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308. THE SQUARE OF THE SUM OF TWO LINES.
1. Study the diagram and observe – D
E (1) That the line AC is the sum of
4 the lines AB and BC.
(2) That the square, 1, is the square of AB. (3) That the rectangle, 2, is as long
2 as AB and as wide as BC. (4) That the rectangle, 3, is as long
B as AB and as wide as BC.
(5) That the square, 4, is the square of BC.
(6) That the square, ACED, is the square of the sum of AB and BC.
2. Since a similar diagram may be drawn with any two lines as a base, the following general statement may be made :
The square of the sum of two lines is equivalent to the square of the first plus twice the rectangle of the two lines plus the square of the second.
3. If the line AB is 10 inches and the line BC, 5 inches, how many square inches in each part of the diagram and how many in the sum of the parts ?
4. Consider the line AB 10 inches and the line BC 3 inches and find the area of each part of the diagram.
5. Suppose the line AB is equal to the line BC; what is the shape of 2 and 3? 6. In the light of the above diagram study the following:
142 = 196. (10 + 4) = 10% + 2 (10 x 4) + 4% = 196.
309. MISCELLANEOUS REVIEW.
1. What is the square root of a' b? ?
What is the square root of 3 x 3 x 5 x 5 ?
2. What is the cube root of a’ 63 ?
What is the cube root of 2 x 2 x 2 x 7 x 7 x 7 ?
3. What is the square root of a’ b*?
What is the square root of 52 x 34 ?
4. What is the cube root of a 60 ?
What is the cube root of 36 x 56 ?
5. The area of a certain square floor is 784 square
feet. How many feet in the perimeter of the floor?
6. The area of a certain square field is 40 acres. How many rods of fence will be required to enclose it?
7. The solid content of a certain cube is 216 cubic inches. How many square inches in one of its faces ?
8. If there are 64 square inches in one face of a cube, how many cubic inches in its solid content ?
9. The square of (30 + 5) is how many more than the square of 30 plus the square of 5 ?
10. The square of (40 + 3) is how many more than the square of 40 plus the square of 3?
11. The square of a is a'; the square of 2 a is 4 a'. The square of two times a number is equal to how many times the square of the number itself?
12. The square of an 8-inch line equals how many times the square of a 4-inch line ?
13. The square of a 6-inch line equals how many times the square of a 2-inch line ?
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, 1-inch squares remaining. From these,
1 2 3 4 additions are to be made to two sides of the square already formed. Each side is 40 5 6 7 inches; hence the additions must be made upon a base line of 80 inches. These addi. 9 | tions can be as many inches wide as 80 is contained times in 195. * 195 · 80
13 14 15 16 The additions are 2 inches wide. These will require 2 times 80, + 2 times 2, 164 square 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 tenih-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.
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.
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.
2. Find the largest square 420 x 2 840 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
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.
(1) 875. (2) 1526. (3) 2754. (4) 4150.
(6) 624.7 (7) 62.47. (8) 6.24.
* The entire divisor is 80 and 2; that is, 82.
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.