AC2 100 CD =5=the versed sine of half the arc, CE 20 20 x 60 -5 x 27=1065=the reserved number. 10 x 2 x 5 x 10 =.9390, and this added to twice the chord 1065 of half the arc gives 20.9390=the length of the arc. 20.9390 x 10 =104.6950=area of the sector OACB. OD=OC=CD=5 the perpendicular height of the tri. angle. AD=V A02–OD=V75=8.6603=1 the chord of the arc. 8.6603 x 5=43.3015=the area of the triangle AOB. 104.6950—43.3015=61.3935= =area of the segment required; it being in this case less than a semicircle. 2. Required the area of a segment whose height is 2, and chord 20. Ans. 26.8783. 3. Required the area of a segment of a circle, the ra. dius being 10, and the chord of the arc 12. Ans. 16.3500. 4. Required the area of a segment of a circle, whose chord is 16, and the diameter of the circle 20. Ans. 44.7195. 5. What is the area of a segment whose arc is a quad rant, the diameter of the circle being 18 feet ? Ans. 23.1174. 6.* What is the area of a segment, whose arc contains 300 degrees, the diameter being 50 ? Ans. 1906.8831. RULE II.T 1. Divide the height, or versed sine, by the diameter and find the quotient in the table of versed sines. 2. Multiply the number on the right hand of the versed sine by the square of the diameter, and the product will be the area. Note 1.—When the diameter or versed sine is not given, it may be found by the notes, page 68 or 74. Note 2.-When the quotient arising from the versed sine divided by the diameter, has a remainder or fraction after the third place of decimals; having taken the area * The chord of the arc will evidently be equal to the radius of the circle. + The table to which this rule refers, is formed of the areas of the segments of a circle whose diameter is 1; and which is supposed to be divided by perpendicular chords into 1000 equal parts. The reason of the rule itself depends upon this property-That the versed sines of similar segments are as the diameters of the circles to which they belong, and the area of those segments as the squares of the diameters; which may be thus demonstrated. Let ADBA and adba be any two similar segments, cut off from the similar sectors ADBCA and adbca, by the chords AB and ab, and let the perpendicular cd bisect them. D Then by similar triangles, DB: db :: BC : bc and DB : db :: Dm : dn; whence, by equality, BC : bc :: Dm: m B or 2BC : 2c :: Dm : dn. cl Again, since similar segments are as the squares of their chords, it will be AB? : ab2 :: seg. ADBA : seg. adba ; but AB?: ab? :: CB2: cb2, whence, by equal. ity, seg. AdBA : seg. adba :: CB2 : cb?, or seg. AdBA : seg. adba :: 4CB? : 4cb2. Q. E. D. C Now, If d be put equal to any diameter, and v the versed sine it will be d:0::1 (diameter in the table):=versed sine of a sinanas d segment in tne table whose area let be called a. • dn, A a n answering to the first three figures, subtract it from the next following area, multiply the remainder by the said fraction, and add the product to the first area, then the súm will be the area for the whole quotient. EXAMPLES. 1. If the chord of a circular segment be 40, its versed sine 10, and the diameter of the circle 50, what is the area? 5.0)1.0 .2=tabular versed sine. 2500=square of 50. 55911500 223646 279.557500=area required. 2. The chord ; the segment is 20, the versed sine 5, what is the area ? Ans. 69.889375. 3. The diameter of a circle is 40, and the versed sine 10; what is the area of the segment ? Ans. 245.6736. 4. If the diameter be 52, and the versed sine 2, what is the area of the segment ? Ans. 26.9197. 5. If the chord of half the arc be 30, and the versed sine 9, what is the area of the segment ? Ans. 350.1100. 6. The diameter of a circle is 100, and the chord of the arc 60 : what is the area of the segment ? Ans. 408.75. Then 12: d2 :: a : ado=area of the segment whose height is v, and diameter d as in the rule. PROBLEM XIV. To find the area of a circular zone, or the space included between any two parallel chords and their intercepted arcs. RULE. From the greater chord subtract half the difference be tween the two, multiply the remainder by the said half dif. ference, divide the product by the breadth of the zone, and add the quotient to the breadth. To the square of this number add the square of the less chord, and the square root of the sum will be the diameter of the circle. Now, having the diameter EG, and the two chords AB and DC, find, by Prob. XIII. Rule II. the areas of the seg. ments ABEA, and DCED, the difference of which will be the area of the zone required. Note 1.-The difference of the tabular segments multi. plied by the square of the circle's diameter will give the area of the zone. Note 2.—When the larger segment AEB is greater than a semicircle, find the areas of the segments AGB, and DCE, and subtract their sum from the area of the whole circle, the remainder will be the area of the zone. * The reason of this rule is too obvious to require a demonstration. Note.When the two parallel sides of the zone are equal, the chord of the small segment will be equal to the breadth of the zone, and the height of this segment will be equal to V4AB? + fasl=AB; as being put for the breadth of the zone. And when one of the sides is the diameter of the circle, the chord of the same segment will be Vas? + D?, and its height=jAB -1102_1as-11 where D=AD-DC. EXAMPLES. J. The greater chord AB is 20, the less DC 15, and their distance Dr 17*: required the area of the zone ABIU. Ans. 395.4388. 20-15 = 2.5=} the difference between the two chords. (20—2.5) x 2.5 17.5+ = 17.5 +2.5=20=DF. 17.5 And ✓ 202 +152=V625=25=the diameter of the cir. The segment AEB being greater than a semicircle, we find by note 1, page 68, the versed sine of DCE=2.5, and that of AGB=5. 2.5 Hence by Prob. XIII. Rule II., =.100 = tabular 25 rersed sine of DEC. 5 And =.200=tabular versed sine of AGB. 25 Now .040875 X 25=area of seg. DEC=25.546875 And .111823 x 252=area of seg. AGB=69.889375 sum 95.43625 By Prob. XI. Rule II. .7854x 252 =490.87500 } |