UC-NRtF suration. ELEMENTS M E N S U R A T I K CONT 1 RULES FOR THE SOLUTION OF THE PRINCIPAL PROBLEMS EXPRESSED IN Til J] MOST CONCISE MANNER, ACCOMPANIED ISY EXPLANATIONS ADAPTED TO THE UXDEKST * NDING OF PUPILS WHO HAVE NOT ";"'V^ i^-' V •■•■ ' T.-i) GEOMETRY TOGEXnER WITH ^iimerotis Examples Illustrating the bartons gules. M. H. RODLriaio, TEACHER OF MATHEMATICS IX THE OIRLS' HIGH ANT) NORMAL SCHOOL OF PHILADELPHIA. PHILADELPHIA : PUBLISHED BY E. II. BUTLEPv & CO. 1862. '/fe» IN MEMORIAM FLORIAN CAJORl ^. (2^-.vn: Rodgers Mensuration. ELEMENTS OP MENSURATION. CONTAINING EULES FOR THE SOLUTION OF THE PRINCIPAL PROBLEMS EXPRESSED IN THE MOST CONCISE MANNER, ACCOMPANIED BY !■ XI'LANATIONS ADAPTED TO THE UNDERSTANDING OF PUPILS WKO HAVE NOT PREVIOUSLY STUDIED GEOMETRY. TOGETHER WITH Jtuntjcrous (Kxampks Illustrating tijc Various glulcs. M. H. KODGEES, XEACHZS OF MATHEMATICS IN THE GIRLS' HIGH AND NORMAL SCHOOL OF PHILADELPHIA. PHILADELPHIA: PUBLISHED BY E. H. BUTLER & CO. 1862. Office op the Contkollers of Public Schools, First School District of Pennstlvania. Philadelphia, February 12th, 1862. At a meeting of the Controllers of Public Schools, First District of Pennsylvania, held at the Controller's Chamber, on Tuesday, February 11th, 1862, the following Besolution was adopted : — Resolved, That Rodgers' Mensuration be Introduced to be used in the Public Schools of this District. EGBERT J. HEMPHILL, Secretary. Entered, according to an Act of Congress, in the year 1862, by M. H RODGERS, in the Clerk's Office of the District Court of the United States for the Eastern District of Pennsylvania. PEEFACE. Perhaps in no branch has the method of instruction, to the young student, been so greatly improved during the last ten years as in Mathematics. Pupils are no longer satisfied with committing mechan- ically rules expressing to them no meaning, because not understood. They no longer receive the words of their text-books, nor the assertions of their teachers, as sufficient proof of the truth of the ivformation imparted to them. Having been taught to inquire the reason why^ they are unwilling to receive a rule without investigating whence it came, and being convinced why it will solve the required problem. In the pursuance of such investigations, the mind, re- fusing to become a mere automaton, must reason, compre- hend, and decide for itself. The dislike evinced by so many of the young for the study of Mathematics, may be traced to the fact that the text-books placed in their hands are frequently too abstruse for their comprehension. The want of simple explanations in treatises on Mensura- (3) IV PREFACE. tion has been greatly felt, more especially by those pupils who have not previously studied Geometry. To such, the solutions of the problems are in a great degree unintelligible, and the retention of the rules rendered doubly difficult, because solely dependent upon the memory. It has been the aim of the Author, in preparing the present work, to supply this deficiency. The wording and explanations of the rules have been made as clear and brief as possible, and explanations have l^een given in all cases where they came within the com- prehension of pupils unacquainted with Geometry and higher Algebra. Particular attention has been paid to the correctness and the arrangement of DEFINITIONS. The EXAMPLES, which are original in construction, have been made very numerous, and illustrative of every variety of application of the different problems. Great pains have been taken in the solutions given of examples to place the dimensions in the figures, an omission common to most authors. The same care has been employed to avoid the improper use of alstract numhers, a fault found in most treatises on this subject, the solutions of examples being performed with abstract numhers, yet producing concrete numhers for results. Hoping this work will accomplish the object for which it was written, the Author commends it to the favorable attention of teachers. CONTENTS. PAQK PREFACE 3 MENSURATION OF SURFACES. PARALLELOGRAMS— Definitions 7 Tables of Long and Square Measure ... 14 Prob. I.— To find the area of a parallelogram 15 II.— To find the sides of a rectangle, the area and the proportion of its sides being given 20 III.— The side of a square given, to find the diagonal ... 21 IV.— The diagonal of a square given, to find the side ... 23 TRIANGLES— Definitions 25 Prob. V.— To find the area of a triangle 27 VI.— To find the base or altitude of a triangle .... 28 VII. — To find the hypothenuse of a triangle 30 VIII. — The hypothenuse and either side of a triangle being given, to find the other 34 IX. — The base and the sum of the other two sides of a right-angled triangle being given, to find those sides .... 36 X. — The three sides of a triangle bL'iiig given, to find a perpen- dicular whiih will divide it into two ri;,'ht-angled triangles 39 XI. — The three sides of a triangle being given, to find the area . 42 XII. — The area of a triiingle and the prt)portlon of its sides being given, to find their length 43 XIII. — The base and perpendicular of a triangle being given, to find the side of the inscribed square . ' 45 XIV. — Two triangles which are alike in one dimension, having their areas and the unequal dimension of the one given, to find the corresponding dimension of the other .... 47 TRAPEZOIDS, TRAPEZIUMS, AND REGULAR POLYGONS HAVING MOIIE THAN FOUR SIDES— Definitions 49 Prob. XV.— To find the area of a trapezoid 50 XVI. — To find the area of a trapezium 52 XVII. — The perimeter and apothem of a regular polygon being given, to find the area 54 XVIII. — One side of a regular polygon being given, to find the area 56 IRREGULAR POLYGONS OF MORE THAN FOUR SIDES— Definition . 58 Prob. XIX. — To find the area of an irregular polygon having moi-e than four sides ... 58 I* (5) VI CONTENTS. PAGK CIRCLES— Definitions 59 Prob. I.— To find the circumference of a circle C2 II. — To find the area of a circle 64 III. — To find the diameter or circumference when the area is given . 67 IV.— To find the area of a circular ring 69 V. — To find the side of a square inscribed in a circle . . .71 ARCS, SECTORS, AND SEGMENTS OF CIRCLES— Definitions . . 72 Prob. VI. — To find the length of a circular arc 74 VII.— To find the arc, when its chord and the chord of half the arc are given 75 VIII.— To find the area of a sector 78 IX. — To find the area of a segment 80 ZONES— Definitions 81 Prob. X.— To find the area of a circular zone 82 THE LUNE— Definitions 87 Prob. XI.— To find the area of a lune 88 THE ELLIPSE— Definitions 90 Prob. XII. — To find the circumference of an ellipse 92 XIII.— To find the area of an ellipse 93 MENSURATION OF SOLIDS. POLYHEDRONS— Definitions 95 CYLINDERS AND CONES— Definitions 99 Table of Cubic or Solid Measure . . .101 Prob. I.— To find the surface of a prism or cylinder .... 102 II. — To find the solidity of a prism or cylinder 106 III. — To find the surface of a cylindrical ring 110 Note. — To find the solidity of a cylindrical ring 112 Peob. IV.— To find the surface of a pyramid or cone .... 113 V. — To fiad the solidity of a pyramid or cone 115 VL— To find the surface of the frustum of a pyramid or cone . 117 VIL— To find the solidity of the frustum of a pyramid or cone . 119 THE WEDGE— Definitions 122 Prob. VIIL-^To fina the solidity of a wedge 122 THE PRISMOID— Definitions 125 Prob. IX.— To find the solidity of a prismoid 125 REGULAR POLYHEDRONS— Definitions 128 Prob. X. — To find the surface of a regular polyhedron .... 130 XI.— To find the solidity of a regular polyhedron .... 131 THE SPHERE-Definitions 132 Prob. XII.— To find the surface of a pphere 136 NoTK. — To find the surface of a spherical segment or zone . . . 136 Peob.XIII.— To find the solidity of a sphere 139 . XIV.— To find the solidity of a spherical segment .... 141 MENSUEATION. ^ Mensuration is practical geometry, or that part of geometry which teaches how to find the lengths of lines, the areas of surfaces, and the volumes of solids. DEFINITIONS. A mathematical point is that which indicates position only, and has no maynitude. The point of a fine needle, or the representation of a point on the black board, is a physical point, perceptible to the senses, and as such possesses magnitude; but a mathematical point, having no magnitude, cannot, strictly speaking, be represented. The beginning and termination of a line are points, but these form no part of the line itself. ' A line is that which has length • only. All lines are either straight, curved, or combinations of these two. ^ A straight line is one that does not change its direction between its extremi- ties, and is the shortest distance between two points. (7) MENSURATION. A curved line is one that changes its direction at every point. A hroken line is one that is made up of a number of limited straight lines. When the ^ .^^^ word line only is used, a straight ^^^^ line is meant. Lines, according to the manner in which they are drawn, are termed 'parallel^ ohlique^ 'perpendicular ^ vertical^ and liorizontal. Parallel lines are those, which being situated in the same plane, if produced to any extent, both ways, never meet. Oblique lines are those which do not maintain the same distance apart, but meet if sufficiently pro- duced. A perpendicular line is one that meets another, making the angles on each side equal. The angles thus formed are called right angles. Thus the line C D is perpendicular to the line A B, and the line A B is perpendicular to C D. '^ d Vertical lines are those which are perpendicular to the horizon or water level. Vertical lines at different points on the earth's surface are not par- allel, but converge towards the centre. Horizontal lines are those which are parallel to the horizon, or water level. DEFINITIONS, A line is intersected when crossed or cut by another. Thus the line A B intersects D C. The line is said to be bisected when it is divided into two equal parts. Thus the line E D is bisected. ANGLES. ' A plane rectilineal angle is the opening or inclination of two straight lines meeting in a -^^ point. The two lines which form it are called its sides, and the point where they meet its vertex. Thus A B G is an ano^le. An angle is read by naming the letters or figures which stand at the vertex^ and at the ends a of the lines which form it, always placing the one at the vertex in the middle. Thus the angle A B C or ^ the angle A B D. If there be but one angle at the vertex, it may be indicated by reading the letter at that point only, as the angle A. Angles are measured by arcs of circles, and their size is expressed in degrees, minutes, and seconds. To measure an angle, make the vertex of the angle the centre of the circle, then with the whole or part of one of its sides for a radius, describe a circle around its centre. Divide the circumference into 360 degrees, and the numher of degrees contained in the arc (or part of the circumfer- ence) between its sides constitutes the measure of the angrle. 10 MENSURATION. The size of an angle is not altered by having its sides produced. The lengthening of its sides increases the circumference of the circle in the same proportion, there- fore it will take the same number of degrees to measure it as before. The size of the degrees is altered, but not their number. A B C is an angle of 45°, as is also D B E. A Angles are divided into right, acute, and obtuse angles. A right angle is formed by a line meeting another peiyendicularlt/ ; as, the angle ABC. It always contains 90 degrees. B An obtuse angle is one greater than a right angle; as, DEC. It may contain any num- d, ber of degrees between 90 and 180. It is called obtuse because its vertex is less pointed than j^ ~ ^ that of a right angle. An acute angle is one less than a right angle; as, B A C. It is called acute because its ver- tex is more pointed than that of a right angle. All angles, except right angles, a are called oblique angles. SURFACES. A surface is that which has length and breadth only. Its boundaries are lines. Surfaces are the boundaries, faces, or liinits of solids ; DEFINITIONS. 11 they must be considered as making no part of the solids themselves, and therefore can have no thickness. A plane surface is one with which a straight line, if laid any direction, will exactly coincide. A curved surface is one that, like a curved line, changes its direction at every point. A concave surface is one which is rounded out on the inside into a spherical form, like the inner surface of a hollow globe. A convex surface is one which swells on the outside into a round or spherical form, like the outside of a globe. POLYGONS. A polygon is any plane figure bounded by straight lines; as the following figures : 12 MENSURATION. The perimeter of a polygon is the sum of the sides which bound it. A regular polygon is one whose sides are all equal. K polygon takes its name from the numher of its sides. A polygon of three sides is called a trigon or triangle. u four (C u tetragon or quadri- lateral. it five pentagon. li six hexagon. a seven heptagon. ic eight octagon. a nine nonagon. a ten decagon. u eleven undecagon. a twelve dodecagon. There are six tetragons or quadrilaterals, four of which are called parallelograms. A. parallelogram is a quadrilateral , whose opposite sides are parallel. A square is a parallelogram whose sides are equal, and whose angles are right angles. A rhombus is a parallelogram whose whose angles are not right angles. The altitude of any parallelo- gram is measured by a straight line drawn from tho top of the figure perpendicular to the base ; as, A E. The base is the line D C on which it stands. are equal, but A B DEFINITIONS. 13 A rectangle is a parallelogram whose opposite sides only are equal, and whose amjles are right angles. A rhomboid is a parallelogram whose opposite sides only are equal, but whose angles are not right angles. A diagonal of a polygon is a straight line joining the vertices of two opposite angles. The area or superficial contents of any figure is the amount of surface which it contains. Similar polygons are those which have the same number of angles, which are equal each to each, and the sides about these angles, taken in the same order, proportional. In similar polygons, the corresponding sides, angles, diagonals, &c., in each, are termed homologous. Similar polygons are to each other as the squares of their homologous sides, or as the squares of their like dimensions. A Problem is a question that requires a solution. An Axiom is a self-evident truth ; as, " Things which are equal to the same thing, are equal to each other." " Things which are doubles of equal things, are equal to each other." "A straight line is the shortest distance between two points," &c. 2 TABLES. LINEAR OR LONG MEASURE. Linear or Long Measure is used to measure distances. 12 inclies (in make 1 foot, ft. 3 feet . u 1 yard, ^t?. 5 J yards, or 16 J feet . 11 1 rod, rd. 4 rods, 22 yards, or 66 feet u 1 chain, ch. 10 chains, or 40 rods a 1 furlong, fur. 8 furlongs . a 1 mile, m. 3 miles . . u 1 league, ha. 69^ miles . a 1 degree, deg. or ° 360 degrees • (( f The circumference 1 of the earth. SURFACE OR SQUARE MEASURE. Square Measure is used in measuring the areas of surfaces. TABLE. 144 square inches (^sq. in.') . make 1 square foot, sq. ft 9 square feet 30i square yards, or 272^^ "| square feet . . . j 16 square rods 2 i square chains, or 40 square "j rods . . . . j 4 roods, 160 square rods, or "j 10 square chains . . j 640 acres .... 1 square yard, sq.7/d. 1 square rod, sq. rd. 1 square chain, sq. ch. 1 rood, R. 1 acre, A. 1 square mile, sq. m. (i-t) MENSURATION OF SURFACES. PARALLELOGRAMS. PROBLEM L To find the area of any parallelogram when the base and altitude are given. Rule. Multiply the base by the altitude. Note. — When the base and altitude are the same, this will be equivalent to squaring the side. EXPLAlfATION. 1. What is the area of a square whose side is 6 feet ? «ft. 6 ft If this square is 6 feet on each side, when the figure is divided into 6 equal parts, through its sides, each strip or divi- sion will be 1 //;. wide. If one of these strips be again divided, in the opposite direction^ into 6 equal parts, these sxdtdivisions will be each 1 ft. long and 1 //. wide, or will con- tain 1 sq.ft. Hence in this strip or section there are 6 sq. ft., and, as there are 6 strips i ft in all, there will be 6 times 6 sq. ft. in the figure, or 36 sq. ft. Res. 36 sq. ft. (15) 16 MENSURATION OF SURFACES. A rhombus and rhomboid may be changed to a rec thus : Let A B C D be a rhombus, then through the altitude A E cut off the part A E D, and apply it to the other side of the figure, it will then become the rectans-le A B D E. o Since the base and altitude of both figures are alike, if their product gives the area of the rectangle, it will also give the area of the rhombus. tangle, EXAMPLES. 2. What is the area of a square whose side is 12 feet? ///^^ 4 3. How many acres in a garden lot 737 feet 6 inches square ? /S-^^o 4. How many acres in a square piece of land whose sides are each 40 rods ? /^ ^. 5. How many square miles in a right-angled field whose sides are each 200 chains ? f" ^^^ ^ . > . '/.-j 6. What is the area of a rectangle whose length is 5 feet, and breadth 7 feet? J>5 >^/. 