inches thick and four string pieces under it the entire length, 4x4 inches Lumber all to be $12.00 per thousand. 52. Find the cost per cubic yard for digging an L shap- ed cellar, 6 feet deep, if $21.60 were paid for the job. Dimensions are as follows, taking the sides in order be, ginning with the base of the L; — 30, 24, 15, 12, 15, 12. :i i '(', ' I 114 vs 53. How many acres in I'^oo feet of street 100 feet wide ? 54. A }4 acre lot is 210 feet in depth. Find its frontage. 55. How many cubic feet, &c., will a wagon-box 10^ fL'et long, 3 J feat wide, and 11 inches deep contain. 56. How many gallons in a trough shaped cistern 7 feet ^eep, the top being 8x12 feet, and the bottom 6x8 teet, the bottom and top being rectangular parallelograms. 57. I want to lay a carpet 27 inches wide to the best advantage on a room 16 by 19 feet. What direction will the strips run, and how many yards will it take? 58. A barn is 40 by 60 feet, the posts 20 feet high. The roof projects i J feet over the end and sides. The roof is 10 feet high above the walls. How many feet of inch hemlock are required to close in the building and sheet the roof ? 59. Find the cost of excavating a circular fish pond 6a feet diameter at top and 30 feet at bottom, 4 feet deep, at 25 cents per cubic yard. 60. There is a conical pile of grain on a floor, 6 feet deep, ID feet in diameter. How many bushels does it con- tain? (^1^ method.) 61. Fmd the cost of lathing the walls and ceiling of an L shaped house of the following dimensions : — the walls of first story are 520 feet long, and walls in second story 60a feet long. First story 1 1 feet high, second story 9 feet high. Cost of lathing 2 J cents per yard. The sides of the house taken consecutively being, 50, 40, 25, 20, 25, and 20 feet. i»5 lirr 62. Find the number of gallons contained in a circular tank, 5 J feet in diameter at bottom, and 4^^^ feet diameter at top and 4 feet high. (Exact answer required.) 63. Find the cost of laying a floor of matched lumber . in a h juse 30 x 40 feet, and kitchen 15 x 15 feet, and two semi-circle bow windows 6 ft. 5 in. in diameter, lumber to be 1} inch thick and costing $30.00 per thousand. 64. Find the cost of covermg a round church steeple at 25 cents per spuare foot if the steeple is 12 ft. 3 inches, . diameter at the base, and 15 ft high. On top of the steeple is a ball one foot in diameter, which is al ••s -v. I20 \\ 2. 3- 4. 5- s a 10. 'I II. 12. ft: 1' ' 13- 14. 15- 16. t 17. Short Rules on Circles, etc. We give these short, handy rules for accurate menfuremtnls in tabu- lar form for reference. They may be cut out and pasted in a memo- randum book and carried in the pocket. I. Diameter of circle x 3. I4i6=:circumference. Radius of circle x 6.283185 = circumference. Sfiuare tlie radius of circle x 3. I4i6 = area. Scjuare the diameter of circle x 0.7854 = area. Square the circumference of circle x 0.07958 = area. 6. Half the circumference of circle x by half its diameter = area. 7. Circumference of circle x o.i59i55 = radius. 