RULES FOR THE APPLICATION OF ARITHMETIC TO GEOMETRY.
The student should make himself familiar with the most common terms of geometry, after which the following rules will present no difficulty. In them all, it must be understood, that when we talk of multiplying one line by another, we mean the repetition of one line as often as there are units of a given kind, as feet or inches, in another. In any other sense, it is absurd to talk of multiplying a quantity by another quantity. All quantities of the same kind should be represented in numbers of the same unit; thus, all the lines should be either feet and decimals of a foot, or inches and decimals of an inch, &c. And in whatever unit a length is represented, a surface is expressed in the corresponding square units, and a solid in the corresponding cubic units. This being understood, the rules apply to all sorts of units.
To find the area of a rectangle. Multiply together the units in two sides which meet, or multiply together two sides which meet; the product is the number of square units in the area. Thus, if 6 feet and 5 feet be the sides, the area is 6 × 5, or 30 square feet. Similarly, the area of a square of 6 feet long is 6 × 6, or 36 square feet (234).
To find the area of a parallelogram. Multiply one side by the perpendicular distance between it and the opposite side; the product is the area required in square units.
To find the area of a trapezium. Multiply either of the two sides which are not parallel by the perpendicular let fall upon it from the middle point of the other.
A four-sided figure, which has two sides parallel, and two sides not parallel.
To find the area of a triangle. Multiply any side by the perpendicular let fall upon it from the opposite vertex, and take half the product. Or, halve the sum of the three sides, subtract the three sides severally from this half sum, multiply the four results together, and find the square root of the product. The result is the number of square units in the area; and twice this, divided by either side, is the perpendicular distance of that side from its opposite vertex.
To find the radius of the internal circle which touches the three sides of a triangle. Divide the area, found in the last paragraph, by half the sum of the sides.
Given the two sides of a right-angled triangle, to find the hypothenuse. Add the squares of the sides, and extract the square root of the sum.
Given the hypothenuse and one of the sides, to find the other side. Multiply the sum of the given lines by their difference, and extract the square root of the product.
To find the circumference of a circle from its radius, very nearly. Multiply twice the radius, or the diameter, by 3·1415927, taking as many decimal places as may be thought necessary. For a rough computation, multiply by 22 and divide by 7. For a very exact computation, in which decimals shall be avoided, multiply by 355 and divide by 113. See (131), last example.
To find the arc of a circular sector, very nearly, knowing the radius and the angle. Turn the angle into seconds, multiply by the radius, and divide the product by 206265. The result will be the number of units in the arc.
The right angle is divided into 90 equal parts called degrees, each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds. Thus, 2° 15′ 40″ means 2 degrees, 15 minutes, and 40 seconds.
To find the area of a circle from its radius, very nearly. Multiply the square of the radius by 3·1415927.
To find the area of a sector, very nearly, knowing the radius and the angle. Turn the angle into seconds, multiply by the square of the radius, and divide by 206265 × 2, or 412530.
To find the solid content of a rectangular parallelopiped. Multiply together three sides which meet: the result is the number of cubic units required. If the figure be not rectangular, multiply the area of one of its planes by the perpendicular distance between it and its opposite plane.
To find the solid content of a pyramid. Multiply the area of the base by the perpendicular let fall from the vertex upon the base, and divide by 3.
To find the solid content of a prism. Multiply the area of the base by the perpendicular distance between the opposite bases.
To find the surface of a sphere. Multiply 4 times the square of the radius by 3·1415927.
To find the solid content of a sphere. Multiply the cube of the radius by 3·1415927 × ⁴/₃, or 4·18879.
To find the surface of a right cone. Take half the product of the circumference of the base and slanting side. To find the solid content, take one-third of the product of the base and the altitude.
To find the surface of a right cylinder. Multiply the circumference of the base by the altitude. To find the solid content, multiply the area of the base by the altitude.
The weight of a body may be found, when its solid content is known, if the weight of one cubic inch or foot of the body be known. But it is usual to form tables, not of the weights of a cubic unit of different bodies, but of the proportion which these weights bear to some one amongst them. The one chosen is usually distilled water, and the proportion just mentioned is called the specific gravity. Thus, the specific gravity of gold is 19·362, or a cubic foot of gold is 19·362 times as heavy as a cubic foot of distilled water. Suppose now the weight of a sphere of gold is required, whose radius is 4 inches. The content of this sphere is 4 × 4 × 4 × 4·1888, or 268·0832 cubic inches; and since, by (217), each cubic inch of water weighs 252·458 grains, each cubic inch of gold weighs 252·458 × 19·362, or 4888·091 grains; so that 268·0832 cubic inches of gold weigh 268·0832 × 4888·091 grains, or 227½ pounds troy nearly. Tables of specific gravities may be found in most works of chemistry and practical mechanics.
The cubic foot of water is 908·8488 troy ounces, 75·7374 troy pounds, 997·1369691 averdupois ounces, and 62·3210606 averdupois pounds. For all rough purposes it will do to consider the cubic foot of water as being 1000 common ounces, which reduces tables of specific gravities to common terms in an obvious way. Thus, when we read of a substance which has the specific gravity 4·1172, we may take it that a cubic foot of the substance weighs 4117 ounces. For greater correctness, diminish this result by 3 parts out of a thousand.
THE END.
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