4. What is the area of a quadrant whose radius is 27 feet 6 inches? Ans. 593 ft. 11 in. 6 pa. 5. What is the area of a sector whose radius is 50 feet, and the length of its arc 60 degrees? PROBLEM XVII. Ans. 1309 feet. To find the area of the segment of a circle. RULE I. Find the area of the sector, having the same arc as the segment; also, find the area of the triangle formed by the chord of the segment and the radii of the sector; then the difference of these areas, when the segment is less than a semicircle, or their sum, when it is greater, will be the area of the segment. EXAMPLES. 1. The radius AD is 25, and the chord AB of the whole arc 40; what is the area of the segment ABC? Here D (252-202)=√(625-400) DE, therefore the versed sine CE 1 = 10. Again, (202 + 102) = ~/(400 + 100) = √500 22.36068 = AC, the chord of half the arc; and (22.36068 8 40) ÷ 3 = (178.88544 40) ÷ 3 = 138.88544 ÷ 3 = 46.295146, the length of the arc ACB; then (46.295146 × 25)2=1157.37865 ÷ 2 = 578.689325, the area of the sector ADBC. Now, AB x DE = 20 × 15 300, the area of the = 278.689325, triangle ADB; hence, 578.689325 — 300 = the area of the segment required. 2. What is the area of the segment of a circle; the chord of the whole arc being 60 feet, and the chord of half the arc 37 feet 6 inches? Ans. 987 ft. 6 in. 3. The chord of the whole arc is 20, and the versed sine 5 feet; what is the area of the segment? Ans. 69 ft. 8 in. 4. What is the area of a segment, the arc of which is a quadrant, whose radius is 24? Ans. 164.3904. 5. What is the area of a segment greater than a semicircle; the chord of the whole arc being 102 feet 6 inches, the chord of half the arc 100 feet, the chord of one quarter of the arc 57 feet 6 inches, and the diameter of the circle 116 feet 6 inches? Ans. 8408.89399 feet. 6. Find the area of a segment whose arc contains 245 degrees 45 minutes; the diameter of the circle being 48 feet 9 inches. Ans. 1545.03853 feet. RULE II. To two-thirds of the product of the chord and height of the segment, add the cube of the height divided by twice the chord, and the sum will be the area of the segment, nearly. Note. When the segment is greate than a semicircle, find the area of the remaining segment, which subtract from the area of the whole circle; and the remainder will be the area required. EXAMPLES. 1. What is the area of a segment of a circle whose chord is 32, and height or versed sine 8? Here (32 × 8) × ÷ + 83 ÷ (32 × 2) = 256 × ÷ + 512 64 170.6666 +8 178.6666, the area required. 2. The chord is 65, and versed sine 15; what is the area of the segment? Ans. 675.96153. 3. What is the area of a segment, greater than a semicircle, whose chord is 30, and height 20? RULE III. Ans. 518.26171875. Divide the height of the segment by the diameter, and find the quotient in the column of heights or versed sines, in the Table at the end of Part II. Take out the corresponding area seg., which multiply by the square of the diameter, and the product will be the area of the segment. Note 1.-If the quotient of the height by the diameter do not terminate in three places of figures, without a fractional remainder, find the area seg. answering to the first three decimals of the quotient; subtract it from the next greater area seg. ; multiply the remainder by the fractional part of the quotient, and the product will be the corresponding proportional part to be added to the first area seg. This method ought to be used when accuracy is required; but for common purposes, the fractional remainder may be omitted. 2. When the area of a segment greater than a semicircle is required, subtract the quotient of the height by the diameter, from 1: find the area seg. corresponding to the remainder, which take from .785398, and the difference will be the area seg. answering to the quotient. 3. The versed sines of similar segments are as the diameters of the circles to which they belong, and the areas of those segments are as the squares of the diameters. EXAMPLES. 1. What is the area of a segment whose chord is 32, the versed sine 8, and the diameter of the circle 40? Here 8.0 40 .2, the quotient or tabular height; and the corresponding area seg. is .111823; hence, .111823 × 402 = .111823 × 1600 = 178.9168, the area of the segment required. 2. What is the area of a segment whose height is 9, and the diameter of the circle 25? Ans. 159.09375. 3. Find the area of a segment whose height is 25, and the diameter of the circle 55. Ans. 1050.60065. 