3. What is the area of a segment, whose arc is 90°, and radius 9 ft.? Ans. 23.117 ft. 4. Required the area of a segment, whose arc contains 280°, the diameter being 25 yds. Ans. 458.729 sq. yds. 15. PROBLEM. The chord and height of an arc of a circle being given, to find the radius and number of degrees in the arc. (See the last figure.) AD2 By Geo. Cor. 1. Art. 61., CD . DF = AD2, .. DF= ; adding CD to each side of this equality, CD Having found the diameter, we now proceed to find the number of degrees in the arc. By Trigo. Art. 9., sin AEC or sin AD AEB= ; hence AE from the table of sines, we find AEC, and .. ZAEB. EXAMPLES. 1. Given the chord AB=8, and the height CD=2, to find the radius of the circle, and the number of degrees in the arc ACB. Here, AD of 8=4; and 2 × DF = 42; .. DF=8, and CF or diameter 8+2= 10, and the radius = 5. From the right-angled triangle ADE, we have, ᎪᎠ ==·8, .. LAEC = 53° 8′ nearly, and sin AEC= = AE LAEB, or degrees in the arc ACB = 53° 8′ x 2 = 106° 16′. 2. Required the same as in the last example, when AB 12, and CD = 4. Ans. radius = 6·5, and arc = 134° 45'. 3. Given the height CD=6, and the chord of the half arc AC=10, to find the radius and number of degrees in the Ans. radius=8}, and arc=147° 28′: arc. Here, AD2 = 102 — 62 — 64; then DF= 64 = 10; and so on, as in the first example. 4. Required the same as in the last example, when CD=9, and the chord of half the arc = 12. Ans. radius = 8, and arc = 194° 22′. 16. PROBLEM. To find the area of any space by means of equidistant ordinates or perpendiculars. RULE. To the two extreme ordinates add twice the sum of the intermediate ordinates, and this sum multiplied by half the common distance between them will be the area. Note. When the curvilineal boundary in the figure is some continuous curve, the area may be more accurately found by Thomas Simpson's rule. EXAMPLES. 1. Required the area of the curvilineal space ABCD, where the extreme ordinates or perpendiculars AD and BC are 4 and 12 ft. respectively, the three intermediate ordinates, taken in order, are 6, 8, and 10 ft., and the common distance between them is 4 ft. A E Here, taking the curve lying between any two consecutive ordinates as a straight line, we may consider the figure as being made up of four trapezoids, whose areas are as follows: area 1st (4+6); area 2d=4 (6+8); area 3d=(8+10); = .. Total area (4+6)+(6+8)+(8+10)+(10+12) = ={4+12+2 (6+8+10)}=128 sq. ft. 2. Required the same as in the last example, when 1 AD = 8 yds., BC= 21 yds., the three intermediate ordinates, taken in order, 11, 14, and 17 yds., and the common distance between them 6 yds. 3. Required the area of the curvilineal space A B ON where the lengths of the equidistant ordinates are as follows: AB=2, CD=3, EF = 5, GH = 6, IK = 9, LM = 10, and NO 10.5 ft.; and their distance apart, AC=CE=&c. = 5 ft. B Ans. 339 sq. yds. 11 D KM O A CE G I L N Ans. 19-625 sq. ft. 17. PROBLEM. To find the area of an ellipse. RULE. Multiply the product of the two diameters by 7854, and the result will be the area. Let A B A B be an ellipse whose major diameter is AA' and minor diameter BB'. Let APA'P' be a circle described upon A A' as a diameter, and let pm and pm be any corresponding ordinates of the circle and the ellipse; then it is one of the most remarkable properties of these curves, that the ratio of pm to pm is the same as the ratio of AA' to BB'. Now if the ellipse and circle be divided into a series of very narrow bands, as shown in the figure, the area of any one of the circular bands will be to the area of the corresponding PB P P elliptical band in the ratio of pm to pm or AA' to BB'; hence it follows that the area of the whole circle will be to the area of the whole ellipse in this same ratio, that is, Area circle .. Area ellipse area ellipse :: A A': BB′, area circle X BB'÷AA' AAX BB × ·7854; which is the analytical expression of the rule. EXAMPLES. 1. What is the area of the ellipse ACBD, whose major diameter AB is 16 ft., and minor diameter CD 12 ft.? Here, area ACBD = 16 x 12 x 7854 = 150·7968 sq. ft. 2. The diameters of an ellipse are the area. 18 and 8 ft.; required Ans. 113-0976 sq. ft. 3. The diameters of an elliptical piece of ground are 3 and 2 chains; required the area. Ans. 1 r. 35-3984 p. MENSURATION OF SOLIDS. 18. PROBLEM. To find the solidity of a rectangular solid or right-angled parallelopiped. RULE. Multiply the length by the breadth, and that product again by the depth or altitude. (See Geo. Art. 90.) Note. To find the surface of a rectangular solid, multiply the perimeter of its end by its length, and the product will be the area of the sides; to which add twice the area of its end, and the sum will be the whole surface. EXAMPLES. 1. Find the number of cubic feet in a rectangular solid ABGE, whose length. AB is 6 ft. 3 in., breadth AE 4 ft. 6 in., and depth or altitude A D 3 ft. 1 in. By fractions. Solid content = 6 × 4 x 31 = 2 x x =8633 c. ft. 2. What is the solidity of a parallelopiped, whose length is 20 ft., breadth 5.75 ft., and depth 3.5 ft.? Ans. 402·5 c. ft. 3. Required the cost of a block of stone 8 ft. 4 in. long, 2 ft. 6 in. broad, and 1 ft. 3 in. deep; at 10d. per cubic foot. Ans. £1. 1s. 81d. 4. What will be the cost of a log of timber 18 ft. long, 1 ft. 8 in. broad, and 1 ft. 6 in. deep; at 2s. 4d. per cubic foot? Ans. £5 5s. 5. What will be the cost of cutting a drain 60 ft. long, 5 ft. 6 in. broad, and 10 ft. 4 in. deep; at 8d. per cubic yard? 6. In the cube ABCE, the side A B or AD or AE is 1.75 ft.; required its solidity. Here, as the length, breadth, and depth are the same, we have, Solidity = 1.75 × 1.75 × 175, or, 1·753 = 5·359 c. ft. |