Page images
PDF
EPUB

Then, 0.0087266 × 73.74 × 20=12.87=arc ACB, nearly.

Again,

and

[blocks in formation]

√EA2-—AD2=√100—36=√64=8=ED; 6x8 48 the area of the triangle EAB. Hence, sect. EACB-EAB=64.35-48=16.35=ACB.

2. Find the area of the segment whose height is 18, the diameter of the circle being 50. Ans. 636.4834. 3. Required the area of the segment whose chord is 16, the diameter being 20. Ans. 44.764.

PROBLEM XIII.

To find the area of a circular ring: that is, the area included between the circumferences of two circles which have a common centre.

RULE.-Take the difference between the areas of the two circles. Or, subtract the square of the less radius from the square of the greater, and multiply the remainder by 3.1416.

[merged small][ocr errors]

Their difference, or the area of the ring, is

[ocr errors][ocr errors][ocr errors][merged small][merged small]

1. The diameters of two concentric circles being 10 and 6, required the area of the ring contained between their circumferences. Ans. 50.2656.

2. What is the area of the ring when the diameters of the circles are 10 and 20? Ans. 235.62.

PROBLEM XIV.

To find the area of an ellipse, or oval.*

RULE.-Multiply the two semi-axes together, and their product by 3.1416.

1. Required the area of an ellipse whose semi-axes AE, EC, are 35 and 25.

Ans. 2748.9.

F EG B

D

* Although this rule, and the one for the following problem, cannot be de monstrated without the aid of principles not yet considered, still it was thought best to insert them, as they complete the rules necessary for the mensuration of planes.

2. Required the area of an ellipse whose axes are 24 and 18 Ans. 339.2928.

PROBLEM XV.

To find the area of any portion of a parabola.

RULE.-Multiply the base by the perpendicular height, and take two-thirds of the product for the required area.

[blocks in formation]

2. Required the area of a parabola, the base being 20 and the altitude 30.

Ans. 400.

MENSURATION OF SOLIDS.

The mensuration of solids is divided into two parts. 1st. The mensuration of their surfaces; and,

2dly. The mensuration of their solidities.

We have already seen, that the unit of measure for plane surfaces is a square whose side is the unit of length.

A curved line which is expressed by numbers is also referred to a unit of length, and its numerical value is the number of times which the line contains its unit. If, then, we suppose the linear unit to be reduced to a right line, and a square constructed on this line, this square will be the unit of measure for curved surfaces.

The unit of solidity is a cube, the face of which is equal to the superficial unit in which the surface of the solid is estimated, and the edge is equal to the linear unit in which the linear dimensions of the solid are expressed (Book VII. Prop. XIII. Sch.).

The following is a table of solid measures:

[blocks in formation]

OF POLYEDRONS, OR SURFACES BOUNDED BY PLANES.

PROBLEM I.

To find the surface of a right prism.

RULE.-Multiply the perimeter of the base by the altitude, and the product will be the convex surface (Book VII. Prop. I.). To this add the area of the two bases, when the entire surface is required.

1. To find the surface of a cube, the length of each side being 20 feet. Ans. 2400 sq. ft. 2. To find the whole surface of a triangular prism, whose base is an equilateral triangle, having each of its sides equal to 18 inches, and altitude 20 feet.

Ans. 91.949.

3. What must be paid for lining a rectangular cistern with lead at 2d. a pound, the thickness of the lead being such as to require 7lbs. for each square foot of surface; the inner dimensions of the cistern being as follows, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and the depth 2 feet 6 inches? Ans. 21. 3s. 10 d.

PROBLEM II.

To find the surface of a regular pyramid.

RULE.-Multiply the perimeter of the base by half the slant height, and the product will be the convex surface (Book VIL Prop. IV.): to this add the area of the base, when the entire surface is required.

1. To find the convex surface of a regular triangular pyra mid, the slant height being 20 feet, and each side of the base 3 feet. Ans. 90 sq. ft. 2. What is the entire surface of a regular pyramid, whose slant height is 15 feet, and the base a pentagon, of which each side is 25 feet? Ans. 2012.798.

PROBLEM III.

To find the convex surface of the frustum of a regular

[ocr errors]

pyramid.

RULE.-Multiply the half-sum of the perimeters of the two bases by the slant height of the frustum, and the product will be the convex surface (Book VII. Prop. IV. Cor.).

1. How many square feet are there in the convex surface of the frustum of a square pyramid, whose slant height is 10 feet, each side of the lower base 3 feet 4 inches, and each side of the upper base 2 feet 2 inches? Ans. 110 sq. ft.

2. What is the convex surface of the frustum of an heptagonal pyramid whose slant height is 55 feet, each side of the lower base 8 feet, and each side of the upper base 4 feet? Ans. 2310 sq. ft.

PROBLEM IV

To find the solidity of a prism.

RULE.-1. Find the area of the base.

2. Multiply the area of the base by the altitude, and the pro duct will be the solidity of the prism (Book VII. Prop. XIV.).

1. What is the solid content of a cube whose side is 24 inches? Ans. 13824. 2. How many cubic feet in a block of marble, of which the length is 3 feet 2 inches, breadth 2 feet 8 inches, and height or thickness 2 feet 6 inches?

[graphic]
[ocr errors]

PROBLEM VI.

To find the solidity of the frustum of a pyramid.

RULE. Add together the areas of the two bases of the frustum and a mean proportional between them, and then multiply the sum by one-third of the altitude (Book VII. Prop. XVIII.).

1. To find the number of solid feet in a piece of timber, whose bases are squares, each side of the lower base being 15 inches, and each side of the upper base 6 inches, the altitude being 24 feet.

Ans. 19.5.

2. Required the solidity of a pentagonal frustum, whose altitude is 5 feet, each side of the lower base 18 inches, and each side of the upper base 6 inches.

Definitions.

1. A wedge is a solid bounded by five planes: viz. a rectangle ABCD, called the base of the wedge; two trapezoids ABHG, DCHG, which are called the sides of the wedge, and which intersect D each other in the edge GH; and the two triangles GDA, HCB, which are called the ends of the wedge.

Ans. 9.31925.

[blocks in formation]

When AB, the length of the base, is equal to GH, the trapezoids ABHG, DCHG, become parallelograms, and the wedge is then one-half the parallelopipedon described on the base ABCD, and having the same altitude with the wedge.

The altitude of the wedge is the perpendicular let fall from any point of the line GH, on the base ABCD.

2. A rectangular prismoid is a solid resembling the frustum of a quadrangular pyramid. The upper and lower bases are rectangles, having their corresponding sides parallel, and the convex surface is made up of four trapezoids. The altitude of. the prismoid is the perpendicular distance between its bases.

PROBLEM VII.

To find the solidity of a wedge.

RULE. To twice the length of the base add the length of the edge. Multiply this sum by the breadth of the base, and then by the altitude of the wedge, and take one-sixth of the product for the solidity.

« PreviousContinue »