EXAMPLE. Take the same example, in which the radius is 10, and the chord AB 12. Then, as before, are found CD=2, and the chord of the half arc AC. 1. Divide the height of the segment by the diameter, and find the quotient in the column of heights or versed sines> in the table of the areas of the segments of a circle. 2. Take out the corresponding area in the next column on the right, and multiply it by the square of the diameter, for the answer. *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, and is at the end of Mensuration. 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 areas of those segments as the squares of the diameters; which may be thus proved: EXAMPLE. The example being the same as before, we have CD equal Co 2, and the diameter 20. Let ADBA and adba be any two similar segments, cut off from the similar sectors ADBCA and adbCa by the chords AB aud ab; and let fall the perpendicular CD. Then, by similar triangles, DB: db :: BC: bC, and DB: db:: Dm: dn; whence, by equality, BC : bC :: Dm: dn, or 2BC: 2AC:: Dm : dn. Again, since similar segments are as the squares of their chords, it will be AB2: ab :: ADBA : adba; but AB2: ab2 :: CB2: Ci2, and therefore, by equality, ADBA adba :: CB2: C52, or ADBA adba: 4CB2 : 4C42. Q. E. D. Now, if d be put equal to any diameter, and v the versed sine, it will be d: v : : 1 (diameter in the table): versed sine of d a similar segment in the table, whose area Jet be called a. OTHER EXAMPLES. 2. What is the area of the segment, whose height is 2, and chord 20? Ans. 26-878787. 3. What is the area of the segment, whose height is 18, and diameter of the circle 50? Ans. 636 375. 4. Required the area of the segment, whose chord is 16, the diameter being 20. Ans. 44 7292. PROBLEM XIV. To find the area of a circular zone* ADCBA. FIGURE 1. V T R P B Then l2 d2 :: a: ad* = area of the segment, whose height is n, and diameter d, as in the rule. *The space included between any two parallel chords and their intercepted arcs. Find the areas of the two segments AEB, DEC, and their difference will be the zone ADCB. Rule 2. To the area of the trapezoid ARDQP add the area of the small segment ADR; and double the sum for the area of the zone ADCB. EXAMPLES. 1. What is the area of the zone, less than a semicircle, figure 1, having the greater chord 16, the less chord 6, and the diameter of the circle 20? Here OQ=VOD-DQ100-9/91-9539392. And OP=OA-AP=100-6436=6. 2. If the greater chord be 96, the less 60, and the distance between them 26; required the area. Here, if R be the middle of the chord AD, and O the cen tre of the circle; then, in figure 1, and by the first rule, DS=26 OQ=OT+TQ=27+13=40 OE=OA VAP2+OP2 =√48 +142=50 EP-OE-OP-50-14=36 100)36(-36, its tab. seg. 25455 Difference '21367 Square Diameter 10000 Answer 2136'7 3. If the greater chord be 40, the less 30, and the distance between them 35; required the area of the zone in figure 2, and by the second rule. |