7. What is the area of a rectangle whose length is 13 chains, and breadth 9 chains ? /' -^ ^ 8. What is the area of a rhomboid whose length is 2 feet 5 inches, and its altitude 2 feet? J^/ ^ d/' 9. What is the area of a rhombus whose base is 13 feet, and altitude 5 feet ? . , 10. What is the area of a rhombus whose base is 15 feet, and altitude 13 feet? /^/-^//J / PARALLELOGRAMS. 17 11. What is the diflference in area between a field 35 rods square, and one of half the dimensions? f/S' /^tir^^ 12. How many small squares, each containing 4 square inches, are contained in a large one which is 3 feet square 'i J2J-'' *■' ■■'■' ' 13. How many squares will it take to equal in area one of 40 rods square, if the smaller ones are each half the dimen- sions of the larger ? <^~y u ^ f ^ 14. What is the area of a square field whose perimeter is 160 rods ? t <^"V -Tt^rij ^^ /^ ''i . 15. What is the area of a square whose perimeter is 140 chains? /Z X StXj, ^ ^^ C^t * ^ ^.^9^^^ 16. If the sum of the base and altitude of a rhomboid is 64 feet, and the base, is to the altitude as 3 to 5, what is the area? " t ■^' .- • 17. If the sum of the base and altitude of a rhomboid is 128 feet, and the base is to the altitude as 7 to 9, what is the area? jiy^^C f f^ 18. The altitude of a rectangle equals 10 times the square root of 256, and the base is to the^ altitude as 1 to 2. What is the area? .U S '^ '-' ^^ - 'i/^.-/?^- 19. How much will it cost to plaster a room, if the length is 25 feet, the width 16 feet, and height 17 feet 5 inches, at 15 cents a square yard ? ^^' / ^ ^ 20. How many yards in length of carpet, which is 3 quarters wide, will it take to cover a floor 22 feet wide, and 30 feet 3 inches long? ;>/ , Sf 7* " *' 21. How many bricks, 8 inches long, and 4 inches wide, will it take to pave a yard which is 16 feet by 25 feet? // 'S 22. How many square yards of paper will it take to cover the walls of a room, which are 16 feet wide, 27 feet long, and 17 feet high, deducting ^^ of the surface for the doors 2* B 18 MENSURATION OF SURFACES. and windows ? How much will it cost to lay it on, at a cent a square yard ? ) f-J- ^ ' ^"^^ ^ /S'- J.^ / Note. — If the area of any parallelogram equals the product of its base and altitude, the area divided by either dimension gives the other ; for if the product of two numbers, and one of them be given to find the other, we divide the product by the given factor. As the area of a square is the product of two equal factors, the square root of its area equals the side. 23. The area of a square field is 663.0625 sq. rods. What is the length of its side 'i Solution. |/663.0625 sq. rods == 25.75 rods. Res. 24. The area of a square is 1600 sq. chains. What is the length of its side ? 25. What is the side of a square field whose area is 4 acres? v' /^ C - 26. What is the side of a square lot whose area is 8 acres, 8 roods, and 25 sq. rods? . ^6' v / ^ */ 27. The diff"erence in area between two squares is 1600 sq. rods, and the area of the smaller one is 900 sq. rods. What is the side of the larger ? ^ , ''{" 28. If it take 64 blocks of stone, whose sides are each 20 inches, to lay a square pavement, how long and wide is the .surface paved ? ^^ 4 ^ *••>-<-- 29. If the area of a rectangle is 149.875 sq. chains, and its length 27.25 chains, what is its breadth? ^^*^/ 80. The area of a rectangle is 2400 sq. rods, and its breadth 40 rods. What is its length ? ^■; 81. A lot 150 feet wide cost $210000, at the rate of $8 a square foot. What was its length ? //^tS^J- ■'^' 82. I wish to saw a square yard from a plank 15 inches wide. How far from the end of the plank shall I commence to cut it '( ^^ ^ ^^ ^ f ARALLELOGRAMS. 19 33. A rectangular field is 50 rods long, and contains 10 acres. How long must another field be, which has the same width, to contain 5 acres ? Z ^ ' 34. What is the difierence between the area of a lot 30 feet square, and that of 2 others, each 15 feet square ? **/'^^ J 35. A man having a field 70 rods square, sold to D 100 square rods, and to E 5 acres. ^ What fractional part of the field remained unsold ? ry'^^U*r\^' tt-^i.*/^ ■'<--/ * 36. A colonel forming his regiment into a square, finds he has 15 men over, but to increase the square, so that it shall contain one more in rank and file, he requires 48 more men. How many men had he ? '^ /^ ^^''• 37. How much ground will be required to enlarge a square lot, 5 feet on .every side, whose area is 225 square feet? //S^ft. Note. — The following examples are given to show that iht same perimeter does not always enclose the same area. That is, if a string 20 inches long be laid on a table in the form of a square, it will enclose an area of 25 sq. inches ; but if it be placed in the form of a rectangle, whose sides are 8 and 2 inches, it will contain only 16 sq. inches, &c. 38. What is the difierence in area between a rectangle which is 80 feet by 20 feet, and a square which has the same perimeter? / -'' ^ ' *" 39. What is the perimeter of a square having the same area as a rectangle, whose length is 20 feet, and breadth 5 feet? --'/ > ^ 40. What must be the perimeter of a square, in order that it may contain the same area as a rectangle of 48 by 12 feet? 41. What is the diS'erence in area between 2 rectangles, the first being ^ feet by 5 feet, and the second 1 foot by 8 feet? ^^J^: 20 MENSURATION OF SURFACES. 42. The perimeters of 2 squares are to each other as 1 to 2. What relation do their areas hold to each other? / ^ J/ PROBLEM 11. The area of a rectangle and the proportion of its sides being given^ to find the length of those sides. EULE. To find the less side, multiply the area by the less num- ber of the proportion, divide the product by the greater, and extract the square root of the quotient. Then find the greater side by simple proportion. EXPLANATION. Let A B C D be a rectangle whose sides are to each other as 3 is to 5. ^ u If we divide it through its longeM side into 5 equal strips, each strip will be 3 times as long as it is wide, and will equal 1 of the rectangle. If one of these divisions is 3 times as long as it is wide, 3 of^ ^ ^ them, that is, | of the rectangle, will equal a square whose side is the less side of the rectangle^ and the square root of these I will be the side. If the sides are to each other as 3 to 5, having the less side, we find the greater by simple proportion. EXAMPLES. 1. The area of a rectangle is 60 sq. inches, and its length is to its breadth as 5 to 3. What are its sides? Solution. | of 60 sq. in. = 36 sq. in. l/36 sq. in. = 6 inches, the less side. 3 : 5 : : 6 in. : 10 in., the greater side. Res. Length 10 inches, breadth 6 inches. PARALLELOGRAMS. 21 2. A rectangle containing 192 square feet is 3 times as ^ long as it is wide. Plow wide is it? ^ \^ ' ^ •<-'•'' ■# 3. What is the breadth of a rectangle containing 150 ^^^^i sq. inches, whose length is to its breadth as 3 is to 2 ? /j'/^ fi 4. A rectangular lot contains 2400 sq. feet, and its length is \\ times the breadth. What is the length? ^'^ jf4 -4 5. A rectangle contains 180 square feet, and the len^fth a is to the breadth as 5 to 4. What are the sides? ^9^' ^ ^ ' / 6. In 2 square fields there are 8586 square rods, and their sides are to each other as 5 to 9. What is the length of a side of each ? J^ f't^U^ fTf ■ *t^<^ Note. — If the sides are to each other as 5 to 9, their areas, being produced by the squares of their sides, will be to each other as 25 : 81. We therefore divide the sum of their areas in the pro- portion of 25 to 81, and extract the square root for the sides. 7 The contents of 2 square fields are 257125 square rods, and their sides are to each other as 6 to 7. What are their sides? Jj^ t^i<^ JVS' trf' 'j 8. The area of 2 squares is 8352 sq. feet, and their sides are to each other as 3 to 7. What are their sides? 3( /^ * f V 9. What are the areas of 2 square lots whose sides are to each other as 7 to 8, the contents of boj;h lots being 1017 kq. feet? >^ -f \i'^ y / S'^ ^^ - 10. What relation do the sides of 2 squares hold to each other whose areas are 81, and 729 sq. feet? / ti J PROBLEM III. The side of a square given to find the diagonal. Rule. Extract the square root of double the area of the square. 22 MENSURATION OF SURFACES. EXPLANATION. Let A B C D be a square, and B J) the diagonal. If we extend tlie line B C until C E equals it, and the line D C until C F equals it, and connect the points B and F, F and E, E and D, by straight lines, we will have described a square on the diagonal. Since this square con- tains four parts, each equal to half the original square, it must be twice as large. Hence the square described on the diagonal equals tivice the square in which it is found. When we have given the side to find the diagonal, the square of the side gives the area of the square in which the diagonal is found, and double this, equals the square on the diagonal; then we have given tiie area of a square to find one side, therefore extract the^ square root. EXAMPLES. 1. The side of a square is 25 feet. What is its diagonal? Solution. 25 ft.^ = 625 sq. feet, the area of the square. Now double this gives the square of the diagonal, and the square root of the square of the diagonal equals the diagonal itself. Hence, |/625 sq. feet X 2 = 35.35 feet. Kes. 35.35 feet. 2. The side of a square is 37 chains. What is its dia- gonal ? 3. The side of a square is 5 feet 7 inches. What is its diagonal ? PARALLELOGRAMS. 23 4. The area of a square field is 6 acres. What is its diagonal ? 5. What is the diagonal of a square whose side is 5 yards 2 feet 3 inches? 6. What is the diagonal of a square whose area is equal to that of a rhombus whose base is 19 feet, and altitude 16 feet? 7. What is the diagonal of a square equal in area to a rectangle, 50 feet by 75 feet ? 8. A cubical room is 16 feet long, of the floor ? 9. A cubical room is 20 feet high, of the floor? ^ ^. H? ^ 2 ) ' 10. The superficial contents of a cubical room are 1350 sq. feet. What is the diagonal of its sides ? Note. — The sides of the room being equal, the contents divided by 6 gives the area of one side, or tlie area of a square of which we are to find the diagonal. 11. The superficial contents of a cubical room are 3750 sq. feet. What is the diagonal of the ceiling? ^ y, J S'^ ^^ PROBLEM IV. The diagonal of a square given to find the side. What is the diagonal What is the diagonal Rule. Extract the square root of half the square of the diagonal. EXPLANATION. The square of the diagonal gives the area of the square described upon it, which is twice the square in which it is found. Therefore the square root of half of this square will be the side. 24 MENSURATION OF SURFACES. EXAMPLES. 1. What is the side of a square whose diagonal is 15 feet? Solution. 15 ft.^ = 225 sq. feet, the square - on the diagonal. 225 sq. feet -4- 2 = 112.5 sq. feet, the area of the square. |/112.5 sq. feet = 10.606 feet. Res. 10.606 feet. 2. The diagonal of a square is 40 rods. What is the length of one side? ^^ . ^' f ;. ■ ■ .' v 3. The diagonal of a square is 55 chains. What is the side? Ji\ p^f dvty. 4. The diagonal of a square field is 75 rods. What is the side ? y V * ^ J J ^ #-«s^ 5. The diagonal of a square is 13 feet 5 inches. What is the length of one side? ^ ,, ^ t€ * 6. The diagonal of a square is 31 chains. What is the area? ^/^. t t-^aJ^^ 7. How many acres in a square Iqt whose diagonal is 16 chains 3 rods and 11 feet? , ? ^j? J' ^'^ ? / 8. In going from the north-east to the south-west corner of a square field containing 4 acres, how much farther will it be to go along its sides than diagonally across ? ^-x ' > t DEFINITIONS. DEFINITIONS. TRIANGLES. A triangle is a plane figure bounded by tlirce straight lines. A B C is a tri- angle. The ha&e of a triangle is generally the side on which it stands ; as, A C. The altitude is mea- sured by a straight line drawn from the vertex opposite the base, per- pendicular to the base. In right-angled tri- angles it equals one of the sides, in acute-angled triangles it is drawn within, and in ohtuse- angled triangles some- times without the figure. Triangles arc divided, according to their sides, into equi- lateralj isosceles, and scalene triangles. An equilateral triangle is one whose sides are all equal. An isosceles triangle is one two of whose sides are equal. 26 MENSURATION OP SURFACES. A scalene triangle is one whose sides are all unequal. Triangles are divided according to their angles into righty acute^ and obtuse- angled triangles. A right-angled triangle is one that has one right angle. The hypothenuse of a right-angled triangle is the longest side, or the side which is opposite to the right angle; as, the side B C. The side A C is called the hase^ and A B the perpen- dicular. An acute-angled triangle is one whose angles are all acute. An ohtuse-angled triangle is one that has one obtuse angle. "When the angles of a triangle are all equal, it is termed equiangular. Similar triangles are those which have all the angles of the one equal to all the angles of the other, each to each, and the sides about the equal angles proportional. Similar triangles are to each other as the squares de- scribed on their homologous sides. Similar triangles have their like dimensions propor- tional. Triangles having the same base are to each other as their altitudes. TRIANGLES. 27 Triangles having the same altitude are to each other as their bases. When two numbeV? are multiplied by the same number, their products are said to be equimultiples of those numbers. Equimultiples of quantities have the same ratio as the quantities themselves. TRIANGLES. PROBLEM V. To find the area of a triangle when the base and altitude are known. RULE. Multiply the base by the altitude, and divide the product by 2. EXPLANATION. Let A B C be a triangle having B C for its base, and A B for its altitude. Add to this, in an inverted position, a triangle having the same base and alti- tude. Since the opposite sides of this figure are equal, it must be a parallelo- gram^ and each triangle equals the half of it; as the hase and altitude are equal in both figures, if the area of the paral- lelogram equals the product of the base and altitude, the area of the triangle will equal half that product. 28 MENSURATION OF SURFACES. EXAMPLES. 1. What is the area of a triangle whose base is 14 inches, and altitude 7 inches ? Solution. 14 in. X 7 in. = 98 sq. in. 98 sq. in. ~ 2 = 49 sq. in. Res. 49 sq. in. 2. What is the area of a triangle whose base is 60 i'eCt, and altitude 40 feet ? 3. What is the area of a triangle whose base is 5 feet 5 inches, and perpendicular 6 feet 3 inches?//'"" / ' " ''^^ ^ ' 4. How many acres in a triangle whose sides, containing the right angle, are 60 rods and 57.38 rods ? /fi4 J^ A^ ^ 5. What is the area of a triangle whose base is 3 rods 5 feet, and altitude 6 yards 2 inches ? /Ti'-fl ^j^ *^f/^jt ^^ ^^/C, 6. How many acres in a triangle whose base is 36.25 chains, and altitude 27.59 chains? ^/^ ^ j4* 7. How many acres in a triangle whose base is 60 rods, and its altitude -| of its base? f^f*i- /^ , // { / 8. What is the area of a triangle which has the same base and altitude as a rectangle,whose sides are 50 feet, and 27 feet 6 inches ? 9. How much was paid an acre for a triangular lot, which cost $375, and whose base is 40, and altitude 60 rods? ^ ^^ 10. What will be the cost of paving a triangular yard with bricks, at 24 cents per square yard, its base being 21 yards, and altitude 15 yards 'iS^^/* p^ PROBLEM VL The area of a triangle and base given to find the altitude. Or, the area and either dimension, to find the other. TRIANGLES. 29 RULE. Double the area, and divide by the given dimension. EXPLANATION. The area is half the product of the base and alt^ude, hence if we double the area we have the product of the base and altitude; then we have given the product of the base and altitude, and one of them to find the other. When we have the product of two factors, and one of them to find the other, we divide the product by the given factor. EXAMPLES. 1. What is the altitude of a triangle whose area is 15 sq. inches, and base 5 inches ? Solution. 15 sq. inches X 2 = 30 sq. inches. 30 sq. inches -~ 5 inches = 6 inches. Result. ,. . , 10 sq. inches ^= ari'a. 2. What is the altitude of a triangle whose area is 708 sq. feet, and whose base is 24 feet ? ^'j^, .{ ," 3. What is the base of a triangle whose area is 5 sq. rods, 6 sq. yards, 3 sq. inches, and whose altitude is 3 rods 2 inches? J, J'^ 4. How many chains in the base of a triangle whose area is 56 sq. chains, and altitude 100 rods? / y » 9 ^ /- 5. What is the altitude of a triangle whose area is 4 acres, and its base 350 rods ? j^ '^ S"-^ t ^ '^^ 6. What is the perpendicular of a triangular lot which cost $500, at the rate of 85 per square rod, if the base is 8 rods? j:i /%- / , • 7. A triangular field contains 600 sq. yards, and its base is to its altitude as 1 to 3. What are its base and altitude? ^ij (jJu , 3 * 30 MENSURATION OF SURFACES. 