PROBLEM IX. To find the area of the Sector of a Circle. ART. 25. Rule.-I. Find the length of the arc by Art. 23. II. Multiply the length of the arc thus found, by half the length of the radius, and the product will be the area. Or, As 360 degrees is to the number of degrees in the arc of the sector, so is the area of the circle to the area of the sector. Ex. 1. If the arc AB be 120 degrees, and the diameter of the circle 226 degrees, what is the area of the sector? Remark, 113 is the diameter, (Art. 23,) and 56 is the radius according to the above rule. OPERATION. A First, .01745x 120x113-236.622 NOTE. It is manifest that the area of the sector has the same ratio to the area of the circle which the number of degrees in the arc has to the number of degrees in the whole circumference. Ex. 2. What is the area of a sector of a circle, in which the radius is 25 and the arc of 26 degrees? (See Art. 23.) Ans. 141.8 Ex. 3. Required the area of a semi-circle, in which the radius is 13. Ans. 265.4143. Ex. 4. What is the area of a circular sector, when the length of the arc is 650 feet, and the radius 325? Ans. 105625 sq. ft. PROBLEM X. To find the area of the Segment of a Circle. ART. 26. Rule.-I. To the chord of the whole arc add of the chord of half the arc. II. Then multiply the sum by the versed sine, or height of the segment, and 1 of the product will be the area of the segment, very nearly. Ex. 1. Required the area of a circular segment whose chord AB=24 feet, and whose radius CA-20 feet? A Ꭰ B OPERATION. First, CA-AP-CP-400-144-16-CP Then, CD-CP-DP-20-16=4-height of segment. Now, AP+PD=AD=✓144+16=12.64911-chord AD And, 24. -the chord of the segment. 12.64911-chord of the segment. 4.21637-3 of the chord of the arc. 40.86548 4-the height of the segment. 163.46192 4 10)653.84768 65.384768-area of the segment. Ex. 2. Required the area of a circular segment whose height is 19.2 and base or chord 70. 35+19.2-4D. Ans. 947.86. Ex. 3. What is the area of a circular segment whose chord is 16, and the diameter of the circle 20? Ans. 44.680. Ex. 4. If the base or chord of a segment be 10 feet, and the radius of the circle 12 feet, what is the area of the segment? Ans. 7.35 sq. ft. NOTE. The following rule is given in Day's Mathematics: Find the area of the sector which has the same arc, and also the area of the triangle formed by the chord of the segment and the radii of the sector. 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. PROBLEM XI. To find the area of a Lune, or Crescent. ART. 27. Rule. Find the difference of the two segments which are between the arcs of the crescent and its chord, for the area. Ex. 1. The chord of two segments AB is 72, and the height of the greater segment HD is 30, and of the less HC 20, what is the area of the D Ex. 2. If the chord AB be 88, the height HC 20, and the height HD 40; what is the area of the crescent, ADB and ACB? Ans. 1478. NOTE. For the demonstration of the above examples consult Art. 26. PROBLEM XII. To find the area of a Circular Zone. ART. 28. Rule.-From the area of the whole circle subtract the areas of the two segments on the sides of the zone. If from the whole circle there be taken the two segments ABC and DFG, there will remain the circular zone ABFD. A D C Ex. 1. What is the area of the zone ABFD, if AB is 7.75, DF, 6.93, and the diameter of the circle 8? OPERATION. 50.26-area of the whole circle. 17.23-area of the segment ABC. And, 50.26-27.05-23.21-area of the zone ABFD. PROBLEM XIII. To find the area of a Ring included between the circumferences of two Concentric Circles. ART. 29. Rule.-I. Square the diameter of each circle, and subtract the square of the less from that of the greater. II. Multiply the difference of the squares by the decimal .7854, and the product will be the area. Or, III. Multiply the product of the sum and difference of the two diameters by .7854. Ex. 1. If the diameter of the outer circle AB be 221, and the inner circle DE, 106, what is the area of the ring? That is, the area of each of these circles is equal to the square of the diameter multiplied by .7854 (Art. 19.) And the difference of these squares is equal to the product of the sum and difference of the diameters. Therefore, the area of the ring is equal to the product of the sum and difference of the two diameters, multipled by .7854. Ex. 2. The diameter of the inner circle is 12 rods, and the outer, 20 rods; required the area of the ring. Ans. 201.06 rods. Ex. 3. Supposing the diameter of Saturn's larger ring to be 205.000 and the smaller one 190.000 miles, required the number of square miles on one side of the ring. Ans. 4.653.495.000. Ex. 4. Two diameters are 21.75 and 9.5; what is the area of the circular ring? Ans. 300.66. Ex. 5. If two diameters are 4 and 74, what is the area of the ring? Ans. 33.80. PROBLEM XIV To find the area of an Ellipse. ART. 30. Rule.-Multiply the longer axis by the shorter, and the product, multiplied by the decimal .7854, will be the area required. |