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Section 49.

EXTRACTION OF THE CUBE ROOT.

A CUBE is a solid, bounded by six equal squares. A number is said to be cubed, when it is multiplied into its square.

To extract the cube root, is to find a number, which, being multiplied into its square, will produce the given number.

The extraction of this root has been illustrated by mathematicians in various ways. But it is believed, that Robert Record, Esquire, of London, in his Arithmetic published in 1673, was among the first, who illustrated this rule by the use of various diagrams and blocks. The same thing, with but little variation, has been done by several arithmeticians in our own country.

The Rule for extracting the root depends on the following

THEOREM.

If any line or number be divided into two parts, the cube of the whole line or number, is equal to the cube of the greater part, plus the square of the greater part multiplied by 3 times the less part, plus the square of the less part multiplied by 3 times the larger part, plus the cube of the less part.

To illustrate this Theorem, let 27 be divided into two parts, 20 and 7. Then, by the hypothesis, the cube of 27 is equal to the cube of 20, plus the square of 20 multiplied by 3 times 7, plus the square of 7 multiplied by 3 times 20, plus the cube of 7.

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Hence the following

RULE.

1. Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure beyond the place of units.

2. Find by the table the greatest cube in the left hand period, and put its root in the quotient.

3. Subtract the cube, thus found, from this period, and to the remainder bring down the next period; call this the dividend.

4. Multiply the square of the quotient by 300, calling it the triple square; multiply also the quotient by 30, calling it the triple quotient; the sum of these call the divisor.

5. Find how many times the divisor is contained in the dividend, and place the result in the quotient.

6. Multiply the triple square by the last quotient figure, and write the product under the dividend; multiply the square of the last quotient figure by the triple quotient, and place this product under the last; under all, set the cube of the last quotient figure, and call their sum the subtrahend.

7. Subtract the subtrahend from the dividend, and to the remainder bring down the next period for a new dividend, with which proceed as before, and so on, till the whole is completed.

NOTE 1. The same rule must be observed for continuing the operation, and pointing for decimals, as in the square root.

NOTE 2. In inquiring how many times the dividend will contain the divisor, we must sometimes make an allowance of two or three units. See National Arithmetic, page 205.

1. What is the cube root of 78402752 ?

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158

APPLICATION OF THE CUBE ROOT. [SECT. 49.

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2. What is the cube root of 74088 ?
3. What is the cube root of 185193 ?
4. What is the cube root of 80621568 ?
5. What is the cube root of 176558481 ?
6. What is the cube root of 257259456 ?
7. What is the cube root of 1860867 ?
8. What is the cube root of 1879080904?
9. What is the cube root of 41673648.563 ?

Ans. 42.

Ans. 57. Ans. 432.

Ans. 561. Ans. 636. Ans. 123. Ans. 1234.

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Spheres are to each other, as the cubes of their diam

eter.

Cones are to each other, as the cubes of their altitudes or bases.

All similar solids are to each other, as the cubes of their homologous sides.

16. If a ball, 4 inches in diameter, weighs 50lbs., what is the weight of a ball 6 inches in diameter ?.

Ans. 168.7 lbs. 17. If a sugar loaf, which is 12 inches in height, weighs 16lbs., how many inches may be broken from the base, that the residue may weigh Slbs. ? Ans. 2.5+ in.

18. If an ox, that weighs 800lbs., girts 6 feet, what is the weight of an ox that girts 7 feet? Ans. 1270.3lbs. 19. If a tree, that is one foot in diameter, make one cord, how many cords are there in a similar tree, whose diameter is two feet? Ans. 8 cords. 20. If a bell, 30 inches high, weighs 1000lbs., what is the weight of a bell 40 inches high? Ans. 2370.3lbs. 21. If an apple, 6 inches in circumference, weighs 16 ounces, what is the weight of an apple 12 inches in circumference? Ans. 128 ounces.

Section 50.

GEOMETRICAL PROBLEMS.

1. To find the area of a square or parallelogram.

RULE. Multiply the length by the breadth, and the product is the superficial contents.

2. To find the area of a rhombus or rhomboid.

RULE. Multiply the length of the base by the perpendicular height.

3. To find the area of a triangle.

RULE. Multiply the base by half the perpendicular height; or, add the three sides together; then take half of that sum, and out of it subtract each side severally; multiply the half of the sum and these remainders together, and the square root of this product will be the area of the triangle.

4. Having the diameter of a circle given, to find the circumference.

RULE. Multiply the diameter by 3.141592, and the product is the circumference.

NOTE. The exact proportion, which the diameter of a circle bears to the circumference, has never been discovered, although some mathematicians, have carried it to 200 places of decimals. If the diameter of a circle be 1 inch, the circumference will be 3.141592653 5897932384626433832795028841971693993751058209749445923078164062 8620899862803482534211706798214808651328230664709384464609550518 22317253594081284802 inches nearly.

5. Having the diameter of a circle given, to find the side of an equal square.

RULE. Multiply the diameter by .886227, and the product is the side of an equal square.

6. Having the diameter of a circle given, to find the side of an equilateral triangle inscribed.

RULE. Multiply the diameter by .707016, and the product is the side of a triangle inscribed.

7. Having the diameter of a circle given, to find the area.

RULE. Multiply the square of the diameter by .785398, and the product is the area. Or, multiply half the diameter by half the circumference, and the product is the area. 8. Having the circumference of a circle given, to find the diameter.

RULE. Multiply the circumference by .31831, and the product is the diameter.

9. Having the circumference of a circle given, to find the side of an equal square.

RULE. Multiply the circumference by .282094, and the product is the side of an equal square.

10. Having the circumference of a circle given, to find the side of an equilateral triangle inscribed.

RULE. Multiply the circumference by .2756646, and the product is the side of an equilateral triangle inscribed. 11. Having the circumference of a circle given, to find the side of an inscribed square.

RULE. Multiply the circumference by .225079, and the product is the side of a square inscribed.

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