last product will be the length of the arc; for, when the radius is 1, half the circumference is 3,14159265, &c. 180 degrees, ,01745329, or ,0174533, nearly which is the length of an arc of 1 degree. Hence CD× ADB× b= the length of the arc, ADB. EXAMPLE. (17) What is the length of the arc, AED, which is 29,5 degrees, and radius 9}. PROBLEM IX. To find the area of any sector of a circle. RULE. Multiply the radius by half the arc of the sector, found by the last problem, and the product will be the area, as in the whole circle. (18) What is the area of a sector, whose radius, CA, is 55, and the length of the arc, AB, 59 ? PROBLEM X. To find the area of the segment of a circle, ADB, whose centre is E. (See Fig. 3.) RULE. Find the area of the triangle ABC, by Prob. III. and of the sector, ADBC, by the last problem: their difference, or sum of these areas, will be that of the segment, according as it is less or greater than a semi-circle. Or, To six times the base AB (see Fig. 8.), add eight times the chord of half the arch AB, viz. DB; multiply the sum by the altitude DE, divide the product by 15, and the quotient will nearly give the area. A TABLE of the segments of circles, whose area is unity or 1, the diameter being divided by parallel chord-lines into 100 equal parts. V. S. Segment. V. S. Segment. V. S. Segment. V. S. Segment. EXAMPLES. (19) Suppose the diameter, FG, of a circle to be 84 inches, and the height of the segment, ED, 30 inches, what will its area be? (20) What is the area of a segment whose are is a quadrant, or contains 90 degrees, and diameter 18 feet? PROBLEM XI. To find the area of a segment of a sector, ABCD, or the front of an arch built with stones of equal length. RULE. Multiply half the sum of the bounding arches, AB and CD, by the distance, AC, and the product will give the area. Fig. 10. AB+CD That is, × AC= the area nearly. 2 B EXAMPLES. (21) What is the area of the front of an arch built with stones 3 feet long, whose upper and lower bounding arches are in length 84 and 72 respectively? (22) What is the area contained between two concentric semicircles, whose diameters are 24 and 16? PROBLEM XII. To find the area of an ellipsis, or oval. RULE. Multiply continually together the two axes, and the number,7854 (b), and the product of these three numbers will express the area. That is, bx AB× CD= the area. Fig. 11. A EXAMPLE. (23) What is the area of an ellipsis whose greatest diameter is 24, and the least diameter 18? OF ARTIFICERS' WORK. I. GLAZIERS' and MASONS' FLAT WORK is measured by the Foot Square. EXAMPLES. (1) What is the content of 12 panes of glass, each measuring 3 feet 10 inches long, and 2 feet 8 inches broad? What will the glazing come to at 8 d. per foot? (2) There is a house with 3 tier of windows, 4 in a tier; the height of the first tier is 6 feet 6 inches, the second 5 feet, and the third 44 feet; the breadth of each window is 3 feet 9 inches. What will the glazing come to at 16d. per foot? (3) What is the price of a marble slab, whose length is 6 feet, and the breadth 34 feet, at 8s. per foot? (4) A looking-glass is 16 inches by 9, and contains a foot of glass. What will the content of the plate be, that bas twice the length, and three times the breadth? II. PAINTING, PLASTERING, PAVING, &c. is measured by the yard square, which is 9 square feet. RULE. Divide the square feet by 9; and the quotient will be the number of square yards. |