APPLICATIONS.

397. A triangle is a plane figure having three sides and three angles, as A, B, C.

A right triangle is a triangle having one right angle, as at C.

The base is the side on which it stands, as AC.

The perpendicular is the side forming a right angle with the base, as BC.

HYPOTHENUSE.

BASE.

The hypothenuse is the side opposite the right angle, as AB.

900

PERPENDICULAR.

398. The following principles relating to right-angled triangles have been established by Geometry:

PRINCIPLES.-I. The square of the hypothenuse of a right-angled triangle is equal to the sum of the squares of the other two sides.

II. The square of the base, or of the perpendicular, of a right-angled triangle is equal to the square of the hypothenuse diminished by the square of the other side.

FORMULAS: 2.

1. √ Base2 + perpendicular2 = hypothenuse. Hypothenuse-base2 = perpendicular. Hypothenuse - perpendicular2 = base.

3.

1. The base of a right-angled triangle is 12, and the perpendicular 16. What is the length of the hypothenuse?

2. The hypothenuse of a right-angled triangle is 35 ft., and the perpendicular 28 ft. What is the base ?

3. The hypothenuse of a right-angled triangle is 53 yd., and the base 84 feet. Find the perpendicular.

4. If the gable end of a house 40 ft. wide is 16 ft. high, what is the length of the rafters ?

5. The foot of a ladder is 15 feet from the base of a building, and the top reaches a window 36 ft. above the base. What is the length of the ladder?

6. A line reaching from the top of a precipice 120 ft. high, on the bank of a river, to the opposite side is 380 feet long. How wide is the river?

CUBE ROOT.

399. The cube root of a number is one of the three equal factors of that number.

Thus, the cube root of 125 is 5, since 5 × 5 × 5 = 125.

PRINCIPLES.-I. The cube of a number contains three times as many figures as the number, or three times as many, less one or two.

Thus,

II. If any perfect cube be separated into periods of three figures each, beginning with units' place, the number of periods will be equal to the number of figures in the cube root of that number.

400. To find the cube root of a number.

1. Find the cube root of 405224.

t3 + 3 × 12 × u + 3 × t × u2 + u3 = 405,224 ( 70 + 4 = 74 t3703

= 343 000

EXPLANATION.-Since 405224 consists of two periods, its cube root will consist of two figures (399, PRIN. II). Since 224 cannot be a part of the cube of the tens of the root, the first figure of the root must be found from the first period, 405. The greatest number of tens whose cube is contained in 405000 is 7. Subtracting the cube of 7 tens from the given number, the remainder is 62224. This remainder is equal to the product of three times the square of the tens of the root by the units, plus three times the product of the tens by the square of the units, plus the cube of the units (400).

If the remainder 62224 be divided by 3 × 702 or 14700, the quotient, which is 4, will be the units' figure of the root or a figure greater than the units' figure. Subtracting 743 from the given number, the result is 0; hence 74 is the required root.

Instead of cubing 74, the parts which make up the remainder 62224 may be formed and added thus:

Or, since 4 is a common factor in the three parts which make up the remainder, these parts may be combined thus:

In this example, 14700 is a partial or trial divisor, and 15556 is a com

plete divisor.

GEOMETRICAL EXPLANATION OF CUBE ROOT.

What is the length of the edge of a cube whose volume is 405224 cubic feet?

Let Fig. 1 represent a cube whose volume is 405224 cu. ft. It is required to find the length of the edge of this cube.

Since the volume of a cube is equal to the cube of one of its edges, an edge may be found by extracting the cube root of the volume.

Since 405224 consists of two periods, its cube root will consist of two figures.

The greatest number of tens whose cube is contained in 405224 is 7. Hence, the length of the edge of the cube is 70 ft., plus the units' figure of the root. Removing the cube whose edge is 70 ft. and whose volume is 343000 cu. ft., there remains a solid whose volume is 62224 cu. ft., Fig. 2. This remainder consists of solids similar to those marked B, C, and D, in Fig. 1 and Fig 2 of Art. 391.

The volume of a rectangular solid is equal to the product of the area of its base by its height or thickness (222); hence, if the volume be divided by the area of the base, the quotient will be the thickness. Now since the three equal rectangular solids, each of which is 70 ft. square, and whose thickness is the units' figure of the root, contain the greater part of the 62224 cu. ft., 3 x 702, or 14700, may be used as a trial divisor to find the thickness. Dividing 62224 by 14700, the quotient is 4. But this quotient may be too large. To test it, find the volume of Fig. 2, considering 4 ft. as the thickness. Thus, 3 x 702 × 4 + 3 x 70 x 42 + 48 = 62224. Hence, 4 ft. is the true thickness, and 74 ft. the length of the edge of a cube whose volume is 405224 cu. ft.

If the cube root contains more than two figures,.it may be found by a similar process, as in the following example, where it will be seen that the partial divisor at each step is equal to three times the square of that part of the root already found.

RULE.-1. Separate the given number into periods of three figures each, beginning at the units' place.

2. Find the greatest number whose cube is contained in the period on the left; this will be the first figure in the root. Subtract the cube of this figure from the period on the left, and to the remainder annex the next period to form a dividend.

3. Divide this dividend by the partial divisor, which is 3 times the square of the root already found, considered as tens; the quotient is the second figure of the root.

4. To the partial divisor add 3 times the product of the second figure of the root by the first considered as tens, also the square of the second figure, the result will be the complete divisor.

5. Multiply the complete divisor by the second figure of the root and subtract the product from the dividend.