UC-NRLF 
 
 il||l!"llUlilil11l 
 
 $B 527 633 
 
m MEMORIAL 
 Irving Strlngham 
 
 J 
 
-^ 
 
A 
 
 PRACTICAL APPLICATION 
 
 OF 
 
 THE PRINCIPLES OF GEOMETRY 
 
 TO THB 
 
 MENSURATION 
 
 or 
 
 SUPERFICIES AND SOLIDS. 
 
 ▲DAmB TO 
 
 THE METHOD OP INSTRUCTION IN SCHOOLS AND ACADEMXEB. 
 
 BY JEREMIAH DAY, D.D. LL.D. 
 
 I.4TS PUMOBBT Or TAU OOLLBVC 
 
 NEW YORK: 
 
 PUBLISHED BY MARK H. NEWMAN <k CO., 
 
 No. 199 BROADWAY. 
 
 1848. 
 
Entered, according to Act of Congress, in the year 1848, by 
 
 JEREMIAH DAY, 
 
 In the Clerk's OfBce of the District Court of the Unit«d States for the 
 Southern District of New York. 
 
 THOMAS B. SMITH, STERKOTTPER, J. D. BEDFORD, PRINTER, 
 
 216 WILLIAM STREET, K. Y. 138 FULTON STREET. 
 
CONTENTS, 
 
 Sectioh L Areas of figures bonnded by right lines, ... 5 
 
 II. The Quadrature of the Circle and its parts, . . 19 
 
 Promiscnoos examples of Areas, 34 
 
 m. Solids bounded by plane surfaces, 37 
 
 rV. The Cylinder, Cone, and Sphere, 66 
 
 Promiscuous examples of Solids, 76 
 
 V. Isoperimetry, 78 
 
 APPENDIX 
 
 Gauging of Casks, 92 
 
 Notes, 99 
 
SECTION I. 
 
 AREAS OF FIGURES BOUNDED BT RIGHT LINES. 
 
 Art. 1. The following definitions, which are nearly the 
 same as in Euclid, are inserted here for the convenience of 
 reference. 
 
 I. Four-sided figures have different names, according to 
 the relative position and length of the sides. A parallelo- 
 gram has its opposite sides equal and parallel, as ABCD. 
 
 1 
 
 D| 1 r ^ |C 
 
 a' ' ' ' 1 
 
 (Fig. 2.) A rectangle, or right parallelogram, has its oppo- 
 site sides equal, and all its angles right angles ; as AC. 
 (Fig. 1.) A square has all its sides equal, and all its angles 
 right angles ; as ABGH. (Fig. 3.) A rhxmibus has all its 
 
 sides equal, and its anL^ 
 rJwmJboid has its opposi \ 
 
 e ; as ABCD. (Fig. 3.) A 
 [ual, and its angles obUque \^ 
 
 as ABCD. (Fig. 2.) A trapezoid has only two of its sides 
 parallel ; as ABCD. (Fig. 4.) Any other four sided figilre 
 is called a trapezium. 
 
6 MENSURATION OF PLANE SUHFACES. 
 
 II. A figure which has more than four sides is called a 
 polygon. A regular polygon has all its sides equal, and all 
 its angles equal. 
 
 in. The height of a triangle 
 is the length of a perpen- 
 dicular, drawn from one of the 
 angles to the opposite side ; as 
 CP. The height of 2kfour sided 
 figure is the perpendicular dis- 
 tance between two of its par- 
 allel sides ; as CP. (Fig. 4.) 
 
 IV. The area or superficial contents of a figure is the 
 space contained within the line or lines by which the figure 
 is bounded. 
 
 2. In calculating areas, some particular portion of surface 
 is fixed upon, as the measuring unit, with which the given 
 figure is to be compared. This is commonly a square ; as a 
 square inch, a square foot, a square rod, &c. For this rea- 
 son, determining the quantity of surface in a figure is called 
 squaring it, or finding its quadrature ; that is, finding a 
 square or number of squares to which it is equal. 
 
 3. The superficial unit has generally the same name, as 
 the linear unit which forms the side of the square. 
 
 The side of a square inch is a linear inch ; 
 of a square foot, a linear foot ; 
 of a square rod, a linear rod, &;c. 
 There are some superficial measures, however, which have 
 no corresponding denominations of length. The acre, for 
 instance, is not a square which has a line of the same name 
 for its side. 
 
 The following tables contain the linear measures in com- 
 mon use, with their corresponding square measures. 
 
MENSURATION OF PLANE SURPAOBS. 
 
 Linear Measures. 
 
 Square Measures. 
 
 12 inches =1 foot. 
 
 144 inches =1 foot. 
 
 3 feet =1 yard. 
 
 9 feet =1 yard. 
 
 6 feet =1 fathom. 
 
 36 feet =1 fathom. 
 
 16i feet =1 rod. 
 
 272i feet =1 rod. 
 
 5i yards =1 rod. 
 
 30-I- yards =1 rod. 
 
 4 rods =1 chain. 
 
 16 rods =1 chain. 
 
 40 rods =1 furlong. 
 
 1600 rods =1 furlong. 
 
 320 rods =1 mile. 
 
 102400" rods =1 mile. 
 
 An acre contains 160 square rods, or 10 square chains. 
 By reducing the denominations of square measure, it will 
 be seen that 
 
 1 sq. mi!e=610 acre8=102400roda=2787»100 feet =4014489600 inchca. 
 1 acre=iachain8=160 rods=:43560 feet=:6272640 inches. 
 
 The fundamental problem in the mensuration of super- 
 ficies is the very simple one of determining the area of a 
 riffht parallelogram. The contents of other figures, partic- 
 ularly those which are rectilinear, may be obtained by find- 
 ing parallelograms which are equal to them, according to the 
 principles laid down in Euclid. 
 
 Problem I. 
 
 To find the area of a parallelogram, square, rhombus, or 
 rliomboid. 
 
 4. .NL . LENGTH BY Tr?r: T.V 
 
 or BREAD! 
 
 It is b evident that the number of 
 square inches in the parallelogram AC 
 is equal to the number of linear 
 inches in the length AB, repeated as 
 many times as there are inches in the 
 breadth BC. For more particular 
 illustration of this see AJg. 886 — 389. 
 
 ••"'VDICULAR llEIGIir 
 
 I1~M.__ 
 
8 
 
 MENSURATION OF PLANE SURFACES. 
 
 The oblique parallelogram or rhomboid ABCD, (Fig. 2.) 
 is equal to the right parallelogram GHCD. (Euc. 36. 1.)* The 
 
 area, therefore, is equal to the length AB multiplied into the 
 perpendicular height HC. And the rhombus ABCD, (Fig. 3.) 
 is equal to ihe parallelogram ABGH. As the sides of a 
 square are all equal, its area is found, by multiplying one of 
 the sides into itself. 
 
 Ex. 1. How many square feet are there in a floor 23|- feet 
 long, and 18 feet broad ? Ans. 23^X18=423. 
 
 2. What are the contents of a piece of ground which is 
 66 feet square ? Ans. 4356 sq. feet=16 sq. rods. 
 
 3. How many square feet are there in the four sides of a 
 room which is 22 feet long, 1*7 feet broad, and 11 feet high ? 
 
 Ans. 858. 
 
 Art. 5. If the sides and angles of a parallelogram are 
 given, the perpendicular height may be easily found by trig- 
 onometry. Thus, CH (Fig. 2.) is the perpendicular of a 
 right angled triangle, of which BC is the hypothenuse. 
 Then, (Trig, 134.) 
 
 R : BC : : sin B : CH. 
 
 The area is obtained by multiplying CH thus found, into 
 the length AB. 
 
 * Thomson's Legendre, 1. 5. 
 
MENSURATION OF PLANE SURFACES. 
 
 Or, to reduce the two operations to one, 
 
 As radius, 
 
 To the sine of any angle of a parallelogram ; 
 
 So is the product of the sides including that angle, 
 
 To the area of the parallelogram. 
 
 For the arm=ABxCH, (Fig. 2.) But CH=^^^^'" ^ 
 
 R 
 Therefore, 
 
 Thearea^^^^^^^''''' ^. Or,R : sinB : : ABxBC : thearea. 
 R 
 
 Ex. If the side AB be 58 rods, BC 42 rods, and the angle 
 B 63°, what is the area of the parallelogram ? 
 
 As radius 10.00000 
 
 To the sine of B 63° 9.94988 
 
 ( So is the product of AB 58 1.76343 
 
 \ Into BC (Trig. 39.) 42 1.62325 
 
 To the area 2170.5 sq. rods 3.33656 
 
 2. If the side of a rhombus is 67 feet, and one of the 
 angles 73**, what is the area ? Ans. 4292.7 feet. 
 
 C. When the dimensions are given in feet and inches, the 
 multiplication may be conveniently performed by the arith- 
 metical rule of Duodecimals ; in which each inferior denom- 
 ination is one-twelfth of the next higher. Considering a foot 
 as the measuring unit, a prime is the twelfth part of a foot ; 
 a second, the twelfth part of a prime, kc. It is to be ob- 
 served, that, in measures of length, inches are primes ; but in 
 superficial measure they are seconds. In both, a prime is -^ 
 of a foot. But iV of a square foot is a parallelogram, a foot 
 long and an inch broad. The twelfth part of this is a square 
 inch, w^hich is i -J y of a square foot. 
 
10 MENSURATION OF PLANE SURFACES. 
 
 Ex. 1. What is the surface of aboard 9 feet 5 inches, by 
 2 feet 1 inches. 
 
 F 
 
 9 5' 
 
 2 '7 
 
 18 10 
 
 5 5 11 
 24 3 11^ or 24 feet 47 inches. 
 
 2. How many feet of glass are there in a window 4 feet 
 11 inches high, and 3 feet 5 inches broad ? 
 
 Ans. 16 F. 9' 1", or 16 feet 115 inches. 
 
 Y. If the area and one side of a parallelogram be given, 
 the other side may be found by dividing the area hy the 
 given side. And if the area of a square be given, the side 
 may be found by extracting the square root of the area. This 
 is merely reversing the rule in Art. 4. See Alg. 520, 521. 
 
 Ex. 1 . What is the breadth of a piece of cloth which is 
 36 yds. long, and which contains 63 square yards. 
 
 Ans. If yds. 
 
 2. What is the side of a square piece of land containing 
 289 square rods ? 
 
 3. How many yards of carpeting 1^ yard wide, will cover 
 a floor 30 feet long and 22^- broad ? 
 
 Ans. 30X22ifeet=10xH='75 yds. And '75-Mi=60. 
 
 4. What is the side of a square which is equal to a par- 
 allelogram 936 feet long and 104 broad ? 
 
 5. How many panes of 8 by 10 glass are there, in a win- 
 dow 5 feet high, and 2 feet 8 inches broad ? 
 
MENSURATION OF PLANE SCRFACES. 
 
 11 
 
 Problem II. 
 
 To find the area of a triangle. 
 
 8. Rule L Multiply onb side by half the perpen- 
 dicular from the opposite angle. Or, multiply half the 
 side by the perpendicular, Or, multiply the whole side by 
 the perpendicular and take half the product. 
 
 The area of the triangle ABC, 
 b equal to ^ PC X AB, because 
 a parallelogram of the same 
 base and height is equal to PC 
 X AB, (Art. 4.) and by Euc. 41, 
 1,* the triangle is half the par- 
 allelogram. 
 
 Ex. 1. If AB be 65 feet, and PC 31.2, what is the area 
 of the triangle? Ans. 1014 square feet. 
 
 2. What is the surface of a triangular board, whose base 
 is 3 feet 2 inches, and perpendicular height 2 feet 9 inches ? 
 Ans. 4 F. 4' 3", or 4 feet 51 inches. 
 
 9. If two sides of a triangle and the included angle, are 
 given, the perpendicular on one of these sides may be easily 
 found by rectangular trigonometry. And the area may be 
 calculated in the same man- 
 ner as the area of a parallel- 
 og^m in Art. 5. In the tri- 
 angle ABC, 
 
 R : BC : : sin B : CH 
 
 And because the triangle is half the parallelogram of the 
 same base and height, 
 
 * Thomson's Legendre, 2. 4. 
 
12 
 
 MENSURATION OF PLANE SURFACES. 
 
 As radius, 
 
 To the sine of any angle of a triangle ; 
 
 So is the product of the sides including that angle, 
 
 To twice the area of the triangle. (Art. 5.) 
 
 Ex. If AC be 39 feet, AB 
 65 feet, and the angle at A 53° 
 7' 48", what is the area of the 
 triangle ? Ans. 1014 square feet. 
 
 9. &. If one side and the angles 
 are given ; then 
 
 As the product of radius and the sine of the angle oppo- 
 site the given side, 
 
 To the product of the sines of the two other angles; 
 So is the square of the given side. 
 To twice the area of the triangle. 
 
 If PC be perpendicular to AB. 
 
 R : sin B : : BC : CP 
 sin ACB : sin A : : AB : BC 
 
 Therefore, (Alg. 351, 345.) 
 R X sin ACB : sin A X sin B : : AB X BC : CP X BC : : 
 AB^ : ABxCP— twice the area of the triangle. 
 
 Ex. If one side of a triangle be 57 feet, and the angles at 
 the ends of this side 50° and 60°, what is the area? 
 
 Ans. 1147 sq. feet. 
 
 10. If the sides only of a triangle are given, an angle may 
 be found, by oblique trigonometry, Case IV, and then the 
 perpendicular and the area may be calculated. But the area 
 may be more directly obtained, by the following method. 
 
 Rule II. When the three sides are given, from half their 
 sum subtract each side severally, multiply together the lialf 
 sum and the three femainders, and extract the square root of 
 the product. 
 
MENSURATION Of PLANE SURFACES. 
 
 13 
 
 If the sides of the triangle are a, h, and c, andif A=half 
 their sum, then 
 
 The area=Vhx{hr—a)x{fir-h)X{h—c) 
 
 Ex. 1. In the triangle ABC, 
 given the sides a 52 feet, 6 39, 
 and c 65 ; to find the side of a 
 square which has the same area 
 
 as tlio Irianrjle. 
 
 ^— a=26 
 
 A— 6=39 
 A— c=13 
 
 Then the area=v78x 26X39X13= 1014 square feet. 
 By logarithms. 
 
 The half sum 
 
 = 78 
 
 1.89209 
 
 First remainder 
 
 = 20 
 
 1.41497 
 
 Second do. 
 
 =39 
 
 1.59106 
 
 Third do. 
 
 = 13 
 
 1.11394 
 
 2)6.01206 
 2)3.00603 
 
 The area required^ =1014 
 
 Side of the square =31.843 (Trig. 47.) 1.50301 
 
 2. If the sides of a triangle are 134, 108, and 80 rods, 
 what is the area? Ans. 4319. 
 
 3. What is the area of a triangle whose sides are 371, 264, 
 and 225 feet ? 
 
 Problem III. 
 To find the area of a trapezoid. 
 21. Multiply half the sum op the parallel sides 
 
 INTO their perpendicular DISTANCE. 
 
14 
 
 MENSURATION OF PLANE SURFACES. 
 
 The area of the trapezoid 
 ABCD, is equal to half the 
 sum of the sides AB and CD, 
 multipHed into the perpendic- 
 ular distance PC or AH. For 
 the whole figure is made up of 
 the two triangles ABC and 
 ADC ; the area of the first of which is equal to the product 
 of half the base AB into the perpendicular PC, (Art. 8.) 
 and the area of the other is equal to the product of half the 
 base DC into the perpendicular AH or PC. 
 
 Ex. If AB be 46 feet, BC 31, DC 38, and the angle B 
 •70°, what is the area of the trapezoid ? 
 
 R : BC : : sin B : PC = 29.13. And 42x29. 13 = 1223^. 
 
 2. What are the contents of a field which has two par- 
 allel sides Q5 and 38 rods, distant from each other 27 rods ? 
 
 Problem IV. 
 To find the area of a trapezium, or of an irregular polygon. 
 
 13. Divide the whole figure into triangles, by draw- 
 ing DIAGONALS, AND FIND THE SUM OF THE AREAS OF THESE 
 
 triangles. (Alg. 394.) 
 
 If the perpendiculars in two triangles fall upon the sa?ne 
 diagonal, the area of the trapezium formed of the two trian- 
 gles, is equal to half the 
 product of the diagonal into 
 the sum of the perpendiculars. 
 
 Thus the area of the trape- 
 zium ABCH, is 
 
 iBHxALH-iBHxCM=iBHx(AL+CM.) 
 Ex. In the irregular polygon ABCDH, 
 
KKNSUBATION OF PLANE SURFACES. 15 
 
 ( BH=36 ( -^^=5.3 
 
 if the diagonals | ' and the perpendiculars ) CM =9.3 
 
 The area=18XU.6 + 16X 7.3=379.6. 
 
 14. If the diagonals of a trapezium are given, the area 
 may be found, nearly in the same manner as the area of a 
 parallelogram in Art. 5, and the area of a triangle in Art. 9. 
 
 In the trapezium ABCD, the sines of the four angles at 
 N, 4he point of intersec- 
 tion of the diagonals, are 
 all equal. For the two 
 acute angles are supple- 
 ments of the other two, 
 and therefore have the 
 same sine. (Trig. 90.) 
 Putting, then, sin N for 
 
 the sine of each of these angles, the areas of the four tri- 
 angles of which the trapezium is composed, are given by the 
 following proportions ; (Art. 9.) 
 
 R : sin N 
 
 fBNxAN : 2 arm ABN" 
 BNxCN : 2ar«xBCN 
 DNxCN : 2 arm CDN 
 
 ^DNxAN : 2 arm ADN 
 
 And by addition, (Alg. 349, Cor. 1.)* 
 
 R : 8inN::BNxAN+BNxCN+DNxCN+DNxAN: 
 2 area ABCD. 
 
 The 3d term=(AN4-CN)x(BN+DN)=ACxBD, by 
 the figure. 
 
 Therefore R : sin N :: AC X BD : 2 arm ABCD. That is, 
 ♦ Euchd, 2, 5. Cor. 
 
16 
 
 MENSURATION OF PLANE SURFACES. 
 
 As Radius, 
 
 To the sine of the angle at the intersection of the 
 
 diagonals of a trapezium ; 
 So is the product of the diagonals. 
 To twice the area of the trapezium. 
 
 It is evident that this rule is applicable to a parallelogram, 
 as well as to a trapezium. 
 
 If the diagonals intersect at right angles, the sine of IST is 
 equal to radius ; (Trig. 95.) and therefore the product of the 
 diagonals is equal to twice the area. (Alg. 356.)* 
 
 Ex. 1. If the two diagonals of a trapezium are 3*7 and 62, 
 
 and if they intersect at an angle of 54°, what is the area of 
 the trapezium ? Ans. 928. 
 
 2. If the diagonals are 85 and 93, and the angle of inter- 
 section 74°, what is the area of the trapezium ? 
 
 Problem V. 
 
 To find the area of a regular polygon. 
 
 15. Multiply one of its sides into half its perpen- 
 dicular DISTANCE FROM THE CENTRE, AND THIS PRODUCT 
 into the number of SIDES. 
 
 A regular polygon contains as many equal triangles as the 
 figure has sides. Thus, the hexagon 
 ABDFGH contains six triangles, each 
 equal to ABC. The area of one of 
 them is equal to the product of the 
 side AB, into half the perpendicular 
 CP. (Art. 8.) The area of the whole, 
 therefore, is equal to this product 
 multiplied into the 7iumler of sides. 
 