8. Square root of area of circle x 0.56419 = radius. 9. Circum''erence of circle X 0.31831 =diimeter. Square root of area of circle x 1. 12838 = diameter. Diameter of circle x o.86 = side of in crib. 'd equilateral triangle. Diameter of circle x 0.7071 = side of inscribed square. Circumference of circle X 0.225 = side of inscribed square. Circumference of circle x 0.282 = side of an equal square. Diameter of ciicle x 0.8862 = side of an equal square. Bise of triangle x by h al itude = area. Multiplying both diameters and .7854 = area of an ellipse. 18. Surface of jiphere x ^ its di? meter -solidity. 19. Circumference of sphere x its diameter = surface. 20. Square the diameter of sphere x 3. 1416 = surface. 21. Squareof circumference of sphere X 0.3183 = surface. 22. Cube the diameter of .'phere x o.5236=soIidity- 23. Cube of radius of sphere x 0.5236= solidity. 24. Cube of circumference of sphere x 0.0 16887= solidity. 25. Square root of surface of sphere X o.564i9=diameter. 26. Square root of surface of sphere x 1.7 7 2454= circumference. 27. Cube root of solidity of sphere x 3. 8978= circumference. 28. Cube root of solidity of sphere x 1.2407= diameter. 29. Radius of sphere x 1.1547= side of inscribed cube. 30. Area ol its base x ^i its altitude = solidity of a cone, or pyra. mid, whether square or triangular. 31. Area of one of its sides X 6=surfaceora cube. r-^ n latu- memo- ANSWERS. a. le. pyra. Page 1 1. 21. 5464 12. 20.78 ^ 22. 19683 13. 11.2101 I. in, 20 23. 2732 14. 37-5 2. 27, 25 24. 1994 . 15. .9128 3- 3, 7 16. .9354 4. 5, 6 Page 136. 17- 1.5411 5- 24, 9 6. 81, 15 T S 1 *K 18. 15.7306 7. 21, 42 8. 54, 33 2' 14 - 25 3- TF 4- f%5 Page 1 Page 13a. , 5- 4 I. 4 6. 5^ 2. 6 I- 33 2. 93.786 + 7- 4,213 8. 81 3- 25 4. 55 3. 25 4- 45 "3 9- i¥r 10. 11 5. 78 6. 196 5- 95 •7 i TT 12^1 7. 97 6. 75 7. 85 •i ;) ;i 12. 95,353 8. 504 9- 145 8. 125 Page 13c. 10. 364 9. 179 II. .45 10. 121 I. .25 12. .046 II. 109 2. '33 13- 2.13 12. 115 3- -035 14. 6.54 13- 307 4. .98 15- 2.34 14. 294 5. .716 16. .25 15- 312 6. .216 17- 758 16. 216 7. 7.924 18. 1^ 17- 351-514 8. 1. 4142 19- .iV 18. 702 9. 1.73205 20. Vk 19. 997 10. 14.004 + 21. 35 20. 1432 1 1. 3.00001 + 22. 1.42 + 122 23- 2 104 ft. lo m. 24. 26.509 + 25. 5.16 + 26. 1494.2+ i 27. 1.25992+ I 28. 1.44224 I 29. 8.572+ I 30. 206.6016+ I Pagre 28. I I. 13 a., 20 rods.; 2. 53:s sq- yds. I 3. 366lt sq. yds. 1 4. 1 5 a. I 5. 10 a. HOC sq.! yds. i 6. 600 a. 7. 28^ 9. i8t a. 3960 sq.| yds ! Pagra 29. ' 1. 44 yds. long. I 2. 403 rods long I 3- 24/.y rds. \ 4. iSfVft- 5. 2Ji yds. 6. 25 ch. 7. 40 rods 8. 5 1.688 + ft. 9. 101.18 rds. 10. 40 rds. 120 rds Page 31. 1. 25 2. 53.814 + 3- 137-295 + 4. 10 ft. 11.48 in. 5. 13ft. 10.811 in. 7. 67 ft. 10.07 in. 8. 22 ft.3.28 in. 9. 96.2 ft. Pagre 32. 1. 528 ft. 2. 180 ft. ' 3. 5 ft. 4-49 if^- 4. 3 ft. 8 in. 5. 