8. What is the base of a lot, which is laid out in the form of a right-angled triangle, and rents for 162.50, at the rate of 25 cents per square rod, the altitude being 10 rods? ^ 9. The base and altitude of a triangle, containing 1211 sq. chains, are to each other in the proportion of 1 to 3. What are the base and altitude? J i" * 10. The area of a triangle is 2832" sq. feet, and its base and altitude are to each other as 7 to 9. What are its base and altitude ? ^ " < ^ ^ f , /' PROBLEM VII. The base and perpendicular of a right-angled triangle being given, to find the hypothenuse. RULE. Extract the square root of the sum of their squares. EXPLANATION. Let us take for an example a triangle whose base is 3 inches, whose perpen- dicular is 4 inches, and hypothenuse 5 inches. Here the square of the hypothenuse will be 25 sq. in., the square of the base 9 sq. in., and that of the perpendicular 16 sq. in. Adding the last two, their sum gives 25 sq. in., which equals the square of the hypothenuse; hence tlie square of tlie Inj-po- tlienuse equals the sum of the squares of the other two sides. Then when we have the base and perpendicular to find the hypothenuse, the sum of their squares gives the square of the TRIANGLES. 31 hypothenuse, and the square root of this square is the hypothenuse. Note, — The solution of this problem, known as the 47th of Euclid, is said to have been discovered by Pythagoras, who was 80 rejoiced thereat that he sacrificed a hundred oxen to the gods, in thankfulness, for enabling him to solve it. The Persians are said to call it The Bride, because it has such a large family or number of propositions dependent upon it. EXAMPLES. 1. What is the hypothenuse of a right-angled triangle whose perpendicular is 2-4 feet, and base 35 feet? Solution. 35 ft.^ =: 1225 sq. ft. 24 ft.^ = 576 sq. ft. 576 sq. ft. + 1225 sq. ft. = 1801 sq. ft. 1/I8OI sq. ft. = 42.43 ft. Res. 42.43 ft. 2. What is the hypothenuse of a triangle whose base is 57.05 rods, and perpendicular 60 rods ? 3. What is the hypothenuse of a triangle whose base is 5 yards 2 feet 6 inches, and perpendicular 7 yards 1 foot 2 inches ? 4. What is the diagonal of a field whose length Is 25 chains,and breadth 23 chains? J" J, f j"* 5. What is the diagonal of the floor of a cubical room whose height is 16 feet? ^^ .^^ /// 6. What is the diagonal of;i cubical room whose length is 18 feet?* J/, ) 7 ff- 7. What is the diagonalof a room 35 feet Ipng, 25 feet wide, and 20 feet high ? J^ /£/h ^^ ft' 8 The area of a rectangular field is 12 acres, and ita length is to its breadth as 3 to 1. What is its diagonal? /^X ^^ 9. What is the longest straight line that can be drawn in a rectangle whose sides are 30 and 40 inches ? * See Key. 32 MENSURATION OF SURFACES. 10. In the centre of a field 40 rods square there is planted a pole 75 feet long. How long will a line be that will reach from the top of the pole to either corner of the garden ? ^ ^^' 11. What is the length of a ladder one end of which rests against a tree 20 feet from the ground, and the other on the ground at a distance of 16 feet 5 inches from its trunk ? 12. What is the length of a ladder one end of which is placed 25 feet from a building, and the other end against the house 30 feet from the ground? 13. What is the length of one of the equal sides in an isosceles triangle whose base is 42 feet, and altitude 30 feet? Note. — The perpendicular of an isosceles or equilateral triangle divides the base into tico equal parts. 14. A ship sailed from port north 50 miles, then west 60 miles, when she stopped to un- j^q ^ ^^^ load her cargo, after which she "^^ | ! ^ sailed still farther north 50 miles ^\ i 3 and west 100 miles. How far was ^"^^^ she then in a direct line from the ^\^^^ % port whence she started ? ^ " 15. A vessel sailed first south 25 miles, then east 75 miles, again south 80 miles, and east 70 miles. How far was she then from the port whence she started ? 16. What is the perimeter of a right-angled triangle whose area is 121.5 sq. chains, and whose base is 3 times the altitude ? 17. What is the perimeter of a triangle whose base of 45 feet is divided by its perpendicular into two parts, which are to each other as 2 to 7, the perpendicular being 50 feet ? 18. What is the perimeter of a right-angled triangle whose area is 2437.5 sq. chains, and whose base is to its altitude as 13 to 15? TRIANGLES. 33 19. What is the longest side of a right-angled triangle whose area is 700 square feet, and whose base is to its alti- tude as 7 to 8 ? Gahle end — the triangular end of a house or other build- ing from the eaves to b Rid-eroie. the top, as A B C. B D is the altitude of the gable end. Ridge pole — the up- per horizontal timber in a roof, against which the rafters pitch. Rafter — a piece of timber that extends from the plate of a building towards the ridge, and serves to support the covering of the roof. Eaves — the edges of the roof of a building, which usually project beyond the face of the walls so as to throw off the water. Beam or plate — the largest or principal piece of timber in a building, that lies across the walls, and serves to sup- port the principal rafters. 20. The distance between the lower ends of two equal rafters is 80 feet, and the perpendicular distance of the ridge pole above the foot of the rafters is 10 feet. How long are the rafters ? 21. The gable ends of a house are 24 feet wide, and the ridge is 15 feet above the eaves. How much will it cost to tin the roof, at 8 cents per sq. foot, if it is 30 feet long ? 22. How wide is the gable end of a house, in which the rafters form a right angle at the top, and are 16 feet long on one side and 12 on the' other, the eaves being at the same height? 23. A house, whose gable ends are 24 feet wide, is 38 feet long, and the height of the ridge above the beam is 10 34 MENSURATION OF SURFACES. feet. The roof projects 1 foot over tlie ends and eaves in all directions. How many shingles will be required to roof it, supposing each shingle to be 4 inches wide, and each course 6 inches? 24. The area of an isosceles triangle is 588 square chains, and its altitude is 42 chains. What is the length of one of its equal sides ? 25. A castle 120 feet high is surrounded by a ditch 40 feet wide. What must be the length of a rope to reach from the outside of the ditch to the top of the castle ? PROBLEM VIII. The hypothenuse and either side of a right-angled tri- angle being given, to find the other side. RULE. Extract the square root of the diflference of their squares. EXPLANATION. If the square of the hypothenuse equals the sum of the squares of the other two sides, the square of the hypothe- nuse minus the square of the given side will equal the square of the required side. When we have the square to find the side, extract the square root. EXAMPLES. 1. The hypothenuse of a triangle is 5 inches, and the base is 3 inches. What is the perpendicular ? TRIANGLES. 35 Solution. 5 in.^ = 25 sq. in. 3 in.^ = 9 sq. in. 25 sq. in. — 9 sq. in. =: 16 sq. in, |/16 sq. in. = 4 in.- Piesult. 4 in. 2. The liypothenuse of a right-angled triangle is 13 feet, and the altitude 11 feet. What is the base ? 3. What is the perpendicular of a right-angled triangle whose hypothenuse is 29 feet, and base 17 feet? 4. The hypothenuse of a triangle is 35 feet, and the per- dicular 20 feet. What is the base ? 5. The hypothenuse of a triangle is 17 feet and 5 inches, and the base 11 feet and 7 inches. What is the perpen- dicular ? 6. One end of a ladder, which is 7 yards long, rests on the ground 13 feet from the trunk of a tree, the other end leans against the tree 5 feet from its top. How high is the tree? 7. A ladder 40 feet long is so placed in the street that it reaches a window 30 feet from the ground, and, whea turned to the opposite side, without changing the position of the foot, reaches another window 25 feet from the ground. How wide is the street ? 8. The diagonal of a rectangular field is 50 rods, and its length 40 rods. What 'is its breadth ? 9. Two boys flying a kite wished to ascertain its height; the one held the string close to the ground, and the other placed himself directly under the kite ; they found the dis- tance between them was 60 feet, and that the length of line out was 95 feet. How high was the kite ? 10. The top of a flag-staff being broken off in a storm, the broken part rested upon the upright, and the top on the ground 30 feet from its foot. The broken part measured 45 feet. How high was the staff? 36 MENSURATION OF SURFACES. 11. What is the ahitude of an equikiteral triangle whose sides are each 40 feet ? 12. The perimeter of a field in the form of an equilateral triangle is 30 chains. How many acres does the field contain ? 13. How wide is the gable of a house, in which the rafters are 25 feet long, and the height of the ridge pole above the eaves 12 feet ? 14. The gable end of a house is 24 feet wide, and the rafters are 16 feet on each side of the roof. What is the perpendicular distance of the ridge pole above the eaves ? 15. How many square feet of boards will it take to close up the gables of a barn whose rafters are 20 feet long, and the distance of the ridge above the foot of the rafters 10 feet? 16. A triangular lot whose hypothenuse is 95 feet, and perpendicular 76 feet, rents for 8300 a year. How much is that a square foot ? 17. A ladder 50 feet long is placed against a house of the same height, so that the end which rests on the ground is 18 feet from the house. How far from the top of the house is it placed ? 18. Two men started from the same place, and travelled, one south at the rate of 4 miles an hour, and the other south-east at the rate of 5 miles an hour. After travelling 11 hours, they turned and travelled directly towards each other, at the same rate as before, until they met. How far did each one travel ? PROBLEM IX. The base and the sum of the other two sides of a right- angled triangle being given, to find those sides. Or, The sum of two numbers and the difference between their squares being given, to find those numbers. TRIANGLES. 37 Since the square of the base equals the difference heticeen the squares of the other two sides, we have the following RULE. Divide the square of the base, or the difference between the squares of two numbers, by their sum, and the quotient will be the difference of those numbers. Add this sum and difference together, and divide the result by 2 for the larger number or hypothenuse. Subtract the hypothenuse from the sum for the perpen- dicular. EXPLANATION. The sum of 5 and 4 = 9. The difference of 5 and 4 = 1. The product of this sum and difference = 9. The difference between the squares of these two numbers, or 5^ — 4^ = 9 ; hence. The difference hetween the squares of two numbers equals the product of two factors^ viz.^ the sum and difference of those n umbers ; therefore, 77ie difference between the squares of tico numbers divided hy their sum gives their difference. When the sum and difference of two numbers are given to find the larger, we add them together and divide by 2. Let A B C be a triangle whose base is 3, ^ and the sum of the other two sides 9. Then the square of the base, or 9, equals the dif- ference between the squares of the other two sides; this, divided by 9, their sum, gives 1, their difference, whence (9 -f- 1) -I- 2 = 5, the larger side, or hypothenuse. 9 — 5 = 4, the smaller side, or perpendicular. 4 38 MENSURATION OF SURFACES. EXAMPLES. 1. The base of a triangle is 18 inches, and the sum of the hypothenuse and perpendicular is 54 inches. What is the length of the hypothenuse ? 2. The base of a triangle is 20 feet, and the sum of the hypothenuse and perpendicular is 40 feet. What is the perpendicular ? 3. The base of a triangle is 210 chains, and the sum of the hypothenuse and perpendicular is 630 chains. What are the hypothenuse and perpendicular ? 4. At what distance above the ground did a tree, which was 90 feet high, break off, when the broken part rests on the upright, and the top on the ground 30 feet from the foot of the tree ? 5. A flag-staff 125 feet high was broken by the wind, the top struck the ground 40 feet from the foot of the staff, and the broken end rested on the upright part. What was the length of the broken piece ? 6. The base of a right-angled triangle is 28 feet, and the difference of the other two sides is 14 feet. What are those sides ? 7. A pole standing in a field was broken by a storm, and fell so that one end rested on the ground 33 feet from the foot, while the other remained attached to the upright part. The difference between the parts of the pole after it was broken was 11 feet. How high was the pole at first? 8 The sum of the hypothenuse and perpendicular of a triangle is 104 feet, and their difference is 26 feet. What is the base ? 9. A field enclosed in the form of a right-angled triangle has 136 rods of fence in its perpendicular and hypothenuse ; TRIANGLES. 39 the diflference of fence in the same two sides is 34 rods. "What is the area of the field ? 10. A started from Philadelphia and travGlled south 20 miles. B started from the same city, and after travelling west for a certain distance, turned and journe3'ed in a straight line until he reached the point at which A had stopped. The distance A travelled was J of the distance they both travelled. How far was B from A when he changed his course ? PROBLEM X. The three sides of a triangle being given, to find a per- pendicular which will divide it into two right-angled tri- angles. Note. — If we consider the longest side to be the base of the tri- angle, and draw from the vertex opposite the base a line perpendicular to it, this line will be the perpendicular of the triangle, and will divide the base into two parts. These parts form the bases of two right-angled triangles (right-angled because the line is perpendicular to the base), whose hypothenuscs are the sides, and whose perpendiculars are the perpendicular of the tri- angle. To find the smaller part, or the hase of the smaller r'ujlit- anglcd triangle, we have the following RULE. From the square of the base subtract the diff'erence of the squares of the other two sides, and divide the remainder by twice the base. We then have the base and hypothenuse of the smaller right-angled triangle to find the perpendicular. 40 MENSURATION OP SURFACES. EXPLANATION. Let A B C be a triangle having 5, 6, and 7 for its sides ; what IS the length of a b perpendicular which will /'^^""""'"^•^-^ divide it into two right- ^^^ 1 ^"^'^^'^-^ angled triangles ? y^ ^ \ 7_a ^""""■^^ Consider A C the base, ^ d 7 c and draw the line B D perpendicular to it. Since the two smaller triangles thus formed are right-angled, we first find A D, in order to obtain B D. Let A D be represented by the letter a, then D C will equal 7 — a. The square of the perpendicular of a right-angled triangle equals the square of the hypothenuse minus the square of the base ; hence the square of B D equals 25 — d\ and for the same reason the square of B D equals 36 — (7 — ay ; therefore these differences must be equal. For things which equal the same thing, are equal to each other ; and we have 25 — a^ = S(j — (7 — ay ) completing the square, 25 — a^ = 86 — 49 -{- 14 a — a^\ cancel and transpose, 49 + 25 — 36 = 14 a. 38 = 14 a. a = 11, or 2f. Hence the rule, from the square of the bnse, or 49, sub- tract the difference of the squares of the other two sides, or 11, and we have 38, which, divided by twice the base, or 14, gives 2 1, which is the value of a, or the length of the Hne A D. We then have A D the base, and A B the hypothenuse, to find the perpendicular. EXAMPLES. 1. The sides of a triangle are 13, 14, and 15 inches. TRIANGLES. 41 What is the length of a perpendicular that will divide it into two right-angled triangles, or what is its altitude ? Solution. 14 in.^ — b 13 in.2 = 27 sq. inches. ^^'^^^^^'^^^ ]5in.2 — 27sq. in.= j^^ 1 ^^^^^^i^L 225 sq. in. — 27 sq. in. y^ ef in. | ^^^^^ = 198 sq. in. '"^ ^ 1^ '''■ ^ 198 sq. in. -- (15 in. X 2) = G| in., the length of A D. l/13~iir.2 _ (3| i,j 2 ^ i/^ip" sq. in. = Hi in., the alti- tude. Res. 11 J- inches. 2. What is the perpendicular of a triangle whose sides are 3, 4, and G inches ? 3. What is the altitude of a triangle whose sides are 15, 18, and 25 inches ? 4. What is the perpendicular of a triangle whose sides are 22, 27, and 40 feet? 5. What is the perpendicular of a triangle whose sides are GO, 70, and 90 chains ? 6. What is the altitude of a scalene triangle whose base is 240 chains, and the other sides 100 and 200 chains ? 7. What is the altitude of a triangle whose base is G8 rods, and its other sides 50 and 25 rods ? 8. What is the perpendicular of a triangle whose sides are 350, 150, and 220 chains? 9. What is the height of the gable end of a house whose width is 28 feet, the rafters being 18 feet long on one side and 25 feet on the other ? 10. What is the altitude of a triangle whose base is 15 'feet, and its other sides 11 and 12 feet? 4* 42 MENSURATION OF SURFACES. PROBLEM XI. The three sides of a triangle being given, to find the area. RULE. First find the altitude by Problem X., Then multiply the base by the altitude, and divide the product by 2. Note. — The rules used in the solution of this problem have been explained. EXAMPLES. 1. What is the area of a triangle whose sides ^y^ ' ^^""""^i?/ are 13, 14, and 15 jf^ \ ^^^^-^^ inches ? ^^-^"- ' ^^^^ A D 15 in. Solution. 14 in.^ — 13 in.^ = 27 sq. inches. 15 in.2 — 27 sq. in. = 225 sq. in. — 27 sq. in., or 198 sq. in. 198 sq. in. ~ (15 in. X 2) = 6| in., the length of A D. |/13 in.