20 FT

20 FT

4

4

and thus may be seen why we add the last figure of the root to double the first. 44 ft. being the length of our additions and 4 ft. being their width, multiplying we have 176 sq. ft. as the area, and this we find exactly absorbs the remaining area given in the power.

We have therefore constructed a square, each side of which is 24 feet long, and the area of which exactly equals the area given, therefore we conclude that 24 feet, the length of one side is the square root of the given area, 576 feet.

From this solution and explanation we have the following:

To Find the Square Root

a. Begin at the right hand and point off in periods of two figures each.

b. Find the greatest square in the left hand period and place its root in the quotient.

c. Subtract the square from the left hand period and bring down the next period.

d. Double the root already found and place this at the left of the dividend for a trial divisor. Find how often this is contained in the dividend, exclusive of the right hand figure, and place the quotient in the result as the second figure of the root.

e. Annex the last figure of the root to the trial divisor for the complete divisor. Multiply the complete divisor by the last figure of the root, write this under the dividend, subtract, bring down the next period if any, and continue as before.

NOTE. The square root of a fraction consists of the square root of its numerator over the square root of its denominator.

Or, the fraction may be reduced to a decimal and then the root extracted.

2. Find the square root of 1024.
3. Find the square root of 1849.
4. Find the square root of 4225.
5. Find the square root of 61504.
6. Find the square root of 444889.

9

847

7. Find the square root of 390625. 8. Find the square root of 1679616. 9. Find the square root of . 10. Find the square root of 301. 11. Find the square root of 1183. 12. Find the square root of 2777. 13. Find the square root of 3. 14. Find the square root of 3. 15. Extract the square root of 3.

16. Extract the square root of 173.

17. Extract the square root of .008836. 18. Extract the square root of .00006561. 19. What is the value of V 3?

20. What is the value of V 125?

APPLICATIONS OF SQUARE ROOT

490. An Angle is the space included between two straight lines which meet. Thus the two lines B A and C A meet and form an angle at A.

491. A Triangle is a figure bounded by three straight lines. The accompanying figure is a triangle.

B

492. A Right-Angle is the space between two lines when one is perpendicular to the other. Thus the angle at B is a right-angle.

493. A Right-Angled Triangle is a triangle having in it a right angle.

494. The Hypotenuse is the longest side, or the slanting side. Thus A C is the hypotenuse.

495. The Base is the side upon which the triangle rests. Thus B C is the base.

496. The Perpendicular is the side which stands at right angles with the base. Thus A B is the perpendicular.

B

It is a known principle, discovered and demonstrated over two thousand years ago by a Greek philosopher and mathematician that the area of the square described on the hypotenuse of a rightangled triangle is equal to the sum of the areas of the squares described on the other two sides.

From this principle we have the following:

To Find the Hypotenuse

a. Add the square of the base and perpendicular together and extract the square root.

To Find Either of the Shorter Sides

Subtract the square of the given side from the square of the hypotenuse, and extract the square root.

1. The base of a right-angled triangle is 30 ft. and the perpendicular is 40 ft. What is the hypotenuse?

2. The base of a right-angled triangle is 18 ft. and the hypotenuse is 22 ft. What is the perpendicular?

3. The perpendicular of a right-angled triangle is 64 inches and the hypotenuse is 82 inches. What is the length of the base? 4. A church steeple is 128 ft. high and stands 72 feet from the opposite side of the street. What is the length of a rope which will reach from the top of the steeple to the opposite side of the street?

5. A window is 16 ft. 4 in. from the ground. What length. of a ladder the foot of which placed 14 ft. 8 in. from the house, will reach to the window?

6. A square field contains 40 A. will be required for one side of it? will be required to entirely enclose it?

How many rods of fence
How many rods of fence

7. A city lot in the form of a rectangle contains 5650 sq. ft. and its length is twice its width. What are its dimensions?

8. A maypole broke off 23 ft. from the ground and in falling the top struck the ground 14 ft. from the bottom of the pole. How high was the pole?

9. A park is one mile square. What is the length of a diag

onal path across it?

10. How many rods of travel does A save who goes diagonally across a field 1 mile square in preference to going by its two sides?

11. A room is 30 feet long, 25 feet wide and 12 feet high. What is the distance from one of the lower corners of the room to the opposite upper corner?

CUBE ROOT

497. The Cube Root of a number is one of three equal factors which produce the number. Thus the cube root of 64 is 4.

By trial we find that the cube of any number consisting of one figure can never exceed three figures, the cube of a number consisting of two figures can never exceed six figures, etc. Hence if we point off the power into periods of three figures each the number of periods will indicate the number of figures in the root. 1. Find the cube root of 91125 cu. in.

=

EXPLANATION.Pointing off the power into periods of three places each, by the principle just given, we see that the root will contain two places. It is required to construct a cube which shall contain 91125 cu. in. We

see by inspection that the cube root of the left period is 4 and since there are to be two figures in the root this is 4 tens or 40. Construct a cube which shall be 40 in. on each side as in the diagram. The solid contents of this cube is 40 X 40 X 40 64000 cu. in. But subtracting the 64000 cu. in. (expressed in the solution as 64 in thousands place) we still have 27125 cu. in. left. Our cube must be increased so as to absorb this, which can be done by adding to three sides of the cube, and still re

tain its cubical form. The surface of one
face of the cube contains 40 X 40
= 1600 sq.
in. and 3 faces will contain 3 times 1600 sq.
in. or 4800 sq. in. This explains why we
"square the root found and multiply it by 300
for a trial divisor," 300 being used instead of
3 because we have used the root found as
units when they are really tens. We find that
our trial divisor is contained in the dividend
5 times and this gives us the thickness of the
additions. But our cube is not yet perfect. We
must add three rectangular pieces at the cor-
ners. These pieces will be 40 in. long and 5 in.
wide and there being three of them their sur-
face will be 40 × 5 × 3 = 600 sq. in. This ex-
plains why we "multiply the last figure of the
root by the last and by 30;" 30 being used in-
stead of 3 because we called the 40 four units
instead of 4 tens, which they really are. But
our cube is still imperfect. We must add a
small block at the corner. This will be 5 in.
long and 5 in. wide and hence its surface will
be 5 × 5 = 25 sq. in. which is also written in
the divisor. Our complete divisor then is
5425 sq. in. which is the surface of all of the
additions that we have made. Now multiply-
ing this by the thickness, 5 in., we have the
solid contents of all of these additions, 27125
cu. in. and this, we find, exactly consumes the
cubical contents given.

From this solution and explanation we have the following:

To Find the Cube Root

a. Begin at the right hand and point off in periods of three places each.

b. Find the greatest cube in the left hand period and place its root in the quotient.

c. Subtract the cube from the left hand period and bring down the next period.