 * Euclid, 14. 5. 
 
XSHSURATION OF PLANE SURVA0K6. 17 
 
 E3C 1. What is the area of a regular octagon, in which the 
 length of a side is 60, and the perpendicular from the centre 
 72.42G? Ans. 17382. 
 
 2. What is the area of a regular decagon whose sides are 
 46 each, and the perpendicular 70.7867 ? 
 
 16. If only the length and number of sides of a regular 
 polygon be given, the perpendicular from the centre may be 
 easily found by trigonometry. The periphery of the circle 
 in which the polygon is inscribed, is divided into as many 
 equal parts as the polygon has sides. (Euc. 16.4. Schol.)* 
 The arc, of which one of the sides is a chord, is therefore 
 known ; and of course, the angle at the centre subtended by 
 this arc. 
 
 Let AB be one side of a regular polygon inscribed in the 
 circle ABDG. The perpendicular CP bisects the Imo AB, 
 and the angle ACB. (Euc. 3. 3.)t Therefore, BCP is the 
 same part of 360°, which BP is of the perimeter of the 
 polygon. Then, in the right angled triangle BCP, if BP be 
 radius, (Trig. 122.) 
 
 R : BP : : cot BCP : CP. That is. 
 
 As Radius, 
 
 To half of one of the sides of the polygon ; 
 So is the cotangent of the opposite angle. 
 To the perpendicular from the centre. 
 
 Ex. 1. If the side of a regular hexagon be 38 inches, what 
 is the area ? 
 
 The angle BCP=-iV of 360^=30°. Then, 
 
 R : 19 : : cot 30*» : 32.909=CP, the perpendicular. 
 
 And the area=10X 32.909X6=3751.6 
 
 ♦ Thomson's Lcgendre, 2. 5. Schol. f Ibid. 6. 2. 
 
18 
 
 MENSURATION OF PLANE SURFACES. 
 
 2. What is the area of a regular decagon whose sides are 
 each 62 feet ? Ans. 29576. 
 
 lY. From the proportion in the preceding article, a table 
 of perpendiculars and areas may be easily formed, for a series 
 of polygons, of which each side is a unit. Putting R=l, 
 (Trig. 100.) and w—the number of sides, the proportion be- 
 comes 
 
 360° 
 1 t -2 * • cot 1 the perpendicular 
 
 So that, the p)erp.^=i^ cot 
 
 2n 
 
 And the area is equal to half the product of the perpen- 
 dicular into the number of sides. (Art. 15.) 
 
 Thus, in the trigon, or equilateral triangle, the perpendic- 
 360° 
 
 ular=^ cot. 
 
 6 
 
 4 cot 60° = 0.2886752. 
 
 And the area=0.4330127. 
 
 In the tetragon, or square, the perpendicular=-^ cot' 
 
 360° 
 
 "8~ 
 =-^ cot 45° = 0.5. And the area=l. 
 
 In this manner, the following table is formed, in which 
 the side of each polygon is supposed to be a unit. 
 
 A TABLE OF RSGULAR POLYGONS. 
 
 Names. 
 
 Sides. Angles. 
 
 Perpendicxilars. 
 
 Areas. 
 
 Trigon, 
 
 3 
 
 60° 
 
 0.2886752 
 
 0.4330127 
 
 Tetragron, 
 
 4 
 
 45° 
 
 0.5000000 
 
 1.0000000 
 
 Pentagon, 
 
 5 
 
 36° 
 
 0.6881910 
 
 1.7204774 
 
 Hexagon, 
 
 6 
 
 30° 
 
 0.8660254 
 
 2.5980762 
 
 Heptagon, 
 
 V 
 
 25-^ 
 
 1.0382601 
 
 3.6339124 
 
 Octagon, 
 
 8 
 
 22i 
 
 1.2071069 
 
 4.8284271 
 
 Nonag'on, 
 
 9 
 
 20° 
 
 1.3737385 
 
 6:1818242 
 
 Decagon, 
 
 10 
 
 18° 
 
 1.5388418 
 
 7.6942088 
 
 Undecagon, 
 
 11 
 
 iQ-h 
 
 1.7028439 
 
 9.3656399 
 
 Dodecagon, 
 
 12 
 
 15° 
 
 1.8660252 
 
 11.1961524 
 
MENSURATION OF THE CIRCLE. ^. 1# 
 
 By this table may be calculated the area of any other reg- 
 ular polygon, of the same number of sides with one of these. 
 For the areas of similar polygons are as the squares of their 
 homologous sides. (Euc. 20. 0.)* 
 
 To find, then, the area of a regular polygon, multiply the 
 square of one of its sides by tlce area of a similar polygon of 
 which the side is a unit. 
 
 Ex. 1. What is the area of a regular decagon whose sides 
 are each 102 rods ? Ans. 80050.5 rods. 
 
 2. What is the area of a regular dodecagon whose sides 
 are each 8Y feet ? 
 
 SECTION II. 
 
 THE QUADRATURE OF THE CIRCLE AND ITS PARTS. 
 
 Art. 18. Definition I. A circle is a plane bounded by a 
 line which is equally distant in ail its parts from a point 
 within called the centre. The bounding line is called the 
 circumference or periphery. An arc is any portion of the 
 circumference. A semi-circle is half, and a quadrant one- 
 fourth of a circle. 
 
 II. A Diameter of a circle is a straight line drawn through 
 the centre, and terminated both ways by the circumference. 
 A Radius is a straight line extending from the centre to the 
 circumference. A Chord is a straidit line which joins the 
 two extremities of an ar 
 
 III. A Circular Sector is a spare coniaincd between an 
 arc and the two radii drawn from the extremities of the arc. 
 
 * Thomson's Legendre 1. 5. Cor. 
 
20 
 
 MENSURATION 01<' THE CIRCLE. 
 
 It may be less than a semi-circle, as 
 AC BO, or greater, as ACBD. 
 
 TV. A Circular Segment is the 
 space contained between an arc and 
 its chord, as ABOorABD. The chord 
 is sometimes called the base of the 
 segment. The lieight of a segment 
 is the perpendicular from the middle 
 of the base to the arc, as PO. 
 
 t V. A Circular Zone is the space 
 between two parallel chords, as 
 AGHB. It is called the middle 
 zone, when the two chords are 
 equal, as GHDE. 
 
 VI. A Circular Ring is the space between the peripheries 
 of two concentric circles, as A A', BB'. (Fig. 13.) 
 
 VII. A Lune or Crescent is the space between two circu- 
 lar arcs which intersect each other, as ACBD. (Fig. 14.) 
 
 19. The Squaring of the Circle is a problem which has 
 exercised the ingenuity of distinguished mathematicians for 
 many centuries. The result of their efforts has been only 
 an approximation to the value of the area. This can be car- 
 ried to a degree of exactness far beyond what is necessary 
 for practical purposes. 
 
MXirSURATION OF TBE CIRCLE. 21 
 
 f 
 
 20. If the circumference of a circle of given diameter 
 were known, its area could be easily found. l\or the area is 
 equal to the product of half the circumference into half the 
 diameter. (Sup. Euc. 5, l.*)t But the circumference of a 
 circle has never been exactly determined. The method of 
 approximating to it is by inscribing and circumscribing poly- 
 gons, or by some process of calculation which is, in principle, 
 the same. The perimeters of the polygons can be easily 
 and exactly determined. That which is circumscribed is 
 greater, and that which is inscribed is less, than the peri- 
 phery of the circle ; and by increasing the number of sides, 
 the difference of the two polygons may be made less than 
 any given quantity. (Sup. Euc. 4, 1.) 
 
 21. The side of a hexagon in- 
 scribed in a circle, as AB, is the 
 chord of an arc of 60°, and there- 
 fore equal to the radius. (Trig. 95.) 
 The chord of half this arc, as BO, 
 is the side of a polygon of 12 equal 
 sides. By repeatedly bisecting the 
 arc, and finding the chord, we may 
 
 obtain the side of a polygon of an immense number of sides. 
 Or we may calculate the sine, which will be half the chord 
 of double the arc, (Trig. 82, cor.,) and the tangent, which 
 will be half the side of a similar circumscribed polygon. 
 Thus the sine AP, is half of AB, a side of the inscribed 
 hexagon ; and the tangent NO is half of NT, a side of the 
 circumscribed hexagon. The difference between the sine 
 and the arc AO is less than the difference between the sine 
 and the tangent. In the section on the computation of the 
 canon, (Trig. 223.) by 12 successive bisections, begi^ping 
 with 60 degrees, an arc is obtained which is the ^ i\ii of 
 the whole circumference. 
 
 * In this manner, the SupplemefU to Play/air's Euclid is referred to in 
 this work. t TtioinMn's Legendre, 11. 5. 
 
22 MENSURATION OF THE CIRCLE. 
 
 The cosine of this, if radius be 1, is found to be .99999996'732 
 The sine is .00025566546 
 
 And the tangent=i^(Trig. 93.) =.00025566347 
 cosine 
 
 The diff. between the sine and tangent is only .00000000001 
 
 And the diflference between the sine and the arc is still less. 
 
 Taking then, .000255663465 for the length of the- arc, 
 multiplying by 24576, and retaining 8 places of decimals, 
 we have 6.28318531 for the whole circumference, the radius 
 being 1. Half of this, 
 
 3.14159265 
 
 is the circumference of a circle whose radius is -^j and diam- 
 eter 1. 
 
 22. If this be multiplied by 7, the product is 21.99+or 
 22 nearly. So that, 
 
 Diam : Circum : : Y : 22, nearly. 
 
 If 3.14159265 be multiplied by 113, the product is 
 354.9999+, or 355, very nearly. So that, 
 
 Diam ; Circum ; : 113 : 355, very nearly. 
 
 The first of these ratios was demonstrated by Archimedes. 
 
 There are various methods, principally by infinite series 
 and fluxions, by which the labor of carrying on the approx- 
 imation to the periphery of a circle may be very much 
 abridged. The calculation has been extended to nearly 150 
 places of decimals. But four or five places are sufl&cient 
 for most practical purposes. 
 
 After determining the ratio between the diameter and the 
 circumference of a circle, the following problems are easily 
 solved. 
 
MENSURATION OF THE CIRCUS. 2S 
 
 Problem I. 
 To find the circitmference of a circle from its diameter. 
 
 23. Multiply THE diameter by 3*1 41 59*^ 
 
 Or, 
 
 Multiply the diameter hy 22 and divide the product hy *l. Or, 
 multiply the diameter by 355, and divide the product by 
 113. (Art. 22.) 
 
 Ex. 1. If the diameter of the earth be ^930 miles, what 
 is the circimiference ? Ans. 249128 miles. 
 
 2. How many miles does the earth move, in revolving 
 round the sun ; supposing the orbii to be a circle whose 
 diameter is 190 million miles ? Ans. 596,902,100. 
 
 8. What is the circumference of a circle whose diameter 
 is V69843 rods ? 
 
 Problem II. 
 To find the diameter of a circle from its circumference, 
 
 24. Divide the circumference by 3.1 41 59* 
 
 Or, 
 
 Multiply the circumference by Y, and divide the product by 
 22. Or, multiply the circumference by 113, and divide 
 the product by 355. (Art. 22.) 
 
 Ex. 1. If the circumference of the sun be 2,800,000 miles, 
 what is his diameter? Ans. 891,267. 
 
 2. What is the diameter of a tree which is 5^ feet round ? 
 
 25. As multiplication is more easily performed than divis- 
 ion, there will be an advantage in exchanging the divisor 
 
 * In many caset, 3.1416 will be sufficiently accurate. 
 
24 MENSURATION OF THE CIRCLE. 
 
 3.14159 for a multiplier which ■will give the same result. 
 In the proportion 
 
 3.14159 : 1 : : Circum : Diam. 
 
 to find the fourth term, we may divide the second by the 
 first, and multiply the quotient into the third. Xow, 1-^ 
 3.14159=0.31831. If, then, the circumference of a circle 
 be multiplied by .31831, the product will be the diameter. 
 
 Ex. 1. If the circumference of the moon be 6850 miles, 
 what is her diameter ? Ans. 2180. 
 
 2. If the whole extent of the orbit of Saturn be 5650 
 million miles, how far is he from the siin ? 
 
 3. If the periphery of a wheel be 4 feet 7 inches, what is 
 its diameter? 
 
 Problem III. 
 To find the length of an arc of a circle. 
 
 26. As 360°, to the number of degrees in the arc ; 
 
 So is the circumference of tlie circle, to the length of the arc. 
 
 The circumference of a circle being divided into 360°jflH|L 
 (Trig. 73.) it is evident that the length of an arc of any less 
 number of degrees must be a proportional part of the whole. 
 
 Ex. What is the length of an arc of 16°, in a circle whose 
 radius is 50 feet ? 
 
 The circumference of the circle is 314.159 feet. (Art. 23.) 
 
 Then 360 : 16 : : 314.159 : 13.96 feet. 
 
 2. If we are 95 millions of miles from the sun, and if the 
 earth revolves round it in 365i days, how far are we carried 
 in 24 hours? Ans. 1 million 634 thousand miles. 
 
 27. The length of an arc may also be found, by multiply- 
 ing the diameter into the number of degrees in the arc, and 
 
MBNSURATION OF THE CIRCIJB. 
 
 25 
 
 this product into .0087266, which is the length of one de- 
 gree, in a circle whose diameter is 1. For 3,14169-r360= 
 0.00872 60. And in different circles, the circumferences, and 
 of course the degrees, are as the diameters. (Sup. Euc. 8, 1.)* 
 
 Ex. 1. What is the length of an arc of 10° 16' in a circle 
 whose radius is 68 rods ? Ans. 12.165 rods. 
 
 2. If the circumference of the earth be 24913 miles, what 
 is the length of a degree at the equator ? 
 
 28. The length of an arc is frequently required, when 
 the number of degrees is not given. But if the radius of the 
 circle, and either the cliord or the fieight of the arc, be 
 known ; the number of degrees 
 may be easily found. 
 
 Let AB be the chord, and PO 
 the height, of the arc AOB. As 
 the angles at P are right angles, and 
 AP is equal to BP; (Art. 18. Def. 
 4.) AO is equal to BO. (Euc. 4, 
 l.)t Then, 
 
 BP is the sine, and CP the cosine, ) ^^ ^^ . ^j^^ ^^ ^^^ 
 OP the versed sine, and BO the chord, ) 
 
 And in the right angled triangle GBP, 
 
 PR ♦ R i ^1* • sin BCP or BO. 
 * i CP : cosBCPorBO. 
 
 Ex. 1. If the radius CO=25, and the chord AB=43.3; 
 what is the length of the arc AOB ? 
 
 CB : R : : BP : sin BCP or BO— 60'' very nearly. 
 
 The circumference of the circle -"3.14159 X50-»167.08. 
 
 Ajid 360° : 60° : : 157.08 : 26.18=OB. Therefore,AOB=62.36. 
 
 * Thomson's Legendre, 10. b. 
 
 t Ibid., 5. 1. 
 
26 
 
 MENSURATION OF THE CIRCLE. 
 
 / 2. What is the length of an arc whose chord is 216^, in a 
 circle whose radius is 126? Ans. 261.8. 
 
 29. If only the chord and the height of an arc be given, 
 the radius of the circle may be 
 found, and then the length of the 
 arc. 
 
 If BA be the chord, and PO the 
 the height of the arc AOB, then 
 (Euc. 35. 3.)* 
 
 DP=Jl_. And DO=OP+DP=OP+S--' 
 OP OP 
 
 That is, the diameter is equal to the height of the arc, + 
 the square of half the chord divided by the height. 
 
 The diameter being found, the length of the arc may be 
 calculated by the two preceding articles. 
 
 Ex. 1. If the chord of an arc be 173.2, and the heischt 50, 
 what is the length of the arc ? 
 
 The diameter =50+5^=200. The arc contains 120^ 
 50 
 
 (Art. 28.) and its length is 209.44. (Art. 26.) 
 
 2. What is the length of an arc whose chord is 120, and 
 height 45? Ans. 160.8. 
 
 Problem IV. 
 
 To find the area of a circle. 
 
 30. Multiply the square of the diameter by the 
 decimals .7854. 
 
 * Thomson's Legendre, 10, 5. 
 
lOBMntATION OF THE CIRCLE. %% 
 
 Or, 
 
 Multiply half the diameter into half the circum- 
 ference. Or, multiply the whole diameter into the whole 
 circumference, and take -^ of the product. 
 
 The area of a circle is equal to the product of half the 
 diameter into half the circumference ; (Sup. Euc, 5, 1.) or, 
 which is the same thing, ^ the product of the diameter and 
 circumference. If the diameter be 1, the circumference is 
 3.14159; (Art. 23.) one-fourth of which is 0.7854 nearly. 
 But the areas of different circles are to each other, as the 
 squares of their diameters. (Sup. Euc. Y, 1.)* The area of 
 any circle, therefore, is equal to the product of the square of 
 its diameter into 0.7854, which is the area of a circle whose 
 diameter is 1. 
 
 Ex. 1. What is the area of a circle whose diameter is 623 
 feet^. Ans. 304836 square feet. 
 
 2. How many acres are there in a circular island whose 
 diameter is 124 rods. Ans. 75 acres, and 76 rods. 
 
 3. If the diameter of a circle be 113, and the circumfer- 
 ence 355, what is the area ? Ans. 10029. 
 
 4. How many square yards are there in a circle whose 
 diaraelcr is 7 feet ? 
 
 81. If the circumference of a circle be given, the area may 
 be obtained, by fir- the diameter ; or, without finding 
 
 the diameter, by mu ^ y - the square of the circumference 
 by .07958. 
 
 For, if the circumference of a circle be 1, tlie diameter = 
 1-^3.14159=0.31831 ; and | the product of this into the 
 circumference is .07958 the area. But the areas of different 
 circles, being as the squares of their diameters, are also as 
 the squares of their circumferences. (Sup. Euc. 8, 1.) 
 
 • Thomson's Legendre, 28. 4. Cor. 
 
28 
 
 MENSURATION OF THE CIRCLE. 
 
 Ex. 1. If the circumference of a circle be 136 feet, what 
 is the area ? Ans. 1472 feet. 
 
 2. What is the surface of a circular fish-pond, which is 
 10 rods in circumference ? 
 
 32. If the area of a circle be given, the diameter may be 
 found, by dividing the area by .7854, and extracting the 
 square root of the quotient. 
 
 This is reversing the rule in Art. 30. 
 
 Ex. 1. What is the diameter of a circle whose area is 
 380.1336 feet? 
 
 Ans. 380.1336H-.'7854 = 484. And V4'84 = 22. 
 
 2. What is the diameter of a circle whose area is 19.635 ? 
 
 33. The area of a circle, is to the area of the circumscribed 
 square ; as .7854 to 1 ; and to that of the inscribed square 
 as .7854 to i. 
 
 Let ABDF be the inscribed 
 square, and LMNO the circum- 
 scribed square, of the circle ABDF. 
 The area of the circle is equal to 
 AD"' X. 7854. (Art. 30.) But the 
 area of the circumscribed square 
 
 (Art. 4.) is equal to ON^=AD^. 
 And the smaller square is half of 
 the larger one. For the latter con- 
 tains 8 equal triangles, of which the former contains only 4. 
 
