7. The sun is 885680 miles in diameter, and the earth 7912 miles; the sun is how many times as large as the earth? Ans. About 1123.

8. There are two balls whose diameters are respectively 3in. and 4in.; what is the diameter of a ball whose contents are equal to them both? Ans. 4.5in. nearly.

SUG.-Cube 3 and 4, take their sum, and then compare this with either of the given balls.

INVOLUTION GEOMETRICALLY.

438. In treating evolution by the Synthetic Method, it is necessary to give a short treatise on involution by the same method.

1. Square 25 by means of a geometrical figure.

SOLUTION.-Let the line AB represent a length of 20 units, and BH, 5 units. Upon AB construct a square, the area will be 202 =400 square units. On the two sides DC and BC construct rectangles, each 20 units long and 5 broad, the area of each will be 520100, and the area of both will be 2100 200 square units. Now add the little square on CG, whose area is 5225 square units, and the sum of the different areas, 400+200+ 25 625, is the area of a square whose side is 25.

D

E

NOTE. When there are three figures after completing the second square as above, we must make additions to it as we did to the first square. When there are four figures there are three additions, etc.

1. Cube 45 by means of the cubical blocks.

SOLUTION.-Let A, Fig. 1, represent a cube whose sides are 40 units, its contents will be 40364000. increase its dimensions by 5 units we must add, 1st, the three rectangular slabs, B, C, D, Fig. 2; 2d, the three corner pieces, E, F, G, Fig. 3; 3d, the little cube H, Fig. 4. The three slabs B, C, D, are 40 units long and wide and 5 units thick; hence their contents are 402X5X3=24000; the contents of the corner pieces, E, F, G, Fig. 3, whose

Fig. 4.

NOTE. When there are three figures in the number, complete the second cube as above, and then make additions and complete the third in the same manner. If there are still some figures and no more blocks to make additions, let the first cube represent the cube already found, and then proceed as at first.

Cube the following numbers geometrically:-

GEOMETRICAL METHOD OF SQUARE ROOT.

439. The Geometrical Method is so called because it makes use of a geometrical figure to explain the

process of

extracting the root.

twice as many places as OPERATION AS IN PRACTICE.

the number itself, or twice as many less one, the square root will consist of two places, and hence will consist of tens and units.

1225(35

3

9

30

Fig. 2.

B

C

30

5

Fig. 3.

B

D

A

The greatest number of tens whose square is contained in the given number is 3 tens. Let A, Fig. 1, represent a square whose sides are 3 tens or 30 units, its area will be the square of 30, or 900. Subtracting 900 from 1225 we find our square is not large enough by 325 square units, we must therefore increase it by 325 units. To do this we add the two rectangles B and C, Fig. 2, each of which is 30 units long, and since they nearly complete the square their area must be nearly 325 units, hence if we divide 325 by their length we will find their width. The length of each is 30, hence the length of both is 30 X 260, and dividing 325 by 60 we find their width to be 5 units. We now complete the square by the addition of the little corner square, D, Fig. 3, whose sides are 5 units, and then the entire length of the additions is 60+5=65 units, and multiplying this by the width we find the whole area of the additions to be 655325 square units. Subtracting and nothing remains, therefore the side of the square whose area is 1225 units is 35 units, hence the square root of 1225 is 35.

NOTE. When there are three or more figures in the root after completing the square as above, we then add rectangles to it and complete a second as we did the first, and so on. With the pupil on the blackboard it is only necessary to use one figure like Fig. 3, omitting Fig. 1 and Fig. 2. In practice we generally abbreviate the work by omitting the ciphers as is shown in the margin.

30

5

The rule is the same as that already given. Let the pupil apply the method to some of the problems under the Analytie Method.

GEOMETRICAL METHOD OF CUBE ROOT.

440. The Geometrical Method of cube root is so called because it makes use of a cube to explain the process.

1. Extract the cube root of 74088.

SOL.-We find the number of figures in the root as before, and then proceed as follows: The greatest number of tens whose cube is contained in the given number is 4 tens. Let A, Fig. 1, represent a cube whose sides are 40, its contents will be 403 = 64000. Subtracting from 74088 we find a remainder of 10088 cubit units, hence the cube A is not large enough to contain 74088 by 10088 cubit units, we will therefore increase it by 10088 cubit units.

To do this we add the three rectangular slabs B, C, D, Fig. 2, each of which is 40 units in length and breadth, and since they nearly complete the cube their contents must be nearly

10088, hence if we

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

OPERATION.

74088(40

4003

64000

2

10088 42

3 X 4024800

3 X 40 X 2: 240
23 -

4

5044 10088

divide 10088 by the sum of the areas of one of their faces as a base, we

can ascertain their thickness.

The area of a face of one slab is 402- 1600, and of the three, 3X16004800, and dividing 10088 by 4800 we have a quotient of 2, hence the thickness of the additions is 2 units. We now add the three corner pieces, E, F, and G, each of which is 40 units long, 2 wide, and 2 thick, hence the surface of a face of each is 40X280 square units, and of the three it is 80X3=240 square units.

We now add the little corner cube H, Fig. 4, whose sides are 2 units, and the surface of a face is 22-4. We now take the sum of the surfaces of the additions, and multiply this by the common thickness, which is 2, and we have their solid contents equal to (4800 +240 +4)

i

X2=10088. Subtracting, nothing remains, hence the cube which contains 74088 cubic units is 40+2 or 42 units on a side.

NOTE. When there are more than three figures we increase the size of the new cube, Fig. 4, as we did the first, or let the first cube, Fig. 1, represent the new cube, and proceed as before.

The rule is the same as that already given. Let the pupil apply this method of explanation to some of the problems given under the Analytic Method. The operation is abbreviated in practice, as is shown in the Analytic Method.

SOME PRINCIPLES OF INVOLUTION AND EVOLUTION.

441. The pupil will now give attention to the following principles of involution and evolution; with quite young pupils they may be deferred until review.

PRIN. I.-The product of any two powers of a number equals a power of the number denoted by the sum of the exponents.

DEM.-If we multiply the square of a number by the cube of the number, we will have the number used five times as a factor, or the 5th power of the number, and the same may be shown in any other case.

EXAMPLES FOR PRACTICE.

1. Find the fifth power of 4.

SOLUTION.-Cube 4, which is 64, and multiply this by the square of 4, which is 16, and we will have the fifth power of 4.

2. Find the fourth power of 7; sixth power of 12; seventh power of 8; eighth power of 9.

Find the value of the following:

5. 123×123.

6. (1)2×(1)3.

Ans 128. 9. (2.5)*X(2.5). Ans. (25)10. Ans. (4)3. | 10. (3.3)2×(3.3)3 Ans. (3.3)§.

PRIN. II.-A power of a number raised to any power equals a power of the number denoted by the product of the exponents.

DEM.-If we square the cube of a number we will use the number as a factor two times 3 times, or 6 times; thus, (43)243× 43=4o, and the same can be shown in any other case.