11.324 yds. 6. I mi. 37.77 rds. 7. 108. 16 + rds. 8. $133625 9. I mi. 245 j^ rds. 10. 210 ft. 11. 51 ^t. 5.03 in. 12. 18 ft. 8.8 in. 13. 40 ft. 4 in. 14. 21.23 + ft. 15. 1620 sq. ft. 16. 31I ac. Pag:es 37-8 1. 150 2. 176.156 3. 76.681 \. 10.79 5. 9.921 6. 30-195 7. 1054352. o8siq. yds. 8. 18060 sq. yds. 9- 5515650 sq ft. 10. 2310, 6930 11. 7sq.ft. i8f5sq.in 21 II 56 n Pages 34.-36. 1. 36 sq. ft 2. 18 sq. ft. 31 sq. in. 3. 20 sq. yds. isq. ft. 72 sq. in. 4. 67 a. 2982 sq. yds. 2 sq. ft. 5- 4a. 6. 2400 sq ft. 7. 96 sq. ft. 8. 384 sq ft. 9. 1 7 100 sq. it. 10. 7a. 80 sq. rds. ! i. 1650 sq. ft 12. 13- 14. I. 2. 3- 4- I. 2. 3- 4- 5- 35 " 93:' " Pag-e 40. I a. 99.8 sq rds. 2 10.444 + sq. ft 779.41 sq yds. 84.44 acres. Page 42. 2160 sq. ft. 6 a. * 75 a. 20 rods. 40 rods Page 43. 11. 1040 sq. 12. 30 a. ft. 13- 80 ft. 14. 14 a. losq. rds. 2. 20 a. 3. 10 sq. 4. 20 sq. 5- 5 a- ft. ft. 123 A*' 8 I. 2. 3- 4- 5- 6. 7 Page 45. 132 ft. 3} 396 yds. $43.20 5- 25'42i: 25132I Pag:e 46. 1. 28 ft. 2. 77 ft. 3. 283 yds 4 7ft.; 7iTr4Vff ^ 5. 84 rds. 6. 70 nil-^. Page 47. I A ft 5544 sq. ft. 154 sq. yds. $44-55- $69.30 246400. 246301.44. 8662! sq. ft. $1-54 8. $27.72. 9. 231 sq. in. lo- 3t; 3- 1416. 1 1, 12. $6 16c 96: Page 48. 1. 280 ft. 2. 21 ft. 3. 40 rds. Page 49. 1. 770 sq. rds. 2. 9 a. 131'^ sq, rds. 3. 6285^ sq. rds. Pages 50—1. 1. 96 2^^ sq. rds. 2. 1350 sq.ft. 3. 6.547 a. 4. 924 sq.ft. 5. 21.4 ft. 6. 1280 sq. ft. 7. 7200 sq. ft. 8. i43y\ deg. 9. 57tt rfeg. 10. 4^V a. 11. 120 deg. Pages 52—3. 1. 14.142 2. 22.627 + 3. 15.6 sq. rds. Page 54. 1. 471V i^q. ft. 2. 165942!^ cq. ft. 3. 69696 sq. ft. Page 55. 1. 7260 2. 16 a. 80 rds. 3. 4466 sq. yds. Page 56 1. 66 ft. 2. 187 ft. Page 56—7. I. 3147 2. 201142857I 3- 2245/1 4. 2464 Page 57, I. 1440 sq. ft. 2- 35 sq- yds. 3. 220 sq. yds. Page 59. 1. 192 sq. ft. 2. 288 sq. ft. 3. 2 a. 52 sq. yds. Page 60. T. 512 cub. ft. 2. 125 cub. ft. 3. 125 cub. yds. 4. 24 ft. 5. 8 pieces Page 62. 1. 160 cub. yds. 2. $60 3. lojif cub. ft. 4. 78 cub. ft. 810 cub. in. 5- 384- 6. 7. 7862} gals. 8. 4 ft. 6 in. 9. 4 ft. Page 64. 1. 48 cub. ft. 2. iS']}y cub. in. 124 3. 194 cub. ft. 363' cub. in. 4. 3 ft. 8 in. e. 748.244 cub. in Pajje 65. 1 . II cub. ft. 2. 440 cub. ft. 3. 9i.63cub. ft. 4 15-55 i"- ^" edge. 5. 2ii"j cub. ft. 6. 14I7 cub. in. 7. I3mls4578i\ft 8. 25454.4 ft ^ 9. 8250 cub. ft. TO. 2650% Page 67. 1. 48^ cub. in. 2. 3267 cub. in. Page 67. 1. i^ cub. ft. 2. 4I in. Pages 69 & 70. 1. ioo5f cub. fc. 2. 960 cub ft. 3. 19.6 cub. ft. 4. 1800 cub. ft. 5. 39 cub. ft. 608 10. 3968626cu!).ft. 11. 3466 I 45 I } cub yds. 12. 1408.157 cub. ft. 13. 1248 cub. ft. Pages 71 & 72. 1. 780 cub. ft. 2. 556 cub. ft. 3. 10780 cub. ft. 4. 304 cub. ft. 5. 76.2 cub. ft. 6. 2389.5 gal. 7. 3 1 20 cub. ft. 8. 5501.9 gals. 9. 280 cub. ft. 10. 10.99. tons. Pages 73 & 74. 1. 905^ cub. in. 2. 1,462,114,401 523,809,523,8 09H cub yds. 3. 14I cub. ft. 4. 1767^ cub. in 5. 2145.524 cub, cub. in, 235t ^" b. ft 7. 