 Ex. What is the area of a square inscribed in a circle 
 whose area is 159 ? Ans. .7854 : i : : 159 : 101.22. 
 
 c \^ 
 
 12 
 
 Problem V. 
 To find the area of a sector of a circle. 
 34. Multiply the radius into half the length of 
 
 THE ARC. 
 
MENSURATION OF THE CIRCLE. 
 
 Or, 
 
 As 360, TO THE WUMBKR OF DEGREES IN THE ARC ,* 
 So IS THE AREA OF THE CIRCLE, TO THE AREA OF THE 
 SECTOR. 
 
 It is evident, that the area of the sector has the same 
 ratio to the area of the circle, which the length of the arc 
 has to the length of the whole circumference ; or which the 
 number of degrees in the arc has to the number of degrees 
 in the circumference. 
 
 Ex. 1. If the arc AOB be 120°, 
 and the diameter of the circle 226 ; 
 what is the area of the sector 
 AOBC? 
 
 The area of the whole circle is 
 40116. (Art. 30.) 
 
 And 360° : 120° : : 40115 : 
 133 71 1, the area of the sector. 
 
 2. AVhat is the area of a quadrant whose radius is 621 ? 
 
 3. What is the area of a semi-circle, whose diameter is 
 328? 
 
 4. What is the area of a sector which is less than a semi- 
 circle, if the radius be 15, and the chord of its arc 12 ? 
 
 Half the chord is the sine of 23° 34|' nearly. (Art. 28.) 
 
 The whole arc, then, is 47° 9^' 
 
 The area of the circle is 706.86 
 
 And 360° : 47° 9i' : : 706.86 : 92.6 the area of the sector. 
 
 5. If the arc'ADB be 240 degrees, and the radius of the 
 circle 113, what is the area of the sector ADBC ? 
 
 Problem VI. 
 To finl the area of a segment of a circle. 
 2o. Find the aeba of the sector which has the 
 
80 MENSURATION" OF THE CIRCLE. 
 
 SAME ARC, AND ALSO THE AREA OF A TRIANGLE FORMED BY 
 THE CHORD OF THE SEGMENT AND THE RADII OF THE SEC- 
 TOR. 
 
 Then, if the segment be less than a semi-circle, 
 subtract the area of the triangle from the area of 
 the sector. but, if the segment be greater than a 
 
 SEMI-CIRCLE, ADD THE AREA OF THE TRIANGLE TO THE AREA 
 OF THE SECTOR. 
 
 If the triangle ABC, be taken 
 from the sector AOBC, it is evi- 
 dent the difference will be the seg- 
 ment AOBP, less than a semi-cir- 
 cle. And if the same triangle be 
 added to the sector ADBC, the 
 sum will be the segment ADBP, 
 greater than a semi-circle. 
 
 The area of the triangle (Art. 8.) 
 is equal to the product of half the chord AB into CP, which 
 is the difference between the radius and PO the height of the 
 segment. Or CP is the cosine of half the arc BOA. If this 
 cosine and the chord of the segment are not given, they 
 may be foimd from the arc and the radius. 
 
 Ex. 1. If the arc AOB be 120°, and the radius of the 
 circle be 113 feet, what is the area of the segment AOBP ? 
 
 In the right angled triangle BCP, 
 
 R : BC : : sin BCO : BP=97.86, half the chord. (Art. 28.) 
 
 The cosine PC=i CO (Trig. 96, Cor.) =56.5 
 
 The area of the sector AOBC (Art. 34.) =13371.67 
 
 The area of the triangle ABC=BPxPC = 5528.97 
 
 The area of the segment, therefore, = 7842.7 
 
 2. If the base of a segment, less than a semi-circle, be 10 
 
MENSURATION OF THE CIRCLE. $X 
 
 feet, and the radius of tlie circle 12 feet, what is the area 
 of the segment ? 
 
 The arc of the segment contains 49-J- degrees. (Art. 28.) 
 The area of the sector =61.89 (Art. 34.) 
 
 The area of the triangle =54.54 
 
 And the area of the segment = 7.35 square feet. 
 
 3. What is the area of a circular segment, whose height 
 is 19.2 and base 70 ? Ans. 947.86. 
 
 4. What is the area of the segment ADBP, (Fig. 9.) if 
 the base AB be 195.7, and the height PD 169.5 ? 
 
 Ans. 32272. j 
 
 36. The area of any figure which is bounded partly by 
 arcs of circles, and partly by right lines, may be calculated, 
 by finding the areas of the segments under the arcs, and then 
 the area of the rectilinear space between the chords of the 
 arcs and the other right lines. 
 
 Thus, the Gothic arch ACB, 
 contains the two segments 
 AC II, BCD, and the plane tri- 
 angle ABC. 
 
 Ex. If AB be 110, each of 
 the fines AC and BC 100, and 
 the height of each of the seg- 
 ments ACH, BCD 10.435; 
 what is the area of the whole figure ? 
 
 The areas of the two segments are 1404 
 
 The area of the triangle ABC is 4593.4 
 
 And the whole figure is 5997.4 
 
32 
 
 MENSURATION OF THE CIRCLE. 
 
 Problem VII. 
 To find the area of a circular zone, 
 37. From the area of the whole circle, subtract 
 
 THE TWO segments ON THE SIDES OF THE ZONE. 
 
 If from the whole circle there be taken the two segments 
 ABC and DFH, there will remain 
 the zone ACDH. 
 
 Or, the area of the zone may be 
 found by subtracting the segment 
 ABC from the segment HBD : Or, 
 by adding the two small segments 
 GAH and VDC, to the trapezoid 
 ACDH. (See Art. 36.) 
 
 The latter method is rather the 
 most expeditious in practice, as the two segments at the end 
 of the zone are equal. 
 
 Ex. 1. What is the area of the zone ACDH, if AC is 
 7.75, DH 6.93, and the diameter of the circle 8 ? 
 
 The area of the whole circle is 
 of the segment ABC 
 of the segment DFH 
 of the zone ACDH 
 
 17.32 
 
 9.82 
 
 50.26 
 
 23.12 
 
 2. What is the area of a zone, one side of which is 23.25, 
 and the other side 20.8, in a circle whose diameter is 24 ? 
 
 Ans. 208. 
 
 38. If the diameter of the circle is not given, it may be 
 found from the sides and the breadth of the zone. 
 
 Let the centre of the circle be at 0. Draw ON perpen- 
 dicular to AH, NM perpendicular to LR, and HP perpen- 
 dicular to AL. Then, 
 
MENSURATION OF THE CIRCLE. 33 
 
 AN=iAH, (Euc. 3. 3.)* MN=KLA+RH) 
 LM=iLR, (Euc. 2. 6.)t PA=LA— RH. 
 
 The triangles API! and OMN are siniilar, because the 
 sides of one are perpendicular to those of the other, each to 
 each. Therefore, 
 
 PII : PA : : MN : MO 
 MO being found, we have ML — MO=OL. 
 
 And the radius CO=vOL»+CL*. (Euc. 41. l.)l 
 
 Ex. If the breadth of the zone ACDH (Fig. 12.) be 6.4, 
 and the sides 6.8 and 6 ; what is the radius of the circle ? 
 
 PA=3.4— 3=0.4. And, MN=i(3.4+3)=3.2. 
 Then, 6.4 : 0.4 : : 3.2 : 0.2=M,O. And, 3.2— 0.2=3=OL 
 And the radius CO=V3'4-(3.4)'=4.534. 
 
 ^^ Problem VIII. 
 
 To find the area of a lune or crescent. 
 39. Find the difference of the two segments which 
 
 ARE between the ARCS OF THE CRESCENT AND ITS CHORD. 
 
 If the segment ABC, be 
 taken from the segment ABD ; 
 there will remain the lune or 
 crescent ACBD. 
 
 Ex. If the chord AB be 8B, 
 the height CH 20, and the 
 height DH 40 ; what is the 
 area of the crescent ACBD ? 
 
 The area of the segment ABD is 2698 
 
 of the segment ABC 1220 
 
 of the crescent ACBD 1478 
 
 • Thomson's Lcgendre, 6. 2. f ^^ 1^- 4. t ^^'^ ^l- 4* 
 2» 
 
84 MENSURATION OF THE CIRCLE. 
 
 Problem IX. 
 
 To find the area of a ring, included between the periphe- 
 ries of two concentric circles. 
 
 40. Find tsk difference op the areas of the two 
 
 CIRCLES. 
 
 Or, 
 
 Multiply the product of the sum and diflference of the two 
 diameters by .TSS^. 
 
 The area of the ring is evidently equal to the diflference 
 between the areas of the two cir- 
 cles AB and A'B'. 
 
 But the area of each circle is equal 
 to the square of its diameter multi- 
 plied into .78.54. (Art. 30.) And 
 the difference of these squares is equal 
 to the product of the sum and dif- 
 ference of the diameters. (Alg. 191.) Therefore the area of 
 the ring is equal to the product of the sum and diflference 
 of the two diameters multiplied by .'7854. 
 
 Ex. 1. If AB be 221, and A'B' 106, what is the area of 
 the ring? Ans. (22p X. 7854)— (106= X.'7854) = 29535. 
 
 2. If the diameters of Saturn's larger ring be 205,000 
 and 190,000 miles, how many square miles are there on one 
 side of the ring ? ^ 
 
 Ans. 395000X15000X. 7854=4,653,495,000. 
 
 PROMISCUOUS EXAMPLES OF AREAS. 
 
 Ex. 1. What is the expense of paving a street 20 rods 
 long and 2 rods wide, at 5 cents for a square foot ? 
 
 Ans. 544-^ dollars. 
 
MKN8URATI0N OF THE CIRCLE. 35 
 
 2. If an equilateral triangle contains as many square feet 
 as there are inches in one of its sides ; what is the area of 
 the trijingle ? 
 
 Let ;2;=the number of square feet in the area. 
 
 X 
 
 Then— •= the number of linear feet in one of the sides. 
 
 And, (Art. 11.) ar=i/-:^yx V3=_^XV3. 
 \12/ 676 
 
 Reducing the equation, a;=_ =332.55 the area. 
 V3 
 
 3. What is the side of a square whose area is equal to that 
 of a circle 452 feet in diameter? 
 
 Ans. V(452)»X. 7854 = 400.574. (Arts. 30 and 7.) 
 
 4. What is the diameter of a circle which is equal to a 
 square whflie side is 36 feet ? 
 
 Ans. V('307ToT7854^4O.62l7. (Arts. 4 and 32.) 
 
 5. What is the area of a square inscribed in a circle whose 
 diameter is 132 feet ? 
 
 /gf^^' 8712 square feet. (Art. 33.) 
 
 6. How much carpeting, a yard wide, will be necessary to 
 cover the floor of a room which is a regular octagon, the 
 sides being eight feet each ? Ans. 34f yards. 
 
 7. If the diagonal of a square be 16 feet, what is the 
 area? Ans. 128 feet. (Art. 14.) 
 
 8. If a carriage-wheel four feet in diameter revolve 300 
 times, in going round a circular green ; what is the area of 
 the green ? 
 
 Ans. 4164i sq. rods, or 26 acres, 3 qrs. and 34^ rods. 
 
 9. What will be the expense of papering the sides of a 
 room, at 10 cents a square yard ; if the room be 21 feet long, 
 
3G 
 
 * MENSURATION OF THE CIRCLE. 
 
 4 
 
 18 feet broad, and 12 feet high ; and if there be deducted 3 
 windows, each 5 feet by 3, two doors 8 feet by 4^, and one 
 fire-place 6 feet by 4i ? Ans. 8 dollars 80 cents. 
 
 10. If a circular pond of water 10 rods in diameter be 
 surrounded by a gravelled walk 8} feet wide ; what is the 
 area of the walk? Ans. 16^ sq. rods. (Art. 40.) 
 
 11. If CD, the base of the 
 isosceles triangle VCD, be 60 
 feet, and the area 1200 feet; 
 and if there be cut off, by the 
 line LG parallel to CD, the tri- 
 angle VLG, whose area is 432 
 feet ; what are the sides of the 
 latter triangle ? 
 
 Ans. 30, 30, and 36 feet. 
 
 12. What is the area of an equilateral triangle inscribed 
 in a circle whose diameter is 52 feet ? 
 
 Ans. 878.15 sq. ft. 
 
 13. If a circular piece of land is inclosed by a fence, in 
 which 10 rails make a rod in length; and if the field con- 
 tains as many square rods, as there are rails in the fence ; 
 what is the value of the land at 120 dollars an acre ? 
 
 Ans. 942.48 dollars. 
 
 14. If the area of the equilat- 
 eral triangle ABD be 219.5375 
 feet ; what is the area of the cir- 
 cle OBDA, in which the triangle 
 is inscribed? 
 The sides of the triangle are each 
 
 22.5167. (Art. 11.) 
 
 And the area of the circle is 630.93. 
 
MENSURATION OF SOLIDS. * 87 
 
 15. If 6 concentric circles are so drawn, that the space 
 between the least or 1st, and the 2d is 21.2058, 
 
 between the 2d and the 3d is 35.343, 
 
 between the 3d and the 4th is 49.4802,' 
 
 between the 4th and the 5th is 63.6174, 
 
 between the 5th and the 6th is V7.Y546; 
 
 what Ihre the several diameters, supposing the longest to be 
 
 equal to 6 times the shortest ? 
 
 Aus. 3, 6, 0, 12, 15, and 18. 
 
 16. If the area between two concentric circles be 1202.64 
 square inches, and the diameter of the lesser circle be 19 
 uiche8»t>what is the diameter of the other ? 
 
 17. What is the area of a circular segment, whose height 
 is 9, and base 24? 
 
 SECTION III. 
 
 SOLIDS BOUNDED BY PLANE SURFACES. 
 
 Abt. 41. DEFINr^(0||^ I. A prism is a solid bounded by 
 plane figures or faces, two of which are parallel, similar, and 
 equal ; and the others are parallelograms. 
 
 II. The parallel planes are sometimes called the bases or 
 ends; and the other figures the sides of the prism. The 
 latter taken together constitute the lateral surface, 
 
 III. A prism is riffht or oblique, according as the sides are 
 perpendicular or oblique to the bases. 
 
 IV. The height of a prism is the perpendicular distance 
 between the planes of the bases. In a right prism, there- 
 fore, tLe height is equal to the length of one of the sides. 
 
 V. A Parallelopiped is a |)ri8ni whose bases are parallelo- 
 grams. 
 
38 
 
 MENSURATION OF SOLIDS. 
 
 VI. A Cuhe is a solid bounded by six equal squares. It 
 is a right prism whose sides and bases are all equal. 
 
 VII. A Pyramid is a solid bounded by a plane figure 
 called the base, and several triangular planes, proceeding from 
 the sides of the base, and all terminating in a single point. 
 These triangles taken together constitute the lateral surface. 
 
 VIII. A pyramid is reguUar, if its base is a regular poly- 
 gon, and if a line from the centre of the base to the vertex 
 of the pyramid is perpendicular to the base. This line is 
 called the axis of the pyramid. 
 
 IX. The height of a pyramid is the perpendicular distance 
 from the summit to the plane of the base. In a regular pyr- 
 amid, it is the length of the axis. 
 
 X. The slant-height of a regular pyramid, is the distance 
 from the summit to the middle of one of the sides of the base. 
 
 XI. A frustum or trunk of a pyramid is a portion of the 
 solid next the base, cut off by a plane parallel to the base. 
 The height of the frustum is the perpendicular distance 
 of the two parallel planes. The slant-height of a frus- 
 tum of a regular pyramid, is the distance from the middle of 
 one of the sides of the base, to the middle of the corres- 
 ponding side in the plane above. It is a line passing on 
 the surface of the frustum, through the middle of one of 
 its sides. 
 
 XII. A Wedge is a soUd of five sides, viz. a rectangular 
 base, two rhomboidal 
 
 sides meeting in an 
 edge, and two tri- 
 angular ends ; as 
 ABHG. The base 
 is ABCD, the sides 
 are ABHG and 
 DCHG, meeting in 
 the edge GH, and 
 the ends are BOH and ADG. The height of the wedge is a 
 
MSNSURATION OF SOLIDS. 39 
 
 perpendicular drawn from any point in the edge, to the plane 
 of the base, as GP. 
 
 XIII. A Prismoid is a solid whose ends or bases are par- 
 allel, but not similar, and whose sides are quadrilateral. It 
 differs from a prism or a frustum of a pyramid, in having its 
 ends dissimilar. It is a rectangular prismoid, when its ends 
 are right parallelograms. 
 
 XIV. A linear side or edge of a solid is the line of intersec- 
 tion of two of the planes which form the surface. 
 
 42. The common measuring unit of solids is a cube, whose 
 sides are squares of the same name. The sides of a cubic 
 inch are square inches ; of a cubic foot, square feet, &c. 
 Finding the capacity, solidity* or solid contents of a body, is 
 fmding the number of cubic measures, of some given de- 
 nomination contained in the body. 
 
 In solid mectsure. 
 
 1728 cubic incha =1 cubic foot, 
 2*7 cubic feet =1 cubic yard, 
 4492i cubic feet =1 cubic rod, 
 32768000 cubic rods =1 cubic mile, 
 282 cubic inches =1 ale gallon, 
 231 cubic inches =1 wine gallon, 
 2150.42 cubic inches =1 bushel, 
 
 1 cubic foot of pure water weighs 1000 
 avoirdupois ounces, or 62i pounds. 
 
 Problkm I. 
 To find the soLmnr of a prism. 
 
 48. MlTLTIPLY THE AREA OF THE BASE BY THE HEIGHT. 
 
 This is a general rule, applicable to parallelopipeds 
 whether right or obhque, cubes, triangular prisms, <fec. 
 
 ♦ Sec note A. 
 
40 MENSURATION OF SOLIDS. 
 
 As surfaces are measured, by comparing them with a right 
 parallelogram (Art. 3.) ; so solids are measured, by com- 
 paring them with a right parallelepiped. 
 
 If ABCD be the base of a right ^ 
 parallelopiped, as a stick of timber 
 standing erect, it is evident that the 
 number of cubic feet contained in one 
 foot of the height, is equal to the 
 mimber of square feet in the area of 
 the base. And if the sohd be of any 
 
 other height, instead of one foot, the contents must have the 
 same ratio. For parallelepipeds of the same base are to 
 each other as their heights. (Sup. Euc. 9. 3.)* The solidity 
 of a right parallelopiped, therefore, is equal to the product 
 of its length, breadth, and thickness. See Alg. 397. 
 
 And an oblique parallelopiped being equal to a right one 
 of the same base and altitude, (Sup. Euc. 7. 3)f is equal to 
 the area of the base multiplied into the perpendicular height. 
 This is true also of prisms, whatever be the form of their 
 bases. (Sup. Euc. 2. Cor. to 8, 3. Thomson's Legendre, 12. 7.) 
 
 44. As the sides of a cube are all equal, the solidity is 
 found by cubing one of its edges. On the other hand, if the 
 solid contents be given, the length of the edges may be 
 found, by extracting the cube root. 
 
 45. When solid measure is cast by Duodecimals, it is to 
 be observed that inches are not primes of feet, but thirds. 
 If the unit is a cubic foot, a solid which is an inch thick and 
 a foot square is a prime ; a parallelopiped a foot long, an 
 inch broad, and an inch thick is a second, or the twelfth part 
 of a prime ; and a cubic inch is a third, or the twelfth part 
 of a second. A linear inch is -i\ of a foot, a square inch -ri t 
 of a foot, and a cubic inch tt^-s of a foot. 
 