=^ — 6| in.2 = VHl^ ^^' i^~= ¥' i"- = ^H inches, the length of B D. Hi in. X y in. = 84 sq. inches. lies. 84 square inches. 2. What is the area of a triangle whose sides are 5, 6, and 7 inches ? 8. What is the area of a triangle whose sides are 3, 4, and 6 inches ? 4. What is the area of a triangle whose sides are 15, 18, and 25 inches ? 5. What is the area of a triangle whose sides are 25, 30, and 46 inches ? 6. How many acres in a field whose sides are 25, 30, and 60 rods ? TRIANGLES. 43 7. How many square yards in a lot whose sides are 76, 43 J, and 35 yards? 8. What is the rent of a field whose sides are 80, 90, and 125 rods, at the rate of $2.00 per acre ? 9. How many square feet of boards will it take to board up the gable ends of a house, the rafters being 18 feet long on one side and 20 feet on the other, and the gables 30 feet wide? 10. What ratio docs a triangle whose sides are G5, 70, and 75 feet, hold to another whose sides are 13, 14, and 15 feet ? 11. What is the area of an equilateral triangle whose sides are each 20 inches ? 12. What is the area of an isosceles triangle whose base is 26 inches, and each of the equal sides 38 inches ? PROBLEM XIL The area of a triangle and the proportion of its sides being given, to find their length. RULE. Find the area of a triangle whose sides equal the numbers expressing the given proportion. Then as this area is to the given area, so is the square of either side to the square of the required side. EXPLANATION. Since these triangles have the j^^oportion, of their sides the samey they are similar, and will therefore be to each other as the squares of their like dimensions. Having found one of the required sides, the rest can be ascertained by simple proportion. 44 MENSURATION OF SURFACES. EXAMPLES. 1. The area of a triangle is 756 square inches, and its sides are to each other ,/?^-^^ as 13, 14, and 15. What ^J/ i ^^^-<^ is the length of each y^ 756|.sq.in. ^->...^^^^^^^^ side ? ' 15 Solution. 15 in.^ — (14 in.^ ~ 13 in.^) = 198 sq. in. 198 sq. in. -^30 in. = 6| in., the smaller base. 13 in.^ — 6| in.* = ^]|-S sq. in. \/^~\%~ sq. in. = K^ in., the altitude. ^-^ in. X 2^ in. = 84 sq. in., the area of a triangle whose sides are to each other as 13, 14, and 15. 84 sq. in. : 756 sq. in. : : 13 in.^ : square of the similar side, or 1521 sq. in. l/1521 sq. in. = 13 in., the side. 13 : 14 : : 39 in. : 42 in. 13 : 15 : : 39 in. : 45 in. 39, 42, and 45 in. Res. ' 2. The area of a triangle is 756 square feet, and its sides are to each other as 13, 14, and 15 feet. AVhat is the length of each side ? 3. The area of a triangle is 1.781 acres, and its sides are to each other in the proportion of 5, 6, and 10 rods. What are those sides ? 4. What are the sides of a triangle containing 486 sq. chains, if they are in the proportion of 3, 4, and 5 chains? 5. The sides of a triangular plot of ground are in the proportion of 2, 3, and 4 feet, and it contains 418.282 sq. feet. What is the length of each side ? 6. If a triangular piece of ground containing 27 acres measures 60 rods on one side, what would be the corres- ponding side of a similar triangle containing 3 acres ? TRIANGLES. 45 7. The area of a triangle is G acres, and its base is 10 chains. What is the area of a similar triangle whose base is 30 chains ? 8. What relation does a triangle whose base and altitude are 5 and 7 feet, hold to one whose area is 105 square feet? 9. If the perpendicular and base of a right-angled triangle are 18 and 24 chains, what will be the sides of a triangle containing 54 sq. chains, which I cut off from it parallel to its base ? 10. I wish to enclose 338 square rods in the form of a triangle whose sides shall be to each other as 13, 14, and 15. What must be the length of each side ? PROBLEM XIIL The base and perpendicular of a triangle being given, to find the side of the inscribed square. RULE. Divide their product by their sum. EXPLANATION. Since the line F G is parallel to A C, the triangles F B G and A B C are similar, » and therefore have their like sides proportional, that is, 8 the base of the larger : 6 its alti- tude : : the side of the square, or base of the smaller : 6 — that side, or its altitude. In every proportion the pro- a ducts of the extremes and means are equal, therefore 48 — 8 46 MENSURATION OP SURFACES. times the side of the square equals 6 times its side ; hence 48 equals 14 times the side of the square, and the side will be -j|, or 3|, whence the rule, Divide the product of the base and perpendicular, or 48, by their sum, or 14. EXAMPLES. 1. The base of a right-angled triangle is 4 feet, and the perpendicular 9 feet. What is the side of the inscribed square ? 2. The base of an isosceles triangle is 20 chains, and the perpendicular 30 chains. What is the side of the inscribed square ? 3. The base of a right-angled triangle is 75 feet, and its perpendicular 50 feet. What is the area of the inscribed square ? 4. A triangle contains 10 acres, and its base is 50 rods. What is the side of the inscribed square ? 5. The equal sides of an isosceles triangle are eacn 20 feet, and its base is 30 feet. What is the area of the greatest square that can be drawn within it? 6. The hypothenuse of a triangle is 45 feet, and its base is 27 feet. What is the side of the inscribed square ? 7. Having inscribed a square in a triangle whose base is 16 feet, and altitude 26 feet, I j&nd I have also formed 3 smaller triangles. What is the area of the one similar to the original triangle ? 8. Having inscribed a square in a triangle Avhose base is 50, and altitude 70 feet, I wnsh to know the area of the 2 small right-angled triangles whose altitudes form sides of the square ? 9. The sides of a scalene triangle are 13, 14, and 15 feet. TRIANGLES. 47 What is the side of the greatest square that can be drawn within it? 10. A triangular farm, whose base was 400 and altitude 320 rods, was divided among 3 children as follows: the eldest received the greatest square that could be drawn within it, the youngest the 2 right-angled triangles left after the eldest had received his portion, and the second the remainder. How many acres did each one's share contain ? PROBLEM XIV. Two triangles which are alike in one dimension, having their areas and the unequal dimension of the one given, to find the corresponding dimension of the other. RULE. As the area of the one whose dimension is given, is to the one whose dimension is required, so is the given to the required dimension. EXPLANATION. Triangles which are alike in one dimension are to each other as their unequal dimensions, that is, when their bases are equal they are to each other as their altitudes, and when their altitudes are equal they are to each other as their bases. The area of a triangle is the product of two factors, the base and half the altitude, hence the areas of two triangles having the same altitude are equimultiples of their bases, but equimultiples of quantities have the same ratio as the quantities themselves, whence the rule. The same reasoning applies when their bases are equal. 48 MENSURATION OF SURFACES. EXAMrLES. 1. A triangle whose base is 10 feet contains 120 square y/ feet. What is the base of a yX triangle having the same al- y^ titude, whose area is 6 square y^ feet? A DC ABC = 120 sq.ft. A B D = 6 sq. ft. A D C = 10 ft. Solution. — 120 sq. ft. : 6 sq. ft. : : 10 ft. : required base or \ foot. Res. \ of a foot or 6 inches. 2. A triangle whose base is 13 feet contains 65 square feet. What is the base of a triangle having the same altitude, whose area is 25 square feet ? 3. The areas of two triangles, having equal bases, are 5 and 37 acres, and the altitude of the smaller is 40 rods. What is the altitude of the larger ? 4. Two triangles having the same altitude contain 50 and 75 square rods. What is the base of the larger if that of the smaller is 5 rods ? 5. What relation do the areas of 2 triangles hold to each other, whose altitudes are 15 and 25 feet, their bases being the same ? 6. Given the area of the triangle ABC equals 2 acres, the triangle A B D equals 15 sq. chains, and the line A C equals 10 chains, to find DC? 7. I have a triangular piece of land whose area is 19 acres and 2 sq. chains, the base of which is 24 chains. I desire to divide it into 3 triangles holding the relation to each other of 1, 2, and 3. What must be the base of each, if they all retain the same altitude as the original figure ? DEFINITIONS. 49 DEFINITIONS. THE TRAPEZOID, TRAPEZIUM, AND REGULAR POLYGONS HAVING MORE THAN FOUR SIDES. A trapezoid is ^ a quadrilateral, ;ht line E F, making B E equal F D, ^^ we have two equal trape- zoids. The altitude of either equals the altitude of the rectangle^ and the parallel ^ F » sides of either equal its hnse. The area of the rectangle equals the base multiplied by the altitude, or, which is the same thing, the sum of the parallel sides of either trapezoid multipled by the altitude. Since the trapezoids are halves THE TRAPEZOID. 51 of the rectangle, each will equal halfih.e same product, or half the sum of the parallel sides multiplied by the altitude. EXAMPLES. 1. What is the area of a trapezoid ^ y^^- whose parallel sides are 6 and 8 yards, and whose perpendicular is 5 yards ? Solution. (6 yards -|- 8 yards) -^2 = 7 yards, or half the sum of its s yds. parallel sides. 7 yards X 5 yards = 35 sq. yards. Kes. 35 sq. yards. 2. What is the area of a trapezoid whose parallel sides are 30 and 20 chains, and whose altitude is 26 chains ? 3. One side of a quadrilateral, having 2 right angles, is 20 feet, and the side parallel to it is 24 feet. What is its area, if the perpendicular is 9 yards ? 4. What is the area of a trapezoid, having 2 right angles, if the sides containing these angles are 4, 6, and 10 feet ; the longest side being the perpendicular ? 5. How many square feet are contained in a plank which is 16 inches wide at one end, and 12 inches at the other; the length being 28 inches ? 6. A farmer has a field in the form of a trapezoid, which he wishes to sell for \ as many dollars per acre as there are acres in it. What is its value if its parallel sides are 600 and 424 rods, and its perpendicular 100 rods ? Note. — Since the area of a trapezoid equals half the product of its parallel sides by its altitude, double the area divided by the altitude will give its parallel sides, &c. 7. The parallel sides of a trapezoid are 60 and 40 chains, and its area is 150 acres. What is its altitude ? 52 MENSURATION OF SURFACES. 8. The area of a trapezoid is 150 acres, and its altitude is 30 chains. What is the sum of its parallel sides ? 9. I have a plank 16 inches long, which contains 540 square inches. Two of its ends are parallel, and hold the same relation to each other as 2 to 3. What is the length of each end ? 10. From one side of a rectangular field 80 rods long, I wish to cut off a trapezoid of 13 acres, whose parallel sides shall be to each other as 5 to 8. What will be the length of each side ? THE TRAPEZIUM. PEOBLEM XVI. The diagonal and perpendiculars of a trapezium being given, to find the area. RULE. Multiply the diagonal by half the sum of the perpen- diculars. EXPLANATION. The diagonal of a trapezium divides it into two triangles, and at the same time forms their bases. The perpen- diculars of the trapezium are the altitudes of the tri- angles. Hence we find the area of the trapezium by finding the areas of the two triangles whose bases and THE TRAPEZIUM. 53 altitudes are known, or, which is the same thing, multiply the diagonal by half the sum of the two perpendiculars. Note. — If we multiply a number by two different multipliers and add the results, we will have the same as if the number were multi- plied by the sum of the multipliers. EXAMPLES. 1. What is the area of a /^' ^'^^"^"^-^^ trapezium whose diagonal is y^ ^j lert^^^^^^ 16 feet, and whose perpen- \^ j / diculars to this diagonal are 6 ^^^ j! / and 8 feet? ^s.^^ | / Solution. (16 ft. x 6 ft.) -^- 2 = 48 sq. ft. (16 ft. X 8 ft.) -V- 2 =:r 64 sq. ft. 112 sq.ir Or, («J^JJ!:)xl0ft. = 112sq.ft. Res. 112 sq. feet. 2. What is the area of a trapezium whose diagonal is 80 feet, and whose perpendiculars to this diagonal are 24 and 20 feet? 3. What is the number of square chains in a trapezium whose diagonal is 40 rods, and whose perpendiculars are 15 and 20 rods ? 4. How many square yards of paving are there in a quadrilateral whose diagonal is 60 feet, and perpendiculars 20 and 30 J feet? 5. How many acres In a quadrangular field whose diago- nal is 50 chains, and perpendiculars 21 chains 3 rods; and 13 chains? 6. The diagonal of a trapezium, separating the shorter from the longer sides, is 80 rods, and the sides are 40, 50, 60, and 70 rods. How many acres does it contain? 5* 54 MENSURATION OF SURFACES. 7. The diagonal of a trapezium, separating the shorter from the longer sides, is 80 rods, and the sides are 40, 50, 60, and 70 rods. What are the lengths of the two perpen- diculars to this diagonal ? 8. What is the area of the trapezium A B C D, in which A C is IGO chains, A B is 70 chains, D C is 80 chains, A E 50 chains, and F C 60 chains? 9. In the trapezium AB C D the perpendiculars BE and FD are 40 and 80 rods, and the sides A B and B C are 60 and 100 rods. What is its area ? 10. In the trapezium A B C D the diagonal A C is 100 feet, the sides A D and DC are 80 and 50 feet, and the perpendicular B E is 30 feet. What is the area ? Note. — In performing the last three examples, draw the figure and place the dimensions, the solution will then become evident. EEGULAB POLYGONS OF MORE THAN FOUR SIDES. PROBLEM XVII. The perimeter and apothem of a regular polygon being given, to find the area. RULE. Multiply half the perimeter by the apothem. EXPLANATION. If straight lines be drawn from the centre of a polygon REGULAR POLYGONS OF MORE THAN FOUR SIDES. 5^) to the vertices of all its angles, we will have as many equal trlamjles as the poli/goii has sides. Since these triangles hav3 the pc?-i- meter^ or sum of the sides, of tJie 'poly- gon for their bases, and its apothem for their altitude, the area of any one of them will equal half of one of the sides multiplied by the apothem, and all of them, or the polygon, will equal half the perimeter multiplied by the apothem. EXAMPLES. 1. What is the area of a regular pentagon whose perimeter is 60 inches, and apothem, or perpendicular to one of its sides, 8.258 inches? Solution. (60 in. -r- 2) = 30 in., or half the perimeter. 30 in. X 8.258 in. = 247-74 sq. in. Res. 247.74 sq. in. 12 m. 2. What is the area of a regular pentagon whose peri- meter is 40 inches, and apothem 5.505 inches ? 3. What is the area of a regular hexagon whose side is 7 feet, and whose perpendicular from the centre to one ot its sides is 6.062 feet ? 4. How many acres in a regular octagon whose side is 8 chains, and apothem 9.656 chains ? 5. What is the area of a resrular nonajxon whose side is 5 feet, and apothem 6.868 feet ? 6. What is the perimeter of a regular dodecagon whose 56 MENSURATION OF SURFACES. area is 279.9 sq. inches^ and whose apotheni is 9.33 inclies ? 7. What is the apothem of a regular tid decagon whose area is 149.842 square feet, and whose sides are each 4 feet? 8. What is the apothem of a regular heptagon whose area is 232.568 square feet^ and whose perimeter is 56 feet ? 9. What is the area of a hexagon whose side is 6 feet, and whose diagonal passing through the centre is 12 feet? 10. What is the area of a regular octagon whose diagonals passing through the centre are 20.904 inches^ and whose sides are 8 inches ? PROBLEM XVIII. One side of a regular polygon being given^ to find the area. RULE. First ascertain, by the table, the area of a similar polygon whose side is 1. As the square of its side is to the square of the given Bide, so is its area to the required area. Note. — Since the first term of this proportion is always the 1^, dividing by it will not affect the product of the second and third terms. EXPLANATION. All similar polygons are to each other as the squares of tbeir like dimensions The following table consists of the areas of regular poly- gons whose sides are 1. REGULAR POLYGONS OF MORE THAN FOUR SIDES. 57 Xumber of Sides. Names of Polygons. A reas. 3 Trigon or triangle 0.433013 4 Tetragon or quadrilateral 1.000000 5 Pentagon 1.720477 6 Hexagon 2.598076 7 Heptagon 3.633912 8 Octagon 4.828427 9 Nonagon 6.181824 10 Decagon 7.694209 11 Un decagon 9.365640 12 Dodecagon 11.196152 EXAMPLES. 1. "What is the area of a regular pentagon whose side is 5 inches ? •Solution. 1 in.^ : 5 in.^ : : 1.720477 sq. in. : the re- quired area, or 43.011925 sq. in. Res. 43.011925 sq. in. 2. How many acres in a regular nonagon whose side is 17 rods ? 3. How many triangles, each containing .433013 square inches, can be formed from a regular decagon whose side is 5 inches ? 4. How many squares of 9 square inches each are con- tained in a regular hexagon whose side is 3 inches ? 5. What is the apothem of a regular octagon whose side is 8 chains ? 