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SQUARE ROOT. 310. TO FIND THE Approximate Square Root OF NUMBERS

THAT ARE NOT PERFECT SQUARES.

Find the square root of 1795. Regard the number as representing 1795 1-inch squares. These are to be arranged in the form of a square, and the length of its side noted. 100 1-inch squares = 1 10-inch square. 1700 1-inch squares

17 10-inch squares. But 16 of the 17 10-inch squares, can be arranged in a square that is 4 by 4; that is, 40 inches by 40 inches. See diagram.

After making this square (40 inches by 40 inches) there are (1700 — 1600 + 95) 195, l-inch squares rémaining. From these,

1

4 additions are to be made to two sides of the square already formed. Each side is 40

56 inches; hence the additions must be made upon a base line of 80 inches. These addi. 9 10 | 11 | 12 tions can be as many inches wide as 80 is contained times in 195. * 195 80 = 2 +.

13 14 15 16 The additions are 2 inches wide. These will require 2 times 80, + 2 times 2,

inches. After making this square (42 in. by 42 in.) there are (195 — 164) 31 square inches remaining. If further additions are to be made to the square, the 31 square inches must be changed to tenth-inch squares. In each 1-inch square there are 100 tenth-inch squares; in 31 square inches there are 3100 tenth-inch squares. From these, additions are to be made upon two sides of the 42-inch square. 42 inches equal 420 tenth-inches. The additions must be made upon a base line (420 x 2) 840 tenth-inches long. These additions can be as many tenth-inches wide as 840 is contained times in 3100. 3100 = 840 = 3+. The additions are 3 tenth-inches wide. These will require 3 times 840, + 3 times 3, = 2529 tenth-inch squares.

After making this square (42.3 by 42.3) there are (3100 — 2529) 571 tenth-inch squares remaining. (If further additions are to be made to the square, the 571 tenth-inch squares must be changed to hundredthinch squares.) The square root of 1795, true to tenths, is 42.3.

* Allowance must be made for filling the little square shown at the upper right hand corner of the diagram.

164 square

Square Root. NOTE.-Pupils who have mastered the work on the preceding page will have no difficulty in discovering that the same result may be obtained by the following process:

Find the square root of 1795.
Operation.

Rule.
1795.'(42.36

1. Beginning with the 16

decimal point, group the fig40 x 2 = 80 * 195

ures as far as possible into 2 164

periods of two figures each. 420 x 2

2. Find the largest square 840 3100 3 2529

in the left-hand period and 4230 x 2 8460 57100

place its root at the right as 6 50796 the first figure of the com6304

plete root.
3. Subtract the

square from the left-hand period and to the difference annex the next period. Regard this as a dividend.

4. Take 2 times 10 times the root already found as a trial divisor, and find how many times it is contained in the dividend. Write the quotient as the second figure of the root, and also as a part of the divisor. Multiply the entire divisor by the second

figure of the root, subtract the product from the dividend and proceed as before.

NOTE.-If, in applying the foregoing rule, a dividend is found that will not contain the divisor, annex a zero to the root, a zero to the trial divisor, a new period to the dividend, and proceed as before.

PROBLEMS.
Find the approximate square root (true to tenths):

(1) 875. (2) 1526. (3) 2754. (4) 4150.
(5) 624.
(6) 624.7

(7) 62.47. (8) 6.24.
(a) Find the sum of the eight results.
* The entire divisor is 80 and 2; that is, 82.
+ The left-hand period may consist of either one or two figures.

Square Root. 311. To FIND THE APPROXIMATE SQUARE Root OF DECI

MALS THAT ARE NOT PERFECT SQUARES.

Find the square root of .6 Regard the number as representing .6 of a 1-inch square. .6 of a 1-inch square 60 tenthinch squares.

But 49 of the 60 tenth-inch squares can be arranged in a square that is 7 by 7; that is, 7 tenths of an inch by 7 tenths of an inch.

After making this square there are (60 — 49) 11 tenth-inch squares remaining. If additions are to be made to the square, the 11 tenth-inch squares must be changed to hundredth-inch squares. In each tenth-inch square there are 100 hundredth-inch squares ; in 11 tenth-inch squares there are 1100 hundredth-inch squares. From these, additions are to be made upon two sides of the .7-inch square.

.7 70 hundredths. The additions must be made upon a base line (70 X 2) 140 hundredth-inches long. These additions can be as many hundredth-inches wide as 140 is contained times in 1100. 1100 - 140 = 7 +. The additions are 7 hundredthinches wide. These will require 7 times 140, + 7 times 7, 1029 hundredth-inch squares.

After making this square (.77 by .77) there are (1100 — 1029) 71 hundredth-inch squares remaining. (If further additions are to be made to the square the 77 hundredth-inch squares must be changed to thousandth-inch squares.) The square root of .6, true to hundredths, is .77.

Note.-The work on this page should be first presented orally by the teacher. It must be given very slowly. Great care must be taken that pupils image each magnitude when its word-symbol is spoken by the teacher. Any attempt to move forward more rapidly than this can be done by the slowest pupil, will result in failure so far as that pupil is concerned. The great skill of many pupils in the Illinois School for the Blind in such work as this, is to be attributed mainly to their practice in creating imaginary magnitudes. When the teacher says tenth-inch square. they “think a tenth-inch square.” Its image comes immediately into consciousness. Teachers of pupils who have sight, may obtain invaluable suggestion from the mathematical ability of these pupils who have opportunity for comparatively little sense-perception of magnitudes, but who, from necessity, are constantly trained to "see with the mind's eye." See preface of this book, page 3.

Square Root. Note.—Pupils who have mastered the work on the preceding page will readily understand the following process. See rule on page 212.

1. Find the square root of .6.

49

Operation.

Observe- 46000(.774

1. That in grouping deci. mals for the purpose of extract

ing the square root it is 70 x 2 = 140 | 1100 7 1029

necessary to begin at the deci

mal point. 770 x 2 1540 7100 4 6176

2. That the square root of 924

any number of hundredths is a

number of tenths, the square root of any number of ten-thousandths is a number of hundredths, etc.

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2. Find the square root of 54264.25.

54264. 25(232.946

4
20 x 2 = 40 142

3 129
230 x 2 460 1364

2

924 2320 x 2 4640 440.25

9.418.41 23290 x 2 = 46580 21.8400

4 18.6336 232940 x 2 = 465880 3.206400

6 2.795316

.411084 Observe that the trial divisor is always 2 timès 10 times the part of the root already found.

Square Root.

312. The following numbers are perfect squares. Find their square roots by both the factor method and the method given on the four preceding pages.

(1) 6889
(2) 841

(3) 71824
(4) 1849

(5) 729

(6) 60516 (a) Find the sum of the six results.

62

(10) 436
(11) 2013

(12)
(b) Find the sum of the six results.

(13) .81
(14) .0625

(15) .04
(16) 1.21
(17) .7921

(18) .0004
(c) Find the sum of the six results.

313. MISCELLANEOUS.

or

or

or

1. The square of a number represented by one digit gives a number represented by

digits. 2. The square of a number represented by two digits gives a number represented by

digits. 3. The square of a number represented by three digits gives a number represented by

digits. 4. The square root of a perfect square represented by one or two digits is a number represented by digit.

5. The square root of a perfect square represented by three or four digits is a number represented by digits.

6. The square root of a perfect square represented by five or six digits is a number represented by — digits.

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