 * Thomson's Legendre, 9. 7. f Ibid., 7. 7. 
 
MKNSURATION OP SOLIDS. 41 
 
 Ex. 1. What are the solid contents of a stick of tlmbe^ 
 which is 31 feet long, 1 foot 3 inches broad, and 9 inches 
 thick? Ans. 29 feet 9", or 29 feet 108 inches. 
 
 2. What is the solidity of a wall which is 22 feet long, 12 
 feet high, and 2 feet 6 inches thick ? 
 
 Ans. 660 cubic feet. 
 
 3. What is the capacity of a cubical vessel which is 2 feet 
 3 inches deep ? 
 
 Ans. 11 F. 4' 8" 3'", or 11 feet 6*75 inches. 
 
 4. If the base of a prism be 108 square inches, and the 
 height 30 feet, what are the solid contents ? 
 
 Ans. 27 cubic feet. 
 
 6. If the height of a square prism be 2\ feet, and each 
 side of the base 10^ feet, what is the solidity ? 
 
 The area of the base = 10ixl0i==106f sq. feet. 
 And the solid contents = 106iX 2-i-=240i cubic feet. 
 
 6. If the height of a prism be 23 feet, and its base a reg- 
 ular pentagon, whose perimeter is 18 feet, what is the so- 
 lidity ? Ans. 512.84 cubic feet. 
 
 46. The number of gallons or bushels which a vessel will 
 contain may be found, by calculating the capacity in inches, 
 and then dividing by the number of inches in 1 gallon or 
 bushel. 
 
 The weight of water in a vessel of given dimensions is 
 easily calculated ; as it is found by experiment, that a cubic 
 foot of pure water weighs 1000 ounces avoirdupois. For 
 the weight in ounces, then, multiply the cubic feet by 1000 ; 
 or for the weight in pounds, multiply by 62^. 
 
 Ex. 1. How many ale gallons are there in a cistern whicli 
 is 1 1 feet 9 inches deep, and whose base is 4 feet 2 inches 
 square? 
 
42 MENSURATION OF SOLIDS. 
 
 •" The cistern contains 352500 cubic inches; 
 
 And 352500-7-282 = 1250. 
 
 2. How many wine gallons will fill a ditch 3 feet 11 inches 
 wide, 3 feet deep, and 462 feet long ? Ans. 40608. 
 
 3. What weight of water can be put into a cubical vessel 
 4 feet deep ? Ans. 4000 lbs. 
 
 Problem II. 
 To find the lateral surface of a right prism. 
 4*7. Multiply the length into the perimeter of the 
 
 BASE. 
 
 Each of the sides of the prism is a right parallelogram, 
 whose area is the product of its length and breadth. But 
 the breadth is one side of the base; and therefore, the sum 
 of the breadths is equal to the perimeter of the base. 
 
 Ex. 1. If the base of a right prism be a regular hex- 
 agon whose sides are each 2 feet 3 inches, and if the height 
 be 16 feet, what is the lateral surface ? 
 
 Ans. 216 square feet. 
 
 If the areas of the two ends be added to the lateral sur- 
 face, the sum will be the whole surface of the prism. And 
 the superficies of any solid bounded by planes, is evidently 
 equal to the areas of all its sides. 
 
 2. If the base of a prism be an equilateral triangle 
 whose perimeter is 6 feet, and if the height be 17 feet, what 
 is the surface ? 
 
 The area of the triangle is 1.732. {Art. 11.) 
 And the whole surface is 105.464. 
 
MENSURATION OF SOLIDS. 49 
 
 Problem III. 
 To find the soLiDirr of a pyramid. 
 48. Multiply the area of the base into \ of the 
 
 HEIOUT. 
 
 The solidity of a prism is equal to the product of the 
 area of the base into the height. (Art. 43.) And a pyramid 
 is -^ of a prism of the same base and altitude. (Sup. Euc. 
 15, 3. Cor. 1.)* Therefore the solidity of a pyramid 
 whether right or oblique, is equal to the product of the base 
 into \ of the perpendicular height. 
 
 Ex. 1. What is the solidity of a triangular pyramid, 
 whose height is 60, and each side of whose base is 4 ? 
 
 The area of the base is 6.928 
 And the solidity is 138.56. 
 
 2. Let ABC be one side of an oblique 
 pyramid whose base is 6 feet square ; 
 let BC be 20 feet, and make an angle 
 of 70 degrees with the plane of the 
 base ; and let CP be perpendicular to this 
 plane. What is the solidity of the pyr- 
 amid ? 
 
 In the right angled triangle BCP, (Trig. 134.) 
 
 R : BC : : sinB :: PC = 18.'79. 
 
 And the solidity of the pyramid is 225.48 feet. 
 
 3. What is the solidity of a pyramid whose perpendicular 
 height is 72, and the sides of whose base are 67, 54, and 
 40? Ans. 25920. 
 
 * Thomson's Legendre, 15 and 18. 7. 
 
44 MENSURATION OP SOLIDS. 
 
 Problem IV. 
 To find the lateral surface of a regular pyramid. 
 
 49. Multiply half the slant-height into the perim- 
 eter OF the base. 
 
 Let the triangle ABC be one of 
 the sides of a regular pyramid. As 
 the sides AC and BC are equal, the 
 angles A and B are equal. Therefore 
 a line drawn from the vertex C to the 
 middle of AB is 'perpendicular to AB. 
 The area of the triangle is equal to 
 the product of half this perpendicular into AB. (Art. 8.) 
 The perimeter of the base is the sum of its sides, each of 
 which is equal to AB. And the areas of all the equal tri- 
 angles which constitute the lateral surface of the pyramid, 
 are together equal to the product of the perimeter into half 
 the slant-height CP. 
 
 The slant-height is the hypothenuse of a right angled tri- 
 angle, whose legs are the axis of the pyramid, and the dis- 
 tance from the centre of the base to the middle of one of the 
 sides. See Def. 10. 
 
 Ex. 1. What is the lateral surface of a regular hexagonal 
 pyramid, whose axis is 20 feet, and the sides of whose base 
 are each 8 feet ? 
 
 The square of the distance from the centre of the base to 
 one of the sides. (Art. 16.) =48. 
 
 The slant-height (Euc. 47. l.)*=:V48-f (20)'=21.16 
 
 And the lateral surface==21.16X 4X6=507.84 sq. feet. 
 
 2. What is the whole surface of a regular triangular pyr^ 
 
 * Thomson's Legendre, 11. 4. 
 
ITION OP SOLIDS. 
 
 4ft. 
 
 amid whose axis is 8, and the sides of whose base are each 
 
 20.78? 
 
 The lateral surface is 312 
 
 Tlie area of the base is 18 Y 
 
 And the whole surface is 499 
 
 3. What is the lateral surface of a regular pyramid 
 whose axis is 12 feet, and whose base is 18 feet square ? 
 
 Ans. 540 square feet. 
 
 The lateral surface of an obliqiie pyramid may be found, 
 by takmg the sum of the areas of the unequal triangles 
 which form its sides. 
 
 Problem V. 
 To Jind the soLiDrrr of a fbustttm of a pyramid. 
 50. Add together the areas of the two ends, and 
 
 THE square root OF THE PRODUCT OF THESE AREAS ; AND 
 MULTIPLY THE SUM BY -^ OF THE PERPENDICULAR HEIGHT 
 OF THE SOLID. 
 
 Let CDGL be a vertical 
 section, through the middle 
 of a frustum of a right pyr- 
 amid CDV, whose base is a 
 square. 
 
 Let CD=a, LG=6, RN=A. 
 
 By similar triangles, 
 LG : CD ::RV : NV 
 
 Subtracting the antecedents, (Alg. 349.) 
 LG : CD— LG : : RV : NV— RV=RN. 
 RNxLG hb 
 
 l/ * 
 
 
 G 
 
 /.■ 
 
 
 \ 
 
 C ] 
 
 H i 
 
 ^ D 
 
 Therefore RV: 
 
 CD— LG""a— 6 
 
46 MENSURATION OF SOLIDS. 
 
 The square of CD is the base 
 of the pyramid CDV ; 
 
 And the square of LG is 
 the base of the small pyr- 
 amid LGV. 
 
 V 
 
 i 
 
 Therefore, the solidity of 
 the larger pyramid (Art. 
 48) is 
 
 p 
 
 \ a — 0/ 3a — 36 
 
 And the solidity of the smaller pyramid is equal to 
 
 LG^XiRV=2''X 
 
 3a— 36 3a— 36 
 
 If the smaller pyramid be taken from the larger, there 
 will remain the frustum CDLG, whose solidity is equal to 
 
 ^t::|^';=x^X^^=iAxK+a6+6^) (Alg.l94. a.) 
 oa — do a — 
 
 Or, because Va'6''=a6. (Alg. 210. a.) 
 
 iAX(a^+6'+Va^F) 
 
 Here A, the height of the frustum, is multiplied into a' 
 and W, the areas of the two ends, and into Va"6'* the square 
 root of the products of these areas. 
 
 In this demonstration the pyramid is supposed to be 
 square. But the rule is equally applicable to a pyramid of 
 any other form. For the solid contents of pyramids are 
 equal, when they have equal heights and bases, whatever be 
 the Jiffure of their bases. (Sup. Euc. 14. 3.)* And the sec- 
 
 * Thomson's Legendre, 14, 7. 
 
MENSURATION OF SOLIDS. W 
 
 tions parallel to the bases, and at equal distances, are equal 
 to one another. (Sup. Euc. 12. 3. Cor. 2.)* 
 
 Ex. 1. If one end of the frustum of a pyramid be 9 feet 
 square, the other end 6 feet square, and the height 36 feet, 
 what is the sohdity ? 
 
 The areas of the two ends are 81 and 36. 
 The square root of their product is 54. 
 And the solidity of the frustum=(81+36+64)xl2=2062. 
 
 2. If the height of a frustum of {f pyramid be 24, and 
 the areas of the two ends 441 and 121 ; what is the solid- 
 ity? Ans. 6344. 
 
 8. If the height of a frustum of a hexagonal pyramid be 
 48, each side of one end 26, and each side of the other end 
 16 ; what is the solidity ? Ans. 56034. 
 
 Problem VI. 
 
 To find the lateral surface of a frustum of a regular 
 pyramid, 
 
 51. Multiply half Ihe slant-heioht by t^k sum of 
 THE perimeters of the two ends. 
 
 Each side of a frustum of a regular pyramid is a trapezoid, 
 as ABCD. The slant-height HP, (Def. 
 11.) though it is oblique to the base of 
 the soHd, is perpendicular to the line AB. 
 The area of the trapezoid is equal to the 
 product of half this perpendicular into 
 the sum of the parallel sides AB and DC. 
 (Art. 12.) Therefore the area of all the 
 equal trapezoids which form tho lateral 
 surface of the frustum, is equal to the 
 
 ♦ Thomson's Legendre, 13, 7, Cor, 
 
48 
 
 MENSURATION OF SOLIDS. 
 
 product of half the slant-height into the sum of the peri- 
 meters of the ends. 
 
 Ex. If the slant-height of a frustum of a regular octag- 
 onal pyramid be 42 feet, the sides of one end 5 feet each, and 
 the sides of the other end 3 feet each ; what is the lateral 
 surface? Ans. 1344 square feet. 
 
 62. If the slant-height be not given, it may be obtained 
 from the perpendicular 
 
 height and the dimensions 
 of the two ends. Let GD 
 be the slant-height of the 
 frustum CDGL, RN or GP 
 the perpendicular height, 
 ND and RG the radii of the 
 circles inscribed in the pe- 
 rimeters of the two ends. ^ 
 Then. PD is the difference of the two radii : 
 
 17 
 
 And the slant-height GD = v(GP'-fPD=^). 
 
 Ex. If the perpendicular height 
 of a frustum of a regular hexagonal 
 pyramid be 24, the sides of one end 
 13 each, and the sides of the other 
 end 8 each ; what is the whole sur- 
 face? 
 
 V(BC'^— BP') = CP, that is, V(l 3'— 6^5^=11.258 
 And V8^— 4= = 6.928 
 The difference of the two radii is, therefore 4.33 
 
 Theslant-height=V(24»+4.33')=24.3875. 
 TJie lateral surface is 1536.4 
 
 And the whole surface, 2 141. 7 5. 
 
USV8URATI0N OF SOLIDS. 4i 
 
 The height of the whole pyramid may be calculated from 
 the dimensions of the frustum. Let VN (Fig. 17.) be the 
 height of the pyramid, RN or GP the height of the frus- 
 tum, ND and RG the radii of the circles inscribed in the 
 perimeters of the ends of the frustum. 
 
 Then, in the similar triangles GPD and VND, 
 
 DP : GP : : DN ; VN. 
 
 The height of the frustum subtracted from VK, gives VR 
 the height of the small pyramid VLG. The solidity and 
 lateral surface of the frustum may then be found, by sub- 
 tracting from the whole pyramid, the part which is above 
 the cutting plane. This method may serve to verify the cal- 
 culations which are made by the rules in Arts. 50 and 51. 
 
 Ex. If one end of the frustum CDGL (Fig. 17.) be 90 feet 
 square, the other end 60 feet square, and the height RN 36 
 feet ; what is the height of the whole pyramid VCD : and 
 what arc the solidity and lateral surface of the frustum ? 
 
 DP=DN—GR=45— 30=15. And, GP=RN=36. 
 
 Then, 15 : 36 : : 45 : 108 =VN, the height of the whole 
 pyramid. 
 
 And, 108— 36=72 =VR, the height of the part VLG. 
 
 The solidity of the large pyramid is 291600 (Art. 48.) 
 of the small pyramid 86400 
 
 of the frustum CDGL 205200 
 
 The lateral surface of the large pyramid is 21060 (Art. 49.) 
 of the small pyramid 9360 
 
 of the frustum 11700 
 
60 
 
 MENSURATION OF SOLIDS. 
 
 Problem VII. 
 To find the solidity of a wedge. 
 64. Add the length of the edge to twice the 
 
 product of the height of the wedge and the breadth 
 of the base. 
 
 Let L = AB the 
 
 length of the base. 
 Let Z=GH the length 
 
 of the edge. 
 Let 6=BC the breadth 
 
 of the base. 
 Let A=PG the height 
 
 of the wedge. 
 Then, L 
 
 A M 
 
 Z=AB— GH=AM 
 
 If the length of the base and the edge be equal, as BM 
 and GH, the wedge MBHG is ha,lf a parallelepiped of the 
 same base and height. And the soHdity (Art. 43.) is equal 
 to half the product of the height, into the length and breadth 
 of the base ; that is i &AZ. 
 
 If the length of the base be greater than that of the edge, 
 as ABGH ; let a section be made by the plane GMN, par- 
 allel to HBC. This will divide the whole wedge into two 
 parts MBH<T and AMG. The latter is a pyramid, whose 
 solidity (Art. 48.) is i hhx{L—l) 
 
 The solidity of the parts together, is, therefore, 
 
 i6A?+,i&AX (L— Z)=iM3Z+i&A2L— i6A2Z=i6A X (2L+?) 
 
 If the length of the base be less than that of the edge, it 
 is evident that the pyi-amid is to be subtracted from half the 
 parallelepiped, which is equal in height and breadth to the 
 wedge, and equal in length to the edge. 
 
MENSURATION OF SOLIDS. 51 
 
 The solidity of the wedge is, therefore, 
 
 Ex. 1. If the base of a wedge be 35 by 15, the edge 55, 
 and the perpendicular height 12.4 ; what is the solidity ? 
 
 Ans. (70 + 55)X^^^^^ = 3875. 
 
 2. If the base of a wedge be 27 by 8, the edge 30, and 
 the perpendicular height 42 ; what is the solidity? 
 
 Ans. 5040. 
 
 Problem VIII. 
 To find the solidity of a rectangular prismoid. 
 
 55. To the areas of the two ends, add four times 
 the area of a parallel section equally distant from 
 the ends, and multiply the sum by ^ of the height. 
 
 Let L and B be the length and 
 
 breadth of one end, 
 Let I and h be the length and breadth 
 
 of the other end. 
 Let M and m be the length and 
 
 breadth of the section in the middle. 
 And h be the height of the pris- 
 
 inoid. 
 
 The solid may be divided into two wedges whose bases are 
 the ends of the prismoid, and whose edges are L and /. The 
 solidity of the whole, by the preceding article is, 
 
 ii?Ax(2L+/)+i*AX(2/+L)=-iA(2BL-f-B^+2W+6L) 
 
 As M is equally distant from L and I, 
 
 2M=«L4-/,2/w-»B+J,and4Mm— {L+/)(B+2')=BL+B;+ 
 
 [bh+lb. 
 
52 MENSURATION OF SOLIDS. 
 
 Substituting 4 Mm for its value, in the preceding ex- 
 pression for the sohdity, we have 
 
 ih(BL+hl+AMm) 
 
 That is, the soHdity of the prismoid is equal to -J- of the 
 height, multiplied into the areas of the two ends, and 4 
 times the area of the section in the middle. 
 
 This rule may be applied to prismoids of other forms. 
 For, whatever be the figure of the two ends, there may be 
 drawn in each, such a number of small rectangles, that the 
 sum of them shall differ less, than by any given quantity, 
 from the figure in which they are contained. And the solids 
 between these rectangles will be rectangular prismoids. 
 
 Ex. 1. If one end of a rectangular prismoid be 44 feet by 
 23, the other end 36 by 21, and the perpendicular heiglit 
 72 ; what is the solidity ? 
 
 The area of the larger end =44X23 = 1012 
 of the smaller end =36X21= 156 
 of the middle section =40X22= 880 
 And the solidity= (101 2+756 +4 X 880) X 12 = 63456 feet. 
 
 2. What is the solidity of a stick of hewn timber, whose 
 ends are 30 inches by 27, and 24 by 18, and whose length 
 is 48 feet ? Ans. 204 feet. 
 
 Other solids not treated of in this section, if they be 
 bounded by plane surfaces, may be measured by supposing 
 them to be divided into prisms, pyramids, and wedges. And, 
 indeed, every such solid may be considered as made up of 
 triangular pyramids. 
 
1. 
 
 The Tetraedron, 
 
 
 2. 
 
 The Ilexaedron or cube, 
 
 whose 
 sides are 
 
 3. 
 4. 
 
 The Octaedron, 
 The Dodecaedran, 
 
 5. 
 
 The Icosaedron, 
 
 
 MENSURATION OP REGULAR SOLIDS. 53 
 
 THE FIVE REGULAR SOLIDS. 
 
 66. A SOLID IS SAID TO BE REGULAR, WHEN ALL ITS 
 SOLID ANGLES ARE EQUAL, AND ALL ITS 6IDKS ARE EQUAL 
 AND REGULAR POLYGONS. 
 
 The following figures are of this description ; 
 
 four triangles ; 
 six squares ; 
 eight triangles ; 
 twelve pentagons; 
 twenty triangles.* 
 
 Besides these five there can be no other regular solids. 
 The only plane figures which can form such solids, are tri- 
 angles, squares, and pentagons. For the plane angles which 
 contain any sohd angle, are together less than four right an- 
 gles or 360°. (Sup. Euc. 21, 2.) And the least number 
 which can form a solid angle is three. (Sup. Euc. Def. 8, 2.) 
 If they are angles of equilateral triangles, each is 60°. The 
 sum of three of them is 180°, of four 240°, oijlve 300°, and 
 of six 360°. The latter number is too great for a solid angle. 
 