6. What is the perimeter of a regular trigon whose area is 3.897117 square feet? 7. What is the diagonal passing through the centre of a regular hexagon whose side is 6 feet, and apothem 5.196 feet? 8. What is the diagonal passing through the centre of a regular octagon whose perimeter is 50 feet, and apothem 7.543 feet? 58 MENSURATION OF SURFACES. IRREGULAE POLYGONS OF MORE THAN FOUR SIDES. Irregular polygons are those whose sides are unequal. PROBLEM XIX. To find tlie area of an irregular polygon having more than four sides. RULE. Draw straight lines dividing the polygon into trapezoids, trapeziums, and triangles, as may be most convenient. Then find the areas of the figures thus formed, and add them together. EXPLANATION. If the polygon can be divided into a number of figures, whose area can be readily found, it is evident it will equal the sum of their areas. EXAMPLES. 1. What is the area of the polygon ABODE, whose dimensions are as follows : c B A B C 14 feet, 13 feet. E F = 10 feet, E D ^ 9 feet, B m = 3 feet, B 8 = 6 feet, B w = 14 feet, B o = 16 feet, BE = 20 feet? DEFINITIONS. 59 2. What is the area of a polygon whose dimensions are one-half of those in the first example ? 3. How many acres in an irregular tract of land having the following dimensions : D B E = 95 chains, B D = 75 chains, A E = 80 chains, A I = 28 chains, D h==SO chains, C <7 = 15 chains, F ^ = 20 chains ? A F 4. What is the area of a tract of land whose dimensions are one-half of those in the third example ? DEFINITIONS. CIRCLES. A circle is a plane figure bounded by a curved line which is everywhere equidistant from a point within called the centre. The circumference or 'periphery of a circle is the curved line which bounds it. The diameter of a circle is a straight line passing through the centre, and terminating at both ends in the circumfer- ence ; as, A B. The radius of a circle is a straight line drawn from the centre to the circumference ; as, C D. 60 MENSURATION OF SURFACES. Since the circumference is everywhere equidistant from the centre, all the radii of a circle are equal. Since the diameters measure the distance from the centre to the circumference twice, they are also equal^ and each is douhle a radius. A diameter is the longest straight line that can be drawn in a circle, and divides the circle and circumference into two equal parts — called semi-circle and semi-circumference. Circles are divided into 360 equal parts called degrees, each degree is divided into GO minutes, and each minute into 60 seconds. Circles whether great or small contain the saine nuniher of degrees. The difference consists in the size of the degrees, not in their number. Quadrants, sextants, and octants receive their names from the number of degrees they contain being aliquot parts of 360°. A quadrant is J of a circle, and contains 90°. A sextant is i of a circle, and con- tains 60°. An octant is ^ of a circle, and con- tains 45°. A tangent is a straight line which touches the circumference at one point, without cutting it; as, C D. The point where it touches is called the point of contact. CIRCLES. 61 A secant is a straight line which intersects the circum- ference in two points, and lies partly within, and partly without the circle ; as, A B.. Concentric circles are those which have the same centre but unequal radii. Their circumferences form parallel curves. A polygon is described upon a circle when each of its sides is tangent to the circumference A polygon is inscribed in a ci?rle when the vertices of all its angles are in the circumference. The circle is then said to be circumscribed about the polygon. A circle includes a greater area than any polygon having the same perimeter. All circles are similar, because their circumferences always hold the same relation to their radii. Circles therefore are to each other as the squares of their like dimensions. The circumferences of two circles are to each other as their radii. A square described upon a circle is double the inscribed square, for its side equals the diagonal of the inscribed square, and the square of the diagonal equals twice the square in which it is found. 6 62 MENSURATION OP SURFACES. CIRCLES. PROBLEM I. The diameter of a circle being given, to find the circum- ference. RULE. Multiply the diameter by 3.1416. EXPLANATION, Mathematicians have never been able to find the exact ratio of the diameter of a circle to the circumference, nor have they accomplished the squaring of the circle, that is, they have never succeeded in drawing a square or other polygon having the same area as any given circle. The ratio of the diameter to the circumference was shown by Metius to be as 113 to 355, which is sufficiently accurate for practical use. 355 -~ 113 = 3.141592-}-, but for convenience we extend the decimal only to 4 places, calling the last 6. Hence the rule, multiply the diameter by 3.1416. EXAMPLES. 1. What is the circumference of a circle whose diameter is 11 inches ? ^ --^^ Solution. 11 in. X 3.1416 = 34.5576 / \ in. 34.5576 in. ==z 2 feet, 10.5576 (" ruu." j , inches. \ / CIRCLES. bd Note. — If the circumference is the product of two factors, viz., the diameter and 3.1416, the circumference divided by 3.1416 gives the diameter. 2. What is the circumference of a cart-wheel whose diameter is 6 feet ? 3. What length of tire will it take to band a carriage- wheel 5 feet 7 inches in diameter ? 4. What is the diameter of a circle whose circumference is 2 feet 10.5576 inches ? 5. What is the thickness of a round tree whose girt is 15 feet? 6. The dial plate of a clock is 2 feet in circumference. What is the length of the minute hand ? 7. At what rate per hour does the city of Quito move from west to east if the equatorial diameter of the earth is 7926 miles, and it turns once on its axis in 24 hours ? 8. What is the circumference of a circle whose diameter equals the diagonal of a square containing 3 acres? 9. If the minute hand of a watch is 1 inch long, over how much of the circumference of the face does it pass in 20 minutes ? 10. What is the diameter of a circle, having the same perimeter as a right-angled triangle, whose base and perpen- dicular are 18 and 24 feet? 11. The base of a right-angled triangle is 28 chains, and the sum of the other two sides is 56 chains. What is the diameter of a circle having the same perimeter ? 12. What is the circumference of the greatest circle which I can draw upon a blackboard with a string of 7 Inches ? 13. How many times will a wheel, 5 feet in diameter, 64 MENSURATION OF SURFACES. turn round in going 5 miles, 2 furlongs, 3 chains, 2 rods, 3 yards, 1 foot, and 5 inches ? 14. A horse is fiistened in a meadow, by a rope 30 feet long, to the top of a post 6 feet high. What is the circum- ference of the greatest circle over which he can graze ? 15. What is the diameter of a wheel which makes 336 revolutions in a minute, when the cars are going 30 miles an hour? PROBLEM II. To find the area of a circle, the diameter or circumference being given. RULE. Multiply the square of the diameter by .7854, or the square of the circumference by .07958. EXPLANATION. The square of the diameter gives the area of the square described upon the circle, which holds i — —;=— the same relation to the circle as 1 to / .7854 J therefore, as [ 1 : .7854 : : the area of the square described upon the circle : the circle. \ Since the first term of the proportion ' — — ^:s- is 1, we simply multiply the area of the square, or the square of the diameter, by .7854. The circumference is 3.1416 times the diameter, there- fore its square is (3.1416)^ or 9.86965056 times the square of the diameter, and must be multiplied by -5755^55055 ^^ .7854, or .07958. CIRCLES. 65 Note.— The area of a circle also equals J the product of the circumference by the radius ; for if in the circle we inscribe a regular polygon, and draw its apothem, the polygon equals ^ the product of the perimeter by the apothem. If the number of the sides of the polygon be indefinitely increased, until they are mere points, the perimeter will become the cir- cumference, the apothem the radius, and the polygon the circle ; therefore the circle will equal ^ the product of the circumfer- ence by the radius, or \ the product of the circumference by the diameter, which is (diam. X ^Ham. X 3.141G) ^ 4. Since dividing one of several factors, before multiplying them together, is the same as to divide their product, we divide 3.1416 by 4, and have the work reduced to (diam. X tliam. X -7854). Now the square of the diameter = the square described on the circle, and since the square of the diameter X "854 = the circle, the square described upon the circle : circle : : 1 : .7854, as stated in the explanation to the rule. EXAMPLES. 1. What is the area of a circle whose diameter is 7 feet ? Solution. 7 ft.^ x .7854 = 38.4846 sq. feet. Ees. 38.4846 sq. feet. 2. What is the area of a circle whose diameter is 11 inches ? 3. What is the area of a circle whose circumference is 15 feet 2 inches ? 4. What is the area of a circle described with a string of 11 inches? 6* E 66 MENSURATION OF SURFACES. 5. How many square yards are there in a circle whose diameter is 27 feet 5 inches ? 6. What is the value of a circular piece of ground whose circumference is 200 chains, at $50 an acre ? 7. What is the diiference in area between a circle whose circumference is 60 feet; and an equilateral triangle having the same perimeter? 8. What is the area of a circle whose diameter corres- ponds to the diagonal of a rectangle 24 feet long, and 18 feet wide ? 9. What is the area of a circular race course, which a horse, at the rate of a mile in 3 minutes, can trot around in 5 minutes? 10. AVhat is the difference in area between a circle 5 chains in diameter and a square described upon it ? 11. How many acres in a semi-circular lot whose radius is 20 rods ? 12. A horse is fastened in a meadow, by a rope 30 feet long, to the top of a post 6 feet high. What is the area of the greatest circle over which he can graze ? 13. What is the difference in area between a circle 60 rods in circumference and a rectangle, having the same perimeter, whose length is twice its breadth ? 14. What relation will the quantity of water that can be forced into a basin in 1 hour, through a pipe 3 feet in diameter, hold to that which can be forced through 3 pipes, each 1 foot in diameter, in the same length of time ? 15. W'hat is the area of a circle which contains as many square feet as its circumference numbers long feet ? CIRCLES. 67 PROBLEM IIL To find the diameter or circumference when the area is given. RULE. To find the diameter — divide the area by .7854 and extract the square root of the quotient. To find the circumference — divide the area by .07958 and extract the square root of the quotient. EXPLANATION. This rule is simply the reverse of the preceding one. EXAMPLES. 1. What is the diameter of a circle containing 490.875 square feet? Solution. 490.875 sq. feet -=- .7854 y^ "^ = 625 sq. feet. / c-- \ |/625 sq. feet = 25 feet. -^^- j Res. 25 feet. \ # / 2. "What is the diameter of a circle containing 78.54 square feet ? 3. What is the circumference of a circle containing 1.9895 square feet? 4. What is the diameter of a circular acre ? 5. What is the circumference of a wheel vrliich turns 200 times in running around a circular bowling-green, whose area is 795.8 square rods? 6. What is the diameter of a circular fish-pond Laving the same area as a square one whose side is 20 rods ? 68 MENSURATION OF SURFACES. 7. What is the circumference of a circle having the same area as a triangle whose sides are 13, 14, and 15 chains ? 8. What is the diameter of a circle having the same area as a rectangle whose diagonal is 10 rods, and length 6 rods ? 9. The area of a circular park is 4 square miles. How long will it take to drive around it, at the rate of 5 miles per hour ? 10. There is a circular garden containing 3216.9984 square feet, in the centre of which stands a pole 64 feet high. The pole being broken by the wind, one end of the broken part rested on the upright, and the other on the ground at the extremity of the garden. What was the length of the broken part ? Note. — Similar surfaces are to each other as the squares of their like dimensions. If the areas of circles are produced by multiplying the squares of the diameters by .7854, or the squares of the circumferences by .07958 — these areas must be equimultiples of the squares of the diameters and circumferences, and as equimultiples of quantities have the same ratio as the quantities themselves, the areas are to each other as the squares of their diameters or circumferences. 11. If the diameter of a wheel is 18 inches, what is the circumference of one 3 times as large ? 12. If a rope 4 inches in circumference is composed or 200 threads, how many threads will be required to make one 9 inches in circumference ? 13. If the wheels of a car which are 2i feet in diameter make 7 revolutions per second, how many revolutions would a wheel 5 feet in diameter make, if it passes over the same distance in the same length of time ? 14. If a pipe 1 foot in diameter will fill a cistern in 6 hours, how large a pipe will it take to fill a cistern 3 times as large in the same time ? CIRCLES. 69 PROBLEM IV. To find the areas of circular rings formed by concentric circles. RULE. Find the difference between the areas of the circles forming the rings. EXPLANATION. If two circles, of unequal radii, be drawn around a common centre, the excess of the larger over the smaller forms a circular ring, therefore the area of this ring equals the difference between the areas of the two circles. EXAMPLES. 1. "What is the area of a circular ring formed by 2 con- centric circles whose diameters are 4 and 2 feet ? Solution. 4ft.2x .7854r:=:12.5G64 sq. ft, the area of the circle C D. 2 ft.2 X .7854 r= 3.141G sq. ft., the area of the circle A B. 12.5664 sq. ft. — 3.1416 sq. ft. = 9.4248 sq. ft., the area of the circular ring. Res. 9.4248 sq. ft. 2. What is the area of a circular ring formed by 2 con- centric circles whose diameters are 4 and 8 feet ? 3. What is the area of a circular ring formed by 2 con- centric circles whose circumferences are 25.1328 feet and 18.8496 feet? 4. I have a circle 4 inches in diameter. What is the 70 MENSURATION OF SURFACES. surface of the ring formed by putting it in the centre of a circle twice as large ? 5. The radii of 3 concentric circles are 3, 4, and 5 feet. What are the areas of the 2 circular rin2;s thus formed i* 6. A bricklayer is to pave a walk, 3 feet in width, around a circular grass-plat 15 feet in diameter. How many bricks will it take if they are 8 inches long and 4 wide, making no allowance for waste ? 7. Within a circular acre is a pond 5 rods in diameter. What fractional part of the acre is not covered by the pond ? 8. A mason is to curb a cylindrical well, at 1 shilling per square foot; the breadth of the curb is to be 1 foot 6 inches. How much will it cost, if the diameter within the curb is 3 feet ? 9. If the minute hand of a clock is 6 inches, and the hour hand 5 inches long, what is the difference of the surfaces over which they travel from sunrise to sunset at the time of the vernal equinox, when the days and nights are equal ? 10. A circular garden, containing 1 acre, is bordered by a gravel walk of uniform width which takes up i of its area. What is the width of the walk ? 11. Three men bought a grindstone 3 feet in diameter, for which they paid equally. What part of the diameter must each grind down for his share ? 12. Four men purchased a grindstone 70 inches in diameter, towards which the first contributed $2.50, the second $2.00, the third $1.50, and the fourth $1.00. What part of the diameter must each grind down for his share, if the one who contributed $2.50 grinds first, and the rest follow according to the amounts they paid ? CIRCLES. 71 PROBLEM V. The diameter of a circle being given, to find the side of the inscribed square. RULE. Extract the square root of half the square of the diameter. EXPLANATION. The diameter of the circle forms the diagonal of the inscribed square. Hence we have the diagonal of a square given to find its side, as in Problem IV. of Polygons. When the circumference is given, first find the diameter. EXAMPLES. 1. What is the side of a square in- scribed in a circle whose diameter is 5 inches 1 Solution. 5 in.* = 25 sq. in. l/25lq. inr^2 = 3.535 in. Res. 3.535 inches. 2. What is the side of a square inscribed in a circle whose diameter is 2 feet 3 inches ? 3. What is the side of a square inscribed in a circle •whose radius is 2 inches ? 4. What is the diagonal of a square inscribed in a circle whose circumference is 20.4204 feet? 5. What is the area of a circle circumscribed about a square whose side is 5 chains 1 72 MENSURATION OF SURFACES. 6. How much more land in a circle SO rods in diameter than in its inscribed square ? 7. An eccentric father bequeathed a circular portion of his estate, containing 800 acres, to his wife, son, and four daughters, in the following manner : the wife and son were to have the two largest isosceles triangles that could be inscribed in the circle on its diameter, and each daughter i of the remainder. How many acres did each receive ? 8. How many squares inscribed in a circle equal one described upon it ? Explain why this is so. 9. Why does multiplying the diameter by .7071, or the circumference by .2251, produce the side of the inscribed square ? 