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EXAMPLES.

1. Required the square root of 133225.

133225(365 root.
9

first divisor 66 | 432 first dividend. 6 396 first product

second divisor 725 | 3625 second dividend. 5 3625 second product.

2. Required the square root of 54990.25 ?
3. Required the square root of .2916?
4. Required the square root of 3?

5. Required the square root of 7?

6. Required the square root of 10?

7. Required the square root of .001225?
8. Required the square root of .00032754 ?

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.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 ?
11. Required the square root of.4 ?
12. Required the square root of 1320.i ?
13. Required the square root of 138.518 ?

VULGAR FRACTIONS.

Ans.

.6

36.3

11.769389

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RULE-Reduce the fraction to its lowest terms, then extract the square root of its numerator and denominator, for a new numerator and denominator; but if the fraction be a surd reduce it to a decimal, and then extract the root. mixed number may be reduced to an improper fraction, or decimal.

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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?

Ans.

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 is 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?

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1. To find a mean proportional between any two given numbers. RULE-The square root of the product of the given numbers, 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 sought.

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 find the diameter.

RULE AS 355: 452: or, as 1: 1.273239:: so is the area to the square of the diameter: or multiply the square root of the area by 1.12837nd the product will be the di

aineter.

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;

I've taken just two acres,

To tether my ass :

Then how long must the cord be,

That feeding all round;

He mayn't graze less or more, than

Two acres of ground.

Ans. 55 yards.

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 circumference.

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. Any two sides of a right angled triangle, to find the third side. 1st. The base and perpendicular 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 hy pothenuse.

2. The hypothenuse and one side given to find the other side.

RULE-From the square of the hypothenuse, subtract the square of the given side, the square root of the remainder is the side required.

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 height, 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 vards,

d from the top of the rock to the said ace, to be 140 yards: Required the height of this stupendouf work.

Ans. 122.07 yards.

129.07

Hypothenuse,

33.

⚫ 21.

Base.

28

Perpendicular.

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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?

36. 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 fathoms; and from the top of the building 425: the height of the edifice is required.

Ans. 56.64 feet.

27

23.

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Ans. 287.80027 fathoms the height of the rock, 2.93155 fath om 17 feet, 7 inches, the height of the light-house."

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39. The height of an Elm, growing in the middle of a cifcular island 30 feet in diameter, plumbs 53 feet, and a line stretched,from the top of the tree straight to

the hither edge

112

of the water,

112 feet: What

then is the breadth of the moat, supposing the land on the other side the water to be level.

CUBE ROOT.

Ans. 83 ft. 8 in.

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.

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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.

24. Multiply the former quotient by 30, and the product by the figure last-put in the quotient for the second part of the divisor.

4th.-Square the last figure in the quotient for the remainJeg part of the divisor; the sum of these three parts will be divisor complete, which multiply by the figure last put byshire quotient, subtract the product from the dividend, bring osities the next period, and proceed as before.

the

EXAMPLES.

to a Required the cube root of 15625.

Ast part of the divis.

20 part of the divis.

15625(25 root.

8

2x2x300=1200
2x30x5 = 300

7625 the dividend

= 25

3d part of the divis.=5×5

The divisor complete=15.5×5-7625 the product

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