 The angles of squares are 90° each. The sum of three of 
 these is 270°, of four 300°, and of any other greater number, 
 still more. 
 
 The angles of regular pentaf/ons are 108° each. The sum 
 of three of them is 324° ; of four, or any other greater num- 
 ber, more than 360°. The angles of all other regular poly- 
 gons are still greater. 
 
 In a regular solid, then, each solid angle must be con- 
 tained by three, four, or five equilateral triangles, by three 
 squares, or by three regular pentagons. 
 
 * For the geometrical construction of these solids, see Legendre's 
 Geometry ; Appendix to Books VI. aojl VII., or Thomson's Legendre, 
 p. 214. 
 
54 MENSURATION OF REGULAR SOLIDS. 
 
 67. As the sides of a regular solid are similar and equal, 
 and the angles are also alike ; it is evident that the sides 
 are all equally distant from a central point in the solid. If 
 then, planes be supposed to proceed from the several edges 
 to the centre, they will divide the solid into as many equal 
 'pyramids, as it has sides. The base of each pyramid will be 
 one of the sides ; their common vertex will be the central 
 point; and their height will be a perpendicular from the 
 centre to one of the sides. 
 
 Problem IX. 
 
 To find the surface of a regular solid. 
 
 68. Multiply the area of one of the sides by the 
 number of sides. 
 
 Or, 
 
 Multiply the square of one of the edges, by the 
 surface of a similar solid whose edges are 1. 
 
 As all the sides are equal, it is evident that the area of 
 one of them, multiplied by the number of sides, will give the 
 area of the whole. 
 
 Or, if a table is prepared, containing the surfaces of the 
 several regular solids whose linear edges are unity ; this may- 
 be used for other regular solids, upon the principle, that the 
 areas of similar polygons are as the squares of their homolo- 
 gous sides. (Euc. 20. 6.)* Such a table is easily formed, by 
 multiplying the area of one of the sides, as given in Art. 17, 
 by the number of sides. Thus, the area of an equilateral 
 triangle whose side is 1, is 0.4330127. Therefore, the siu:- 
 face 
 
 * Thomson's Legendre, 27. 4. 
 
MEN SI RATION OP REGULAR SOLIDS. 5ff 
 
 Of aregular tetraedron =.433012'7X4 =1.7320508. 
 Of a regular octaedron =.4330127x8 =3.4641016. 
 Of a regular icosaedron =.4330127x20=8.0602540. 
 
 See the table in the following article. 
 
 Ex. 1 . What is the surface of a regular dodecaedron whose 
 edges are each 25 inches ? 
 
 The area of one of the sides is 1075.3 
 And the surface of the whole solid =1075.3X12=12903.6. 
 
 2. What is the surface of a regular icosaedron whose 
 edges are each 102? Ans. 90101.3. 
 
 Problem X. 
 . 7b ^rw? /^€ solidity o/* a REGULAR solid. 
 59. Multiply the surface by ^ of the perpendicular 
 
 DISTANCE from THE CENTRE TO ONE OF THE SIDES. 
 
 Or, 
 
 McLTIPLY the cube of ONE OF THE EDGES, BY THE 
 SOLIDITY OF A SIMILAR SOLID WHOSE EDGES ARE 1. 
 
 As the solid is made up of a number of equal pyramids, 
 whose bases are the sides, and whose height is the perpendic- 
 ular distance of the sides from the centre (Art. 57.) ; the 
 solidity of the whole must be equal to the areas of all the 
 sides midtiplied into ^ of this perpendicular. (Art. 48.) 
 
 If the contents pf the several regular solids whose edges 
 are 1, be inserted in a table, this may be used to measure 
 other similar .solids. For two similar regular solids contain 
 the same number of similar pyramids ; and the.se are to each 
 other as the cubes of their linear sides or edges. (Sup. Euc. 
 15. 3. Cor. 3.)* 
 
 • Thomson's Legendre, 20. 7. 
 
56 
 
 MENSURATION OF THE CYLINDER. 
 
 A TABLE OF REGULAR SOLIDS WHOSE EDGES ARE 1. 
 
 Names. 
 
 No. of sides. 
 
 Surfaces. | Solidities. | 
 
 Tetraedron 
 
 Hexaedron 
 
 Octaedron 
 
 Dodecaedron 
 
 Icosaedron 
 
 4 
 
 6 
 
 8 
 
 12 
 
 20 
 
 1.7320508 
 6.0000000 
 3.4641016 
 20.6457288 
 8.6602540 
 
 0.1178513 
 1.0000000 
 0.4714045 
 7.6631189 
 2.1816950 
 
 For the method of calculating the last column of this table, 
 see Hutton's Mensuration, Part. III. Sec. 2. 
 
 Ex. What is the solidity of a regular octaedron whose 
 edges are each 32 inches ? Ans. 15447 inches. 
 
 SECTION IV. 
 
 THE CYLINDER, CONE, AND SPHERE. 
 
 Art. 61. Definition I. A right cylinder is a solid de- 
 scribed by the revolution of a rectangle about one of its 
 sides. The ends or hases are evidently equal and parallel 
 circles. And the axis, which is a line passing through the 
 middle of the cylinder, is perpendicular to the bases. 
 
 The ends of an oblique cylinder are also equal and paral- 
 lel circles ; but they are not perpendicular to the axis. The 
 height of a cylinder is the perpendicular distance from one 
 base to the plane of the other. In a right cylinder, it is the 
 length of the axis. 
 
 II. A right cone is a solid described by the revolution of 
 a right angled triangle about one of the sides which contain 
 the right angle. The hase is a circle, and is perpendicular to 
 
MENSX7RATI0N OF TUB CVLINDER. 
 
 «lr 
 
 the axiSf which proceeds from the middle of the base to the 
 vertex. 
 
 The base of an oblique cone is also a circle, but is not per- 
 pendicular to the axis. The height of a cone is the perpen- 
 dicular distiince from the vertex to the plane of the base. In 
 a right cone, it is the length of the axis. The slant-height 
 of a right cone is the distance from the vertex to the circum- 
 ference of the base. 
 
 III. A frustum of a cone is a portion cut oflf by a plane 
 parallel to the base. The height of the 
 
 frustum is the perpendicular distance of 
 the two ends. The slant-height of a 
 frustum of a right cone, is the distance 
 between the peripheries of the two 
 ends, measured on the outside of the 
 solid ; as AD. 
 
 IV. A sphere or globe is a solid which 
 has a centre equally distant from every 
 
 part of the surface. It may be described by the revolution 
 of a semicircle about a diameter. A radius of the sphere is 
 a line drawn from the centre to any part of the surface. A 
 diameter is a line passing through the centre, and terminated 
 at both ends by the surface. The circumference is the same 
 as the circumference of a circle whose plane passes through 
 the centre of the sphere. Such a circle is called a great 
 circle. 
 
 V. A segment oi a sphere is a part cut off by any plane. 
 The height of the segment is a per- 
 pendicular from the middle of the 
 base to the convex surface, as LB. 
 
 VI. A spherical zone or frustum is 
 a part of the sphere included be- 
 tween two parallel planes. It is 
 called the middle zone, if the planes 
 are equally distant from the centre. 
 
 3* 
 
58 MENSURATION OF THE CYLINDER. 
 
 The height of a zone is the distance of the two planes, as 
 LR.* 
 
 VII. A spherical sector is a solid produced by a circular 
 sector, revolving in tlie same manner 
 as the semicircle which describes the 
 whole sphere. Thus a spherical sec- 
 tor is described by the circular sec- 
 tor ACP or GCE revolving on the 
 axis CP. 
 
 VIII. A solid described by the 
 revolution of any figure about a fixed 
 axis, is called a solid of revolution. 
 
 Problem I. 
 To find the convex surface of a right cylinder. 
 62. Multiply the length into the circumference of 
 
 THE BASE. 
 
 If a right cylinder be covered with a thin substance like 
 paper, which can be spread out into a plane ; it is evident 
 that the plane will be a parallelogram, whose length and 
 breadth will be equal to the length and circumference of the 
 cylinder. The area must, therefore, be equal to the length 
 multiplied into the circumference. (Art. 4.) 
 
 Ex. 1. What is the convex surface of a right cylinder 
 which is 42 feet long, and 15 inches in diameter? 
 
 Ans. 42X1.25X3.14159 = 164.933 sq. feet. 
 
 2. What is the whole surface of a right cylinder, which 
 is 2 feet in diameter and 36 feet long ? 
 
 * According to some writers, a spherical segment is either a solid 
 which is cut off from the sphere by a single plane, or one which is in- 
 cluded between two planes : and a zone is the surface of either of these. 
 In this sense, the term zone is commonly used in geography. 
 
MENSURATION OF THB CYLINDER. 56 
 
 The convex surface is 226.1945 
 
 The area of the two ends (Art. 30.) is 6.2832 
 
 The whole surface is 232.4777 
 
 3. What is the whole surface of a right cylinder whose 
 axis is 82, and circumference 71 ? Ans. 6024.32. 
 
 63. It will be observed that the rules for the prism and 
 pyramid in the preceding section, are substantially the same, 
 as the rules for the cylinder and cone in this. There may be 
 some advantage, however, in considering the latter by them- 
 selves. 
 
 In the base of a cylinder, there may be inscribed a poly- 
 gon, which shall differ from it less than by any given space. 
 (Sup. Euc. 6. 1. Cor.)* If the polygon be the base of a 
 prism, of the same height as the cylinder, the two solids 
 may differ less than by any given quantity. In the same 
 manner, the base of a pyramid may be a polygon of so many 
 sides, as to differ less than by any given quantity, from the 
 base of a cone in which it is inscribed. A cyhnder is there- 
 fore considered, by many writers, as a prisr^ of an infinite 
 number of sides ; and a cone, as a pyramid of an infinite 
 number of sides. (For the meaning of the term " infinite," 
 when used in the mathematical sense, see Alg. Sec. XV.) 
 
 Problem II. 
 To find the solidity of a cyundkr. 
 
 64. Multiply the area op the base by the height. 
 
 The solidity ^f a parallelopiped is equal to the product of 
 the base into the perpendicular altitude. (Art. 43.) And a 
 parallelopiped and a cylinder which have equal bases and 
 altitudes are equal to each other. (Sup. Euc. 17. 3.)f 
 
 • Thomson's Legendre, 9. 5. t ^^^^y 2* 8. 
 
60 
 
 MENSURATION OF THE CYLINDER. 
 
 Ex. 1. What is the solidity of a cylinder, whose height is 
 121, and diameter 45.2 ? 
 
 Ans. 45.2'X.V854X 121 = 194156.6. 
 
 2. What is the solidity of a cylinder, whose height is 424, 
 and circumference 213 ? Ans. 1530837. 
 
 3. If the side AC of an oblique 
 cylinder be 27, and the area of the base 
 32.61, and if the side make an angle 
 of 62° 44' with the base, what is the 
 sohdity ? 
 
 E : AC : : sin A : BC = 24 the per- 
 pendicular height. 
 
 And the soHdity is 782.64. 
 
 4. The Winchester bushel is a hollow cylinder, 18^ inches 
 in diameter, and 8 inches deep. What is its capacity ? 
 The area of the base=(18. 5)' X. 7853982 = 268.8025. 
 And the "^^apacity is 2150.42 cubic inches. See the 
 table in Art. 42. 
 
 Problem III. 
 
 To find the convex surface of a right cone. 
 
 65. Multiply half the slant-height into the cir- 
 cumference OF the base. 
 
 If the convex surface of a right cone be spread out into a 
 plane, it will evidently form a sector of a circle whose radius 
 is equal to the slant-height of the cone. But the area of the 
 sector is equal to the product of half the radius into the 
 length of the arc. (Art. 34.) Or if the cone be considered 
 as a pyramid of an infinite number of sides, its lateral sur- 
 
MENSURATION OF THE CONE. 61 
 
 face is equal to the product of half the slant-height into the 
 perimeter of the base. (Art. 49.) 
 
 Ex. 1. If the slant-height of a right cone be 82, and the 
 diameter of the base 24, what is the convex surface ? 
 
 Ans. 41X24X3.14159=8091.3 square feet. 
 
 2. If the axis of a right cone be 48, and the diameter of 
 le base 72, what is the whole surface ? 
 
 The slant-height = V(36'+48')=60. (Euc. 47. 1.) 
 The convex surface is 6786 
 
 The area of the base 4071.6 
 
 And the whole^urfaco 10857.6 
 
 3. If the axis of a right cone be 16, and the circumfer- 
 ence of the base 75.4 ; what is the whole surface ? 
 
 Ans. 1206.4. 
 
 Problem IV. 
 To find the sounrrY of a cone. 
 66. Multiply the area op the base into -^ of the 
 
 HEIGHT. 
 
 The solidity of a cylinder is equal to the product of the 
 base into the perpendicular height. (Art. 64.) And if a cone 
 and a cylinder have the same base and altitude, the cone is 
 \ of the cylinder. (Sup. Euc. 18. 3.)f Or if a cone be con- 
 sidercd as a pyramid of an infinite number of sides, the so- 
 lidity is equal to the product of the base into \ of the height, 
 by Art. 48. 
 
 Ex. 1. What is the solidity of a right cone whose height 
 is 663, and the diameter of whose base is 101 ? 
 
 Ans. ioPx. 7854X221 = 1770622. 
 
 * Thomcon's Legendre, 4. 8. Cor. 
 
62 MENSURATION OF THE CONE. 
 
 2. If the axis of an oblique cone be 738, and make an 
 angle of 30° with the plane of the base ; and if the circum- 
 ference of the base be 355, what is the solidity ? 
 
 Ans. 1233536. 
 
 Problem V.. 
 To find the convex surface of a frustum of a right cone. 
 67. Multiply half the slant-height by the sum of 
 
 THE peripheries OF THE TWO ENDS. 
 
 This is the rule for a frustum of a pyramid ; (Art. 51.) 
 and is equally applicable to a frustum of a cone, if a cone be 
 considered as a pyramid of an infinite number of sides. 
 (Art. 63.) 
 
 Or thus, 
 
 Let the sector ABV represent the 
 convex surface of a right cone, (Art. 
 65.) andyDCV the surface of a portion 
 of the cone, cat off by a plane parallel 
 to the base. Then will ABCD be tbe 
 surface of the frustum. 
 
 Let AB=a, DC=6, YI>=d, AD=h. '^ 
 
 Then the area ABV =iaX{h-\-d)=iah+iad. (Art. 34.) 
 And the area 'DCY=^bd. 
 Subtracting the one from the other. 
 
 The area ABDC=iaA+iac/ — ^bd. 
 
 But d : d-\-h :\h : a. (Sup. Euc. 8. 1.)* Therefore ^ad^ 
 ibd=ibh. 
 
 The surface of the frustum then, is equal to 
 ^ah+^bh. or^hx(a-{-b) 
 
 * Thomson's Legendre, 10, 5. Cor, 
 
MENSURATION OF THE CONE. 63 
 
 Cor. The surface of the frustum is equal to the product 
 of the slant-height into the circumference of a circle which 
 is equally distant from the two ends. Thus, the surface 
 ABCD is equal to the product of AD mto MN. For MN is 
 equal to half the sum of AB and DC. 
 
 Ex. 1. What is the convex surface of a frustum of a right 
 cone, if the diameters of the two ends be 44 and 33, and 
 the slant-height 84 ? ^ Ans. 10159.8. 
 
 2. If the perpendicular height of a frustum of a right 
 cone be 24, and the diameters of the two ends 80 and 44, 
 what is the whole surface ? 
 
 * Half the difference of the diameters is 18. 
 
 And V 18'+24'=30, the slant-height, (Art. 52.) 
 The convex sur&ce of the frustum is 5843 
 The sum of the areas of the two ends is 6547 
 
 And the whole surface is 12390 
 
 Problem VI. 
 
 To find the solidity of a frustum of a cone. 
 
 68. Add toosther the areas of the two ends, and 
 the square root of the product of these areas ; and 
 multiply the sum by \ of the perpendicular height. 
 
 This rule, which was given for the frustum of a pyramid, 
 (Art. 60.) is equally applicable to the frustum of a cone ; be- 
 cause a cone and a pyramid which have equal bases and alti- 
 tudes are equal to each other. 
 
 Ex. 1. What is the solidity of a mast which is 72 feet 
 long, 2 feet in diameter at one end, and 18 inches at the 
 other? Ans. 174.36 cubic feet. 
 
64 
 
 MENSURATION OF THE SPHERE. 
 
 2. What is tlie capacity of a conical cistern which is 9 
 feet deep, 4 feet in diameter at the bottom, and 3 feet at the 
 top ? Ans. 87.18 cubic feet— 652.15 wine gallons. 
 
 3. How many gallons of ale can be put into a vat in the 
 form of a conic frustum, if the larger diameter be 7 feet, the 
 smaller diameter 6 feet, and the depth 8 feet ? 
 
 Problem VII. 
 
 To find the surface of a sphere. 
 
 69. Multiply the diameter by the circumference. 
 
 Let a hemisphere be described by the quadrant CPD, 
 revolving on the line CD. Let 
 AB be the side of a regular poly- 
 gon inscribed in the circle of 
 which DBF is an arc. Draw AO 
 and BN perpendicular to CD, 
 and BH perpendicular to AO. 
 Extend AB till it meets CD con- 
 tinued. The triangle AOY, re- 
 volving on OV as an axis, will 
 describe a right cone. (Defin. 2.) 
 AB will be the slant-height of a 
 frustum of this cone extending 
 from AO to BN. From G the middle of AB, draw GM 
 parallel to AO. The surface of the frustum described by 
 AB. (Art. 67. Cor.) is equal to 
 
 ABxc^VcGM.* 
 
 From the centre C draw CG, which will be perpendicular 
 to AB, (Euc. 3. 3.) and the radius of a circle inscribed in 
 
 * By drc GM is meant the circumference of a circle the radius of 
 which is GM. 
 
MENSURATION OF THE SPHERE. 05 
 
 the polygon. The triangles ABH and CGM are similar, be- 
 cause the sides are perpendicular, each to each. Therefore, 
 
 HB or ON : AB : : GM : GC : : circ GM : circ GC. 
 
 So that ONxciVc GC=ABx«>c GM, that is, the sur- 
 face of the frustum is equal to the product of ON the per- 
 pendicular height, into circ GC, the perpendicular distance 
 from the centre of the polygon to one of the sides. 
 
 In the same manner it may l>e proved, that the surfaces 
 produced by the revolution of the lines BD and AP about 
 the axis DC, are equal to 
 
 ND X circ GC, and CO X circ GC. 
 
 The surface of the whole solid, therefore, (Euc. 1. 2.) is equal to 
 
 CDxcirc GC. 
 
 The demonstration is applicable to a solid produced by 
 the revolution of a polygon of any number of sides. But a 
 polygon may be supposed which shall differ less than by 
 any given quantity from the circle in which it is inscribed ; 
 (Sup. Euc. 4. 1.)* and in which the perpendicular GC shall 
 differ less than by any given quantity from the radius of the 
 circle. Therefore, the surface of a hemisphere is equal to 
 the product of its radius into the circumference of its base ; 
 and the surface of a sphere is equal to the product of its 
 diameter into its circumference. 
 