10. Why does multiplying the diameter of a circle by .8862, or the circumference by .2821, give the side of a square having the same area as the circle ? DEFINITIONS. ARCS, SECTORS, AND SEGMENTS OF CIRCLES. A An arc of a circle is any part of its circumference } as, A B. The chord or subtense of an arc is a straight line joining its extremities; as, C D. c E D Half the chord of the arc is the u; ~? half of the line C D ; as, C E or E D. \^^^., / The chord of half the arc is a ^ straight line joining the extremities of half the arc; as, F C. DEFINITIONS. 78 The chord of half the arc is the hypofhenuse of a right- angled triangle, whose perpendicular is half the chord of the whole arc, as may be seen by joining the points E and F. The sine of an arc is a straight line drawn from one of its extremities perpendicular to a diameter passing through the other extremity ; thus, C B is the sine of the arc A B. The versed sine of an arc is that part of the diameter which is inter- cepted between the arc and its sine ; thus, C A is the versed sine of the arc A B. The cosine of an arc is that part of the diameter which is intercepted between the centre of the circle and the sine of the arc ; as, D C. It always equals the radius minus the versed sine. A sector of a circle is that part in- cluded between an arc and two radii drawn to its extremities. A segment of a circle is that part included between an arc and its chord. The difference between a sector and segment^ luiving the same arc, is a tri- angle whose base is the chord of the arc, and altitude the cosine of half the arc ; as, A B C. ^ Two sectors, segments, or arcs of circles are similar they correspond to equal angles at the centre. 7 c when 74 MENSURATION OP SURFACES. CIRCULAR AND ANGULAR MEASURE. This measure is applied to the measurement of circles and angles. 60 seconds ('') .... make 1 minute, ' 60 minutes . . . . . "1 degree, ° 30 degrees ....." 1 sign, S. 12 signs, or 360 degrees . . "1 circle, 0. ARCS OF CIRCLES. PROBLEM VI. The number of degrees in a circular arc and the radius of the circle being given, to find the length of the arc. RULE. First find the circumference of the circle whose radius is given. Then as 360 degrees is to the number of degrees in the arc, so is the circumference to the required arc. EXPLANATION. The circumference is the length of 860 degrees, and we wish to find the length of the number of degrees in the arc^ therefore as an arc is less than a circumference, we have by simple proportion as 360° : number of ° in the arc : : circumference : arc. EXAMPLES. 1. What is the length of an arc of 95° whose diameter is 5 inches ? ARCS OF CIRCLES. 75 Solution. 5 in. X 3.1416 = 15.708 in., the circum- ference. 360° : 95° : : 15.708 in. : 4.145 in., tlie arc. Res. 4.145 inches. 2. What is the length of an arc of 30°, the radius of the circle being 7 feet ? 8. What is the length of an arc of 30° 5' 7", the radius of the circle being 20 feet ? 4. What is the length of a degree of the earth's circum- ference, if its equatorial diameter is 7"926 miles ? 5. How far is an inhabitant of any place on the equator carried in 5 hours, the diameter of the earth being 7926 miles ? PROBLEM VIL The chord of the arc and the chord of half the arc being given, to find the length of the arc. RULE. From 8 times the chord of half the arc, subtract the chord of the whole arc, and divide the remainder by 3. Note. — When the chord of the arc and the chord of half the arc are not given, but other terms from which to find them, they can generally be obtained from those terms, by a knowledge of the properties of a right-angled triangle, as will be seen by drawing the figure and placing the given dimensions. EXAMPLES. 1. What is the length of the arc, if the versed sine of half the arc is 2 feet, and the radius of the circle is 5 feet? .258 ft. 76 MENSURATION OF SURFACES. Solution. If the radius is 5 feet and the versed sine of half the arc is 2 feet, 5 feet — 2 feet = 3 feet, the cosine of the same. Placing these dimensions in the figure, we have the hypothenuse and perpendicular of the triangle D B E, therefore its base, or half the chord of the whole arc, equals the |/5~ft.=^'— B'ft.' = 4 ft., the length of E B, also of E A. Therefore A B, the chord of the arc = 8 feet. "VVe now have the perpendicular and base of the right-angled triangle B C E to get the hypothenuse B C, or chord of half the arc. Hence |/2lt7=^+ 4 ft.'^ •== 4.472 ft. or C B. From 8 times the chord of half the arc, subtract, &c. 4.472 ft. X 8 = 35.776 ft. 35.776ft-8ft.^^,^3^^ o Res. 9.258 feet. 2. What is the length of the arc, if the versed sine of half the arc is 6 feet, and the radius of the circle is 10 feet? 3. What is the length of the arc, if the versed sine of half the arc is 2 feet, and the radius of the circle is 6 feet? 4. What is the arc, if the versed sine of half the arc is 2 feet, and the cosine of the same is 8 feet ? 5. What is the length of an arc whose chord is 26 yards, and the diameter of the circle 100 yards ? 6. What is the length of an arc whose chord is 30 chains, and the versed sine of half the arc is 8 chains ? 7. What is the length of the arc, when the diameter of the circle is 36 feet, and the chord of half the arc is 12 feet? ARCS OF CIRCLES. 77 Noj'K. — The chord of half the arc always equals the square root of the product of two factors, viz., the diameter and versed sine of half the arc, therefore the square of the chord of half the arc equals their product, and being divided by either gives the other; that is, The sq. of the chord of lialf the arc -v- by the diam. = the versed sine of half the arc. The sq. of the cord of half the arc -f- by the versed sine of half the arc = the diameter. Let the versed sine = v. Let the radius = r. Let the chord of half the arc = c. Let the cosine of half the arc = r Then from the 2 right-angled tri- angles C B E and D B E we have c^ — ^2 __ tijg square of the line E B. Also r' — (r — v)''' = the square of the line E B. Hence c^ — v- = r^ — (r — vy, for things which equal the same thing are equal to each other. Completing the multiplication in the equation, c^ — ■Tp.2-=v* — >^* + 2ry — V^ cancelling, c"^ = 2rt>, or, since 2r :== the diameter, c2 = do, or the square of the chord of half the arc ^^ the pro- duct of the diameter and versed sine, and c'-^d = V. c'^-i- v = d. Solution of the 7tii Example. 12 ft.'^ -^ 36 ft. = 4 ft., the versed sine of half the arc. |/12 ft.*'' — 4 ft.2 z:= 11.813 ft., half of the whole chord. 11.313 ft. X 2 = 22.626 ft., the chord. 12 ft. X 8 — 22.626 ft. 24.458 ft = 24.458 ft., the arc. Res. 24.458 feet. 78 MENSURATION OF SURFACES. 8. What is the length of the arc when the diameter of the circle is 50 yards, and the chord of half the arc is 10 yards ? 9. What is the arc, if its chord is 8 feet, and the versed sine of half the arc is 3 feet ? 10. What is the arc, if the chord of half the arc is 16 feet, and the diameter of the circle is 32 feet ? SECTORS OF CIRCLES. PROBLEM VIII. To find the area of a sector. RULE. Multiply the arc by the radius, and divide the product by 2. EXPLANATION. Draw a triangle in the sector having the radii and chord for its sides. The area of the triangle equals ^ the product of its base and altitude, but if the number of triangles^ drawn in the sector, be indefinitely increased, their bases will equal the arc of the sector, their altitudes its radius^ and the triamjles the sector. Hence the rule, multiply the arc by the radius and divide the product by 2. EXAMPLES. 1. What is the area of a sector, if the chord of its arc is 16 feet, and the radius of the circle 10 feet? SECTORS OP CIRCLES. 7S Solution. |/1U ft;^ — « It.^ =: 6 ft., the cosine of J the arc. 10 ft. _ G ft. = 4 ft., the versed sine of } the arc. y'S It.^ -I- 4 lt.2 = 8.944 ft., chord of ^ the arc. 8.944 ft. X 8 — IG ft 3 18.517 ft. X 10 ft. = 18.517 ft., the arc =r 92.585 sq. ft., the sector. Res. 92.585 sq. ft. 2. What is the area of a sector whose arc is 75 feet, and the radius of the circle 30 feet ? 3. What is the area of a sector, the chord of whose arc is 18 feet, and the diameter of the circle 30 feet ? 4. What is the area of a sector, if the chord of half the arc is 5 feet, and its versed sine is 3 feet ? 5. What is the area of a sector, if the diameter of the circle is 20 feet, and the versed sine of half the arc 2 feet ? 6. What is the area of a sector whose arc is 90°, if the diameter of the circle is 2 feet 3 inches ? 7. What is the area of a sextant, the diameter of the circle being 20 yards ? 8. What is the area of an octant, the circumference of the circle being 25.1328 feet? 9. What is the area of a sector whose arc is a quadrant, the diameter of the circle being 5 feet ? 10. If a sector of a circle, whose diameter is 6 feet, contain 3.5343 sq. feet, what will be the area of a similar sector, in a circle whose diameter is 10 feet ? 80 MENSURATION OF SURFACES. SEGMENTS OF CIRCLES. PROBLEM IX. To find the area of a segment. RULE. First find the area of a sector having the same arc. From this subtract the area of the triangle whose base is the chord of the arc, and whose altitude is the cosine of half the arc. EXPLANATION. Since a sector exceeds a segment havin< by the area of a triangle whose base is the chord of the arc, and altitude the cosine of half the arc, the difference be- tween that sector and triangle will give the segment. EXAMPLES. the same arc, 1. What is the area of a segment the chord of whose arc is 16 feet, and the chord of half the arc 10 feet? Solution. -/lOft.^ — 8 ft.2 = 6 ft., the versed side of ^ the arc. 6 ft. = 16? ft., the 100 sq. ft. -~ diameter. 16| ft. -- 2 8-J ft., the radius. ^ ft. 6 ft. sine of A the arc. 10 ft. X 8 — 16 ft, 21 ft., the CO- v = J^^ 21 j ft. = 21-J- ft., the arc. 211 ft. X U ft. ^-^ g—^ — = 88f sq. ft., the sector. DEFINITIONS. 81 16 ft. X 2.1 ft. = 18| sq. ft., the triangle. 88| sq. ft. — 18 5 sq. ft. = 70| sq. ft., the segment. Res. 70| sq. feet. 2. What is the area of a segment, the chord of wliose arc is 24 yards, and the chord of half the arc 15 yards? 3. What is the area of a segment, the versed sine of half the arc being 1 foot, and the radius of the circle 5 feet ? 4. What is the area of a segment, the chord of whose arc is 8 feet, and the cosine of half the arc 3 feet ? 5. Compute, by the rules for segments, the area of 1 of the 4 segments, formed by inscribing a square in a circle whose diameter is 40 rods ? DEFINITIONS. ZONES. A circular zone is a part of a circle included between two parallel chords, and their inter- cepted arcs. The breadth of the zone is that / ! \ part of the diameter contained be- tween the two parallel chords. The chords may lie on the same or different sides of the diameter, and one of them may form the diameter. They may be equal or unequal, but if equal the diameter passes through the middle of the zone. 82 MENSURATION OF SURFACES. ZONES. PROBLEM X. To find the area of a circular zone. RULE. First draw the chords of the intercepted arcs. Then to twice the area of one of the segments thus formed, add the area of the trapezoid or rectangle formed at the same time. When one of the chords is the diameter of the circle, take from the semi-circle the segment formed by the smaller chord of the zone and its arc. Note. — The rules for trapezoids, rectangles, segments, &c., have already been explained. To find the diameter of the circle, Divide the difference between the square of ^ the larger chord, and the sum of the square of the breadth plus the square of I the smaller chord by twice the breadth, and the quotient will be the hase of a right-angled triangle whose perpendicidar is i the larger chord, and whose hypothenuse is the radius of the circle. Double the radius for the diameter. EXPLANATION. If two parallel chords of a zone are 96 and 60, and its breadth 26, what is the diameter ? Let 26 -|- a: = distance from centre to shortest chord Let X = distance from centre to longest chord. ZONES. Then from tlie properties of a right-angled triangle, 30^ + (26 + xy = radius^ ; also, 48^ -\- x^ = radius^; hence, 30^ -f (26 + xy = 48^ -f x\ or 900 + 676 + 52x + ai» = 2304 +. ^\ cancel and collect, 1576 + 52x = 2304. 52a: = 2304 — 1576. 52x = 728. x = 14, the base, of which 48 is the perpendicular, and the radius the hypothenuse. Then l/U^* + 48=* = 50, the radius. 50 X 2 = 100, the diameter. Since from the equation 52a; = 2304 — 1576, we obtain Xj we have the rule, Divide the difference between the square of J the larger chord, and the sum of the square of the breadth plus the square of i the smaller chord, by twice the breadth, to find the base of a right-angled triangle, whose perpendicular is \ the larger chord, and whose hypothenuse is the radius. Note. — If the square of h the larger chord exceed the sum of the square of the breadth plus the square of J the smaller chord, the chords are on the same side of the diameter ; but if it be less than the sum, they are on different sides. EXAMPLES. 1. What is the area of a circular zone, whose parallel chords are 96 and 60 feet, and whose breadth is 26 feet ? Solution. To perform this example we will first find the area of the trapezoid, then the area of one of the segments, 84 MENSURATION OF SURFACES. the double of which added to the trapezoid will equal the zone. (60 ft. + 96 ft.) X 26 ft. ^^ ' — ^-^ = 2028 sq. ft., the trapezoid. Li To find the segment we can obtain the chord of its arc and also the diameter of the circle, which will, by the rules for segments, be sufficient. To find the diameter, divide the diff'erence between the square of \ the larger chord, and the sum of the square of the breadth plus the square of \ the smaller chord by twice the breadth, and we will have the base of a right-angled triangle, whose perpendicular is \ the larger chord, and whose hypothenuse is the radius. 48 ft.^ — (30 ft.2 + 26 ft.O ^ = 728 sq. feet. 728 sq.ft. — 52 ft. = 14 ft. l/i4 ft.2-i-48 ft.2 = 50 ft., the radius. 50 ft. X 2 = 100 ft., the diameter. To find the chord of the arc of the segment, if from \ of the larger chord of the zone we take \ of the smaller, the remainder will be the base of the right-angled triangle B E D, whose perpendicular is 26 ft, the breadth of the zone, and whose hypothenuse is the chord of the arc of the seg- ment, hence l/18 ft.2 -f 26 ft.=^ == 31.6228 ft., the chord B D of the ZONES. 85 The dimensions of this segment can be more readily placed by drawing a segment to represent it in another circle. If we retain the same dimensions the area of the segment will be the same. 32.1748 ft. Note. — Always place the dimensions as soon as they are found. y/50ft.=^~—"T5^."8ir3 "ft? =47.4341 ft., the cosine of half the arc. 50 ft. — 47.4341 ft. = 2.5659 ft., the versed sine of half the arc. Since the square root of the (diameter X versed sine of half the arc) = chord of half the arc, ^100 ft. X 2.5659 ft. = 16.0184 ft., the chord of half the arc. From 8 times the chord of half the arc subtract Che chord of the arc, and -—- the remainder by 3. 16.0184 ft. X 8 — 31.6227 ft. oo ^^ao ^ x-u ^ u ^^s = 32.1748 ft., the length o of the arc. Half the product of radius and arc = sector. 32.1748 ft. X 50 ft. „^, ,,^ ^ , = 804.37 sq. ft., the area of sector. Half the product of the cosine of half the arc and the ol 8 chord of the arc = the triangle. 8C MENSURATION OF SURFACES. 31.6227 ft. X 47.4341 ft. ■ = 749.9971 sq. ft. the triangle. Sector 804.37 sq. ft. — 749.9971 sq. ft., triangle r= 64.3729 sq. ft, segment. 54.3729 sq. ft. X 2 = 108.7458 sq. ft., the two segments. Trapezoid. Segments. Zone. 2028 sq. ft. + 108.7458 sq. ft. = 2136.7458 sq. ft. Res. 2136.7458 sq. ft. 2. What is the area of a circular zone whose parallel chords are 18 and 24 inches, and whose breadth is 3 inches ? 3. What is the area of a circular zone whose breadth is 7 feet, and whose parallel chords are 42 and 56 feet ? 4. What is the area of a circular zone whose parallel chords are 18 and 24 yards, and whose breadth is 21 yards? 5. What is the area of a circular zone whose parallel chords are 24 and 32 rods, and whose breadth is 28 rods ? 6. What is the area of a circular zone whose breadth is 48 feet, and whose parallel chords are each 36 feet ? 7. What is the area of a circular zone whose smaller chord is 10 inches, and whose larger chord is 20 inches, being the diameter of the circle ? 8. What is the area of a circular zone whose parallel chords are each 12 feet, and whose breadth is 16 feet? 9. What is the area of a circular zone whose larger chord being the diameter of the circle is 10 feet, and whose breadth is 4 feet ? DEFINITIONS. 87 DEFINITIONS. THE LUNE. A lune is the space included be- tween the intersecting arcs of two eccentric circles. These arcs with their chord form two segments, whose difference consti- tutes the lune. The first curvilinear figure whose surface was exactly calculated was the lune of Hippocrates. This lune is formed by drawing semi-cir- cles on the sides of a right- angled triangle ; thus, if we describe semi-circles on the d/ sides C B, G A, and A B, we have the lunes C D A and AEB. c These lunes equal the triangle CAB, for, the largest semi-circle =^ \ the C B '^ X .7854, and the smaller ones = i the (C A » + A B X .7854. But from the properties of a right-angled triangle we have CA^-fAB^^ziCB^; therefore, the larger semi-circle equals the two smaller ones. If from these equals we subtract the segments C F A and A G B, we have left the triangle CAB equal to the two lunes. For if equals be taken from equals, the remainders will be equal. If the perpendicular and base are equal, the lunes will be equal, and each will equal \ of a square inscribed in a circle whose diameter is the hypothenflse. 