15. The short method of ascertaining the convex surface should not be given until 18. Each of the convex faces is a trapezoid, whose parallel sides measure 4 inches and 8 inches, respectively, the altitude being 10 inches. 16. Two inches apart, draw two parallel lines, AB and CD (Fig. 47), measuring 1 inch and 2 inches, respectively, a 2-inch perpendicular, XY, connecting the middle point of each. Draw AC and BD, and produce them until they meet in M. With this as a center, and MC as a radius, draw an arc; on which three other chords equal to CD are laid off, as in Fig. 46. With M as a center and a radius MA, lay off another arc, on which chords are laid off equal to AB; etc. See Arithmetic, Art. 1296. À L1 in 1 in.) 17. The entire surface = convex surface + Fig. 47. 4 sq. ft. + 9 sq. ft. 18. The pupils can readily understand this rule, using the frustums of problems 15 and 17 as illustrations. 19. The pupil should first locate the apex of the cone. This he can do by following the method shown in 16 ; making AB (Fig. 47) 21 inches; CD, 1} inches; and AC, 2 inches. This makes MA 5 inches, the slant height of the whole cone. The circumference of the base of the cone=217. The slant height, 5 inches, is the radius of the required sector, its circumference being 10. 247, the length of the arc of the sector, being onefourth of 107, shows that the required sector is a quadrant. 20. The number of square inches in the convex surface = [circumference (perimeter) of upper base (9 x 3.1416) + circumference of lower base (6 x 3.1416)] x 1 slant height (2). Adding to this, the area of the bottom (32) 9 sq. in. x 3.1416 (Art. 1124, 2) gives the number of square inches of material required. 22. Circumference of upper base = 6 x 3.1416 Circumference of lower base = 10 x 3.1416 One-half sum = 8 x 3.1416 Multiplying by slant height gives 48 X 3.1416 Add to this the area of the upper base, 9 x 3.1416 And the area of the lower base, 25 x 3.1416 Total in square yards, 82 x 3.1416 or ([(3 + 5) x 6] +9+25) 3.1416. 23. EA: EC ::6:8 :x+9::6:8 Ans. 27 in., 24 ft. The slant height of the whole cone=27 in. +9 in. = 3 ft. 24. The convex surface of the whole cone=į (8 x 3.1416 x 3) sq. ft.; of the part cut off=1 (6 x 3.1416 x 24) sq. ft. 1287. A sphere (a croquet ball, for instance) and a hemisphere should be used to illustrate these problems. On the plane face of the latter can be drawn the lines AD, FG, HI, CI, etc.; while on the curved face can be drawn HYI, FXG, etc. 25. 1 of 25,000 miles. 26. IH=chord of 60° of the great circle = radius of the great circle= 4000 miles. IB= 1 of IH=2000 miles. 27. The diameter, HI, of the small circle is į diameter FG of the great circle; the circumference HYI=1 of 25,000 miles, or 12,500 miles. 28. The length of a degree of longitude on the 60th parallel is about one-half of the length of a degree on the equator. (See Art. 995, Problem 10.) A 29. On the plane face of the hemisphere suggested above (Art. 1287), draw diameters AD and FG at right angles; and 45° from G, a chord NM parallel to FG. (This chord will not bisect AC.) As MCG :/ = 45°, WCM=45°; and the triangle Y WCM is a right-angled isosceles tri- FH angle, and WM'={CM”; WM= .7071CM. (See Art. 995, Problem 12.) If CG measures 4000 miles, CM=4000 miles, and WM= V8,000,000= 2828.4 miles. Fig. 48. 1289. The pupils have already learned that the volume of a rectangular prism is equal to the area of the base multiplied by its altitude; these problems are intended to show that the same is true of all prisms and of the cylinder (6). 1292. 8. The volume of the frustum is obtained by deducting the volume of the part cut off from the volume of the whole pyramid (Problem 7). The rule is given later. 