 Cor. 1. From this demonstration it follows, that the sur- 
 face of any segment or zone of a sphere is equal to the 
 product of the height of the segment or zone into the cir- 
 cumference of the sphere. The surface of the zone pro- 
 duced by the revolution of the arc AB about ON, is equal 
 to ON X«rc CP. And the surface of the segment pro- 
 
 ♦ Thomson's Legendre, 9. 5. 
 
66 
 
 MENSURATION OF THE SPHERE. 
 
 duced by the revolution of BD about DN is equal to DNx 
 ci7'c CP. 
 
 Cor. 2. The surface of a sphere is equal to four times tlie 
 area of a circle of the same diameter ; and therefore, the 
 convex surface of a hemisphere is equal to twice the area of 
 its base. For the area of a circle is equal to the product of 
 half the diameter into half the circumference ; (Art. 30.) 
 that is, to ^ the product of the diameter and circumference. 
 
 Cor. 3. The surface of a sphere, or the convex surface 
 of any spherical segment or zone, 
 is equal to that of the circum- 
 scribing cylinder. A hemis- 
 phere described by the revolu- 
 tion of the arc DBP, is cir- 
 cumscribed by a cylinder pro- 
 duced by the revolution of the 
 parallelogram Dc?CP. The con- 
 vex surface of the cylinder is 
 equal to its height multiplied 
 by its circumference. (Art. 62.) 
 And this is also the surface of 
 the hemisphere. 
 
 So the surface produced by the revolution of AB is equal 
 to that produced by the revolution of ab. And the surface 
 produced by BD is equal to that produced by hd. 
 
 Ex. 1. Considering the earth as a sphere 7 930 miles in 
 diameter, how many square miles are there on its surface ? 
 
 Ans. 197,558,500. 
 
 2. If the circumference of the sun be 2,800,000, what is 
 his surface? Ans. 2,495,547,600,000 sq. miles. 
 
 3. How many square feet of lead will it require, to cover 
 a hemispherical dome whose base is 13 feet across ? 
 
 Ans. 265i-. 
 
UBNSITBATIOK OF THE SrUERE. 67 
 
 Problem VIII. 
 To find the solidiit of a sphere. 
 70. 1. Multiply tse cube of the diameter by •5236. 
 
 Or, 
 
 2. Multiply the square of the diameter by \ of the 
 circumfekence. 
 
 Or, 
 
 3. Multiply the' surface by \ of the diameter, 
 
 1. A sphere is two-thirds of its circumscribing cylinder. 
 (Sup. Euc. 21. 3.)* The height and diameter of the cylin- 
 der are each equal to the diameter of the sphere. The solid- 
 ity of the cylinder is equal to its height multiplied into the 
 area of its base, (Art. 64.) that is putting D for the diam- 
 eter, 
 
 DxD'X.7854 or D'x.7854. 
 
 And the solidity of the sphere, being f of this, is 
 D'X.6236. 
 
 2. The base of the circumscribing cylinder is equal to half 
 the circumference multiplied into half the diameter ; (Art. 
 30.) that is, if C be put for the circumference, 
 
 iC X D ; and the solidity is \Q X D'. 
 
 Therefore, the solidity of the sphere is 
 iofiCxD«=D«XiC. 
 
 8. In the last expression, which is the same as CxDxiD, 
 • ThomMa's Legendre, 12. 8. 
 
68 MENSURATION OF THE SPHERE. 
 
 we may substitute S, the surface, for CxD. (Art. 69.) We 
 then have the solidity of the sphere equal to 
 
 SxiD. 
 
 Or, the sphere may be supposed to be filled with small 
 pyramids, standing on the surface of the sphere, and having 
 their common vertex in the centre. The number of these 
 may be such, that the difference between their sum and the 
 sphere shall be less than any given quantity. The solidity 
 of each pyramid is equal to the product of its base into -|- 
 of its height. (Art. 48.) The solidity of the whole, there- 
 fore, is equal to the product of the surface of the sphere 
 into "i of its radius, or ^ of its diameter. 
 
 71. The numbers 3.14159, .7854, .5236, should be made 
 perfectly familiar. The first expresses the ratio of the 
 circumference of a circle to the diameter; (Art. 23.) the 
 second, the ratio of the area of a circle to the square of the 
 diameter (Art. 30.) ; and the third, the ratio of the solidity 
 of a sphere to the cube of the diameter. The secon(J*is -J- 
 of the first, and the third is ^ of the first. 
 . As these numbers are frequently occurring in mathemat- 
 ical investigations, it is common to represent the first of them 
 by the Greek letter n. According to this notation, 
 
 7r=3.14159, i7r=.7854, i^=.5236. 
 
 If D=the diameter, and R=:the radius of any circle or 
 sphere ; 
 
 Then, D = 2R D^=4R^ D'r=8R^ 
 
 And nD ) ^,^^^ -j^^ i-D= ) =the area oi-nW ) ^^^^ 
 Or, 27rR j ^ ^ or ttR^ S the circ. or f^iR^ \ 
 
 solidity of the sphere. 
 
 Ex 1. What is the solidity of the earth, if it be a sphere 
 7930 miles in diameter ? 
 
 Ans. 261,107,000,000 cubic miles. 
 
MENSURATION OF THE SPHERE. 60 
 
 2, How many wine gajlons will fill a hollow sphere 4 feet 
 in diameter ? 
 
 Ans. The capacity is 33.5104 fect=250f gallons. 
 
 3. If the diameter of the moon be 2180 miles, what is its 
 sohdity? Ans. 6,424,600,000 miles. 
 
 72. If the solidity of a sphere be ^Iven, the diameter may 
 be found by reversing the first rule in the preceding article ; 
 that is, dividing hy .5236 and extracting the cube root of the 
 quotient, 
 
 Ex. 1. What is the diameter of a sphere whose solidity is 
 65.45 cubic feet ? Ans. 5 feet. 
 
 2. What must be the diameter of a globe to contain 16755 
 poimds of water? Ans. 8 feet. 
 
 Problem IX. 
 
 To find the convex surface of a segment or zone of a 
 spJiere. 
 
 73. Multiply the height of the segment or zone 
 
 INTO the circumference OF THE SPHERE. 
 
 For the demonstration of this rule, see Art. 69. 
 
 Ex. 1. If the earth be considered a perfect sphere 7930 
 miles in diameter, and if the polar circle be 23° 28' from the 
 pole, how many square miles are there in one of the frigid 
 zones ? 
 
 If PQOE be a meridian on the 
 earth, ADB one of the polar circles, 
 and P the pole ; then the frigid zone 
 is a spherical segment described by 
 the revolution of the arc APB about 
 PD. The angle ACD subtended by 
 the arc AP is 23° 28'. And in the 
 rififht angled triangle ACD, 
 
10 
 
 MENSURATION OF THE SPHERE. 
 
 R : AC : : COS ACD ; CD=3C37. 
 
 Then, CP~CD=3965— 363'7=328=PD the height of 
 the segment. 
 
 And 328X^930X3.14159=8] 71400 the surface. 
 
 2. If the diameter of the earth be 7930 miles, what is the 
 surface of the torrid zone, extending 
 23° 28' on each side of the equator ? 
 
 If EQ be the equator, and GH one 
 of the tropics, then the angle ECG is 
 23° 28'. And in the right angled 
 triangle GCM, 
 
 R : CG : : sin ECG : GM=C]Sr=15'78.9 the height of 
 half the zone. 
 
 The surface of the whole zone is 78669700. 
 
 3. What is the surface of each of the temperate zones ? 
 
 The height DN=CP— CN— PD = 2058.1 
 
 And the surface of the zone is 51273000. 
 
 The surface of the two temperate zones is 102,546,000 
 
 of the two frigid zones 16,342,800 
 
 of the torrid zone 78,669,700 
 
 of the whole globe 197,558,500 
 
 Problem X. 
 
 To find the solidity of a spherical sector. 
 
 74. Multiply the spherical surface by ^ of the 
 radius of the sphere. 
 
 The spherical sector produced by the revolution of ACBD 
 
laNSURATION OF THE SPHERfi. 
 
 a 
 
 about CD, may be supposed to be filled 
 
 with small pyramids, standing on the 
 
 spherical surface ADB, and temainating 
 
 in the point C. Their number may be 
 
 so great, that the height of each shall 
 
 differ less than by any given length 
 
 from the radius CD, and the sum of their 
 
 bases shall differ less than by any given 
 
 quantity from the surface ABD. The 
 
 solidity of each is equal to the product of its base into \ of 
 
 the radius CD. (Art. 48.) Therefore, the solidity of all of 
 
 them, that is, of the sector ADBC, is equal to the product 
 
 of the spherical surface into i of the radius. 
 
 Ex. Supposing the earth to be a 
 sphere 7930 miles in diameter, and 
 the polar circle ADB to be 23° 28' 
 from the pole ; what is the solidity of 
 the spherical sector ACBP ? 
 
 Ans. 10,799,867,000 miles. 
 
 Problem XI. 
 
 To find the solidity of a spherical segment. 
 
 75, Multiply half the heioht of the segment into 
 the area of the base, and the cube of the height 
 into .5236 ; and add the two products. 
 
 As the circular sector AOBC consists of two parts, the 
 segment AOBP and the triangle 
 ABC ; (Art. 35.) so the spherical 
 sector produced by the revolution of 
 AOC about OC consbts of two parts, 
 the segvient produced by the revolu- 
 tion of AOP, and the cone produced 
 by the revolution of ACP. If then 
 
12 
 
 MENSURATION OF THE SPHERE. 
 
 the cone be subtracted from the sec- 
 tor, the remainder will be the seg- 
 ment. 
 
 Let CO=R, the radius of the sphere, 
 PB=r, the radius of the base of 
 
 the segment. 
 PO=A, the height of the segment, 
 Then PC=R — h, the axis of the cone. 
 
 The sectors 27iRx^XiR. (Arts. '71, 73, Y4.)=f;rAR«. 
 The cone=7r/^xi (R— A) (Arts. 71, 66.)=i7rr"R — \7Tkr'' 
 
 Subtracting the one from the other, 
 
 The segment =-|7rAR2— |;rr^R+|7rAr^ 
 
 ButDOxPO=BO' (Trig. 97.*)=PO^-fPB^ (Euc. 47. 1.) 
 
 That is, 2R7i=:A^-f r^ So that, R=^'+^' 
 
 2h 
 
 And R^= 
 
 A'+7-\= A4-f2AV='+H 
 
 2= /^_j:!_\ = 
 
 V 2A / 
 
 W 
 
 Substituting then, for R and R^, their vahies, and multi- 
 plying the factors, 
 
 The segment=i7r7i=' ^\nhr'' -\-\-j^ — i^Ar"— iy +\n1ir^ 
 Which, by uniting the terms, becomes 
 
 The first term here is ihx^r^, half the height of the seg- 
 ment multiplied into the area of the base ; (Art. 71.) and the 
 other h^Xi^, the cube of the height multiplied into .5236. 
 
 * Euclid 31, 3, and 8, 6. Cor. 
 
MENSURATION OF THE SPHERE. 
 
 IS 
 
 If the segment be greater than a hemisphere, as ABD ; 
 the 'cone ABC must be added to the sector ACBD. 
 
 Let PD=A the height of the segment, 
 Then PC=A — R the axis of the cone. 
 
 The sector ACBD=inAR« 
 
 The cone=nr*Xi(A— R)=Wir' — \rtr*R 
 
 Adding them together, we have as before. 
 
 The segment =inhR'' — \nr'R-^inhr\ 
 
 Cor. The solidity of a spherical segment is equal: to half a 
 cylinder of the same base and height + a sphere whose 
 diameter is the height of the segment. For a cylinder is 
 equal to its height multiplied into the area of its base ; and 
 a sphere is equal to the cube of its diameter multiplied by 
 .5230. 
 
 Thus, if Oy be half Ox, the spher- 
 ical segment produced by the revo- 
 tution, of Oxt is equal to the cylin- 
 der produced by ivt/x -f the sphere 
 produced by Oyxz ; supposing each 
 to revolve on the line Ox. 
 
 Ex. 1. If the height of a spherical segment be 8 feet, and 
 the diameter of its base 25 feet ; what is the solidity ? 
 
 Ans. (25)'X.V854X4+8'X.5236=2231.58 feet. 
 
 2. If the earth be a sphere 7930 miles in diameter, and the 
 polar circle 23° 28' from the pole, what is the solidity of 
 one of the frigid zones ? Ans. 1,303,000,000 miles. 
 
 4 
 
 I 
 
14 
 
 MENSURATION OF THE SPHERE. 
 
 « Problem XII. 
 
 To find the SOLIDITY of a sph rlcal zone or frustum, 
 
 16. From the solidity of the whole sphere, sub- 
 tract THE TWO segments ON THE SIDES OF THE ZONE. 
 
 Or, 
 
 Add together the squares of the radii of the two 
 ends, and ^ the square of their distance ; and multiply 
 the sum by three times this distancc, and the product 
 BY .5236. 
 
 If from the whole sphere, there 
 be taken the two segments ABP and 
 GHO, there will remain the zone 
 or frustum ABGH. 
 
 Or, the zone ABGH is equal to 
 the difference between the segments 
 GHP and ABP. 
 
 i the Jieights of the two segments 
 
 the radii of their bases. 
 
 DP 
 
 GN=R ] 
 AD=r ) 
 
 DN=c?=H — h the distance of the two bases, or the 
 height of the zone. 
 
 Then the larger segment=i7rHR'+^rrH^ ) , . . ^ . 
 And the smaller segment=^rrAr^+-6-^^^ i 
 
 Therefore the zone ABGH=i7r (SHR'^+H^— 3^r'— A^) 
 
 By the properties of the circle, (Euc. 35, 3.) 
 
 0]S^xH=:R^ Therefore, (ON+H)xH=R^+H=' 
 
 R=+H^ 
 
 Or, 0P= 
 
 H 
 
MENSURATION OF THE SPHERE. 76 
 
 In the same manner, OP=!li — 
 
 h 
 
 Therefore, 3Hx(r'+A*)=3Ax(R'+H*.) 
 
 Or, 8Hr«+3lIA'— 3AR'— 3AH'=0. (Alg. l'/8.) 
 
 To reduce the expression for the sohdity of the zone to 
 the required form, without altering its value, let these terms 
 be added to it : and it will become 
 
 in(3HR'+3Hr«— 3AR'— 3Ar»+H«— 3H'A+3HA»— A3) 
 
 Which is equal to 
 
 i^r X 3(H— A) X (R'+r'+i (H-~A)») 
 
 Or, as \n equals .6236 (Art. 71.) and H — h equals d, 
 
 The zone=.6236X3rfx(R'+r2+i<i'.) 
 
 Ex. 1. If the diameter of one end of a spherical zone is 
 24 feet, the diameter of the other end 20 feet, and the dis- 
 tance of the two ends, or the height of the zone 4 feet ; 
 what is the solidity ? Ans. 1566.6 feet. 
 
 2. If the earth be a sphere 7930 miles in diameter, and 
 the obliquity of the ecliptic 23** i?8' ; what is the solidity of 
 one of the temperate zones ? 
 
 Ans. 55,390,500,000 miles. 
 
 3. What is the solidity of the torrid zone ? 
 
 Ans. 147,720,000,000 miles. 
 
 The solidity of the two temperate zones is 110,781,000,000 
 of the two frigid zones 2,606,000,000 
 
 of the torrid zone 147,720,000,000 
 
 of the whole globe 261,107,000,000 
 
 4. What is the convex surface of a spherical zone, whose 
 breadth is 4 feet, on a sphere of 25 feet diameter ? 
 
76 MENSURATION OF SOLIDS. 
 
 5. What is the solidity of a spherical segment, whose 
 teight is 18 feet, and the diameter of its base 40 feet ? 
 
 PROMISCUOUS EXAMPLES OF SOLIDS. 
 
 Ex. 1. How much water can be put into a cubical vessel 
 three feet deep, which has been previously filled with cannon 
 balls of the same size, 2, 4, 6, or 9 inches in diameter, regu- 
 larly arranged in tiers, one directly above another ? 
 
 Ans. 96^ wine gallons. 
 
 2. If a cone or pyramid, whose height is three feet, be 
 divided into three equal portions, by sections parallel to the 
 base ; what will be the heights of the several parts ? 
 
 Ans. 24,961, 6-.488, and 4.551 inches. 
 
 3. What is the solidity of the greatest square prism which 
 can be cut from a cylindrical stick of timber, 2 feet 6 
 inches in diameter and 56 feet long ?* 
 
 Ans. 175 cubic feet. 
 
 4. How many such globes as the earth are equal in bulk 
 to the sun; if the former is 7930 miles in diameter, and the 
 latter 890,000 ? Ans. 1,413,678. 
 
 * The common rule for measuring round timber is to multiply the 
 square of the quarter-girt by the length. The quarter-girt is one-fourth 
 of the circumference. This method does not give the whole solidity. It 
 makes an allowance of about one-fifth, for waste in hewing, bark, &c. 
 The solidity of a cyUnder is equal to the product of the height into the 
 area of the base. 
 
 If C=the circumference, and t=3.14159, then (Art. 31.) 
 
 The area of the base^— =(—--)=:(— —-) 
 4s- \ y/47r/ V3.545/ 
 
 If then the circumference were divided by 3.545, instead of 4, and the 
 quotient squared, the area of the base would be correctly found. See 
 note B. 
 
MENSURATION OF SOLIDS. 77 
 
 5. How many cubic feet of wall are there in a conical 
 tower 66 feet high, if the diameter of the base be 20 feet 
 from outside to outside, and the diameter of the top 8 feet ; 
 the thickness of the wall being 4 feet^t the bottom, and de- 
 creasing regularly, so as to be only two feet at the top ? 
 
 Ans. Viae. 
 
 6. If a metallic globe filled with wine, which cost as much 
 at 5 dollars a gallon, as the globe itself at 20 cents for every 
 square inch of its surface ; what is the diameter of the globe ? 
 
 Ans. 55.44 inches. 
 
 7. If the circumference of the earth be 25,000 miles, 
 what must be the diameter of a metallic globe, which, when 
 drawn into a wire j^ oi an inch in diameter, would reach 
 round the earth ? Ans. 15 feet and 1 inch. 
 
 8. If a conical cistern be 3 feet deep, 7^ feet in diameter 
 at the bottom, and 5 feet at the top ; what will be the depth 
 of a fluid occupying half its capacity ? 
 
 Ans. 14.535 inches. 
 
 9. If a globe 20 inches in diameter, be perforated by a 
 cylinder 16 inches in diameter, the axis of the latter passing 
 through the centre of the former ; what part of the solidity, 
 and the surface of the globe, will be cut away by the cyl- 
 inder ? 
 
 Ans. 3284 inches of the solidity, and 502,655 of the surface. 
 
 10. What is the solidity of the greatest cube which can 
 be cut from a sphere three feet in diameter ? 
 
 Ans. 6f feet. 
 
 11. What i3 the solidity of a conic fhistum, the altitude 
 of which is 36 feet, the greater diameter 16, and the lesser 
 diameter 8 f 
 
 12. What is the solidity of a spherical segment 4 feet 
 high, cut from a sphere 16 feet in diameter ? 
 
VS ISOPERIMETRY. 
 
 SECTION V. 
 