88 MENSURATION OF SURFACES. LUNES. PROBLEM XL To find the area of a lune. RULE. Find the difference between the two segments formed by the arcs of the lune, and their chord. Note. — The reason for this rule is too obvious to require any explanation. EXAMPLES. 1. The chord of the segments forming a lune is 12 feet, .ind the heights of the segments are 4 and 3 feet. What is the area of the lune ? Solution. G ft.^ -f 4 ft.^ = 52 y^ ""\ sq. ft. / \ |/52~sq7'ft.~= 7.2111 ft., the / \ chord of J the arc. 1 s^>'^'''^"^^^« I 52 sq. ft. -f- 4 ft. r^ 13 ft., the W-^^6ft. pj ^^-^^ diameter. \^ pi ^ ^Vj>^J/ 13 ft. -.- 2 = 6.5 ft., radius. TlS^T 6.5 ft. — 4 ft. =: 2.5 ft., the cosine of i the arc. 7.2111 ft. X 8 — 12 ft. 3 15.2296 ft. X 6.5 ft. = 15.2296 ft., the arc. = 49.4962 sq. ft., the sector. 12 ft. X 2.5 ft. f, ,1, ,. 1 — rr= 15 sq. it., the triangle. 49.4962 sq. ft. — 15 sq. ft. = 34.4962 sq. ft., the larger segment. LUNES. 89 6 ft.' + 3 ft.2 45 sq. ft. 6.7082 ft., the 15 ft., the |/45 sq. ft. chord of i the arc. 45 sq. ft. -^ 3 ft diameter. 15 ft. -T- 2 = 7.5 ft., the radius. 7.5 ft. — 3 ft. ==: 4.5 ft., the cosine of i the arc. 6.7082 ft. X 8 — 1 2 ft 3 13.SS86 ft. = 13.8885 ft., the arc. 13.8885 ft. X 7.5 ft. 52.0818 sq. ft., the sector. 12 ft. X 4.5 ft. : 27 sq. ft., the triangle. 27 sq. ft. = 25.0818 sq. ft, the smaller 52.0818 sq. ft. segment. 34.4962 sq. ft. — 25.0818 sq. ft. = 9.4144 sq. ft., the lune. Res. 9.4144 sq. ft. 2. The chord of the segments forming a lune is 1 foot 4 inches, and the heights of the segments are 7 and 5 inches. What is the area of the lune ? 3. The chord of the segments is 32 inches, and the heights of the segments are 12 and 6 inches. "What is the area of the lune ? 4. Two eccentric circles 24 and 18 inches in diameter intersect each other so as to form a lune. What is the area of the lune if the chord of the intersecting arcs is 8 in. ? 5. What is the area of the lunes formed by describing semi-circles on the three sides of a right-angled triangle, if those sides are 6, 8, and 10 inches ? 6. What is the area of a lune formed by describing a semi-circle on the side of a square inscribed in a circle, whose diameter is 10 yards? 90 MENSURATION OF SURFACES. DEFINITIONS. THE ELLIPSE. An ellipse is a section of a cone, generated by a plane being passed througb its slant sides obliquely to the base; as, the curve A B C D. A diameter of an ellipsis is any straight line passing through its centre, and terminating at both ends in the circumference ; as, the lines A C, B D, E F, and G H. The transverse axis of an ellipse is its longest diameter ; as, A C. It is also called the major axis. The conjugate aa:tsof an ellipse is its shortest diameter ; as, B D. This axis is perpendicular to the transverse axis, and is sometimes called the minor axis. Tlie vertices of a diameter are the points in which the diameter meets the circumference : thus, A and C are the vertices of the transverse axis A C ; B and D are the vertices of the conjugate axis B B. DEFINITIONS. 01 The foci of an ellipse are two points in the longest diameter^ from which if two straight lines be drawn meeting each other in the circumference, their sum will equal that diameter; as, the points F and/. To find the foci of an ellipse, take a straight line equal to half the longest diameter, and, having placed one end of it on either vertex of the shortest diameter, let the other end fall where it will on the longest diameter. The points where it meets that diameter on each side of the centre of the ellipse are the ybci. If the lines B F and B / each equal half the diameter A C, the points F and/ are the foci ; for the straight lines B F and B/ drawn from them, meet each other in the circumference, and equal the longest diameter. The centre of an ellipse is the middle point of the straight line which joins the foci; as, E. All the diameters bisect each other at the centre. The eccentricity of an ellipse is the distance from the centre to either focus ; as, E F, or E /. The simplest method of constructing an ellipse when its major and minor axis are given, is to place them so that they bisect each other, and then find the foci. Then take a string the length of the major axis and fasten its ends at the foci. The curve which a pencil will then describe, on both sides of the major axis, by being pressed against the string stretched to its greatest extent, will be the circumference of the ellipse. 92 MENSURATION OF SURFACES. Two ellipses are similar when their axes are respectively? proportional to each other. THE ELLIPSE. PROBLEM XII. The transverse and conjugate axes of an ellipse being given, to find the circumference. RULE. Multiply the square root of half the sum of the squares of the transverse and conjugate axes by 3.1416. EXAMPLES. 1. What is the circumference of an ellipse whose trans- verse and conjugate axes are 4 and 6 feet? Solution. ^^^ ^^J^+6i'-= 5.099 ft. / __ 5.099ft. X3.1416 = 16.0190184ft. T Ees. 16.0190184 feet. \^ 2. What is the circumference of an ellipse whose trans- verse and conjugate axes are 70 and 50 feet? 3. What is the circumference of an ellipse whose semi- axes are 10 and 15 inches ? 4. What is the circumference of an elliptical field whose major and minor axes are 24 and 20 chains ? 5. What is the circumference of an ellipse, if the rectangle which is described upon it is twice as long as it is wide, and contains 18 square yards ? THE ELLIPSE. ^f^ 6. What is the eccentricity of an ellipse whose transverse and conjugate axes are 10 and 6 inches? 7. How far are the foci of an ellipse from the vertices of the transverse diameter, if the transverse and conjugate diameters are 20 and 12 feet ? PROBLEM XIII. The transverse and conjugate axes of an ellipse being given, to find the area. RULE. Multiply the square of a mean proportional between the two axes by .7854. EXPLANATION. It can be proved by geometrical analysis that a mean proportional between the axes of an ellipse gives the dia- meter of an equivalent circle. Therefore to get the area of an ellipse we first find a mean proportional between the axes by extracting the square root of their product. We then have the diameter of a circle equal in area to the ellipse; and since the square of the diameter X by .7854 equals the circle, the square of the mean proportional X by .7854 will equal the ellipse. EXAMPLES. 1. What is the area of an ellipse whose transverse and conjugate axes are GO and 40 feet? _^-y— .^.^^ Solution, yiji) ft. x 4 J ft. = 48.989 ft. 48.989 ft.' X .7854 = 1884.96 sq. ft. Res. 1884.96 sq. feet. 94 MENSURATION OF SURFACES. 2. How many acres are there in an elliptical field whose longest and shortest diameters are 80 and 60 chains ^ 3. What is the area of an elliptical park whose major and minor axes are 92 and 78 rods ? 4. The area of an elliptical fish-pond is 19.635 sq. rods. What is the diameter of a circular one of equal area ? 5. The area of an ellipse is 6.2832 sq. feet, and its con- jugate axis is 2 feet. What is the transverse axis ? 6. The axes of an ellipse containing 808.1766 sq. feet are to each other as 3 to 7. What are the axes ? 7. In finding the area of an ellipse, why is it the same to multiply the square of a mean proportional between the two axes by .7854 as to multiply the product of the two axes by .7854? MENSURATION OF SOLIDS. DEFINITIONS. POLYHEDRONS. A volume, solid, or hocJi/, is a quantity of space having three dimensions, viz., length, breadth, and thickness. The terms solid and lod?/, as generally applied, would infer the existence of matter, whereas the reasoning's of geometry carefully exclude every such idea. For this reason the term volume ia preferable, as it denotes a quantity of space limited in every direction, irrespective of what that space may be filled with, or whether it be entirely void. An empti/ barrel is just as much a solid, mathematically considered, as one filled with lead ; a bod^ capable of con- taining, as that which contains. A polyhedron is a solid or volume bounded by polygons ', these polygons are called the faces, and the straight lines in which the faces meet, the sides or edges of the polyhedron. The solid angles formed by three or more of these faces meeting at a common point are termed polyhedral angles. Polyhedrons, from the number of their faces, are de- nominated tetrahedrons, pentahedrons, hexahedrons, hcpta- hedrons, &c. A regular polyhedron is one whose faces are equal regular polygons, and whose polyhedral angles are all equal. (95) 96 MENSURATION OF SOLIDS. The diagonal of a polyhedron is a straight line joining the vertices of any two polyhedral angles not adjacent. Similar polyhedrons are those which are bounded by an equal number of mutually similar faces, similarly situated. All regular j)olij]iedrons of the same name are similar. The corresponding parts of similar solids are termed homologous. Similar solids or volumes (that is solids or volumes whose dimensions vary proportionally) are to each other as the cubes of their like dimensions. Solids of the same name, having two dimensions alike, are to each other as their third dimensions. Solids of the same name, having one dimension alike in each, are to each other as the products of the other two. Solids, generally, are to each other as the products of their bases and altitudes. The 'principal irregidar polyhedrons are prisms and pi/ramids. A prism is a polyhedron whose ends are two equal par- allel polygons^ and whose sides are right-angled parallelo- grams. The two equal parallel polygons form the upper and loicer bases of the prism, and the right-angled parallelograms its convex surface. The convex surface plus the areas of the bases form the entire surface. The altitude of a prism is the perpendicular distance between its bases. A regular prism is one whose edges are perpendicular to DEFINITIONS 97 its bases. Hence, in a regular prism the edges and altitude are equal. Other prisms are termed ohiique, and in saoh prisms the edges are (jreattr than the altitude. Prisms, from the shapes of their hases^ are classified into trianrjular or trigonal, quadrangular or tetragonal, penta- gonal, hexagonal, heptagonal, octagonal prisms , &c. A parallelopipedon is a prism whose faces are all right- angled parallelograms. When these are all equal the solid is termed a a/6e, otherwise a rectangular parallelopipedon. RectaQguJar ParaUulopipedon. 98 MENSURATION OF SOLIDS. A pyramid is a polyliedron whose sides are triaugles, uniting in a point at the apcx^ and terminating in the edges of a 2wli/e?ulicular, wliich remains fixed. EXAMPLES. 1. What is the surface of the frustum of a pentagonal pyramid, the slant height being 6 inches, each side of the upper base 2 inches, and each side of the lower base 4 inches ? Solution. 2 in. X 5 — 10 in., the dis- tance around the upper base. 4 in. X 5 = 20 in., the distance around the lower base. 10 in. + 20 in. == 30 in., the sum of these distances. , . 30 in. X (6 in. -j- 2) = 90 sq. in., the convex surface. Find the areas of the bases by Prob. XYIII., page 56. 1 in.2 : 2 in.^ : : 1.720477 sq. in. : the upper base, or 6.881908 sq. in. 1 in.2 : 4 in.^ : : 1.720477 sq. in. : the lower base, or 27.527632 sq. in. 90 sq. in. -p 6.881908 sq. in. + 27.527632 sq. in. = 124.40954 sq. in., the surface of the frustum. Res. 124.40954 sq. inches. FRUSTUMS OF PYRAMIDS AND CONES. !!& 2. What is the convex surface of the frustum of a hex- agonal pyramid, the slant height being 8 inches, each side of the upper end 3 inches, and each side of the lower end 6 inches ? 3. How many square yards are contained in the surface of the frustum of an octagonal pyramid whose slant height is 7 feet, each side of its uppeT base 3 feet, and of its lower base 5 feet ? 4. What is the convex surface of a frustum of a cone whose slant height is 8 feet, and the diameters of whose bases are 4 and G feet ? 5. What is the surface of the frustum of a cone described by a trapezoid whose parallel sides are 3 and 6 inches, and whose altitude is 4 inches ? 6. How much will it cost to line a cistern with cement at 10 cents a square foot, if it is 8 feet square at the top, 4 feet at the bottom, and 10 feet deep ? PROBLEM Vir. To find the solidity of the frustum of a right pyramid or cone. RULE. Add together the areas of the two bases and a mean proportional between them, and multiply this sum by one- third the altitude. Note. — To get a mean proportional between two numbers, extract the square root of their product. Note. — The sum of the two bases and the mean proportional between them can be obtained with much less work by multiplying the squares of the diameters plus their product by .7854 when ilie frustum is of a cone, or the squares of the sides plus their product by the number found in the tabular area when the frustum is of a pyramid. For reason, see Key. 120 MENSURATION OF SOLIDS. EXPLANATION. It can be proved by geometrical analysis that the frustum of a regular pyramid is equal in solidity to the sum of three pi/ramids, having for their altitudes the altitude of the frustum, and for their hases the lower base of the frustum^ the upper base of the frustum^ and a mean pnportional heticeen them. Every triangular pyramid equals the product of its base by one-third its altitude, therefore the three p)i/ramids will equal the sum of their bases, multiplied by one-third their altitude, or the sum of the upper and lower bases of the frustum plus a mean proportional between them, multiplied by one-third the altitude of the frustum. Since they equal in solidity the frustum, the solidity of the frustum will also o.j[ual the sum of its upper and lower bases plus a mean proportional between them, multiplied by one-third its altitude. Note. — It has been shown, in the note under Problem VI., that tho rules for the frustums of pyramids and cones arc similar. EXAMPLES. 1. What is the solidity of the frustum of a cone, the diameters of its upper and lower bases being 6 and 10 feet, and its altitude 12 feet? Solution. (6 ft.^ -j- 10 ft.^ -f (G ft. X 0^ 10 ft.)) X -7854 = 153.9384 sq. ft., the i§^k. sum of the bases and a mean proportional mM/% ^mk between them. »//" "^'""'^^^ 153.9384sq.ft.X(12ft.--3) = G15.7536 W^''''"'^W cu. ft., the solidity of the frustum. Kes. G15.7536 cu. ft. 2. What are the solid contents of the frustum of a cone the diameters of whose ends are 3 and 7 inches, and whose altitude is 9 inches ? 3. How many perches of stone are contained in the frustum of a pyramid composed of granite, whose altitude FRUSTUMS OF PYRAMIDS AND CONES. 121 is IG feet and whose bases are squares, the upper one measuring 4 feet on each side and the lower one 5 feet? 4. What will be the cost of a stick of hewn timber which is 2 feet 6 inches square at one end, 1 foot 6 inches square at the other, and 15 feet long, at the rate of 5 cents per cubic foot ? 5. IIow many hogsheads of rain water will a cistern contain, which is 12 feet in diameter at the bottom, 8 feet at the top, and 9 feet deep ? 6. A crucible which is 4 inches in diameter at the top, 2 inches at the bottom, and 3^~§J inches in depth is filled with melted silver. How many cubes, each containing 1 solid inch, can be made from the metal ? 7. How many feet high is the frustum of a tetragonal pyramid which contains 2 cubic yards and 13 cubic feet, if the upper end measures 3 feet on each side and the lower end 7 feet ? 8. What is the altitude of the trapezoid describing the frustum of a cone, if the parallel sides of the trapezoid are 4 and 5 inches, and the frustum contains 1149.8256 cubic inches ? Note. — If the frustums of cones or pyramids have the same altitude, and ihcir ends proportional, their solidities will be to each other as the squares of their diameters or like sides. 9. A marble monument shaped in the form of a frustum of a hexagonal pyramid measures 3 feet at each side of the base, 1 foot at the top, and 15 feet high. What are the sides of a monument containing G75.4997G cubic feet, whose altitude is the same as the former, and whose ends hold the same relation to each other ? 10. If the diameters of the top and bottom of a basket are 10 and 8 inches, and the depth 9 inches, what must be the dimensions of a similarly-shaped basket to contain 8 times as much? 11 122 MENSURATION OF SOLIDS. DEFINITIONS. THE WEDGE. The icedge is a solid bounded by five polygons ; viz. rectangle forming its back or base, two trapezoids forming its faces, and two triangles its ends. In the wedge A B C D E F, tlie rectangle C D E F is the base, the trapezoids A B C F and A B D E are the faces, and the triangles A E F and BBC the ends of the wedge. The edge of a wedge is the straight line in which the trapezoids forming its faces meet; as, A B. The altitude of a wedge is the perpendicular distance from the edge to the base; as, A G. THE WEDGE. PROBLEM VIII. To find the solidity of a wedge. RULE. When the base exceeds the edge in lengthy at each end of the edge pass a plane through the wedge perpendicular to its base. The portion cut off by each plane will be a quad- rangular pyramid^ having half the difference in length between the base and edge, and the breadth of the base for its base, and the altitude of the wedge for its altitude. The middle portion will equal a triangular prism, whose altitude is the edge of the wedge, and whose ends are tri- angles, having for their bases the breadth of the base THE WEDGE. 