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Section LXI.

900

EXTRACTION OF THE SQUARE ROOT.

1. Let it be required to find what number multiplied into itself will produce 1296. OPERATION.

In contemplating this 1296(30 +6= 36 Ans. problem, we perceive

that the root or number

sought must consist of 60 +6= 66)396 396

two figures, since the

product of any two numbers can have at most but as many figures as there are in both factors, and at least but one less. We perceive, also, that the first figure of the root multiplied by itself must give a num. ber not exceeding 12, and as 12 is not the second power of any number, and the second power of 4 is more than 12, we take 3 for the first figure of the root, which multiplied into itself gives 9. Now, the second power of 3, considered as occupying the place of tens, is 900, of which we have the root 30. Taking 900 from 1296, we have a remainder, 396; and having found the root of 900, we are now to seek a number, which, being added to this root (30) and multiplied into itself once, and into 30 twice,* will produce 396. This number is found by dividing 396 by twice 30 plus the number sought.

Q. E. D. Note. — Owing to the fact that the number of figures in the product of any two numbers is always limited as above stated, we ascertain the number of figures in the root of any given second power by putting a dot over the place of units, then over the place of hundreds, and so on. The number of dots gives the number of figures in the root. Thus the square root of 133225 consists of three figures.

2. What is the square root of 576 ? OPERATION. To illustrate this question in a different BOG24 Ans way from the first, we will suppose that we 400

have 576 tiles, each of which is one foot

square, and we wish to know the side of a 44)176

square room whose floor they will pave or 176

cover.

* By adding 6 to 30 and multiplying the sum (36) into itself, we can easily see that we multiply 6 by itself once, and 30' by 6 twice, since 30 is contained in both factors, and in the operation is multiplied by the 6 in each.

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If we find a number which multiplied into itself will produce 576, that number will give the side of the room required. We perceive that as our number (576) consists of three figures, there will be two figures in its root, since the square of no number expressed by a single figure can be so large as 576; and, if the root were supposed to have more than two figures, its square would exceed 576. Dividing the number into periods thus, 576, we now find by trial, or by the table of powers, that the greatest square number or second power in the left-hand period, 5 (hundred), is 4 (hundred), and that its root is 2, which we write in the quotient. (See operation.) As this 2 is in the place of tens, its value must be 20, and its square or second power 400.

Let this be represented by a square whose sides measure 20 feet each, and Fig. 1. whose contents will therefore be 400 square feet. See figure D. We now subtract 400 from 576 and there remain 176 square feet to be arranged on two sides of the figure D, in order that its form may remain square. We therefore double the root 20, one of the sides, and it gives the length of the two sides to be enlarged, viz. 40. We then inquire how many times 40 as a divisor is contained in the dividend (except the right-hand figure) and find it to be 4 times : this we write in the root, and also in the divisor.

This 4 is the breadth of the addi. tion to our square. (See figure 2.)

Fig. 2. And this breadth, multiplied by the

20 length of the two additions (40), gives the contents of the two figures E and F, 160 square feet, which is 80 feet for each.

There now remains the space G, to complete the square, each side of a which is 4 feet; it being equal to the

400 breadth of the additions E and F. Therefore, if we square 4 we have the contents of the last addition G= 16. It is on account of this last addition that the last figure of the root is placed in the divisor. If we now multiply the divisor, 44, by the last figure in the root (4), the product will be

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176, which is equal to the remain. D contains 400 square feet. ing feet after we had formed our first square, and equal to the addi. F. 66 80 66 66 tions E, F, and G, in figure 2. G 1 16 66 66 We therefore perceive that fig. Proof, 576 ure 2 may represent a floor 24 feet square, containing 576 square 24 x 24=576 feet.

RULE. — 1. Distinguish the given number into periods of two figures each, by putting a point over the place of units, another over the place of hundreds, and so on, which points show the number of figures the root will consist of.

2. Find the greatest square number in the first or left-hand period, placing the root of it at the right hand of the given number (after the manner of a quotient in division), for the first figure of the root, and the square number under the period, subtracting it therefrom; and to the remainder bring down the next period for a dividend, always, however, omitting the right-hand figure of this dividend in dividing.

