OPERATION. SIMILAR FIGURES. 819. Similar Figures are those which have the same form. Thus, circles are similar figures; also squares, etc. 820. The Principles of similar figures, derived from geometry, are as follows: PRINCIPLES. 1. The areas of all similar figures are to each other as the squares of their like dimensions. 2. Hence, the like dimensions of similar figures are to each other as the square roots of their areas. EXAMPLES FOR PRACTICE. 1. The area of a rectangle is 648 sq. yd., and one side is 27 yd.; required the area of a similar rectangle whose corresponding side is 36 yd. SOLUTION.—Since the rectangles are similar, their areas Area of 2d : 648 :: 362 : 272 are as the squares of their cor 648 X 362 responding sides ; hence we Area of 2d =1152, Ans. have the proportion in the mar 272 gin. Cancelling and multiplying, we have 1152 sq. yd. 2. The area of a circle whose diameter is 10 meters is 78.54 square meters; what is the diameter of a circle whose area is 1963.5 square meters ? Ans. 50 meters. 3. A farmer has a field 40 rods long and 32 rods wide; required the dimensions of a similar field containing 41 Ans. 30 rd.; 24 rd. 4. A man has two circular gardens; the one is 6.5 meters in diameter, the other 2.6 decameters; the second is how many times the size of the first ? Ans. 16 times. 5. If a horse tied to a post by a rope 1 ch. 784 li. can graze upon an acre, what length of rope would allow it to graze upon 11į acres ? Ans. 5 ch. 943 li. 6. The altitudes of two similar triangles are 18 ft. and 5.4 ft. ; what is the relation of their areas ? Ans. 11) 7. The area of a rectangular building lot is 720 sq. rd.; its sides are as 4 to 5; required the sides. Ans. 24; 30. 8. The sides of a rectangular field are as 3 to 4, and its area is 30 acres; required its dimensions. Ans. 60 rd., 80 rd. . acres. 9. If a pipe of an inch in diameter fill a cistern in-3 hours, what is the diameter of a pipe which will fill it in 1 hour ? Ans. 1.299 in. 10. If a pipe whose diameter is 11 in. fill a cistern in 5 hours, in what time will a pipe whose diameter is 32 inches fill it? Ans. 55 min. 11. If a pipe of 6 inches bore is 4 hours in running off a quantity of water, in what time will three pipes, each 4 inches bore, discharge double the quantity ? Ans. 6 bours. 12. Four men bought a grindstone 40 inches in diameter; how much of the diameter must each grind off so as to share it equally, no allowance being made for the hole ? Ans. 1st, 5.359+; 2d, 6.357; 3d, 8.284; 4th, 20 inches. CUBE ROOT. 821. We give Three Methods of extracting the cube root; the Common Method, a New Method, and Horner's Method. 822. There are Two Methods of explaining the methods of extracting the Cube Root, called the Analytic or Algebraic Method, and the Geometrical Method. 823. The Analytic Method of cube root is so called because it analyzes the number into its elements, and derives the process from the law of involution. 824. The Geometrical Method of cube root is so called because it makes use of a cube to explain the process. COMMON METHOD. 1. Extract the cube root of 91125. ANALYTIC SOLU OPERATION. TION.-Since the cube tu of a number consists of three times as many t: +3t2u+3tu?+u = 91125(40 t = 403 64000 5 places as the number itself, or of three times as 3t2u+3tu?+u;= 27125 45 3t2 many less one or two, 3x 402 = 4800 the cube root of 91125 3tu: 3X 40 X 5 = 600 consists of two places, 52 = 25 or of tens and units, (3t2+3tu+u?u= 5x 5425 27125 and the number itself consists of tens3+3X tens? x units +3x tens x units2 +units8. U2 = The greatest number of tens whose cube is contained in 91125 is 4 tens. Cubing the tens and subtracting, we have 27125, which equals 3x tenso X units +3x tens X units? +units'. Now, since 3x tens” x units is much greater than 3x tens X units+units), 27125 must consist principally of 3 times tens? X units; hence if we divide by 3 times tens”, we can ascertain the units. 3 times tensequals 3X 402=4800; dividing by 4800, we find the units to be 5. We then find 3 times tens X units equal to 3x40x5= 600, and units:=52= 25, and adding these and multiplying by units, we have (3X tens?+3X tens X units+units?) X units, which equals 5425 X 5=27125; subtracting, nothing remains, hence the cube root of 91125 is 45. Fig. 1. Fig. 3. R1 806 Fig. 4. GEOMETRICAL SOLUTION.