123 of the wedge, and for their altitudes the altitude of the wedge. Hence, the sum of the solidUies of the prism and hco pyramith will equal that of the wedge. When the edge exceeds the base in length, add to each end of the wedge a quadrangular pyramid, having half this excess, and the breadth of the base for its base, and the altitude of the wedge for its altitude. The solid thus formed will be a triangular prism, whose altitude is the edge of the wedge, and whose ends are triangles, hav- ing for their bases the breadth of the base of the wedge, and for their altitudes the altitude of the wedge. Hence, the difference between the solidities of the prism and two added pyramids will equal that of the wedge. e A wed^re wliose base exceeds ita edge in length. Tho same wedge divided iuto two quadrangular pyramids, and a tri- angular pritiui. A wedge whose edge exceeds its The same wedse wUh the base in km/th. rangular pyramids added, mn triangular prism. Note.— The rule needs no further explanation. EXAMPLES. 1. What are the solid contents of a wedge whose base is 30 inches long and 10 broad, the length of the edge being 20 inches, and the altitude 18 inches ? 124 MENSURATION OF SOLIDS. Solution. 30 in. — 20 in. = 10 in., the difference in length of the base and edge. 10 in. -f- 2 = 5 in., one side of the base of each quadrangular pyramid. 10 in. X 5 in- X (18 in. --- P>) = 300 cu. in., the solidity of each pyramid. 300 cu. in. X 2 = 600 cu. in., the solidity of both pyramids. 10 in. X (18 in. --2) = 90 sq. in., the area of the triangle forming the base of the prism. 90 sq. in. X 20 in. = 1800 cu. in., the solidity of the triangular prism. 600 cu. in., the pyramids, -\- 1800 cu. in., the prism, =: 2400 cu. in., the solidity of the wedge. Res. 2400 cu. inches. 2. What are the solid contents of a wedge whose base is 3 feet 4 inches long, 1 foot 3 inches broad, the length of the edge being 2 feet 3 inches, and the altitude 2 feet 9 inches? 3. What are the solid contents of a wedge whose base is 1 yard 3 inches long and 26 inches broad, the length of the edge being 31 inches, and the altitude 24 inches ? 4. What is the solidity of a stone, in the form of a wedge, which is 10 feet 2 inches long at the base and 3 feet 3 inches broad, the altitude of the stone being 6 feet, and the length of its edge one-half the length of its base ? 5. What is the difference in volume between two wedges, each being 60 inches in altitude, and 25 inches broad at the base, but the base of one being 70 inches long and its edge 50 inches, while the base of the other is 50 inches long and its ed^-e 70 inches ? THE PRISMOID. 126 DEFINITIONS. THE PRISMOID. A rectangular prismoid is a polyhedron whose ends are two unequal but parallel rectangles, and . . whose sides are trapezoids. Or, it is a / frustum of a wedge formed by passing / a plane through it parallel to its base. The altitude of a prismoid is the per- \ ^i; pendicukr distance between its ends ; as, ' ^' AB. THE PRISMOID. PROBLEM IX. To find the solidity of a rectangular prismoid. RULE. Add the solidity of a wedge whose altitude and base are the altitude and lower base of the prismoid, and whose edge is the length of the upper base, to that of a wedge having the same altitude, but whose base is the upper base of the prismoid, and whose edge is the length of its lower base. EXPLANATION. A prismoid may be considered as composed of two wedges having for their altitudes the altitude of the priamoid, for their bases the hases of the prismoid, and for their edges the lengths of its upper and lower hatie^i. For if through the lines A B and C D we pass the plane A B C D, it will divide the prismoid into the two wedges 11* 126 MENSURATION OF SOLIDS. 16 in. A B C D F E and C D II I B A, having for tlieir alti- tudes the ahitude of the prismoid, for their bases the upper and lower bases of the prismoid, and for their edges the lengths of its upper and lower bases. Hence, the solidities of the two wedges will equal that of the prismoid. EXAMPLES. 1. How many cubic inches are there in a rectangular prismoid, the greater end being 20 by 18 inches, the less 16 by 12 inches, and the altitude 30 inches ? Solution. Divide the prismoid into the w^edges A B C D F E and C D H I B A. 20 in. — 16 in. « • m ^ • = z m. Ihe wea2:e 2 A B C D F E may be divided into two quadrangular pyramids — whose bases are 18 in. by 2 in., whose altitudes are 30 in. — and a trianglar prism whose altitude ^ 20 in. e is 16 in., and whose ends are triangles having 18 and 30 in. for their bases and altitudes. 18 in. X 2 in. X (30 in. -f- 3) = 360 cu. in., the solidity of each pyramid. 360 cu. in. X 2 = 720 cu. in., the solidity of both pyra- mids. 18 in. X (30 in. -^2) =z 270 sq. in., the base of the triangular prism. 270 sq. in. X 16 in. = 4320 cu. in., the soHdity of the prism. 720 cu. in. -j- 4320 cu. in. = 5040 cu. in., the wedge A B C D F E. The wedge C D H I B A has its edge C D 20 in., and its base H I B A 16 in. long. 20 in. — 16 in. ^ . ^ = 2 in. THE PRISMOID. 127 Hence, the quadrangular pyramids necessary to be added to the wedge, in order to change it to a triangular prism having the same altitude as the wedge, have their bases 12 in. by 2 in., and their altitudes 30 in. The wedge, with the addition of these pyramids, equals a prism whose alti- tude is 20 in., and whose ends are triangles having 12 and 30 in. for their bases and altitudes. 12 in. X (30 in. -f- 2) = 180 sq. in., the base of the triangular prism. 180 sq. in. X 20 in. =3600 cu. in., the solidity of the prism. 2 in. X 12 in..X (30 in. -^ 3) = 240 cu. in., the solidity of each pyramid. 240 cu. in. X 2 =480 cu. in., the solidity of both pyramids. Tlie prism. The pyramids. 3G00 cu. in.— 4«0 cu. in. = 3120 cu. in., the wedge C D II I B A. 5040 cu. in -|- 3120 cu. in. =8160 cu. in., the prismoid. Res. 8160 cu inches. 2. How many cubic feet are there in a rectangular pri.«- moid, if the smaller end is 3 feet by 2 feet 6 inches, the greater 4 feet by 3 feet 4 inches, and the altitude 5 feet ? 3. What is the solidity of a block of granite, cut in the shape of a frustum of a wedge, the upper base being 1 yard 4 inches by 2 feet 6 inches, the lower 1 yard 2 feet by 1 yard 1 foot 2 inches, and the altitude 1 yard 1 foot and 10 inches ? 4. How many bushels will a box contain, shaped in the form of a rectangular prismoid, the top being 2 by 2^ feet, the bottom 3 by 3 J feet, and the depth 1^ feet ? 5. What will be the cost of the base of a marble monu- ment, at $1.50 per cubic foot, if it is 12 by 10 feet at the bottom, 8 by 6 feet at the top, and 6 feet high ? 128 MENSURATION OF SOLIDS. DEFINITIONS. THE REGULAR POLYHEDRONS. A REGULAR POLYHEDRON is One whose faces are equal regular polygons, and whose polyhedral angles are all equal. There are jive regular polyhedrons, which derive their names from the number of their s,ide&. They are the tetrahedron^ the hexahedron or cuhe, the octahedron, dodecahedron, and icosahedrop. The tetrahedron is a regular polyhedron bounded by four triangles. The hexahedron or cube is a regular polyhedron bounded by six squares. The octahedron is a regular polyhedron bounded by eight triangles. The dodecahedron is a regular poly- hedron bounded by twelve pentagons. DEFINITIONS. I2d The icosaJiedron is a regular polyhedron bounded by twenty triangles. If the following figures be cut out of pasteboard, and the lines cut partly through, so that the parts may be turned up and glued together, the solids thus formed will represent the five regular polyhedrons. 130 MENSURATION OF SOLIDS. EEGULAIl POLYHEDRONS. PllOBLEM X. To find the surface of a regular polyhedron. RULE. Multiply the square of the given edge hy the surface of a similar polyhedron whose edges are 1. EXPLANATION. Since the faces of a regular polyhedron are eqnal^ regular poJi/ ni.^;=15 in., the ra- dius of the sphere. 15 in.X - = 30 in., the diameter. 30 in. X 3.1416 r= 94.248 in., the circumference of a great circle. 94.248 in. X 3 in. =. 282.744 sq. in., the surface of the zone. Res. 282.744 sq. inches. * For explanation, see Key. f For explanation, see note, p. 82. SPHERES. 1319 13. What is the surface of the torrid zone, if its height is 3150.68 miles, and the diameter of the earth is 7912 miles? PROBLEM Xiri. To find the solidity of a sphere. RULE. Multiply its surface by one-third of its radius. Note — The solidity of a spherical pyramid or sector also equals the product of the surface of the polygon or zone forming the base, by one-third of the radius of the sphere to which it belongs. EXPLANATION. If wc inscribe a regular poli/hedron in a sphere, we may consider the polyhedron to be composed of j^J/^ft^i^'^s, each having for its vertex the centre of the sphere^ and for its base one of the faces of the polyhedron. The solidity of each of these pyramids equals the product of its basq by one-third of its altitude, and the solidity of all the pyramids, or of the polyhedron, equals the product of the sum of their bases, or the surface of the polyhedron, by one-third of their common altitude. Now if we increase the numher of the faces of the poly- hedron, its surface, or the bases of the pyramids, will equal the surface of the sphere, the altitude of the pyramids will equal the radius of the sphere, and the polyhedron will become the sphere. Hence, the solidity of a sphere equals the product of its surface by one-third of its radius. Note. — The solidity of a spherical pyramid or sector is obtained in the same manner as that of a sphere, for either may be con- sidered as composed of an indefinite number of pyramids whose bases form the tjase of the pyramid or sector, and whose vertices, like the vertex of the pyramid and sector, are at the centre of the sphere. EXAMPLES. 1. What is the solidity of a sphere whose diameter is 7 ft.? 140 mensuration of solids. Solution. 7 ft. X 3.141G = 21.C912 ft., the circumference of a great circle. ^1 21.9912 ft. X 7 ft. == 153.9384 sq. ft., the surface of the sphere. 7 ft. -^- 2 = 2" ft., radius. a u = 7 ft. 153.9384 sq. ft. X i^i ft- -^ 3) = 179.5948 cu. ft., the soHdity of the sphere. Res. 179.5948 cu. ft. 2. What is the solidity of a sphere whose diameter is 5 ft. ? 3. What is the solidity of a globe whose radius is 10 rods ? 4. How many cubic miles does the earth contain if its diameter is 7912 miles, supposing it to be a perfect sphere? 5. The diameter of a small circle of a certain globe is 8 ft,, and the distance from the centre of its plane to the centre of the sphere is 3 ft. What are the solid contents of the sphere ? G. What is the diameter of a globe whose solidity is 179.5948 cu. feet? 7. What is the diameter of a globe containing as many cubic foet as a cone 2 feet in diameter, and 3 feet high ? 8. What is the radius of that sphere whose solid contents equal as many cubic feet, as its surface contains square feet ? 9. Why does cubing the diameter of a sphere, and multi- plying it by .5236 give its solidity? 10. What is the solidity of the greatest cube that can be cut from a globe 6 inches in diameter ? 11. If a globe 3 feet in diameter weighs 2G0 lbs., what will be the weight of another whose diameter is G feet ? 12. What is the diameter of a sphere which contains 8 times as much as one 5 feet in diameter ? 13. What is the solidity of a spherical pyramid the area of whose base is 100 sq. feet, and the diameter of the sphero 20 feet? 14. What is the solidity of a spherical sector whose base is a zone 5 feet in altitude, in a sphere 24 feet in diameter? SPHERES. Ul 15. What is the solidity of a spherical sector of the earth, whose base is the north frigid zone, the height of which is 327.19 miles, the diameter of the earth being 7912 miles? PROBLEM XIV. To find the solidity of a spherical segment with one base. RULE. When the spherical segment is less than a hemisphere, from the solidity of a spherical sector having the zone of the segment for its base, subtract the solidity of the cone which forms the difference between the segment and sector. When the spherical segment is greater than a hemisphere, it equals the sum of this sector and cone. Note. — The solidity of a spherical segment having two bases equals the difference between the solidities of two segments, the one extending from the top of the sphere to its upper base, the other from the top of the sphere to its lower base. EXPLANATION. If a spherical sector be divided into two parts by passing a plane through the circumference of its base, the section will be a circle, and the solid will be divided into a cone and a spherical seg- ment; therefore, if we take the cone from the spherical sector the remainder will be the spherical segment. When the spherical segment has two bases, as ABC D, it is evident that the solidity of a segment extending from E to the base C D will exceed the given segment, by a segment extending from E to the base A B. Hence, the difference between the solidities of these two will be the solidity of the required segment. 142 MENSURATION OF SOLIDS. EXAMPLES. 1. What is the solidity of a spherical segment the radius of whose base is 9 inches, and whose height is 3 inches ? Solution. 9 in.2-^ 3 in.2^90sq. in. 90 sq. in. — ■ G in. = 15 in., the ra- dius of the sphere. 15 in. X -^ = 30 in,, the diameter. 30 in. X 3.1416 = 94.248 in., the circumference of a great circle. 94.248 in. X '^ in. = 282.744 sq. in., the convex surface of the spherical segment. 28-2.744 sq. in. X (15 in. ~ 3) = 1413.72 cu. in., the solidity of a spherical sector having the same base as the segment. 18 in.^ X .7854 = 254.4G96 sq. in., the base of the cone forming the difference between the sector and segment. 254.4696 sq. in. X (12 in. -- 3) = 1017.8784 cu. in., the solidity of the cone. 1413.72 cu. in. — 1017.878 cu. in. = 395.8416 cu. in., the solidity of the spherical segment. Res. 395.8416 cu. inches. 2. What is the solidity of a spherical segment the radius of whose base is 21 inches, and whose height is 7 inches?' 3. What is the solidity of a spherical segment, the height of the zone forming the base being 2 feet, and the diameter of the sphere being 20 feet? 4. What is the solidity of a spherical segment whose height is 18 inches, the diameter of the sphere being 30 inches ? SPHERES. 143 5. What is the solidity of a spherical segment the diam- eters of whose bases are 24 and 18 inches, and whose height is 3 inches ? Solution. 12 in.^ — (9 in.^ -|- 3 in.'') = 54 sq. in. 54 sq. in. -T-(3 in. X 2) r== 9 in., the base of a right-angled triangle whose hypothenuse is the radius. y/12 in.=^-f-9 in.^^15 in., the radius of the sphere. 15 in. X - = 30 in,, the diameter. Then from the solidity of the spher- ical segment D E take the segment A E ]i 30 in. X 3.1416 = 94.248 in. circumi'crence of a sreat circle. the 94.248 in. X G in. =: 565.488 sq. in., the convex surface of the spherical segment DEC. 565.488 sq. in. X (15 in. -f- 3) =r 2827.44 cu. in., the f?olidity of the spherical sector, having for its base the base of the spherical segment, and for its vertex the centre of the sphere. 24 in.2 X -7854 = 452.3904 sq in., the ba?e of the cone forming the difference between this spherical sector and 452.3904 sq. in. X (0 in.-^- 3) = 1357.1712 cu. in., the solidity of the cone. 2827.44 cu in. —1357.1712 cu. in = 1470.2688 cu. in., the solidity of the spherical segment D E C. 04.248 in X 3 in. = 282.744 sq. in., the convex surface of the spherical segment A E B. 282.744 in. X (15 in -- 3) = 1413.72 cu. in., the solidity of the spherical sector, having for its base the base 144 MENSURATION OP SOLIDS. of the spherical segment A E B, and for its vertex the centre of the sphere. 18 in.2 X .7854 = 254.4696 sq. in., the base of the cone forniing the difference between this spherical sector and segment. 254.4696 sq. in. X (12 in. -- 3) r= 1017.8784 cu. in., the solidity of the cone. 1413.72 cu. in. ^ 1017.8784 cu. in. = 395.8416 cu. in., the solidity of the spherical segment A E B. 1470.2688 cu. in. — 395 8416 cu. in. == 1074.4272 cu. in., the solidity of the required spherical segment A B C D. Bes. 1074.4272 cu. inches. 6. What is the solidity of a spherical segment the diam- eters of whose bases are 12 and 10 feet, and whose height is 2 feet? 7. What is the solidity of a spherical segment the diam- eters of whose bases are 6 and 10 feet, and whose height is 4 feet ? 8. What is the solidity of a spherical segment the diam- eters of whose bases are 24 and 32 feet, and whose height is 28 feet? THE END. CATALOGUE OF ^pokir Scljaol anir College Sitrf-^oclis. PUBLISHED BY E. H. BUTLER & CO., 137 Soath Fourth St., Philadelphia. Goodrich's Pictorial History of the United States. A Pictorial History of the United States, with notices of other portions of America. By S. G. 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