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2. To find the diameter, having given the circumference. Divide the circumference by 3.1416.

3. To find the area of a circle. Multiply the circumference by one-fourth the diameter. Let

diameter, then D x 3.1416 = circumference,

D and D x 8.X4X6 X = D2 x .7854;

4

D =

.7854

or, multiply the square of the diameter by.7854.

EXAMPLES.

1. Find the circumference of a circle whose diameter is 2 ft.

Ans. 6.2832 ft. 2. Find the diameter of a circle whose circumference is 6.2832 ft.

Ans. 2 ft.

3. Find the area of a circle whose diameter is 2 ft.

2 x 2 x .7854 = 3.1416 sq. ft. PROB. 4.–Find the space between two concentric circles. Find the area of both circles, and the difference of the areas will be the area of the space.

Ex. 4. Two circles have the same centre; the larger one has a diameter of 4 feet and the smaller one of 2 ft. What is the area of the space between them?

4 x 4 = 16
2 x 2 = 4

12 x .7854 = 9.4248 sq. ft.

CUBIC MEASURE.

DEFINITIONS.

1. A Right Prism is a solid which has equal and parallel polygons for its bases, and its edges are perpendicular to the bases.

2. If the bases are squares and each face square and equal to a base, the figure is called a Cube.

3. When the bases are circles, the figure becomes a Cylinder.

4. Figures with a polygon for a base and tapering to a point are called Pyramids.

5. If the upper part is cut off by a plane parallel to the base, the lower part is called the Frustum of a Pyramid.

6. When the base is a circle and the figure tapers to a point, it is called a Cone.

17. If the upper part is cut off by a plane parallel to the base, the lower part is called the Frustum of a Cone.

PROBLEMS.

To find the volume of a prism or cylinder.
Multiply the area of the base by the altitude. (Geom.)
If it be a cube, the cube of an edge will be the volume.

REM.—The lateral surface of a prism or cylinder is the product of the altitude and the perimeter of the base.

EXAMPLES.

1. What is the volume of a prism whose base contains 6 sq. ft. and altitude 5 feet?

6 sq.

ft. x 5 ft. = 30 cu. ft. 2. What is the volume of a cube whose edges are each 3 feet?

3 X3 X3 = 27 cu. ft. 3. What is the volume of a cylinder whose base contains 12 sq. ft. and its altitude is 5 ft.

12 sq. ft. x 5 ft. = 60 cu. ft.

PROBLEM.

To find the volume of a pyramid or cone.
Multiply the area of the base by one-third the altitude.

EXAMPLES.

1. What is the volume of a pyramid whose base contains 15 sq. ft. and altitude 9 ft. ?

15 X 3 = 45 cu. ft. 2. The volume of a pyramid is 90 cu. ft. and the altitudo 9 ft.; what is the area of base ? Ans. 30 sq. ft.

3. What is the volume of a cone whose base contains 36 sq. ft. and its altitude is 18 ft. ?. Ans. 216 cu. ft.

PROBLEM.

To find the volume of the frustum of a pyramid or

cone.

The volume of the frustum is equivalent to three pyramids or cones, one having the lower base for its base,

the second having the upper base for its base, each hav ing for its altitude the altitude of the frustum, and the volume of the third pyramid or cone is a mean proportional between the other two.

Therefore, extract the square root of the product of the areas of the lower and upper bases, and add together this root and the areas of the two bases and multiply their sum by one-third the altitude of the frustum, and this product will be the volume of the frustum.

EX AMPLES.

1. The areas of the lower and upper bases of the frustum of a pyramid are 16 sq. ft. and 9 sq. ft., and the altitude is 12 ft.; what is the volume of the frustum.

16x9 = 7144

= 12

16
9

37 x 4 = 148 cu. ft.

DEFINITIONS.

1. A Sphere is a solid with a curved surface, every part of which is equally distant from a point within called the centre.

2. The Ascis or Diameter of a sphere is a straight line passing through the centre and terminated at both ends by the surface.

3. A Radius is one-half the diameter, or a straight line drawn from the centre to any point of the surface.

PROBLEMS.

I. To find the surface of a sphere.

The surface of a sphere is equal to the product of the circumference and diameter.

TT X D X D = surface. (Geometry.) COR.—The surfaces of spheres are to each other as the squares of their diameters.

EXAMPLES.

What is the surface of a sphere whose diameter is 3 ft. ?

Circumference = 3.1416 x 3.
Surface = 3.1416 x 3 x3 = 28.2744 sq. ft.

II. To find the volume of a sphere.

The product of the surface and the radius or ; the diameter. = 3.1414 and D = diameter.

D
Volume = ax D x D x = .5236 x D3,

6 The volume of a sphere is equal to the cube of the diameter multiplied by .5236.

COR.—The volumes of spheres are to each other as the cubes of their diameters.

EXAMPLES.

1. What is the volume of a sphere whose diameter is 2 feet ? .5236 x 8 = 4.1888 cu. ft.

2. How many shot 1 of an inch in diameter may be made of a ball of lead 6 inches in diameter ? 6 x 6 x 6

= 6 x 6 x 6 x 16 x 16 x 16 = 884736 shot. te x1 10

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