EXAMPLES. 1. The greater chord AB is 20, the less DC 15, and their distance Dr 17: required the area of the zone ABCD. Ans. 395.4388. E n G 20-15 ·=2.5=} the difference between the two chords. 2 17.5+ And 202 +152 √625=25=the diameter of the cir. (20—2.5) × 2.5 = 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., 25 =.100 = tabular Now .040875 × 25=area of seg. DEC=25.546875 By Prob. XI. Rule II. .7854× 252 area of the whole circle. sum 95.43625 =490.87500 Difference area of the zone ABCD=395.43875 2. The greater chord is 96, the lesser 60, and the breadth 26; what is the area of the zone? Ans. 2136.7500. 3. One end of a circular zone is 48, the other end is 30, and the breadth is 13; what is the area of the zone? Ans. 534.1875 PROBLEM XV. space included To find the area of a circular ring, or the RULE.* The difference between the areas of the two circles will be the area of the ring. Or, multiply the sum of diameters by their difference, and this product again by .7854, and it will give the area required. EXAMPLES. 1. The diameters AB and CD are 20 and 15; required the area of the circular ring, or the space included between the circumferences of those circles. E B *Demon. The area of the circle AEBA=AB X.7854, and the area of the small circle CD is=CD2 X.7854; therefore the area of the ring=ab2 × .7854-cD2 x .7854= AB+CDX AB-CDX.7854. Q. E. D. Coroll. If CE be a perpendicular at the point c, the area of the ring will be equal to that of a circle whose radius is CE. Rule 2. Multiply half the sum of the circumferences by half the difference of the diameters, and the product will be the area. This rule will also serve for any part of the ring, using half the sum of the intercepted arcs for half the sum of the circumferences. Here AB+CDx AB-CD=35x5=175; and 175× .7854=137.4450=area of the ring required. 2. The diameters of two concentric circles are 16 and 10; what is the area of the ring formed by those circles? Ans. 122.5224. 3. The two diameters are 21.75 and 9.5, required the area of the circular ring. Ans. 300.6609. 4. Required the area of the ring, the diameters of whose Ans. 15.708. bounding circles are 6 and 4. PROBLEM XVI. To find the areas of lunes, or the spaces between the intersecting arcs of two eccentric circles. RULE.* Find the areas of the two segments from which the lune is formed, and their difference will be the area required. * Whoever wishes to be acquainted with the properties of lunes, and the various theorems arising from them, may consult Mr. Wiston's Commentary on Tacquet's Euclid, where they will find this subject very ingeniously managed. The following property is one of the most curious: If ABC be a right angled triangle; and semicircles be described on the three sides as diameters, then will the said triangle be equal to the two lunes D and F taken together. For the semicircles described on AC and BC=the one described on AB (31.66,) from each take the segments cut off by AC and BC, then will the lunes AFCE and BDCG=the triangle ACB. Q. E D. MENSURATION EXAMPLES. The length of the chord AB is 40, the height DC 10, and DE 4; required the area of the lune ACBEA. By note, page 74, the diameter of the circle of which ACB is a part= 202 +102 10 = 50. And the diameter of the circle of which AEB is a part 202 +42 4 =104. Now having the diameter and versed sines, we find by Prob. XIII. Rule III. The area of seg. ACB=.111823 × 502279.5575 Their difference is the area of the } =171.8842 2. The chord is 20, and the heights of the segment 10 and 2; required the area of the lune. Ans. 130.287. 3. The length of the chord is 48, and the heights of the segments 18 and 7; required the area of the lune. Ans. 408.7057. PROBLEM XVII. To find the area of an irregular polygon, or a figure of any number of sides. RULE.* 1. Divide the figure into triangles and trapeziums, and find the area of each separately. 2. Add these areas together, and the sum will be equal to the area of the whole polygon. EXAMPLES. 1. Required the area of the irregular figure ABCDEFGA, the following lines being given. GB=30.5 An=11.2, CO=6 * When any part of the figure is bounded by a curve, be found as follows: the area may Rule 1. Erect any number of perpendiculars upon the base, at equal distances, and find their lengths. 2. Add the lengths of the perpendiculars, thus found, together, and the sum divided by their number will give the mean breadth. 3. Multiply the mean breadth by the length of the base, and it will give the area of that part of the figure required. H |