 ISOPERIMETRY. 
 
 Art. 77. It is often necessary to compare a number of 
 different figures or solids, for the purpose of ascertaining 
 which has the greatest area, within a given perimeter, or the 
 greatest capacity under a given surface. We may have oc- 
 casion to determine, for instance, what must be the form of 
 a fort, to contain a given number of troops, with the least 
 extent of wall ; or what the shape of a metallic pipe to con- 
 vey a given portion of water, or of a cistern to hold a given 
 quantity of liquor, with the least expense of materials. 
 
 78. Figures which have equal perimeters are called Iso- 
 perimeters. When a quantity is greater than any other of the 
 same class, it is called a maximum. A multitude of straight 
 lines, of different lengths, may be drawn within a circle. 
 But among them all, the diameter is a maximum. Of all 
 sines of angles, which can be drawn in a circle, the sine of 
 90° is a maximum. 
 
 When a quantity is less than any other of the same class, 
 it is called a minimum. Thus, of all straight lines drawn 
 from a given point to a given straight line, that which is per- 
 pendicular to the given line is a minimum. Of all straight 
 lines drawn from a given point in a circle, to the circumfer- 
 ence, the maximum and the minimum are the two parts of 
 the diameter which pass through that point. (Euc. 7, 3.) 
 
 In isoperimetry, the object is to determine, on the one 
 hand, in what cases the area is a maximum, within a given 
 perimeter ; or the capacity a maximum, within a given sur- 
 face : and on the other hand, in what cases the perimeter is 
 
ISOPERIMETRY. Vd 
 
 a minimum for a given area, or the surface a minimumf for a 
 given capacity. ^ 
 
 Proposition I. 
 
 79. An Isosceles Triangle has a greater area than any 
 scalene triangle, of equal base and perimeter. 
 
 If ABC be an isosceles trian- 
 gle whose equal sides are AC and 
 i)C ; and if ABC be a scalene tri- 
 angle on the same base AB, and 
 having AC'-f-BC' = AC+BC; 
 ilien the area of ABC is greater 
 than that of ABC. 
 
 Let perpendiculars be raised 
 from each end of the base, extend 
 AC to D, make CD' equal to AC, join BD, and draw CH 
 and CH' parallel to AB. 
 
 As the angle CAB=ABC, (Euc. 5, 1.) and ABD is a right 
 
 angle, ABC-f CBD=CAB+CDB = ABC+CDB. Therefore 
 
 CBD = CDB, so that CD=CB ; and by constniction, CD'= 
 
 AC. The perpendiculars of the equal right angled triangles 
 
 CHD and CHB are equal; therefore, BH=iBD. In the 
 
 ime manner, AH'=iAD'. The line AD=AC+BC=AC 
 
 HBC'=D'C+BC'. But D'C+BOBD'. (Euc. 20, 1.) 
 
 rhcrefore, AD>BD' ; BD>AD', (Euc. 47, 1.) and i BD> 
 
 i AD'. But iBD, or BH, is the height of the isosceles tri- 
 
 ngle ; (Art. 1.) and |AD' or AH', the height of the scalene 
 
 i iii'^'le ; and the areas of two triangles which have the same 
 
 ire as their heights. (Art. 8.) Therefore the area of 
 
 \ijC is greater than that of ABC. Among all triangles, 
 
 ; lien, of a given perimeter, and upon a given base, the isos 
 
 eles triangle is a maximum. 
 
 Cor. The isosceles triangle has a less perimeter than any 
 
 ■calene triangle of the same base and area. The triangle 
 
80 
 
 ISOPERIMETRY. 
 
 ABC being less than ABC, it is evident the perimeter of the 
 former must be enlarged, to make its area equal to the area 
 of the latter. 
 
 Proposition II. 
 
 80. A triangle in which two given sides make a right 
 ANGLE, has a greater area than any triangle in which the same 
 sides maJce an oblique angle. 
 
 If BC, BC and BC" be equal, 
 and if BC be perpendicular to 
 AB ; then the ric^ht anfyled trian- 
 gle ABC, has a greater area than 
 the acute angled triangle 2VBC', or 
 the oblique angled triangle AB C". 
 
 Let P'C and PC" be perpen- 
 dicular to AP. Then, as the 
 three triangles have the same base AB, their areas are as 
 their heights ; that is, as the perpendiculars BC, P'C, and 
 PC". But BC is equal to BC, and therefore greater than 
 P'C. (Euc. 47. 1.) BC is also equal to BC", and therefore 
 greater than PC". 
 
 Proposition III. 
 
 81. If all the sides except one of a polygon be given, 
 the area will be the greatest, when the given sides are so dis- 
 2)osed that the figure may he inscribed in a semicircle, of 
 which the undetermined side is the diameter. 
 
 If the sides AB, BC, CD, DE, 
 be given, and if their position 
 be such that the area, included 
 between these and another side 
 whose length is not determined, 
 is a nmximum ; the figure may 
 
I 
 
 ISOPERIMETRY. 81 
 
 be inscribed in a semicircle, of which the undetermined side 
 AE is the diameter. 
 
 Draw the lines AD, AC, EB, EC. By varying the angle 
 at D, the triangle ADE may be enlarged or diminished, with- 
 out affecting the area of the other parts of the figure. The 
 whole area, therefore, cannot be a maximum, unless this tri- 
 angle be a maximum, while the sides AD and ED are given. 
 But if the triangle ADE be a maximum, under these con- 
 ditions, the angle ADE is a right angle ; (Art. 80.) and 
 therefore the point D is in the circumference of a circle, of 
 ^vhich AE is the diameter. (Euc. 31,3.) In the same man- 
 ner it may be proved, that the angles ACE and ABE are 
 right angles, and therefore that the points C and B are in 
 tlie circumference of the same circle. 
 
 The term polygon is used in this section to include trian- 
 fjles, and four-sided figures, as well as other right-lined 
 figures. 
 
 82. The area of a polygon, inscribed in a semicircle, in 
 the manner stated above, will not be altered by varying the 
 order of the given sides. 
 
 The sides AB, BC, CD, DE, are the chords of so 
 many arcs. The sum of these arcs, in whatever order 
 they are arranged, will evidently be equal to the semicircum- 
 ference. Audi the segments between the given sides and the 
 arcs will be the same in whatever part of the circle they are 
 -ituated. But the area of the polygon is equal to the area 
 f the semicircle, diminished by the sum of these segments. 
 
 83. If a polygon, of which all the sides except one are 
 given, be inscribed in a semicircle whose diameter is the un- 
 determined side ; a polygon having the same given sides, 
 cannot be inscribed in any other semicircle which is either 
 greater or less than this, and whose diameter is the undeter- 
 mined side. 
 
 The given sides AB, BC, CD, DE, are the chords of arcs 
 whose sum is 180 degrees. But in a larger circle, each 
 
 4* 
 
82 
 
 ISOPERIMETRY. 
 
 would be the chord of a less number of degrees, and there- 
 fore the sum of the .arcs would be less than 180° : and in a 
 smaller circle, each would be the chord of a greater number 
 of degrees, and the sum of the arcs would be greater than 
 180°. 
 
 Proposition IV. 
 
 84. A polygon inscribed in a circle has a greater area, 
 than any polygon of equal perimeter, and the same number of 
 sides, which cannot he inscribed in a circle. 
 
 If in the circle ACHF, (Fig. 30.) there be inscribed a 
 c 
 
 polygon ABCDEFG ; and if another polygon ahcdefg (Fig. 
 31.) be formed of sides which are the same in number and 
 length, but which are so disposed, that the figure cannot be 
 inscribed in a circle; the area of the former polygon is 
 greater than that of the latter. 
 
 Draw the diameter AH, and the chords DH and EH. 
 Upon de make the triangle deh equal and similar to DEH, 
 and join ah. The line ah divides the figure abcdhefg into two 
 parts, of which one at least cannot, by supposition, be in- 
 scribed in a semicircle of which the diameter is AH, nor in 
 any other semicircle of which the diameter is the undeter- 
 mined side. (Art. 83.) It is therefore less than the corres- 
 ponding part of the figure ABCDHEFG. (Art. 81.) And 
 the other part of abcdhefg is not greater than the correspond- 
 
ISOPERIMETRY. 83 
 
 ing part of ABCDHEFiS. Therefore, the whole figure 
 ABCDIIEFG is greater than the whole figure ahcdhefy. If 
 from these there be taken the equal triangles DEU and dch, 
 there will remain the polygon ABCDEFG greater than the 
 polygon ahcdcfg. 
 
 85. A polygon of which all the sides are given in num- 
 ber and length, cannot be inscribed in circles of different 
 diameters. (Art. 83.) And the area of the polygon will not 
 be altered by changing the order of the sides. (Art. 82.) 
 
 Proposition V. 
 
 86. When a polygon has a greater area than any other, of 
 the same number of sides, and of equal perimeter, the sides are 
 
 EQUAL. 
 
 The polygon ABCDF (Fig. 29.) 
 cannot be a maximum, among all 
 polygons of the same number of 
 sides, and of equal perimeters, un- 
 less it be equilateral. For if any 
 two of the sides, as CD and FD, 
 are unequal, let CH and FH be 
 
 lual, and their sum the same as 
 
 lie sum of CD and FD. The 
 
 ')sceles triangle CHF is greater than the scalene triangle 
 L'DF (Art. 19.); and therefore the polygon ABCIIF is 
 greater than the polygon ABCDF ; so that the latter is not 
 a maximum, 
 
 PBOP08inoN*VL 
 
 87. A REQULAB POLraoN has a greater area than any 
 other polygon of equal perimeter, and of the same number of 
 sides. 
 
84 ISOPERIMETRY 
 
 For, by tlie preceding article,. tlie polygon which is a max- 
 imum among others of equal perimeters, and the same num- 
 ber of sides, is equilateral, and by Art. 84, it may be in- 
 scribed in a circle. But if a poly- 
 gon inscribed in a circle is equilat- 
 eral, as ABDFGH, it is also equian- 
 gular. For the sides of the polygon 
 are the bases of so many isosceles 
 triangles, whose common vertex is 
 the centre C. The angles at these 
 bases are all equal ; and two of them, 
 
 as AHC and GHC, are equal to AHG one of the angles of 
 the polygon, ^he polygon, then, being equiangular, as well 
 as equilateral, is a re^z^Zar polygon. (Art. 1. Def. 2.) 
 
 Thus an equilateral triangle has a greater area, than any 
 other triangle of equal perimeter. And a square has a 
 greater area than any other four-sided figure of equal pe- 
 rimeter; 
 
 Cor. A regular polygon has a less perimeter than any 
 other polygon of equal area, and the same number of 
 sides. 
 
 For if, with a given perimeter, the regular polygon is 
 greater than one which is not regular ; it is evident the pe- 
 rimeter of the former must be diminished, to make its area 
 equal to that of the latter. 
 
 Proposition VII. 
 
 88. If a polggon he BESCRiBEB about a circle, the areas 
 of the two figures are as their ^perimeters. 
 
 Let ST be one of the sides of a polygon, either regular or 
 
ISOPERIMETRY. 
 
 85 
 
 T 1 
 
 A 8 
 
 A 
 
 
 /. \ 
 
 V 
 
 y 
 
 not, which is described about the cir- 
 cle LUR. Join OS and OT, and 
 to the point of contact M draw the 
 radius OM,-«|»ich will be perpen- 
 dicular to ST. (Euc. 18, 3.) The 
 triangle OST is equal to half the 
 liase ST multiplied into the radius 
 >M. (Art. 8.) And if lines b« 
 drawn, in the same manner, from 
 
 the centre of the circle, to the extremities of the sev- 
 tral sides of the circumscribed polygon, each of the trian- 
 j;lcs thus formed will be equal to half its base multiplied 
 into the radius of the circle. Therefore t^e area of the 
 ^vllole polygon is equal to half its perimeter multiplied into 
 the radius : and the area of the circle is equal to half its cir- 
 cumference multiplied into the radius. (Art 30.) So that 
 ' he two areas aic to each other as their perimeters. 
 
 Cor. 1. If diflferent polygons are described about the 
 >.ime circle, their areas are to each other as their perimeters. 
 For the area of each is equal to half its perimeter, multi- 
 j)lied into the radius of the inscribed circle. 
 
 Cor. 2. The tangent of an arc is always greater than the 
 arc itself. The triangle OMT is to OMN, as MT to MN. 
 But OMT is greater than OMN, because the former includes 
 the latter. Therefore, the tangent MT is greater than the 
 arc MN. 
 
 Proposition VIII. 
 
 89. A CIRCLE has a greater area than any polygon of equal 
 perimeter. ^ 
 
 If a circle and a regular polygon have the same centre, 
 and equal perimeters ; each of the sides of the polygon 
 must fall partly within the circle. For the area of a circum-» 
 
A A 
 
 1 \/o 1 
 
 V 
 
 A 
 
 A^--^ 
 
 r-^^^D 
 
 86 ISOPERIMETRY. 
 
 scribing polygon is greater tlian the area of the circle, as 
 the one includes the other : and therefore, by the preceding 
 article, the perimeter of the former is greater than that of 
 the latter. 
 
 Let AD then be one side of a 
 regular polygon, whos* perimeter 
 is equal to the circumference of 
 the circle RLN. As this falls 
 partly within the circle, the per- 
 pendicular OP is less than the 
 radius OR. But the area of the 
 
 polygon is equal to half its pe- " ~ 
 
 rimeter multiplied into this per- 
 pendicular (Art. 15.) ; and the area of the circle is equal to 
 half its circumference multiplied into the radius. (Art. 30.) 
 The circle then is greater than the given regular polygon ; 
 and therefore greater than any other polygon of equal pe- 
 rimeter. (Art. 87.) 
 
 Cor. 1. A circle has a less perimeter, than any polygon of 
 equal area. 
 
 Cor. 2. Among regular polygons of a given perimeter, 
 that which has the greatest number of sides, has also the 
 greatest area. For the greater the number of sides, the 
 more nearly does the perimeter of the polygon approach to 
 a coincidence with the circumference of a circle. 
 
 Proposition IX. 
 
 90. A right prism whose bases are regular polygons, has 
 a less surface than any other right prism of the. same solidity , 
 the same altitude, and the same number of sides. 
 
 If the altitude of a prism is given, the area of the base is 
 as the soHdity (Art. 43.) ; and if the number of sides is 
 
isorKiiiM!:juv. 87 
 
 perimeter is & minimurn vf hen the base is a 
 
 i ^ 1. (Art. 87. Cor.) But the lateral surface is 
 
 s the perimeter. (Art- ^y.)^ Of two right prisms, then, 
 
 vhich have the same altitude, the same solidity, and the 
 
 me number of sides, that whose bases are regular polygons 
 
 iias the least lateral surface, while the areas of the ends are 
 
 equal. 
 
 Cor. A right prism whose bases are regular polygons has 
 a greater solidity, than any other right prism of the same 
 surface, the same altitude, and the same number of sides. 
 
 PROPosmoN X. 
 
 91. A right cruNDER has a less surface than any right 
 prism of the same altitude and solidity. 
 
 For if the prism and cylinder have the same altitude and 
 solidity, the areas of their bases are equal. (Art. 64.) But 
 the perimeter of the cylinder is less, than that of the prism 
 (Art. 89. Cor. 1.) ; and therefore its lateral surface is less, 
 while the areas of the ends are equal. 
 
 Cor. A right cylinder has a greater solidity, than any right 
 prism of the same altitude and surface. 
 
 PuorosiTioN XI. 
 
 92. A CUBE has a less surface titan any other right paral- 
 lelepiped of the same solidity. 
 
 A parallelepiped is a prism, any one of whose faces may 
 be considered a base. (Art. 41. Def. I and V.) If these are 
 not all squares, let one which is not a square be taken for a 
 base. The perimeter of this may be diminished, without 
 altering its area (Art. 87. Cor.); and therefore the surface 
 
88 ISOPERIMETRY. 
 
 of tlie solid may be diminislied, without altering its altitude 
 or solidity. (Art. 43, 47.) The same may be proved of 
 each of the other faces which are not squares. The surface 
 is therefore a minimum, when all the faces are squares, that 
 is, when the solid is a cuhc. 
 
 Cor. A cube has a greater solidity than any other right 
 parallelopiped of the same surface. 
 
 Proposition XII. 
 
 ^93. A CUBE has a greater solidity than any other right par- 
 allelojriped, the sum of whose length, breadth and depth, is equal 
 to ike sum of the corresponding dimensions of the cube. 
 
 The solidity is equal to the product of the length, breadth, 
 and depth. If the length and breadth are unequal, the 
 solidity may be increased, without altering the sum of the 
 three dimensions. For the product of two factors whose 
 sum is given, is the greatest when the factors are equal. 
 (Euc. 27. 6.) In the same manner, if* the breadth and 
 depth are unequal, the solidity may be increased, without 
 altering the sum of the three dimensions. Therefore, the 
 solid cannot be a maximum, unless its length, breadth, and 
 depth are equal. 
 
 Proposition XIII. 
 
 94. If a PRISM BE DESCRIBED ABOUT A CYLINDER, the 
 
 cap>acities of the two solids are as their surfaces. 
 
 The capacities of the solids are as the areas of their bases, 
 that is, as the perimeters of their bases. (Art. 88.) But the 
 lateral surfaces are also as the perimeters of the bases. 
 Therefore the whole surfaces are as the solidities. 
 
 Cor. The capacities of different prisms, described about 
 the same right cylinder, are to each other as their surfaces. 
 
ISOrERIMETUY. 80 
 
 Proposition XIV. 
 
 J... J right cylinder whose height is equal to the 
 DIAMETER OF ITS BASE hos a greater solidity t/iak any other 
 right cylinder of equal surface. 
 
 Let C be a rij^ht cylinder whose lieight is equal to the di- 
 ameter of its base ; and C another right cylinder having the 
 same surface, but a different altitude. If a square prism P 
 be described about the former, it will be a cube. But a 
 square prism P' described about the latter will not be a cube. 
 
 Then the surfaces of C and P are as their bases (Art. 47. 
 and 88.) ; which arc as the bases of C and P', (Sup. Euc. 
 7, 1.); 80 that, 
 
 surf C : aurfV : : base C : base V : : base C t base P' : : surfC \ 
 mrfV. 
 
 But the surface of C is, by supposition, equal to the sur- 
 face of C Therefore, (Alg. 396.) the surface of P is equal 
 to the surface of P'. And by the preceding article, 
 
 solid P : solid C : : surfP : sur/C : : siirfV : surfC : : solid 
 V : solid C. 
 
 Cut the solidity of P is greater than that of P'. (Art. 92. 
 Cor.) Therefore the solidity of C is greater than that of C. 
 
 Schol. A right cylinder whose height is equal to the di- 
 ameter of its base, is that which circumscribes a sjihere. It 
 is also called Archimedes* cylinder ; as he discovered the 
 ratio of a sphere to its circumscribing cylinder ; and these 
 are the figures which were put upon his tomb. 
 
 Cor. Archimedes* cylinder has a less surface^ than any 
 other right cylinder of the same capacity. -Mt^'* 
 
90 ISOPERIMETRY. 
 
 Proposition XY. 
 
 96. If a SPHERE BE CIRCUMSCRIBED hy a solid hounded hy 
 'plane surfaces ; the capacities of the two solids are as their 
 surfaces. 
 
