tract the area of the triangle from the area of the sector; and the remainder will be the area of the segment A B C.

Examples.

1. What is the area of the segment A B C, whose arch contains 82°: and its chord AC 17.5 feet: and the perpendicular of its triangle D E 10-4 ft : and semidiameter A D, or DC 13.7 feet?

First, find the area of a circle whose diameter is 27.4 feet.

ft.

As 113 355: 27.4: 86+ circumference. 27.4×86—235645891 area of circle.

To find the area of the sector, say

area.

As 360°: 589-1 :: 82° to 134.1 the area of the sector A B C D.

To find the area of the triangle.

The cord A C 17.5, perpendicular D E 10-4x17·5 18291 feet area of the triangle A D C.

area of the sector A B C D=1341

Area of the triangle A D C

Area of the segment A B C

CASE XIII.

910.

43:1 ft. Ans.

To find the area of a polygon.

of

DEFINITION. A polygon is a figure having equal sides and equal angles, see figures 1. 2. 3. RULE. Multiply half the length of all the sides by the nearest distance from the centre to any the sides; or multiply, the area of one of the triangles by the number of triangles contained in the figure, the product is the answer.

Figures.

B

In measuring figure first, it is evident that mul tiplying the distance from the centre to any one side, by half the length of all the sides is measur ing 5 equal triangles; the sides represent the ba ses, and the distance from the centre, to a side the perpendicular.

Examples.

1. What is the area of figure first AC equal to 8 ft.; B D equal to 5.8 ft.?

8x5.846.4 23.2 area of B A C.

there being 5 triangles contained in this figure the area of one viz. 23:2 multiplied by 5 will give the area of the polygon. 23.2×5 116.0 Ans.

2. What is the area of figure second, the dis tance from the centre to a side is 6 ft.; the sides are 6, and their length 7 ft.? Ans. 126 ft.

3. What is the area of figure third, the distance from the centre to a side 6.2 ft.; number of sides 7; their length 5-9 ft. ? Ans. 128-03 ft.

CASE XIV.

To describe and find the area of an ellipsis, or oval.

First, to describe an oval or elipsis.

RULE-Draw a line and set one foot of the dividers on said line as a centre, and describe a circle, and move the dividers to some other point on the given line (less than the semidiameter) and describe another circle of the same radius, and in the two points where the circle's peripheries intersect, set the dividers and complete the sides of the oval, and through these two points, draw the con jugate diameter, crossing the transverse diameter in the centre of the oval,

Nore. The longest diameter of an oval is called the transverse, and the shortest, the conjugate diameter.

To find the area of an ellipsis.

RULE.-Multiply the transverse by the congugate diameter, and this again by 7854, and the last product is the area; or multiply the two diameters together and of the product is the answer.

Examples.

1. What is the area of figure first, the longest diameter is 17.5 and shortest 13. ?

17 5X13X 7854 178 67850 area Ans.

2. What is the area of figure 2d, the longest diameter being 21 and the shortest 17 ?

By rule first. 21×17×·7854-280·3878 ́area. By rule second. 21×17 357÷1280 Ans. CASE XV.

To find the area of a globe or sphere, DEFINITION. A globe or sphere is bounded by a circumference every way equally distant from a point within called the centre, thus a cannon ball may be called a globe or sphere.

RULE. Multiply the diameter and circumference together, the product is the area.

Examples.

1. What is the superficial area of a globe whose circumference is 44, diameter 14?

44×14-616 area Ans.

2. What is the area of the globe we inhabit, allowing it to contain 360 degrees and 69 miles to a degree on the equator?

360°×6925020 circum. and 7964 diam. nearly. 25020×7964-199259280 area in square miles.

MENSURATION OF SOLIDS.

DEFINITION.-The mensuration of solids includes the measuring of all bodies which have length, breadth and thickness; such as timber, stone and wood, &c.

In solid measure 1728 in. make a foot, that is 12 in. in length, 12 in breadth, and 12 in thickness; thus a solid foot would make 1728 little blocks, one inch square.

GENERAL RULE.

Multiply the length, breadth and thickness together, the last product is the solidity required,

CASE I.

To find the solidity of a prism.

DEFINITION. A prism represents a three cornered file, that is all its length of the same bigness; therefore the ends are triangles.

RULE. Find the area of one end, multiply this area by the length of the prism, the last product will be the solidity.

NOTE. If the area of the end is found in inches, and mul tiplied into the length in inches, the solidity is in solid inches, and must be divided by 1728 to bring it to solid feet; if the area of one end in inches is multiplied into the length in feet, dividing by 144 will give the solidity in solid feet.

Examples.

1. What is the solidity of a prism, the sides of the triangles of which measure 13 in. and the perpendiculars of its triangle 12 in.; and its length 12 feet?

in. in.

13×12÷178×12 feet 936÷144-6,72 Ans.

2. What is the solidity of a prism the base of which is 2 ft. 3 in.; and the perpendicular of which is 1 ft. 10 in.; and length 20 ft. 6 in. ?

Ans. 42 feet 3′ 4" 6"

CASE II.

To find the solidity of any figure that is of equal
width and equal thickness.

RULE.-Multiply the length, breadth and thickness together, the last product is the solidity. Examples.

1. How many solid feet are in a hall, that is 36 ft. 6' long, 14 ft. 6' wide, and 8 ft. 6' high?

36 feet 6'x14 feet 6' x 8 feet 6′ =4498 ft. 7′ 6′′, or 4498 ft. Ans.

2. How many solid feet will be occupied by 21 chests of tea, which are 3 feet 3 inches every way? Ans. 720 ft.

CASE III.

To find the solidity of any figure that has equal thickness, but unequal width.

RULE. Find the mean width by adding together the width of both ends, and taking half the sum for the mean width: the width taken in the middle is also the mean width; multiply the mean width, depth and length together, and the product is the solidity.