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ters, in the different sides of a roof, is 32 feet, and the height of the ridge above the foot of the rafters is 12 feet. What is the length of a rafter?

6. What is the distance measured through the centre of a cube, from one corner to its opposite corner, the cube being 3 feet, or one yard, on a side?

We know, from the principles of geometry, that all similar surfaces, or areas, are to each other as the squares of their like dimensions.

7. Suppose we have two circular pieces of land, the one 100 feet in diameter, the other 20 feet in diameter. How much more land is there in the larger than in the smaller?

By the above principle of geometry it follows, that the quantity of land in the two circles, must be as the squares of the diameters, that is, as 100 to 20, or as 25 to 1. Hence, there is 25 times as much in the one piece, as there is in the other.

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2

8. Two persons, the one 6 feet high, the other 5 feet: now suppose they are both well proportioned in all respects; how much more cloth will it take to make a suit of clothes for the first, than for the second?

Ans.

It will require 1 times as much for the first as for the second.

9. Suppose by observation, it is found that 4 gallons of water flow through a circular orifice of 1 inch in diameter in one minute. How many gallons would, under similar circumstances, be discharged through an orifice of 3 inches in diameter, in the same length of time?

Ans. 36.

10. What must be the circumference of a circular pond, which shall contain part as much surface as a pond 13 miles in circumference ?

Ans. 32 miles..

11. Required the width and depth of a rectangular box, whose length is 3 feet, which shall contain 30000 solid inches; the width being to the depth as 2 to 3.

12. What length of thread is required to wind spirally around a cylinder, 2 feet in circumference and 3 feet in length, so as to go but once around?

It is evident that if the cylinder be developed, or placed upon a plane, and caused to roll once over, that the convex surface of the cylinder will give a rectangle, whose width is 2 feet, and length 3 feet; at the same time the thread will form its diagonal. Hence, the length of the thread is √4+9=√13=3.60555 feet.

13. Seven men purchase a grinding stone, of 60 inches in diameter. What part of the diameter must each grind off, so as to have of the whole stone?

SOLUTION.

In this question, we disregard the thickness of the stone. After the first one has ground off his share, the remaining stone will be of the original stone. Therefore its diameter will be 60√√42=55.54922, nearly.

The diameter, after the 2d one has ground off his share, will be 60√√35=50.70925, nearly.

The diameter, after the 3d one has ground off his share, will be 60√√28=45.35574, nearly.

The diameter, after the 4th one has ground off his share, will be 60√√21=39.27923, nearly.

The diameter, after the 5th one has ground off his share, will be 60√√14=32.07135, nearly.

The diameter, after the 6th one has ground off his share, will be 60√√7=22.67786, nearly.

Hence, the parts of the diameter ground off are as follows: inches, nearly.

The 1st ground off 60.00000-55.54922= 4.45078

2d 66

"55.54922-50.70925= 4.83997

3d 66 4th 66

"50.70925-45.35574= 5.35351

[blocks in formation]
[blocks in formation]

39.27923-32.07135= 7.20788

[blocks in formation]

32.07135-22.67786=
= 9.39349

66

" 22.67786

=22.67786

Proof, 60.00000

EXTRACTION OF THE CUBE ROOT.

79. If we cube 45 by the usual process, we find 45 91125.

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If, instead of 45, we take its equal, 40+5, and then cube it by actual multiplication, as explained under Art. 4, we shall have

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Now, to reverse this process, that is, to extract the cube root of 64000+24000+3000+125, we proceed as follows:

I. We first find the cube root of 64000 to be 40, which we place to the right of the number, in the form of a quotient in division, for the first part of the root sought.

We also place it on the left of the number, in a column headed 1st coL.; we next multiply it into itself, and place the result in a column headed 2d coL.; this last result, also multiplied by 40, gives 64000, which we subtract from the number, and obtain the remainder, 24000+3000+125, which we will call the FIRST Dividend.

II. We obtain the second term of the 1st column by adding the first term to itself, the result being multiplied by this first term, and added to the first term of the 2d column, gives its second term. Again, adding this first term to the second term of the 1st column, we get its third term.

III. We seek how many times the second term of the 2d column, is contained in the first dividend, or simply how many times it is contained in its firstpart, 24000, which gives 5, for the second part of the root.

IV. Finally, we add this 5 to the last term of the 1st column, whose result, multiplied by 5 and added to the last term of the 2d column, gives its third term; which, multiplied by 5, gives 27125=24000+3000+125.

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This work can be written in a more condensed form, as

follows, where the ciphers upon the right have been omitted.

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CASE I.

From the preceding operation we may draw the following rule for extracting the cube root of a whole number.

RULE.

I. Since the cube of any number can not have more than three times as many places of figures as the number, we must separate the number into periods of three figures each, counting from the unit's place towards the left. When the number of figures is not divisible by 3, the left-hand period will contain less than 3 figures.

II. Seek the greatest cube of the first, or left-hand period, place its root at the right of the number, after the manner of a quotient in division; also place it to the left of the number, for the first term of a column, marked 1ST COL. Then multiply it into itself, and place the product for the first term of a column, marked 2D COL. Again, multiply this last result by the same figure, and subtract the product from the number, and to the remainder annex the next period, and it will give the FIRST DIVIThis same figure must be added to the first term of the 1st column; the sum will be its second term, which must be multiplied by the same figure, and the product added to the first term of the 2d column; the sum will be its second term, which we shall name the FIRST TRIAL DIVISOR.

DEND.

The same figure of the root must be added to the second term of the 1st column, to form its third term.

III. See how many times the trial divisor, with two ciphers annexed, is contained in the dividend; the quotient figure will be the second figure of the root, which must be placed at the right of the first figure; also annex it to the third term of the 1st column, and multiply the result by this second figure, and add

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