2. Find the first figure of the root by trial or inspection into a table of powers, and subtract its power from the left hand period of the given number. 3. To the remainder bring down the first figure in the next period, and call it a dividend. 4. Involve the root, already obtained, to the next inferior power (one less power) to that which is given, and multiply it by the number denoting the given power, for a divisor. 5. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the root. 6. Involve the whole root to the given power, and subtract it always from as many periods of the given number as you have found periods in the root. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor, as before, and in like manner proceed till the whole be finished. 8. If remainders should occur, the root may be continued at pleasure, by annexing ciphers to the given powers. NOTE 1.-There must be as many whole numbers and decimals, respectively, in the root as there are periods of whole numbers and decimals in the power. NOTE 2.-Any two sides of a right-angled triangle being given, the third side may be readily obtained, as the square of the hypothenuse, or longest side, equals the sum of the squares of the other two sides, and consequently the difference of the squares of the longest side and either of the other sides, is the square of the remaining side. NOTE 3.-When the base and perpendicular, or the two shortest sides, are given, to find the hypothenuse, or longest side. RULE Square the two sides, from their sum extract the square root, and it will be the hypothenuse, or longest side. The longest side and one of the shortest being given, to find the other. RULE-From the square of the longest side, subtract the square of the other given side, and the square root of the remainder will be the side required. 51 x 51 x 3 7803)31457= 2d dividend 514×514 × 514=135796744 cube of the quotient 3. What is the biquadrate or 4th root of 19987173376? 19987173376(376 root 3×3×3×3=81-4th power of the quotient 3×3×3×4=108)1188=1st dividend 37 x 37 x 37 x 37=1874161=4th power of quo. 37 × 37 × 37×4=202612)1245563=2d dividend 376 × 376 × 376 × 376 = 19987173376=4th power of Ans. 365. Ans. 2.23606+ Ans.. 7. Required the square root of 133225. 1331 Ans. 01809+ Ans. 4.145392 13. I require the 7th root of 21035.8 14. A general has an army of 4096 men-how many must he place in rank and file to form them into a square? Ans. 64. 15. If 1936 trees be planted in a square orchard, how y must be in a row each way? Ans. 44. 16. Two vessels sail from the same port; one goes due north 30 leagues, and the other due west 40 leagues-how far are they from each other? 17. A line of 50 feet in length extends from the top of a wall to a point 40 feet from its base-what was its height? 18. A ladder 50 feet in length being placed in the street, will reach a window 40 feet high on one side, and without moving the foot will reach a window 30 feet high on the opposite side-how wide is the street? Ans. 70 feet. Multiply one of the given numbers by the other, and extract the square root of the product, and the root will be the mean proportional required. NOTE. When the first number is as many times greater than the second, as the second is times greater than the third, the second number is called a mean proportional between the other two. Ex. What is the mean proportional between 36 and 144? 36x144-5184, and √5184-72 Ans. PROBLEM 2. To find the side of a square equal in area to any given superficies whatever. RULE. Find the area, and the square root is the side of the square sought. Ex. If the area of a triangle be 160, what is the side of a square equal in area thereto? √160=12.649+ Ans. PROBLEM 3. If a pipe of 6 inches bore, will be 4 hours in running off a certain quantity of water, in what time will 3 pipes, each 4 inches bore, be in discharging double the quantity? 6×6=36 Then, as 48: 36 :: 4h 4× 4=16, and 16 x 3=48 : 3h. and as 1w. : 2w. :: 3h: 6h. ans. PROBLEM 4. Having the area of a circle to find the diameter. RULE. Multiply the square root of the area by 1.12837, and the product will be the diameter. Or multiply the area by 1.2732, and take the square root of the product. Ex. 1. Required the diameter of a circle whose area is 82 feet 8 inches. Ans. 10 feet 3.13 inches. 2. I have driven a stake in my meadow, to which I wish to tie my horse by a rope of such a length as that he may graze exactly two acres-how long must the rope be? Ans. 55.5+ yards. G PROBLEM 5. The area of a circle given, to find its circumference. RULE. Multiply the square root of the area by 3.5449, and the product will be the circumference. Ex. 1. Required the circumference of that circle whose area Ans. 12.279+ is 12. |