2X40=80 18'66'24(432 16 3 83 249 2 862 2X430=860 17 24 17 24 Explanation: First find the largest number whose square is equal to or less than 18, the first period. This is 4, since 52 is more than 18. Write the 4 to the right for the first figure of the root just as the quotient is put down in long division. The first figure of the root is 4. Square the 4 and write its square (16) under the first period (18) and subtract, leaving 2. Bring down the next period (66) and annex it to the remainder, giving 266 for what is called the dividend. Annex a cipher to the part of the root already found (4) giving 40; then multiply this by 2, making 80, which is called the trial divisor. Set this off to the left. Divide the dividend (266) by the trial divisor (80). We obtain 3, which is probably the next figure of the root. Write this 3 in the root as the second figure and also add it to the trial divisor, giving 83, which is the final divisor. Multiply this by the figure of the root just found (3) giving 249. Subtract this from the dividend (266) leaving 17. Bring down the next period (24) and annex to the 17, giving a new dividend 1724. Repeat the preceding process as follows: Annex a cipher to the part of the root already found (43) giving 430; and multiply by 2, giving 860, the trial divisor. Divide the dividend by this divisor and obtain 2 as the next figure of the root. Put this down as the third figure of the root and also add it to the trial divisor, giving 862 as the final divisor. Multiply this by the 2 and obtain 1724, which leaves no remainder when subtracted from the dividend: As there are no more periods in the original number, the root is complete. 77. Square Roots of Mixed Numbers.—If it is required to find the square root of a number composed of a whole number and a decimal, begin at the decimal point and point off periods to right and left. Then find the root as before. Example: Find the square root of 257.8623 2'57.86'23'00(16.058+, Answer. 1 20 1 57 6 26 1 56 3200 5 3205 32100 8 32108 1 86 23 (a) 1 60 25 25 98 00 25 68 64 29 36 Explanation: 1 is the largest number whose square is equal to or less than 2, the first period. Proceeding as before, we get 6 for the second figure. After subtracting the second time (at a) we find that the trial divisor 320 is larger than the dividend 186. In this case, we place a cipher in the root, annex another cipher to 320 making 3200, annex the next period, 23, to the dividend and then proceed as before. If the root proves, as in this case, to be an interminable decimal (one that does not end) continue for two or three decimal places and put a + sign after the root as in division. In this example the decimal point comes after 16, because there must be two figures in the whole number part of the root since there are two periods in the whole number part of our original number. 78. Square Roots of Decimals.-Sometimes, in the case of a decimal, one or more periods are composed entirely of ciphers. The root will then contain one cipher following the decimal point for each full period of ciphers in the number. Example: Take .0007856 as an example. Beginning at the decimal point and pointing off into periods of two figures each, we have .00'07'85'60. Hence, the first figure of the root must be a cipher. To obtain the rest of the root we proceed as before. It will be noticed that the square root of a decimal will always be a decimal. If we square a fraction, we will get a smaller fraction for its square, (); or as a decimal, .252.0625. Therefore, the opposite is true; that, if we take the square root of a number entirely a decimal, we get a decimal, but it will be larger than the one of which it is the square root. Notice the example just given: .0007856 is less than its square root .028. 79. Rules for Square Root.-From the preceding examples the following rules may be deduced: 1. Beginning at the decimal point separate the number into periods of two figures each. If there is no decimal point begin with the figure farthest to the right. 2. Find the greatest whole number whose square is contained in the first or left-hand period. Write this number as the first figure in the root; subtract the square of this number from the first period, and annex the second period to the remainder. 3. Annex a cipher to the part of the root already found and multiply by 2; this gives the trial divisor. Divide the dividend by the trial divisor for the second figure of the root and add this figure to the trial divisor for the complete divisor. Multiply the complete divisor by the second figure in the root and subtract this result from the dividend. (If this result is larger than the dividend, a smaller number must be tried for the second figure of the root.) Bring down the third period and annex it to the last remainder for the new dividend. 4. Repeat rule 3 until the last period is used, after which, if any additional decimal places are required, annex cipher periods and continue as before. If the last period in the decimal should contain but one figure, annex a cipher to make a full period. 5. If at any time the trial divisor is not contained in the dividend, place a cipher in the root, annex a cipher to the trial divisor and bring down another period. 6. To locate the decimal point, remember that there will be as many figures in the root to the left of the decimal point as there were periods to the left of the decimal point in our original number. 80. The Law of Right Triangles.-One of the most useful laws of geometry is that relating to the sides of a right angled triangle. Fig. 28 shows a right angled triangle, or "right triangle," so A FIG. 28. B U FIG. 29. called because one of its angles (the one at C) is a right angle, or 90°. The longest side (c) is called the hypotenuse. "In any right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides." Written as a formula this would read c2=b2+a2. This can be illustrated by drawing squares on each side, as in Fig. 29, and noting that the area of the square on the hypotenuse is equal to the sum of the areas of the other two. In using this rule, however, we do not care anything about these areas and seldom think of them except as being the squares of numbers. It is used to find one side of such a triangle when the. other two are known. Examples: 1. If the trolley pole in Fig. 30 is 24 ft. high, and the guy wire is anchored 7 ft. from the base of the pole, what is the length of the guy wire? The guy wire is the hypotenuse of a right triangle whose sides are 24 ft. and 7 ft. c2= b2+a2 =576+49=-625 c=625-25 ft., Answer. 2. If the triangle of Fig. 31 is a right triangle having the hypotenuse c-13 in. and the side a=5 in., what is the length of the side b? c2=b2+ a2 Hence, b2c2-a2 b2-132-52 =169-25-144 b =√144-12 in., Answer. This property of right triangles is also useful in laying out right angles on a large scale more accurately than it can be done with a square. This is done by using three strings, wires, or chains of such lengths that when stretched they form a right triangle. A useful set of numbers that will give this are 3, 4, and 5, since 32+42=52 (9+16=25). Any three other numbers having the same ratios as 3, 4, and 5 can be used if desired. 6, 8, and 10; 9, 12, and 15; 12, 16, and 20; 15, 20, and 25; any of these sets of numbers can be used. A surveyor will often use lengths of 15 ft., 20 ft., and 25 ft. on his chain to lay out a square corner; this method can also be used in aligning engines, shafting, etc. 81. Dimensions of Squares and Circles.-Square Root must be used in getting the dimensions of a square or a circle to have a given area. If the area of a square is given, the length of one side can be obtained by extracting the square root of the area. If we wish to know the diameter of a circle which shall have a certain area, we can find it by the following process: The area is .7854× the square of the diameter or, briefly, A.7854XD2 If we divide the given area by .7854, we will get the area of the square constructed around the circle (see Fig. 19). One side of this square is the same as the diameter of the circle and is equal to the square root of the area of the square. Then, to find the diameter of a circle to have a given area: Divide the given area by .7854 and extract the square root of the quotient. 82. Dimensions of Rectangles.-Occasionally one encounters a problem in which he wants a rectangle of a certain area and knows only that the two dimensions must be in some ratio. It may be that a factory building is to cover, say 40,000 sq. ft. of ground and is to be four times as long as it is wide, or some problem of a similar nature. Suppose we take the case of this factory and see how we would proceed to find the dimensions of the building. Example: Wanted a factory building to cover 40,000 sq. ft. of ground. Ratio of length to breadth, 4:1. Find the dimensions. 1=4b FIG. 32. 40000÷4=10000 b2=10000 7=4X100=400. |