of the sector ABCD, in square inches; and 2315.76 2150.42 1.0768, the area in malt bushels. To find the area of the segment of a circle. of the height divided by twice the chord; then 2730.666 +128=2858.666, the area in square inches; whence, 2858.666282 = 10.137, the area in ale gallons; and 2858.666231 = 12.375, the area in wine gallons. EXAM. 3. 302 900 By Note 2, we have~+40= + 40 40 <=225 + 40 = 62.5, the diameter CD; then 2.5 x .78543906.25 x .78543067.96875, square inches, the area of the whole circle. Again, 62.5 - 40=22.5, the versed sine CE; and 2700 22.5 x 60 x = 1350 x = =900, two-thirds 3 of the product of the chord and height of the remain22.5 11390.625 ing segment ABC; also 60×2 = 120 = = 94.921, the cube of the height divided by twice the chord; then 900 + 94.921 = 994.921 square inches, the area of the segment ABC. Now, 3067.96875 - 994.921 = 2073.04775, the area of the required segment ADB, in square inches; then 2073.047282 = 7.351, the area in ale gallons; 2073.047 ÷ 231 = 8.974, the area in wine gallons; and 2073.04775 ÷ 2150.42 = .964, the area in malt bushels. Here 32.0 RULE II. EXAM. 2. =.2, the quotient or tabular height; and the corresponding Area Seg. is .111823; hence .111823 x 1602.111823 x 25600 = 2862.6688, the area of the segment, in square inches; then 2862.668 282 10.151, the area in ale gallons; and 2862.668 12.392, the area in wine gallons. 231 Here, = 40.0 ЕХАМ. 3. = .64, the tabular height; then 1.64 62.5 .36; and the Area Seg. answering to .36, is .254550, which being taken from .785398, leaves .530848, the Area Seg. corresponding to .64; hence .530848 x 62.5 .530848 x 3906.252073.625, the area of the segment in square inches; then 2073.625282 = 7.353, the area in ale gallons; 2073.625231 8.976, the area in wine gallons; and 2073.6252150.42 = .9642, the area in malt bushels. (See Note 2.) PROBLEM XVI. To find the area of an ellipse. EXAM. 2. Ale gallons. 10.8 depth. 171736 21467 231.8436 Ans. EXAM. 3. Here 85.9 x 63.8294.125480.42294.12 = 18.633, the area in wine gallons. EXAM. 4. Here 96.8 x 73.2 ÷ 2738 = 7085.762738 = 2.587, the area in malt bushels; and 2.587 × 34.7 = 89.7689, the content in malt bushels. EXAM. 5. Here 96.4 x 82.3359.057933.72359.05 = 22.096, the area in ale gallons; and 22.096 x 53.8 = 1188.7648, the content in ale gallons. PROBLEM XVIL Tfnd the cres of an elliptical segment, the base of enci is parallel to either of the diameters of the EXAM. 2. By Prob. XVI, we have 120.8 x 75.4294.12 = STAR ÷ 3412 = 30.9680 wine gallons, the area di the whole ellipse. By the last example, the area of the segment ACB is 181 3.3298 square inches; hence 1813.3258231 = THUR the area in wine gallons; then 30.9680 —-3400—23.1181 wine gallons, the area of the sgment ADR. 180 Here EXAM. 3. (See the figure in the Gauging.) +27=90+27=117 = DG, the height of the segment; then 117 180.650, the tabular beight; and by Prob. XV., Rule 2, Note 2, we have 1 - .650 = 350; and the Area Seg. answering to 350, is 244980, which being taken from .785398, leaves 540418, the Area Seg. corresponding to .650; bence 180 x 120 x .540418 = 21600 × 540418= 11673.0288 square inches, the area of the segment ADB; and 11673.02882150.42 = 5.4282, the area in malt bushels. PROBLEM XVIII. To find the area of a parabola, its base and height being inches; and 2620.88 ÷ 231 = 11.3458, the area in wine gallons. PROBLEM XIX. To find the area of compound figures. 75, the half the sum of the sides; then, 150 - 75 = first remainder; 150 100 50, the second remaind er; 150 12525, the third remainder; whence 150 × 75 × 50 × 25 = √14062500 = 3750, the area of the triangle in square inches. By Prob. XV., we have 125 × 26 × 6500 =2166.666, two thirds of the 3 chord and versed sine; and 263 = 3250 × 3 = product of the 17576 = 70.304, 125 × 2 250 the cube of the height divided by twice the chord; then 2166.666 + 70.304 = 2236.97, the area of the segment in square inches. Now, 3750+2236.97 5986.97, the area of the whole compound figure; hence 5986.97 282 = 21.23, the area in ale gallons; and, 21.23 × 86 = 182.578, the content in ale gallons. PROBLEM XX. To find the area of any curvilineal figure, by means of equidistant ordinates. |