Multiply this completed divisor by this last root figure; subtract the product, if possible, from the dividend, and to the remainder annex the next period for a new dividend. Double the whole root found, for a new trial divisor, and proceed as before, till all the periods have been used. NOTE 1.-When the product is greater than the dividend, erase the root figure which produced it, and put a figure of less value in both the root and trial divisor, till the product is small enough for subtraction. NOTE 2.--When the trial divisor is not contained in all of the dividend except its right-hand figure, annex a cipher both to the root and trial divisor, bring down the next period, and proceed as usual. NOTE 3.-If, in marking off, the last decimal place has but one figure, annex a cipher to fill the period. NOTE 4.-In surds, decimal periods of ciphers may be used to any sufficient degree of approximation. NOTE 5.-Point off from the right of the root as many figures for decimals as there are decimal periods in the operation. NOTE 6.-In large numbers, after finding more than the first half of the root, we can lessen labor by constantly dividing the last remainder by the last divisor except its right-hand figure. (See Example 3.) NOTE 7.-The square root of a perfect square may be found by resolving it into its prime factors, and finding the product of one of each two equal factors. Thus, 1764 (2 × 2) × (3 × 3) × (7 × 7): hence ✓1764 =2X 3X7 = 42. EXAMPLES FOR PRACTICE. 3. Find the square root of 579. FULL PROCESS. (See Note 4.) 579.0000000000(24.062418 4 Ans. 24.062418+. By Note 6. 579(24.062418 44)179 176 4 44)179 176 Art. 435. To extract the square root of common fractions and of mixed numbers. Ex. 1. Find the square root of 18. FIRST METHOD. 1/16 V/25 The EXPLANATION. Ans.. square root of § is one of the two equal component factors of. Therefore is the product of two equal fractions, and the numerator 16 is the square of the numerator of those fractions, and the denominator 25 is the square of the denominator of those fractions. Hence the numerator of those fractions is the square root of 16, and their denominator is the square root of 25. SECOND METHOD.-9.64: v.64.81% = {. FIRST METHOD.-50% = Ans. 23. 478: √478 = 44 = 2% = 23. SECOND METHOD.-51% = 5.76: 5.76 2.4 = 23. Rules.-I. For a fraction. Make the square root of the numerator a new numerator, and that of the denominator a new denominator; or Reduce it to a decimal, and find its square root. II. For a mixed number. Reduce it to an improper fraction, or to a mixed decimal, and find its square root in this form. APPLICATIONS OF THE SQUARE ROOT. Art. 436. A plane triangle is a plane figure bounded by three straight lines called its sides. Perpendicular. Hypotenuse. Base. A right-angled triangle is a triangle, two of whose sides are perpendicular to each other. The hypotenuse is the side opposite to the right angle. It is the longest side. The perpendicular sides are the two sides which form the right angle. The base is that side on which the figure is supposed to rest. Any side of a triangle may be considered the base. Art. 437. The square of the hypotenuse is equal to the sum of the squares of the perpendicular sides. On inspecting the figure in the margin, it ILLUSTRATION. is seen that the square Of the side 3 units long And that the sum, 9 sq. units. 66 =16" 66 25 sq. units, is equal to the square of the hypotenuse, which is 5 units long. This example is only one instance of a general truth, proved in Geometry. Rules.-I. To find the hypotenuse. Extract the square root of the sum of the squares of the other two sides. II. To find one perpendicular side. Take the square of the given perpendicular from the square of the hypotenuse, and extract the square root of the remainder. III. To find the side of a square equal to a given area. Find the square root of that area. EXAMPLES FOR PRACTICE. Find the hypotenuse of a triangle whose 1. Sides are 315 ft. and 420 ft. Ans. 525 ft. 2. Sides are 12 ft. 6 in. and 6 ft. 8 in. Ans. 14 ft. 2 in. Find the other perpendicular of a triangle whose 3. Perp. is 24 ft. and hypotenuse 50 ft. Ans. 43 ft. 10.35+ in. 4. Perp. is 15 ft. and hypotenuse 23 ft. 4 in. Find the side of a square 5. Containing 1444804 sq. in. 7. Containing 40 A. Ans. 100 ft. 2 in. 8. Two men start from the same point, and one travels due east 7.5 miles, the other due south 10 miles; how far are they apart? 9. The top of a pole standing 30 ft. from the shore of a river, is 80 ft. above the water, and 300 ft. in a straight line from the opposite shore; how wide is the river? Ans. 259.13+ ft. 10. A ladder 45 ft. long reaches from a spot in the street 28 ft. up a house on one side, and 30 ft. up a house on the other side. What is the width of the street? Ans. 68.768+ ft. 11. What is the distance from the lower corner to the opposite upper corner of a room 33 ft. long, 24 ft. wide, and 12 ft. high? Ans. 42.53+ ft. Art. 438. The areas of similar plane figures are to each other as the squares of the like dimensions of the figures. COROLLARY.-The like dimensions of similar plane figures are to each other as the square roots of the areas of these figures. NOTE.-These propositions are proved in Geometry. EXAMPLES FOR PRACTICE. 1. How much larger is a circle 12 inches in diameter than a circle 4 inches in diameter ? Ans. 9 times. 2. How much larger is a lot 40 rods square than a lot 8 rods square? Ans. 25 times. 3. How much larger is a water-pipe 20 inches in diameter than one 6 inches in diameter ? Ans. 11 times. 4. If a pipe 1 in. in diameter fill a cistern containing 48 gal. in a given time, what is the capacity of a cistern that a pipe 2 in. in diameter, will fill in the same time? Ans. 1683 gal. 5. If a pipe in. in diameter fill a cistern in 8 hours, in what time will a pipe 2 in. in diameter fill the same cistern? Ans. & of an hour. 6. A farmer has a field 60 rd. long and 48 rd. wide; what are the dimensions of a similar field containing 50 acres? Ans. 100 rd. by 80 rd. 7. How much larger is a hole bored by a 2-inch bit than one bored by a g-inch bit? Ans. 256 times. 8. If a pipe 11⁄2 inches in diameter will fill a cistern in 2 hr. 42 min., what must be the diameter of a pipe that will fill the same cistern in 2 hr. 8 min.? EXTRACTION OF THE CUBE ROOT. Art. 439. To extract the cube root of a number is to resolve that number into three equal factors. CUBES WHOSE CUBE ROOTS ARE SINGLE INTEGRAL FIGURES. CUBES.-1, 8, 27, 64, 125, 216, 343, 512, 729. ROOTS.-1, 2, 3, 4, 5, 6, 7, 8, 9. |