56 



MENSURATION. 



Mentura- NOTE. This rule also applies to a sector of a circle, 

 lion of ROLE 2. Multiply the square of the diameter by the 

 Plune Ft- nun ,i, er ,7854, the product is the area. 

 x U 7AL/ The first rule has been demonstrated in GEOMETRY, 

 Prop. VI I. Sect. 5. Part 1. 



If the diameter be supposed = 1, the circumference 

 will be 3.1415927 (10.5), and the area, 



=5 4 diaro. X $ circum. = .7853982 = .7854 nearly. 

 And since (8.5) circles are as the squares of their dia. 

 meters, 1 X 1, or 1, will be to the square of the diame- 

 ter of any circle as .7854 to the diameter of that circle : 

 Hence Rule 2. is formed. 



Ex. Find the area of a circle whose diameter is 12 

 feet. 



In this case half the circumference is 3.1416 X 6 = 

 18.8496 feet, and the area, by rule 1, 



18.8496 X 6 = 113.0976 sq.fcet the answer. 



Or, by Rule 2, the area is .7854 x 144=113.0976 

 sq. feet. 



PROBLEM X. 



To find the area of any sector of a circle. 



RULE I. Multiply the radius, or half the diameter, 

 by half the arc of the sector, and the product will be 

 the area, as in the whole circle. 



RULE II. As 360" is to the degrees in the arc of the 

 sector, so is the area of the whole circle to the area of 

 the sector. 



The first of these rules is contained in the last pro- 

 blem, and the truth of the second is sufficiently evi- 

 dent. 



fr'g- ' Ex. Required the area of a sector CADB, the angle 



ACB at the centre being 18, and the diameter three 

 feet. 



By Prob. 7, 3.1416 X 3 = 9.4248 the whole circumfe- 

 rence ; and by Prob. 8, 360 : 18 : : 9.4248 : : 47124 the 

 arc of sector: Hence, by Rule 1, .47124 x .75 = 

 .35343, the area of the sector. Otherwise, by Prob. 9, 

 the area of the circle = .7854 X 9 = 7.0686 ; and by 

 Rule 2, 360: 18 : : 7-0686: .35343, the area of the sec. 

 tor. 



PROBLEM XI. 



To find the area of the segment of a circle. 



RULE. Find by the last problem the area of the sec- 

 tor having the same arc as the segment, also the area 

 contained by the chord of the arc, and the two radii of 

 the sector, their sum, or difference, according as the 

 segment is greater or less than a semicircle, will be 

 the area of the sector. 



The reason of this rule is sufficiently evident. 

 Fig. 9. i Ex. Find the area of the segment ADB, which is 

 less than a semicircle, its chord AB being 12, and the 

 radius AC or BC 10. 



By TRIO. AC : AP : : rad. : sin. ACP = 36 52'.2 = 

 S6 .87, the degrees in the Angle ACD ; and their 

 double, 73.74 = the degrees in the arc ACB. Now, 

 .7854 x 400 = 314.16, the area of the whole circle : 

 Therefore 360 : 36 87 :: 314.16 : 64.3504 = area of 

 the sector CADB. Again, the three sides of the tri- 

 angle ACB being 10, 10, and 12 ; its area (Prob. 2, 

 Rule 3.) will be ^(16 X 6 X 6 x 4) = 48. Therefore 

 the area of the segment = 64.3504 48 == 16.3504. 



PROBLEM XII. 



To find the area of any segment of a Parabola. 

 RULE. Multiply the base of the segment by its height, 



and take of the product for the area. 



9 



The truth of this rule is proved in CONIC SECTIONS, 

 Sect. 7, Prop. 1. 



Ex. The base AB of a parabolic segment ACB is 

 10, and its altitude CD is 6. Hence, by the rule, 



The area = f x 10 x 6 = 40. Fi S- 



PROBLEM XIII. 



To find the area of an ellipse. 



RULE. Multiply the product of the two axes by the 

 number .7854 for the area. 



Let a denote the transverse axis AB, and b the con- Fig. II. 

 jugate axis CD ; the area of a circle that has a for its 

 diameter is .7854 a 1 (Prob. 9-) And a is to 6 as the 

 area of this circle to the area of the ellipse, (CoNic 

 SECTIONS, Sect. VII. Prop. 3.) that is, 



a : b : : .7854 a' : area of ellipse. 



Hence area of ellipse = .7854 a b. 



Ex. The area of an ellipse, whose axes are 10 and 8 

 feet, is required. 



10 X 8 X .7854 = 62.832 sq.feel. 



Nole. 1. Rules for computing the areas of elliptic and 

 hyperbolic sectors, may be derived from the investiga- 

 tions given in CONIC SECTIONS, Sect. VII. Prop. 3, 4, 

 and 5. 



Note 2. Rules for hyperbolic areas are also investi- 

 gated in FLUXIONS, Art. 150. 



PROBLEM XIV. 



To find nearly the area of a figure bounded by a 

 straight line BQ, two straight lines BA, QP perpendi- 

 cular to BQ, and any curve line A. a a' a" .... P. 



RULE. Let BQ, the base of the figure, be divided into Fig. if. 

 any number of equal parts by the perpendiculars b a, 

 V a', b" a", &c. which meet the curve in a a! a'', &c. 

 Let F denote the first perpendicular. 

 L .... the last. 



"the sum of the remaining even perpen- 

 E . . . \ diculars, viz. a b, a" b'', &c. the 2d, 



the 4th, &c. 



"the sum of the remaining odd perpen- 

 O . . . diculars, viz. a' b', a'" It"', &c. the 



i _ 3d, the 5th, &c. 



D . . . . the common distance between the per. 

 pendiculsrs. 



The area of the figure will be nearly equal to 



And the approximation will be so much the more accu 

 rate, according as the number of the perpendiculars is 

 the greater. 



To prove this rule, join the tops of the first and third 

 perpendiculars by the line A a, meeting the second per- 

 pendicular in E, and through a draw CD parallel to 

 A a', meeting AB and a' b' in C and D. The space 

 bounded by the curve A a a', and the straight lines 

 AB, B b', b'a' is made up of the trapezoid AB b'a' and 

 the space contained by the arc A a a', and its chord 

 A a' ; now if the arc be small, it may be considered as 

 a parabolic arc, and then the curvilineal space between 

 the arc and its chord will be ^ of the parallelogram 

 ACD a'. (CoNic SECTIONS, Sect. vii. Prop. I.) There- 

 fore the space A a a'b' B will be nearly the sum of the 

 trapezoid AB b'a', and f of the parallelogram ACD a' ; 

 that is, the sum of j of the trapezoid AB 6' a', and | of 

 the trapezoid CBi'D. 



Now, Trap. AB b'a' = (AB+a'i') B b (GtoM. Part I. 

 Sect. 4. Prop. 7-) 