4. What is the area of a segment, greater than a semicircle, whose height is 66 feet, and the chord of the whole arc 60 feet 10 inches? Ans. 4435 ft. 8 in. 11 pa. 5. The base of a stone column is the greater segment of a circle whose chord measures 2 feet, and versed sine 1 foot 4 inches; what is its area ? PROBLEM XVIII. Ans. 2.304 feet. To find the area of a circular zone, or the space included between any two parallel chords and their intercepted arcs. RULE. Find the area of that part of the zone forming the trapezoid ABCD, to which add twice the area of the segment AED; and the sum will be the area of the zone required. (See the next figure.) Note 1.-When great accuracy is required, the area of the segment should be found by Rule 3, Problem XVII. 2. The chord AD and versed sine Em may be found from the parallel sides AB, DC, and the perpendicular distance DF, by the help of Theorem 12, Part I.; much calculation may, however, be saved by measuring the chord and versed sine in taking the dimensions of the zone. (See the Key to the Mensuration, page 22.) 3. When the parallel sides and their perpendicular distance are given, the zone may be constructed in the following manner: draw the side AB; make AF equal to half the difference between AD and DC, and erect the perpendicular FD. From the point D, draw DC parallel to AB, and join AD. Bisect AB and AD with the perpendiculars mo, no; and o will be the centre of the circle of which the zone is a part; thus you will determine whether the centre of the circle falls within or without the zone. EXAMPLES. 1. The greater side AB measures 40 feet, the less DC 30 feet, the perpendicular FD 35 feet, the chord AD E 35 feet 3 inches, and the versed sine Em 7 feet 3 inches; what is the area of the zone ABCD? By Problem VIII. we have (40 + 30) × 35 = 70 × 35 = 2450; and 2450 ÷ 2 = 1225, the area of the trapezoid ABCD. Also, by Problem XVII. Rule 2, we have (35.25 × 7.25) × + 7.253 ÷ (35.25 × 2) =255.5625 × ÷ + 381.078125 70.5 = 170.375 + 5.40536 = 175.78036, the area of the segment AED; hence, 1225+ (175.78036 x 2) = 1225 + 351.560721576.56072 feet 1576 ft. 6 in. 8 pa., the area of the zone ABCD, as required. = 2. The greater side is 120 feet, the less 75 feet, the chord 39 feet 6 inches, and the versed sine 3 feet 3 inches; what is the area of the zone? Ans. 3337 ft. 4 in. 10 pa. 3. What is the area of a zone whose greater side measures 72 feet, the less 45 feet, and its breadth 19 feet 6 inches? Ans. 1201 ft. 11 in. 1 pa. 4. The greater side is 80, the less 60, and their distance 70; what is the area of the zone? Ans. 6326.96. PROBLEM XIX. To find the area of a circular ring, or the space included between the circumferences of two concentric circles. RULE. Multiply the sum of the diameters by their difference, and this product by .7854; and it will give the area required. Or, the difference of the areas of the two circles will be the area of the ring. Note 1. The area of a circular ring may also be found by multiplying half the sum of the circumference by half the difference of the diameters. 2. The area of part of a ring, or the segment of a sector, may be found by multiplying half the sum of the bounding arcs by the nearest distance between them. EXAMPLES. 1. The diameters AB and CD are 30 and 20; what is the area of the circular ring? A ӨӨ Here (30 +20) × (30 — 20) × .7854 .7854 500 x .7854 = quired. B = 50 x 10 x 392.7, the area of the ring re 2. The diameters of two concentric circles are 35 and 23; what is the area of the ring formed by the circumference of those circles? Ans. 546.6384. 3. The inner diameter of a circular building is 73 feet 3 inches, and the thickness of the wall 1 foot 9 inches; how many square feet of ground does the wall occupy? Ans. 412.335 feet. 4. What was the expense of making a moat round a circular island, at 2s. 6d. per square yard; the diameter of the island being 525 feet, and the breadth of the moat 21 feet 6 inches? Ans. £512. 13s. 74d. 5. What is the area of the front of a circular arch, built with stones, each 3 feet 6 inches long; the length of the upper bounding arc being 35 feet 3 inches, and the length of the lower 24 feet 9 inches? Ans. 105 feet. PROBLEM XX. To find the area of a lune, or the space included between the intersecting arcs of two eccentric circles. RULE. Find the areas of the two segments forming the lune, and their difference will be the area required. Several other curious properties or lunes may be seen in Dr. Hutton's Recreations, and Mathematical Dictionary. |