 If planes be supposed to be drawn from the centre of the 
 sphere, to each of the edges of the circumscribing soHd, 
 they will divide it into as many pyramids as the solid has 
 faces. The base of each pyramid will be one of the faces ; 
 and the height will be the radius of the sphere. The 
 capacity of the pyramid will be equal, therefore, to its base 
 multiplied into -^ of the radius (Art. 48.) ; and the capacity 
 of the whole circumscribing solid, must be equal to its whole 
 surface multiplied into ^ of the radius. But the capacity of 
 the sphere is also equal to its surface multiplied into ^ of its 
 radius. (Art. '70.) 
 
 Cor. The capacities of different solids circumscribing the 
 same sphere, are as their surfaces. 
 
 Proposition XVI. 
 
 97. A SPHERE has a greater solidity than any 7'egularpoly- 
 edron of equal suyface. 
 
 If a sphere and a regular polyedron have the same centre, 
 and equal surfaces ; each of the faces of the polyedron must 
 fall partly 2vithin the sphere. For the solidity of a circum- 
 scriUng solid is greater than the solidity of the sphere, as 
 the one includes the other : and therefore, by the preceding 
 article, the surface of the former is greater than that of the 
 latter. 
 
 But if the faces of the polyedron fall partly within the 
 sphere, their perpendicular distance from the centre must be 
 less than the radius. And therefore, if the surface of the 
 
180PERIMETRY. 91 
 
 polyedron be only equal to that of the sphere, its solidity 
 must be less. For the solidity of the polyedron is equal to 
 its surface multiplied into -J of the distance from the centre. 
 (Art. 59.) And the solidity of the sphere is equal to its 
 surface multiplied into -^ of the radius. 
 
 Cor. A sphere has a less surface than any regular poly- 
 edron of the same capacity. 
 
APPENDIX 
 
 GAUGING OF CASKS. 
 
 Art. 119. Gauging is a practical art, which does not ad- 
 mit of being treated in a very scientific manner. Casks are 
 not commonly constructed in exact conformity with any reg- 
 ular mathematical figure. By most writers on the subject, 
 however, they are considered r as nearly coinciding with one 
 of the following forms : 
 
 1. ? ^, .-,,,/. ( of a spheroid, 
 
 2. 1 ^^^ ^^^^^" ^^^'*^^ i of a parabolic spindle. 
 3- I rri.. ..„„! ^.„..,„.. S ^f a paraboloid. 
 
 .' f The equal frustums ] . 
 
 4. ) ^ ( of a cone. 
 
 The second of these varieties agrees more nearly than any 
 of the others, with the forms of casks, as they are com- 
 monly made. The first is too much curved, the third too 
 little, and the fourth not at all, from the head to the bung. 
 
 120. Rules have already been given, for finding the capa- 
 city of each of the four varieties of casks. (Arts. 68, 110, 
 112, 118.) As the dimensions are taken in mcAes, these rules 
 will give the contents in cubic inches. To abridge the com- 
 putation, and adapt it to the particular measures used in 
 gauging, the factor .'7854 is divided by 282 or 231 ; and 
 the quotient is used instead of .7854, for finding the capa- 
 city in ale gallons or wine gallons. 
 
GAUGING. 9S 
 
 Now ^^rr^ =.002785, or .0028 nearly ; 
 
 And -l^^.OOSi 
 231 
 
 If then .0028 and .0034 be substituted for .'7854, in the 
 rules referred to above; the contents of the cask will be 
 given in ale gallons and wine gallons. These numbers are 
 to eacli other nearly as 9 to 11. 
 
 Problem I. 
 
 To calculate the contents of a cask, in the form of a middle 
 frustum of a spheroid. 
 
 121. Add together the square of the head diameter, and 
 twice the square of the bung diameter : multiply the sum by 
 ^ of the length, and the product by .0028 for ale gallons, or 
 by .0034 for wine gallons. 
 
 If D and fl?=the two diameters, and /=the length ; 
 
 The capacity in inches=(2D'+c?')Xi/X.'7854. (Art. 110.) 
 
 And by substituting .0028 or .0034 for .7854, we have 
 the capacity in ale gallons or wine gallons. 
 
 Ex. What is the capacity of a cask of the first form, 
 whose length is 30 inches, its head diameter 18, and its bung 
 diameter 24 ? 
 
 Ans. 41.3 ale gallons, or 50.2 wine gallons. 
 
 Problem II. 
 
 To calculate t/u: contents of a cask, in the form of tlie mid- 
 dle frustum of a parabouc spixdle. 
 
 122. Add together the square of the head diameter, and 
 twice the square of the bung diameter, and from the sum 
 
94 GAUGING. 
 
 subtract | of the square of the difference of the diaraeters ; 
 multiply the remainder by -^ of the length, and the product 
 by .0028 for ale gallons, or .0034 for wine gallons. 
 
 The capacity in inches ={2J)^+cP—l (D—dy)x¥X 
 .Y854. (Art. 118.) 
 
 Ex. What is the capacity of a cask of the second form, 
 whose length is 30 inches, its head diameter 18, and its 
 bung diameter 24 ? 
 
 Ans. 40.9 ale gallons, or 49.7 wine gallons. 
 
 Problem III. 
 
 To calculate the contents of a cask, in the form of two equal 
 frustums of a paraboloid. 
 
 123. Add together the square of the head diameter, and 
 the square of the bung diameter ; multiply the sum by half 
 the length, and the product by .0028 for ale gallons, or 
 .0034 for wine gallons. 
 
 The capacity in inches =(D'+d')xilX.'7854. (Art. 112 
 Cor.) 
 
 Ex. What is the capacity of a cask of the third form, 
 ■whose dimensions are, as before, 30, 18, and 24 ? 
 
 Ans. 37.8 ale gallons, or 45.9 wine gallons. 
 
 Problem IV. 
 
 To calculate the contents of a cash, in the form of tivo equal 
 frustums of a coine. 
 
 124. Add together the square of the head diameter, the 
 square of the bung diameter, and the product of the two 
 diameters ; multiply the sum by \ of the length, and the 
 product by.0028 for ale gallons, or .0034 for wine gallons. 
 The capacity in inches=(D-+c^'+Dc?)x-J^X.7854. (Art. 68.) 
 
GAUGINO. 95 
 
 Ex. What is the capacity of a cask of the fourth form, 
 whoso length is 30, and its diameters 18 and 24 ? 
 
 Ans. 37.3 .ale gallons, or 45.3 wine gallons. 
 
 12o. The precedini^ rules, though correct in theory, are 
 not very well adapted to practice, as they suppose the form 
 of the cask to be Jcnoion^ The two following rules, taken 
 from Hut ton's Mensuration, may be used for casks of the 
 usual forms. For the first, three dimensions are required, 
 the length, the head diameter, and the bung diameter. It 
 is evident tliat no allowance is made by this, for different 
 degrees of curvature from the head to the bung. If the 
 cask is more or less curved than usual, the following rule is 
 to be preferred, for which /owr dimensions are required, the 
 head and bung diameters, and a third diameter taken in the 
 middle between the bung and the head. For the demon- 
 stration of these rules, see Hutton's Mensuration, Part V. 
 Sec. 2. Ch. 5 and 7. 
 
 Problem V. 
 
 To calculate the contents of any common cask, from three 
 dimcnsums. 
 
 126. Add together 
 
 25 times the square of the head diameter, 
 39 times the square of the bung diameter, and 
 20 times the product of the two diameters ; 
 Multiply the sum by the length, divide the product by 90, 
 and multiply the quotient by .0028 for ale gallons, or .0034 
 for wine gallons. 
 
 The capacity in inches=(39 D'-f25cr'4-26D(/)x - X. 7854. 
 
 Ex. Wliat is the capacity of a cjisk whose length is 30 
 inches, the head diameter 18, and the bung diameter 24? 
 Ans, 39 ale gallons, or 47^ wine gallons. 
 
96 GAUGING. 
 
 Problem VI. 
 
 2h calculate the contents of a cask from four dimensions, the 
 length, the head and hung diameters, and a diameter taken 
 in the middle between the head and the hung. 
 
 127. Add together the squares of the head diameter, of 
 the bung diameter, and of double the middle diameter ; 
 multiply the sum by -^ of the length, and the product by 
 .0028 for ale gallons, or .0034 for wine gallons. 
 
 If D=the bung diameter, c:?=the head diameter, m=the 
 middle diameter, and Z=the length ; 
 
 The capacity in inches=(D'-|-c?'+2m^)Xi?X.'7854. 
 
 Ex. What is the capacity of a cask, whose length is 30 
 inches, the head diameter 18, the bung diameter 24, and the 
 middle diameter 22^ ? 
 
 Ans. 41 ale gallons, or 49| wine gallons. 
 
 128. In making the calculations in gauging, according to 
 the preceding rules, the multiplications and divisions are fre- 
 quently performed by means of a Sliding Rule, on which 
 are placed a number of logarithmic lines, similar to those on 
 Gunter's Scale. See Trigonometry, Sec. VI., and Note C. 
 p. 149. 
 
 Another instniment commonly used in gauging is the 
 Diagonal Rod. By this, the capacity of a cask is very ex- 
 peditiously found, from a single dimension, the distance from 
 the bung to the intersection of the opposite stave with the 
 head ; but this process is not considered sufficiently accurate 
 for casks of a capacity exceeding 40 gallons. The measure 
 is taken by extending the rod through the cask, from the 
 bung to the most distant part of the head. The number of 
 gallons corresponding to the length of the line thus found, is 
 marked on the rod. The logarithmic lines on the gauging 
 
GAUGING. 97 
 
 rod are to be used in the same manner, as on the sliding 
 rule. 
 
 ULLAGE OF CASKS. 
 
 129. When a cask is partly filled, the whole capacity is 
 divided, by the surface of the liquor, into two portions ; the 
 least of which, whether full or empty, is called the ullage. 
 In finding the ullage, t]«e cask is supposed to be in one of 
 two positions ; either standinr/, with its axis perpendicular to 
 the horizon ; oj- lyi7i(/, with its axis parallel to the horizon. 
 The rules for ullage which are exact, particularly those for 
 lying casks, are too complicated for common use. The fol- 
 lowing are considered as sufficiently near approximations. 
 See Hutton's Mensuration. 
 
 Problem VII. 
 
 To calculate the ullage of a standing cask. 
 
 130. Add together the squares of the diameter at the sur- 
 face of the liquor, of the diameter of the nearest end, and 
 of double the diameter in the middle between the other two ; 
 multiply the sum by \ of the distance between the surface 
 and the nearest end, and the product by .0028 for ale gal- 
 lons, or .0034 for wine gallons. 
 
 If D=the diameter of the surface of the liquor, 
 rf=the diameter of the nearest end, 
 m=the middle diameter, and 
 
 /=the distance between the surface and the nearest end ; 
 The ullage in inches=(D'+rf'+2m^)Xi/X-V8o4. 
 
 Ex. If the diameter at the surface of the liquor, in a stand- 
 ing cask, be 32 inches, the diameter of the nearest end 24, 
 the middle diameter 29, and the distance between the sur- 
 
 5 
 
98 GAUGING. 
 
 face of the liquor and the nearest end 1 2 ; what is the ul- 
 lage? Ans. 2li ale gallons, or 33^ wine gallons. 
 
 Problem VIII. 
 To calculate the ullage of a lying cask. 
 
 131. Divide the distance from the buna: to the surface of 
 the liquor, by the whole bung diameter, find the quotient in 
 the column of heights or versed sines in a table of circular 
 segments, take out the corresponding segment, and multiply 
 it by the whole capacity of the cask, and the product by 1-^ 
 for the part which is empty. 
 
 If the cask be not half full, divide the depth of the liquor 
 by the whole bung diameter, take out the segment, multiply, 
 &c., for the contents of the part which is full. 
 
 Ex. If the whole capacity of a lying cask be 41 ale gal- 
 lons, or 49-f wine gallons, the bung diameter 24 inches and 
 the distance from the bung to the surface of the liquor 6 
 inches ; what is the ullage ? 
 
 Ans. Vf- ale gallons, or 9^ Avine gallons. 
 
NOTES 
 
 Note A. p. 39. , 
 
 The term solidity i» used here in the customary sense, to 
 express the magnitude of any geometrical quantity of three 
 dimensions, length, breadth, and thickness ; whether it be a 
 solid body, or a fluid, or even a portion of empty space. This 
 use of the word, however, is not altogether free from objec- 
 tion. The same term is applied to one of the general prop- 
 erties of matter ; and also to that peculiar quality by which 
 certain substances are distinguished from fluids. There 
 seems to be an impropriety in speaking of the solidity of a 
 body of water^ or of a vessel which is ejnpty. Some writers 
 have therefore substituted the word volume for solidity. But 
 the latter term, if it be properly defined, may be retained 
 without danger of leading to mistake. 
 
 Note B. p. 76. 
 
 The following simple rule for the solidity of round timber, 
 or of any cylinder, is nearly exact : 
 
 Multiply the length into twice the square of ^ of the circum,' 
 fereiice. 
 
 If C=the circumference of a cylinder; 
 
 The area of the base=-^=— ^^— But 2(^\ =-^1- 
 4^ 12.566 \5/ 12.5 
 
 It is common to mea.siire heton timber, by multiplying the 
 length into the square of the quarter-ffirt. This gives ex- 
 
100 NOTES. 
 
 actly the solidity of a parallelopiped, if tLc ends are squares. 
 But if the ends are parallelograms, the area of each is less 
 than the square of the quarter-girt. (Euc. 27. 6.) 
 
 Timber which is taiKring may be exactly measured by the 
 rule for the frustum of a pyramid or cone (Art. 50, 68.) ; 
 or, if the ends are not similar figures, by the rule for a pris- 
 moid. (Art. 55.) But for common purposes, it will be suf- 
 ficient to multiply the length by the area of a section in the 
 middle between the two ends. 
 
^ 
 
 4 
 
 
 
 NATURAL 
 
 TAXCKN'TS. 
 
 
 
 
 
 M 
 
 
 
 44 Degrees. 
 
 M 
 
 60 
 
 31 
 
 44 Degrees. 
 
 M 
 
 29 
 
 
 N.Tan. 
 1)0509 
 
 N.Cot. 
 1.03553 
 
 N. Til 11. 
 
 N. Cot. 
 
 
 98327 
 
 1.01702 
 
 
 
 1 
 
 90025 
 
 1.03493 
 
 59 
 
 .32 
 
 98384 
 
 1.01642 
 
 28 
 
 
 
 2 
 
 90081 
 
 1.03433 
 
 58 
 
 33 
 
 98441 
 
 1.01583 
 
 27 
 
 
 
 3 
 
 90738 
 
 1.03372 
 
 57 
 
 34 
 
 98499 
 
 1.01524 
 
 26 
 
 
 
 4 
 
 9671)4 
 
 1.03312 
 
 56 
 
 35 
 
 98556 
 
 1.01405 
 
 25 
 
 
 
 5 
 
 96850 
 
 1.03252 
 
 55 
 
 36 
 
 98613 
 
 1.01406 
 
 24 
 
 
 
 6 
 
 90907 
 
 1.03192 
 
 54 
 
 37 
 
 98071 
 
 1.01347 
 
 23 
 
 
 
 7 
 
 90963 
 
 1.03132 
 
 53 
 
 38 
 
 98728 
 
 1.01288 
 
 22 
 
 
 
 8 
 
 97020 
 
 1.03072 
 
 52 
 
 39 
 
 98780 
 
 1 .01229 
 
 21 
 
 
 
 9 
 
 97076 
 
 1.03012 
 
 51 
 
 40 
 
 98843 
 
 1.01170 
 
 20 
 
 
 
 10 
 
 97133 
 
 1.02952 
 
 50 
 
 41 
 
 98901 
 
 1.01112 
 
 19 
 
 
 
 11 
 
 97189 
 
 1.02892 
 
 49 
 
 42 
 
 98958 
 
 1.01053 
 
 18 
 
 
 
 11' 
 
 97246 
 
 1.05832 
 
 48 
 
 43 
 
 99016 
 
 1.00994 
 
 17 
 
 
 
 •^■»02 
 
 1.02772 
 
 47 
 
 44 
 
 99073 
 
 1.00935 
 
 16 
 
 
 
 14 
 
 97359 
 
 1.02713 
 
 40 
 
 45 
 
 99131 
 
 1.00876 
 
 15 
 
 
 
 15 
 
 97416 
 
 1.02053 
 
 45 
 
 46 
 
 99189 
 
 1.00818 
 
 14 
 
 
 
 16 
 
 97472 
 
 1.02.193 
 
 44 
 
 47 
 
 99247 
 
 1.00759 
 
 13 
 
 
 
 17 
 
 97529 
 
 1.02533 
 
 43 
 
 48 
 
 99304 
 
 1.00701 
 
 12 
 
 
 
 18 
 
 97586 
 
 1.02474 
 
 42 
 
 49 
 
 99302 
 
 1.00042 
 
 11 
 
 
 ... 1 
 
 «. 19 
 
 97043. 
 
 1.02414 
 
 41 
 
 50 
 
 99420 
 
 1.00583 
 
 10 
 
 
 
 "■ 20 
 
 •97700- 
 
 1.02355 
 
 40 
 
 51 
 
 99478 
 
 1.00525 
 
 9' 
 
 
 
 21 
 
 97756 
 
 1.02295 
 
 39 
 
 52 1 
 
 W^^Q 
 
 
 8 
 
 
 
 • 22 
 23 
 
 a:8i3 
 
 1.02236 
 ^1.02176 
 
 38 
 37 
 
 5? 
 
 99594 ,' 
 ' 99b'52 
 
 'l-oosl 
 
 7 . 
 6 
 
 
 
 24 
 
 9792'? 
 
 1.02117 
 
 36 
 
 55 
 
 99710 
 
 l! 00291 
 
 5 
 
 
 
 25 
 
 97984 
 
 1.02057 
 
 35 
 
 56 
 
 99708 
 
 1.00233 
 
 4 
 
 
 
 26 
 
 98041 
 
 1.01998 
 
 34 
 
 57 
 
 99326 
 
 1.00175 
 
 3 
 
 
 
 27 
 
 98098 
 
 1.01939 
 
 33 
 
 58 
 
 99884 
 
 1.00116 
 
 2 
 
 
 
 28 
 
 98155 
 
 1.01879 
 
 32 
 
 59 
 
 99942 
 
 1.00058 
 
 1 
 
 
 
 29 
 
 98213 
 
 1.0L820 
 
 31 
 
 CO 
 
 10000 
 
 1.00000 
 
 
 
 
 
 - ?^. 
 
 9§?70 
 
 1.01^ 
 
 30 
 
 
 
 
 M 38 
 
 
 
 M 
 
 N. Cot. 
 
 N. Tan. • 
 
 M 
 
 HT 
 
 N.Cot. 
 
 N. Tan. 
 
 
 45 Degrees. 
 
 ^5D^ 
 
 grees. 
 
 
 
 -■ 
 
 ■ 
 
 
 
 
 
 
 
 
 
 
 
 i 
 
14 DAY USE 
 
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 This book is due on the last date stamped below, or 
 
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 ' NQV 13 1957 
 
 I7MaroCA, 
 
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 260ct59FC 
 
 MAii-^es--4m 
 
 REC'D LD 
 
 OCT 2 6 1959 
 
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 NOV 1 .> 1959 
 
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 General Library 
 
 University of California 
 
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