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EXAMPLES. 1. Required the square root of 133225.
first divisor 66 | 432 first dividend.
6 / 396 first product
1825 2190 1095
second divisor 725 | 3625 second dividend.
53625 second product.
2. Required the square root of 54990.25? Ans. 234.5 3. Required the square root of .2916 ?
.54 4. Required the square root of 3 ?
1.73205 5. Required the square root of 7 ?
2.645751 6. Required the square root of 10?
3.162277 7. Required the square root of .001225?
.035 8. Required the square root of .00032754 ? .01809 9. Required the square root of 4.000067121?- 2.000016 REPEATING AND CIRCULATING DECIMALS.
RULE-Instead of ciphers, annex periods of the repeating or circulating figures.
EXAMPLES. 10. Required the square root of .i ? Ans.
.3 11. Required the square root of 4 ?
. 12. Required the square root of 1320.i ?
36-3 13. Required the square root of 138.518 ? 11.769389
VULGAR FRACTIONS. RULE-Reduce the fraction to its lowest terms, then extract the square root of its numerator and denominator, for a jew numerator and denominator ; but if the fraction be a. surd reduce it to a decimal, and then extract the root. A mixed number may be reduced to an improper fraction, or decimal.
EXAMPLES. 14. Required the square root of ?
Ans. 15. Required the square root of 92.16 16. Required the square root of 357
.86602 17. Required the square root of 18
.93308 18. Required the square root of 5131 19. Required the square root of 945?
34. 20. Required the square root of 8 ?
2.9519. 21. Required the square root of cí?
2 704 4225
QUESTIONS FOR EXERCISE. 22. A certain number of men gave $789 61 cents for a charitable purpose, each man giving as many cents as there were men : quere their number?
Ans. 281 23. If 1369 fruit trees be planted in a square orchard, how many must be in a row each way?
37. 24. There are two numbers, whereof the lesser is 3456, their difference is 293392 : what is the greater, and what the square root of their sum!
Ans. the greater
296848, and the square root of their sum 548.
25. There is an army consisting of a certain number of men who are placed rank and file that is in the form of a square, each side having 432 men : required the number?
PROBLEMS. 1. To find a mean proportional between any two given numbers.
Rule-The square root of the product of the given num. bers, is the mean proportional sought.
EXAMPLES. 26. Required the mean proportional between 3 and 12 ?
Ans. 6. 27. Required the mean proportional between 4276 and 842 ?
Ans. 1897.469. II. To find the side of a square equal in area to any given
superfices. RULE--The square root of the given superfices, is the side sougiit.
EXAMPLES. 28. If the content of a given circle be 160, required the side of the square equal thereto ? Ans.
12.64911. 29. Suppose I have an elliptical fish pond containing 9 acres, 2 roods, 15 poles, and would have a square one of the same content : Required the length of each side ?
Ans. 215.484918 yards. III. Having the are of a circle to findthe diameter. RULE-As 355 : 452 3: or, as 1 : 1.273239 : : so is the area : to the square of the diameter : or multiply the square root of the arexby 1.12837, and the product will be the diaineter.
EXAMPLES. 30. Required the diameter of that circle whose area is feet 9 inches?
Ans. 10 feet. 6 inch.
31. In the midst of a meadow,
Well stored with grass ;,
To tether my ass :
That feeding all round ;
Ans. 55) zards. IV. The area of a circle given to find the circumference. RULE-As 113 : 1420, or, as 1 : 12.56637 : : the area : to the square of the circumference, or periphery : or, multiply the square root of the area by 3.5449, and the product is the circuinference.
EXAMPLES. 32. Required the circumference of that circle whose area is 12?
Ans. 12.2798. 33. When the area is 160 perches, required the circumference?
Ans. 44.839. V. 4ny two sides of a right angled triangle
, to fined the third side. 1st. The base and perpendicuları given
to find the hypothenuse. Rule—The square root of the sum of the squares of the base and perpendicular, is the length of the hypöthenuse. 21. The hypothenuse and one side given
to find the other side. Rule-From the square of the hypotheni'se, subtract the square of the
21. given side, the square root of the remurinder is the side required.
Base EXAMPLES. 34. At Matlock, near the Peak in Derbyshire, where are many surprising curiosities in nature, is a rock by the side of the 'river Derwent, rising perpendicular to a wonderful heigt, which being inaccessible, I endeavoured to measure, and and by a mathematical method, that the distance between the place fobservation, and the foot of the rock to be jords,
d from the top of the rock to che said hace, to be 140? yai's : Required this 1.ight vi this sitpead ou wok.
Ans. 242.07 yards.
35. A ladder 40 feet long may
be 'so planted, that it shall reach a window 33 feet from the ground on one side the street, and, without moving it at the foot, will do the same by a window 21 feet high on the other side : the breadth of the street is required ? Ans.
56.64 feet. 35. A line 27 yards long, will exactly reach from the top of a fort, on the opposite bank of a river known to be 23 yards broad: the height of the wall is required. Ans. 14.142 yds.
37. Suppose' a light-house built on the top of a rock, the distance between the place of observation and that part of the rock level with the eye, and directly under the building, is given 310 fathoms; the distance from the top of the rock to the place of observation is 423 fuboms; and from the top of the building 425 : the height of the edifice is required.
310 Ans, 287.80027 fathems the height of the rock, 2.93155 fath
oms=17 feet, 7 inches, the height of the light-house." 93. Two ships set sail from the
84, samo port, one of them sails. due east 84 leagues, the other due
iesser, south 50 : How
lesser far are they assunder.
Ans. 97.75= 97: lea. the dist. squired.
39. The height ef an Elm, growing in the middle of a cit cular island 30 feet in diame ter, plumbs 53 feet, and a line stretched from the top of the tree straight to the hither edge of the waters 112 feet : What then is the breadth of the moat, supposing the land on the other side the water to be level.
Ans. 83 ft. 8 in.
CUBE ROOT. To extract the Cube Root is to find out a number, which being multiplied into itself, and then into that product produceth the given number.
Rule 1st.- Divide the given number into periods of three figures, beginning at the units place, then find the highest cube to the first period, and subtract it therefrom, put the root in the quotient, and bring down the next period to the remainder for a dividend.
2d. Square the quotient, and multiply it by 300 for a trial divisor. Find how often it is contained in the dividend, and put the result in the quotient.
1. Multiply the former quotient by 30, and the product by the figure last-put' in the quotient for the second part of Die dirisor.
4th.--Square the last figure in the quotient for the remaini pitit of the divisor ; the sum of these three parts will be
risor complete, which multiply by the figure last put byshire,uotient, subtraet the product from the dividend, bring osities the next period, and proceed as before. the
EXAMPLES. 102 Required the cube root of 15625.
1st part of the divis.=2X2X300=1200 7625 the dividend 20 part of the divis.=2X30x5 = 300 so part of the divis.=:5X5
25 The divisor complete=11.5X5=7625 the product