10. Fig. 49 gives the method of calculating the slant height. The illustration shows a section of the frustum formed by passing a plane through the center of the frustum perpendicular to the base. The center of the upper base of the frustum of a right pyramid is directly above the center of the lower base, so that a perpendicular let fall from A will fall on CD at a Fig. 49. point X, 5 in. from C. In the right-angled triangle AXC, AX= altitude = 12 in. AC = AX + XOʻ = 144+ 25 = 169; AC= V169 = 13. 5 in. 18 in. 12 in. 5 ini. 1306. A person that contracts to receive a rate of interest greater than is permitted by law, is liable to a penalty, except in Connecticut. In Delaware, Minnesota, etc., the penalty is the forfeiture of the contract; in New York, the forfeiture of the contract, $ 1000 fine, and 6 months' imprisonment; in Indiana, Kansas, Kentucky, Maryland, Michigan, Mississippi, Ohio, Pennsylvania, Tennessee, Vermont, Virginia, and West Virginia, the forfeiture of the excess of interest; in North Carolina, the forfeiture of double the amount of interest ; in Georgia and New Hampshire, the forfeiture of three times the excess of interest; in Alabama, Delaware, Florida, Illinois, Iowa, Louisiana, Missouri, Nebraska, New Jersey, South Carolina, Texas, and Wisconsin, the forfeiture of all interest ; in Arkansas and Oregon, the forfeiture of principal and interest. 1307. 3. Amount of $ 1000, June 1, 1896, to June 1, 1897 Amount of $150, Sept. 16, 1896, to June 1, 1897, 8} mo.. . Due June 1, 1897 . . . . . Interest to settlement, Apr. 16, 1898, 101 mo. . . . . Amount Apr. 16, 1898 Balance due . . . . . . . . 4. Amount of $500, July 25, 1896, to Apr. 1, 1897, 8 mo. 6 da. Amount of $ 100, Sept. 18, 1896, to Apr: 1, 1897, 6 mo. 13 da. . . . Amount of $200, Feb. 5, 1897, to Apr. 1, 1897, 1 mo. 26 da. . . . . . . . .. Balance due . . . . . . 5. Amount of $870.50, Jan. 2, 1894, to March 18, 1896, 2 yr. 2 mo. 16 da. . . . . . . . . $985.99 EN . . $ 750.99 Interest March 18, 1896, to Jan. 2, 1897, 9 mo. 14 da. . . . 35.54 Amount . . ini . . . . . . . $786.53 Amount of $ 250, Aug. 24, 1896, to Jan. 2, 1897, 4 mo. 8 da. . 255.33 Due at maturity . . . . . . . $531.20 Below will be found answers to the partial payments examples of Chapters XIII and XIV, according to the Connecticut rule. NOTE. — Although the legal rate in Connecticut is 6%, the rates given in the examples should be used. Art. 1008. $70.51. Art. 1009. 1. $224.22. 2. $261.21. 3. $1278.15. Art. 1011. 4. $771.24. 5. $899.32 (see Art. 1307, 3). Art. 1013. 5. $ 1088.31. Art. 1015. 7. $ 700.50. Art. 1023. 7. $1232.26. Art. 1051. 7. $77.07. Art. 1090. 3. $649.13. 4. $ 224.64. Art. 1107. 2. $ 1089.64. The time in each of the preceding examples was found by compound subtraction, taking 30 days to each month. In the examples in Art. 1110, the Connecticut rule has been followed. In these, the time is taken in days. (See Art. 1111.) 1308. 7. Principal . . . . . . $ 700.00 Interest to June 15, 1897, 2 yr. . . . . . 84.00 1 year's interest on $ 42, unpaid interest . . . 2.52 Amount June 15, 1897 . . . $786.52 Amount of $20, Nov. 15, 1895, to June 15, 1897. . $21.90 Amount of $80, Feb. 15, 1897, to June 15, 1897. . 81.60 103.50 New principal June 15, 1897 . . $683.02 Interest to Oct. 15, 1899, 2 yr. 4 mo. . . 95.62 Interest for 1 yr. 4 mo. on $ 40.98, unpaid interest. 3.28 Interest for 4 mo. on $ 40.98, unpaid interest . . . .82 Amount Oct. 15, 1899 . . . . . $782.74 Amount of $15, Sept. 15, 1898, to Oct. 15, 1899 . 15.98 Due Oct. 15, 1899 . . . . . $766.76 NOTE. — If four places of decimals are used, the answer to 6 is $367.6442, or $367.64 +; to 7, $ 766.7658, or $ 766.77 – . |