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

Fig. 1 A C-8.

Examples.

BD 5.8X823.2 area of A B C, and there being 5 triangles contained in this figure, the area of one is 23.2 | 5=116-0 area of figure 1-and is the same as the distance B D multiplied by half the length of all the sides, viz.

8+8+8+8+8=40÷=20 × 5·8 116·0 area as before. Figure 2. The distance from the centre is 6; the sides are 6 in number, and their length 7:

6X7=42=21×6=126 area of figure 2.

Figure 3 has 7 sides 5:9 long; and the distance from the centre to any side is

5.9X7 41.3 ÷

6•2

20·65X62128-03 area, figure $.

CASE XIV,

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

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 fine (less than the semi diameter) 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 conjugate diameter, crossing the transverse diameter in the centre of the oval*

RULE. Multiply the transverse by the congugate diameter, and this again by 0-7854, and the last product is the area?

N. B.

Examples.

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

17.5X13X0-7854-178.67850 area.

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

21 X17X0-7854-280-3878 area. by rule 2d 21 x 17X1280 area.

14

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

CASE XV.

To find the area of 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 X 14 616 area Ans.

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

360° 69=25020 circum. and 7964 diam. nearly.
25020×7964—199259280 area in square miles.

MENSURATIONS OF SOLIDS.

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

In solid measure 1728 inches make a foot, that is 12 inches 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.

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

* A globe or sphere is a round ball.

RULE. Find the area of one end in inches, 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 multiplied 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 inches, and the perpendiculars of its triangle 12 inches; and the length 12 feet?

in. in.

13×1278×12 feet-936-÷144-6,

Ans.

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

CASE II.

Ans. 42 feet 3' 4" 6'

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 feet 6' long 14 feet 6' wide and 8 feet 6' high? 36 feet 6' x 14 feet 6' x8 feet 6' er 44985 ft. Ans.

4498 ft. 7′6′′,

2. How many solid feet will be occupied by 21 chests of tea, which are 3 feet 3 inches every way? Ans. 72057 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.

Examples.

1. How many solid feet are in a stone that is 21 feet long, 2 feet wide at one end, and 3 feet at the other, and 1 foot 6 inches thick?

235 2.5X21X1.5=78.75, Ans.

2. How many solid feet in a stone wall that is 51 feet long, and 7 feet high, and mean thickness 2 feet Ans. 892 feet.

CASE IV.

To find the solidity of a cylinder.

NOTE. A cylinder is a long round body, all of its length of equal bigness.

RULE. Multiply half its circumference by half its diameter, (the product is the area of one end) and this product by the length; the last product will be the solidity required.

Example.

What is the solidity of a cylinder that has a diameter 22 inches and is 20 feet long?

22÷—11 and 69.1434.57 × 11 ×20=7605·4÷144 cqual to 52·81† ft. Ans.