3. Place the double of the root already found on the left hand of the dividend for a divisor.

4. Find how often the divisor is contained in the dividend (omitting the right-hand figure), placing the answer in the root for the second figure of it, and likewise on the right hand of the divisor.* Multiply the divisor with the figure last annexed by the figure last placed in the root, and subtract the product from the dividend. To the remainder join the next period for a new dividend.

5. Double the figures already found in the root for a new divisor (or bring down the last divisor for a new one, doubling the right-hand figure of it), and from these find the next figure in the root as last directed, and continue the operation in the same manner, till you have brought down all the periods.

Note 1.- If, when the given power is pointed off as the case requires, the left-hand period should be deficient, it must nevertheless stand as the first period.

Note 2. — If there be decimals in the given number, it must be pointed both ways from the place of units. If, when there are integers, the first period in the decimals be deficient, it may be completed by annexing so many ciphers as the power requires. And the root must be made to consist of so many whole numbers and decimals as there are periods belonging to each ; and when the periods belonging to the given numbers are exhausted, the operation may be continued at pleasure by annexing ciphers.

* One or two units are generally to be allowed on account of other deficiencies in enlarging the square.

3. What is the square root of 278784 ?

278784(528 Answer.

25

102)287

204 1048)8384

8384 4. What is the square root of 776161 ? Ans. 881. 5. What is the square root of 998001 ? Ans. 999. 6. What is the square root of 10342656 ? Ans. 3216. 7. What is the square root of 9645192360241 ?

Ans. 3105671. 8. Extract the square root of 234.09.

Ans. 15.3. 9. Extract the square root of .000729. Ans. .027. 10. Required the square root of 17.3056. Ans. 4.16. 11. Required the square root of 373. Ans. 19.3132079+. 12. Required the square root of 8.93. Ans. 2.98831055+. 13. What is the square root of 1.96 ?

Ans. 1.4. 14. Extract the square root of 3.15. Ans. 1.77482393+. 15. What is the square root of 572199960721 ?

Ans. 756439. If it be required to extract the square root of a vulgar fraction, reduce the fraction to its lowest terms; then extract the square root of the numerator for a new numerator, and of the denominator for a new denominator; or reduce the vulgar fraction to a decimal, and extract its root. 16. What is the square root of 13417 ?

Ans. 1 17. What is the square root of 383?

Ans. Tai 18. What is the square root of 421 ?

Ans. 64. 19. What is the square root of 525 ?

Ans. 71. 20. What is the square root of 9516?

Ans. 93. 21. What is the square root of 363751 ? Ans. 1919. 22. Extract the square root of 6. Ans. 2.52987. 23. Extract the square root of 89. Ans. 2.9519+. 24. Required the square root of 2. Ans. 1.41421+.

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APPLICATION OF THE SQUARE ROOT.

DEFINITIONS. 1. A circle is a figure bounded by a line equally distant from a point called the centre.

2. A triangle is a figure of three sides. 3. An equilateral triangle is that whose three sides are equal.

4. An isosceles triangle is that which has two sides equal, and a perpendicular from the angle where the equal sides meet bisects the base.

5. A right-angled triangle is a figure of three sides and three angles, one of which is a right angle. See figure 1.

Fig. 1.

The longest side is called the hypothenuse, the horizontal side the base, and the other side is called the perpendicular.

Hypothenuse.

Perpendicular.

Base.
The following propositions are demonstrably true.
In a right-angled tri-

Fig. 2.
angle, the square of the
longest side is equal to
the sum of the squares
of the other two sides.

In all similar triangles, that is, in all triangles whose correspond. ing angles are equal, the sides about the equal angles are in direct proportion to each other ; that is, as the longest side of one triangle is to the longest side of the other, so is either of the other sides of the former triangle to the corresponding side of the latter. Let A BC and DEF be

Fig. 3.

Fig. 4. two similar triangles, and let A B be 6 feet, B C, 8 feet, and A C, 10 feet. Again, let D E be 12 feet, E F, 16 feet, and D F will 2016 be 20 feet. That is, BC

10/ 8 will be to E F as A B to DE. Then 8 feet : 16

E ALB

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