- Let OPERATION. Fig. 1 represent the cube which 91125(40 contains 91125 cubic units, then our 40% = 64000 5 object is to find the number of linear 27125 45 units in its edge. The number of terms in the root, found as before, is 3X 402 4800 3 X 40 X 5 600 two. The greatest number of tens whose cube is contained in the given 52 = 25 number is 4 tens. Let A, Fig. 1, 5425 27125 represent a cube whose sides are 40, its contents will be 403 = 64000. Subtracting 64000 from 91125, we find a remainder of 27125 cubic units, which, by removing the cube A from Fig. 1, leaves a solid represented by Fig. 2. Inspecting this solid, we perceive that the greater part of it consists of the three rectangular slabs, B, C, and D, each of which is 40 units in length and breadth; hence if we divide 27125 by the sum of the areas of one face of each regarded as a base, we can ascertain their thickness, SHOWN BY LETTERS. The area of a face of one slab is 402 =1600, and of the three, 3x1600 =4800, and dividing 27125 by 4800 we have a quotient of 5, hence the thickness of the slab is 5 units. Removing the rectangular slabs, there remain three other rectangular solids, E, F, G, as shown in Fig. 3, each of which is 40 units long and 5 units thick, hence the surface of a face of each is 40x5=200 square units, and of the three, 3X40X5= 600 square units. Finally, removing E, F, and G, there remains only the little corner cube H, Fig. 4, whose sides are 5 units, and the surface of one of its faces, 52 = 25 square units. We now take the sum of the surfaces of the solids remaining after the removal of the cube A, and multiply this by the common thickness, which is 5, and we have their solid contents equal to (4800+600+25) X5=27125 cubic units, which, subtracted from the number of cubic units remaining after the removal of A, leaves no remainder. Hence the cube which contains 91125 cubic units is 40+5, or 45 units on a side. NOTE.—This can also be explained by building up the cube instead of separating it into its parts, for which see Manual. 825. When there are three figures in the root, the solution by the analytic method is as follows: OPERATION AS IN PRACTICE. htu 14:706·125(245 14706125(245 23=8 23=2003= 8000000 22 X 300=1200/6706 3h2= 3X 2002=1200006706125 2X4 X 30= 240 3ht=3X 200 X 40= 24000 42 = 16 t2= 402= 1600 1456 5824 145600/5824000 1882125 24 X 5X 30= 3600 52= 25 176425 882125 Notes.-1. By the geometric method, when there are more than two figures we remove the first cube, rectangular slabs and solids, and small cube, and we have remaining three slabs, three solids, and a small cube, as before. 2. The method employed in actual practice is derived from the other by omitting ciphers, using parts of the number instead of the whole number each time we obtain a figure of the root, etc. It will also be seen that by separating the number into periods of 3 figures each, we have the number of places in the root, the part of the number used in obtaining each figure of the root, etc. Rule.-I. Begin at units and separate the number into periods of three fiyures each. II. Find the greatest number whose cube is contained in the left hand period, write it for the first term of the root, subtract its cube from the left hand period, and annex the next period to this remainder for a dividend. = III. Multiply the square of the first term of the root by 300 for a TRIAL DIVISOR; divide the dividend by it, and the result will be the second term of the root. IV. To the trial divisor add 30 times the product of the second term of the root by the first term, and also the square of the second term; their sum will be the TRUE DIVISOR. V. Multiply the true divisor by the second term of the root, subtract the product from the dividend, and annex the next period for another dividend. Square the root now found, multiply by 300, and find the third figure as before, and thus continue until all the periods have been used. Notes.–1. If the product of the true divisor by the term of the root exceeds the dividend, the root must be diminished by a unit. 2. When a dividend will not contain a trial divisor, place a cipher in the root and two ciphers at the right of the trial divisor, bring down the next period, and proceed as before. 3. To find the cube root of a common fraction, extract the cube root of both terms. When these are not perfect cubes, reduce to a decimal and then extract the root. 4. By cubing 1, .1, .01, etc., we see that the cube of a decimal contains three times as many dccimal places as 13 1 the decimal ; hence, to extract the cube root of a deci- .13 = .001 mal, we point off the decimal in periods of three figures .013 = .000001 each, counting from the decimal point. Find the cube root of 1. 42875. Ans. 35. 16. 343319 Ans. 7.002+. 2. 166375. Ans. 55. 17. (33 +43 +53). Ans. 6. 3. 185193. Ans. 57. 18. (83 +483 +643). Ans. 72. 4. 262144. Ans. 64. 19. (243 +323 +403) Ans. 48. 5. 438976. Ans. 76. 20. 8998912. Ans. 208. 6. 614125. Ans. 85. 21. 629412793. Ans. 857. 7. 941192. Ans. 98. 22. 1879080,904. Ans. 1234. 8. 14886936. Ans. 246. 23. 16,348384872. Ans. 2538. 9. 48228544. Ans. 364. 24. 8427392875. Ans. 2035. 10. 105154048. Ans. 472. 25. 46967731712. Ans. 3608. 11. 2.370. Ans. 1.3. 26. 17040.727103. Ans. 25.73. 12. 1.953125. Ans. 11. 27. 16503.467336. Ans. 25.46. 13. 1.587962. Ans. 11. 28. 46928.689543. Ans. 36.07 14. 129554216. Ans. 506. 29. 8625.214936512. Ans. 20.508. 15. 10117 Ans. 4. 30. 8421182563625. Ans. 20345. |