467.64 cub. ft. 8. 528.8 + bus. 9. 20565 9 gals. in. 6. 26.524 lt)s. 7. 185.55 tt)s. 8. 9.55 ft. dia. 9. 90029.761 lt)S. 10. 12.04 ft. Page 78. I. 11.547 m Is. 4.782 mis. 3 671 mis. 10 8 ft. Pages 80- 82. ^^ 1. 2^ cords. 2. 7! cords. 3. 14 cords. 4. 129 cords 88 cub. ft. 5. 21 cords 84 cub. ft. 6. $356.65. 7. 4A ft. 8. 7l ft^ 9- 5iVft- 10. 3tV ft- ^ 11. 37H cords 12. 112J cords 13. 60 loads 14. $271.69 15. $4.69 16. i2ift. 17. 21^ ft. 18. 7^1 ft- 19. 20. 21^ in. 21. i6| cords 22. $3.12 Pages 83— 85. i|i cords 1 144 ft. 2. 24 fr., 36 't. 42 [2 ft. ft. 60 ft. 5. 720 ft. 6. 933^ ft- 7. 256 ft. 8. 300 ft. 9- 6753^ ft- $49.05. 125 lo. 224 ft. II. 252! ft. 12. 1200 ft. 13. 987T ft. 14. 31. 15- 576 ft. 16. $14.48. 17. $1257.795. 18. 4200 ft. 19. $77.86. 20. 84 ft. 21. 80 ft. 22. {a) 24025 ft. (/^)$2 10.02. Pages 86 & 87. 1. 160 rods .sq. 2. $11520. 3. $1.50. 4- 576. 5- 7ia. 6. 40 a. 7. {a) 15 a. (<^)nearly 49 rds 8. 3 a. 136 rds. 9. 22.413 a. 10. 266| rds. 11. 51'f'ft. 12. 2,^1 a.. 13- 33-47 rds. 14. 1. 001 a. Page 89. 1. 90 sq. in. 2. 31] sq. ft. 3. 1040 laths. 4. 5234 5- 105- 6. 24 bunches. 7. 4493- 8. {a) 4993- (^) 1 00 bunches 9. $17.58. 10. 360. Pages 90 & 91. I. 2. 88«. 102 " !•• 3- 24IlT- 4. 294^. 5. $42.25. 6. $5.04 7. $16.00. 8. $10.44. 9. $11.70. 0. $15.68. 1. $15.98. 2. 62.55. 3. $107.90. 4. $8.31. 5. $6.46. 6. $6.12. 7. $171.72. Page 92. 1. $27.50. 2. $14,70. 3. $11.25. 4. $145- 5- $84.87. Page 93. 1. 52 strips. 2. 28 II 3. 28 n 4. 3 " 5. 20 rolls. 6. 3 double or 5 single rolls. 7. 17 rolls. 8. 9 rolls. 9. 23^ yds. 10. $6.30. Pages 94—5. 1. 5 strips. 2. 8 strips. 3. 9 in. 4. 38J yds. 5. 68^ yds. 6. $2.41 7. $1.83 8. $2.99 9- $8.68 lo. $12.48 ii. $17.11 T2. $22.66 I3 $24.00 14. 48 yds. if laid lengthwise or 45^ when laid across. Pages 96—7. I. 2Io''Tr Cds. 2. 28|- cds. 3. 156] cds. 4. 56} yds. sand 112^ bu. lime. 5. 7,:^ loads 6. 40 loads 7. $76.70 8. 12/^ cds. 9. 24H yds. sand 48#i bu. lime 126 lo. $31. 60 II. 12. 13- 14. $186.50 50-573 $198.00 $122.51, $49, $122.51. Page 102. Pages 107&108 w I. 5 cub. ft. 1 2 16 Page 98. 6. $40480 1. 100 posts 2. 2376 feet 3. 14784 feet . . 4. $218.35 i 3- 9'9gals. 5. 45760 feet , 4- K332-4S 2. 2 ::ub. in 4-5 gal^ :als. I. 2. 3- 4- 5- 6. 6000 16666 4 yds. sand, 18 bu. lime $33-33 49000 32!^ yds. sand 147 bu. lime 7. 64934 bricks. 8. $259-74 Pages 99 & 100. 111^, iq. rt. Pages 103 & 104. Page 108. 138^ SI- ft- 720 8 thousand 53^!^ 1. 2. 3- 4- 5- 6. $10.75 7- 38707 8. $101.38 Pages 100—1. 1. 768 feet 2. 1400 feet 3. 2143 feel 4. $44-45 5. 25 studs 6. 494 feet 7. 684 feet 8. $205.21 I. 30 feet. 2. 112, 3. 10. 1 feet. 8. ft. 1. 6,V in. 2. iiff in 3. 5I in. 4. 197.88 cub. ft Pages 109-119 5- 1 585 ft., $237-1 I- 7680 acres 75- 6. 29958^ ft. 7- $4677-33^- Page 104. 2. 110592 qrs 3. 22 miles 4- i46ii yd 5- ^SA n rds. 6. 6 inch pipes 1. 502.85 cub. ft. i 7. 300 pail 2. 1057.67 gals. ' 8. $97-20 3. 1495-7 g'lls. 9. 77,/eet 4. 