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Let E represent a corner piece which is 20 × 5 × 5, or 20 × 5o. It is seen that there are three of these, whose contents are 3 × 20 × 52. Finally, a small cube, O, which is 5 units on a side and whose contents are 53, will complete the new cube.

We shall then have

203 + 3(202 × 5) + 3(20 × 52) + 53,

or 15625 for the contents of a cube whose side is 25.

Hence, the cube of any number consisting of two figures is equal to the cube of the tens, plus three times the tens squared into the units, plus three times the tens into the units squared, plus the units cubed.

Cube the following numbers geometrically:

54. 24.

57. 54.

55. 35.

58. 67.

56. 43.

59. 72.

51. 12.

52. 15.

53. 18.

60. 84.

61. 92.

62. 95.

EVOLUTION.

214. Evolution is the process of finding a root of a number.

215. A Root of a number is one of the equal factors of that number.

A Root is indicated by the symbol V, called the Radical Sign.

Thus,

25 =

5 × 5, hence 1/25 = 5, the square root of 25. 64 4 X 4X4, hence 1/64 4, the cube root of 64. 813 X 3 X 3 X 3, hence 813, the fourth root of 81.

216. The Index of the root is the number written in the angle of the radical sign to indicate what root of the number is required.

217. A Perfect Power is a number of which the required

root can be found exactly.

Thus, 25 is a perfect square, since 52 = 25.

And 27 is a perfect cube, since 33 = 27.

ORAL EXERCISES.

The following table should be committed to memory.

Square Root.

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3

3

=

3

3

Find the value

1. Of 100.

13. Of 1/400.

2. Of √121. 3. Of √169. 4. Of √144.

3

3

5. Of/1000.
6. Of V.09.
7. Of V.16.

14. Of 1/64000.
15. Of .008.
16. Of V4900.
17. Of .027.
18. Of/512000.
19. Of 1900.
20. Of 1/729000.
21. Of/.343.
22. Of V.81.

3

8. Of V.25.
9. Of/8000.

3

10. Of V.36.

3

11. Of/27000. 23. Of .064.

12. Of V.49.

24. Of V.64.

8

27

3

64 = 4

125 = = 5

PROCESS.

/125 3 512

=

2

=

Cube Root.

Ans.

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25. Of.125.

26. Of343000.
27. Of 18100.
28. Of.512.
29. Of 1/6400.
30. Of729.
31. Of.216.
32. Of 1600.

3

33. Of 125000. V

34. Of 1/2500.

35. Of 216000. 36. Of 13600.

8

9

[blocks in formation]

5)3025

5)605

11)121

11

43. Of V
44. Of v12.

100
169.
27

3

45. Of V

3

46. Of v

Hence 1/3025

EVOLUTION BY FACTORING.

218. When the number is a perfect power and the factors are readily found, the root may be obtained as follows:

64

343

47. Of V.

81 100

WRITTEN EXERCISES.

1. Find the square root of 3025, and the cube root of 3375.

PROCESS.

2. V/441.
3.2744.
4. V1225.
5.9261.
6. V/6561.

=

· 5 × 11 = 55, Ans.

Hence/3375

48. Of 1.
49. Of Vt.

50. Of V121.

51. Of 47.

52. Of v61.

7.970299.

8.50625.
9. 7776.

5

10.1728.
11.15625.

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RULE.

Resolve the number into its prime factors, and take the product of one of every two equal factors for the square root; the product of one of every three equal factors for the cube root, etc.

Find the

12.

13.

[blocks in formation]

2401.

5

V 59049.

6

6

14. 46656.

15.

65536.

262144.

16.

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102

2 × 10

SQUARE ROOT.

=

The

square of 1, the smallest number of one figure, is 1; the square of 9, the largest number of one figure, is 81; hence the square of a number consisting of one figure must contain one figure or two figures.

The square of 10, the smallest number of two figures, is 100; the square of 99, the largest number of two figures, is 9801; hence the square of a number of two figures must contain three figures or four figures that is, twice two, or twice two less one.

1. Find the square root of 196.

PROCESS.

102 = 100.

992 = 9801.

The same may be shown for the square of a number consisting of any number of figures. Hence the

220. Principle.-The square of a number contains twice as many figures as the number itself, or twice as many less one.

10 1/96 4

20 96

4

24 96

WRITTEN EXERCISES.

= 1 00 14, Ans.

1002

10000.

9992 998001.

B

10

=

A

=

10

ILLUSTRATION.

D

10

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Pointing the number off into periods of two figures each, by ART. 219 we find that there are two figures in its square root; hence the root will consist of tens and units.

The greatest number of tens whose square is contained in 196 is 1 ten, cr 10 units. This is shown in the illustration by the uare A, whose sides are 10 units and whose area is 102, or 100 square units. Subtracting 100 from 196, the remainder is 96. This we see consists principally of two rectangles B and C, each of which is 10 units long. Since they nearly complete the square, their area is nearly 96 square units. If we divide 96 by their length, we shall find their width. The length of both rectangles is 2 × 10 20. Dividing 96 by 20, we find the width to be 4 units. Adding the length of the small square D, whose sides are 4 units, the entire length of the remaining surface is 24 units. Multiplying this by the width, 4, the area is 24 × 4 = 96 square units. Subtracting, nothing remains. Hence the square root of 196 is 14.

In practice we omit the ciphers and condense the process, as in the next example.

2. Find the square root of 8464.

FULL FORM.

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IN PRACTICE.

| 84/64 92, Ans.

81

182 364
364

RULE.

Begin at the unit's place and separate the number into periods of two figures each.

Find the largest number whose square is contained in the lefthand period; write it as the first figure of the root, subtract its square from the left-hand period, and annex the next period to the remainder.

Double the root found, for a trial divisor; divide the remainder, omitting the last figure, by this divisor, and annex the quotient to the root, and also to the trial divisor.

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