1527-12 96.56 10. II gal. I r. Iff 5. 1196.56. II. 360 iiy 6. 2343.86 gals. ; 12. 4046 sq. yd 13. 25.4 m Page 105. L^; 5 o^. ft iches 500 bu. 2. i40i bu. 3- 303125. 4- 530tV- Page 106. 1. 20 tons. 2. 55 11 tons. 3. 19.56. 4- 5-43- 5. 21.522 15. 48.06 yds. 16. 24 cub. ft. 17- 23.44 rds. 18. 13.85 ft. 19. 48 rods long 20. 35ii^ feet. 21. 2 2 8."88i lbs. 22. 6.78 + ft. side ■^j- 4 It 24. 1980000 gal. 25. 168 ll)s. 127 26. 2.42 in. 53- 2.86 in. 54. 3.72 in. 55- 9 in. 56. 27. 6::- feet 57- 28. $396.00 29. 2715'i cub. ft. 53 30- 23213 shingles 59. 31- 20 tt)S. 60. 32. $4320. 61. 33- 15 acres 62. 34- 4io6fj loads 63- 35- $22.00 64. 36. 758I feet 65- 37- 66.9 66. 3»- 126,2 +l)ush. 67. 39- 7.3 + ft. 69. 40. $57.60 70. 41. $13305.60 71- 42. $37-81 43- 1 880. 28 gal. 44. 20.845 cords 45- 386.39 bush. 1 48. 6.616 cub. yd. 72. 49. $250.00 5°- 1089 feet. 73- 5^- $196.35 74 52. 18 cents 3.673 acres io3i ft. 31.28 cub. ft. 3081.01 4- gals lay cross-wise 48 yds. 7406.36 ft. 1 22.77 bush. $39-2 2f,. 540.6 gals. 54-65- $72.97- $80.56. 1 100 scj. yds. 925-2. 678.857. $6000. 792 cords. 3168 yds. ot sand. 1584 hush, of lime. 9621.60. .un"!b.$i6per M 249-53 cub. ft. 10000 ft. lumb. 100.000 shing. 75. n.49. 76. 5584 laths. 77. Lengthwise. $30-55- 78. 40 rds. breadth 80 rds. length. 79. 6.18 ac. 80. 50 90. 81. 16384 bricks. 82. $38.00. 86. 3133-805?. 87. 111.86. 88. 9081; s(i. yds. 89. I3-1907- 90. 156] bush. 91. 11.28 yds. 92. 2560 11)S. 9;;. 2]'^ acres. 94- $3-47-, 95. 2320 ft. 96. II iV tons. $76.88. 97. 19.20+ ft. 98. 14248 shingles 102. 23.44 rods. 99. 4.14 in. 100. 1385 ft. I 101. 72 rds. I I « t -Me TI3:£: 4- iTOKTHBRiT sN- BMSiNess eotteee. OWEN SOUND, ONT. J C A. P^lcnrjincT - Principal. The most practical Business Training Scliool in Canada. Circulars giving information, Course of Study, &c., free to any address. PUBLICATIONS BY C. A. FLEMING. Expert Book-keeping, $2.00. The Laws of Business, $1.50. How to Write a Business Letter, cloth 75 cenf;. Practical Mensuration, boards 50 cents ; cloth 60 cents. Fleming's Self-Instructor in Business and Ornamental Penmanship, $1.00 Thirty Lessons in Punctuation, with ruled paper with each lesson for exercises, 25 cents. Physician's Complete Account Book, containing Register of Visits, Ledger, Obstetric Record, &c., in Office and Pocket Size. Prices according to size and binding. Church Collection Register. Family Record, 22 x 17 inches, suitable for framing, 25 cents. OTHER WORKS IN PREPARATION. DO YOU . . LIKE A GOOD PEN? W-W' — Dse Fleming's 'Excelsior' Pens IK YOU WISH TO LEARN TO WRITE. Price : Sample Quarter Gross, 30c. ; i gross $1, by mail, post paid. P«iih older # I Is a patented device for holding tlie Pen so that the points will press evenly on the paper. A fine pen will run smoothly tvhen in this holder and can be used on any kind of work. Price by mail, 15 cents each. Address, C. A. FLEMING, Principal, Northern Bus. College, Owen Sound, Ont. *