Ex. How many square feet are contained in a board which is 12 feet 6 inches long, 15 inches wide at the greater end, and 11 inches wide at the less end?

Here (15+11)÷2-13 inches, the mean width; and 12ft. 6in. 150 inches, the length. Then, 150×13-1950 sq. in. 13 sq. feet, Ans.

PROBLEM VI.-To find the area of any Trapezium.

RULE. Divide the trapezium into two triangles by a diagonal; then find the areas of these triangles, and add them together for the whole area.

Or thus: Let fall two perpendiculars on the diagonal from the other two opposite angles; then multiply the sum of these two perpendiculars by the diagonal, and half of the product will be the area of the trapezium.

Ex. 1. To find the area of a trapezium whose diagonal is 42 chains, and the two perpendiculars on it, 16 and 18 chains.

Here 16+18=34, the sum of the two perpendiculars. Then, 42×34-1428, and 1428÷2=714 sq. chains, Ans.

2. A certain tract of land, in form of a trapezium, consists of two triangles-the sides of one triangle are 30, 40, and 50 rods; and the sides of the other, 29, 37, and 50 rods. Required the area of the tract.

Ans. 7 acres, 11.57+ sq. rods. PROBLEM VII. To find the area of a Regular Polygon. RULE I-Multiply together the length of one side, the number of sides, and the length of a perpendicular drawn from the centre of the figure to the middle of one of its sides, and half of the last product will be the area sought.

Ex 1. To find the area of a regular pentagon, each of the 5 sides being 25 feet, and the perpendicular from the centre on each side 17.2048 feet.

Here 17.2048×5×25=2150.6000

And 2150.6+2=1075.3sq. feet, Ans.

RULE II.*--Square one of the sides of the polygon; then multiply that square by the multiplier set against the name

*This rule depends on the principle, that similar figures are to each other as the squares of their like sides. The multipliers in the table are the areas of the respective figures to the side 1. Whence the rule is manifest,

of the figure in the following table, and the product will

4

Tetragon* 1.

9 Nonagon

6.18182

5

7.69421

6

9.36564

Pentagon 1.72048 10 Decagon
Hexagon 2.59808 11 Undecagon

7 Heptagon 3.63391 12 Duodecagon 11.19615

Taking here the same example as before, viz. a pentagon, whose side is 25 feet: Then, 25×25-625; and the multiplier for a pentagon being 1.72048, we have 1.72048 X625-1075.3 sq. feet, the area, as before.

2. What is the area of a trigon, or an equilateral triangle, whose side is 20 rods? Ans. 173.204 sq. rods.

3. What is the area of a hexagon whose side is 5 yards? Ans. 64.952 sq. yards.

PROBLEM VIII. To find the area of an Irregular Pol

ygon.

RULE.--Divide the polygon (or suppose it to be divided) into triangles, by diagonal lines. Then find the areas of all the triangles, separately, and the sum of these areas will be the area of the whole polygon.

Ex. A certain irregular polygon contains three triangles-the base of the first triangle measures 55 rods, and the perpendicular 18 rods; the base of the second triangle is 55 rods, and the perpendicular 13 rods; the base of the third is 44 rods, and the perpendicular is 22 rods: What is the area of the polygon?

Here 55x18-990, and 55x13-715, and 44x22-968. Then, 990+715+968=2673 sq. rods, double the area of the polygon. Then, 2673-2-1336 sq. rods, 8 acres, 1 rood, 16 sq. rods, the Ans.

These two figures are not polygous. A trigon is an equilateral triangle, and a tetragon is a square.

PROBLEM IX.-The Diameter of a Circle being given, to find the Circumference.

RULE. AS 7 is to 22, so is the diameter to the circumference, nearly: Or, more exactly, as 113 is to 355, so is the diameter to the circumference.* Or, multiply the diameter by 3.1416,† and the product will be the circumference, very nearly.

Note. By the converse of this rule the diameter of a circle may be found, when the circumference is known. Ex. 1. What is the circumference of a circle whose diameter is 226 feet?

As 7: 22 :: 226: 710 ft. Ans.

Or, as 113: 355 :: 226 : 710ft., the Ans.

Or, 226×3.1416-710.0016ft., the Ans.

2. If the circumference of the earth be 24912 miles, what is its diameter ?

As 355: 113 :: 24912 : 79293 miles, Ans.

PROBLEM X.-To find the Area of a Circle.

RULE I.-Multiply the circumference by the diameter, and of the product will be the area : Or, multiply one of these dimensions by of the other, and the product will be the area.

RULE II.--Multiply the square of the diameter by .7854,‡` for the area.

RULE III.-Multiply the square of the circumference by .07958.

Ex. What is the area of a circle whose diameter is 22.6, and circumference 71 ?

The answer, by the first rule, is 401.15; by the second rule, 401.150904; and by the third, 401.16278

Note. A circular ring, is the figure contained between the peripheries of two concentric circles; and hence the area of a circular ring must be the difference of the areas

*The first of these proportions will give the circumference true to 3 or 4 places of figures, and the second to 7 places.

This multiplier, correct to fifteen places of decimals, is 3.141592653689793; but 3.1416 is sufficiently exact for common use.

This multiplier is 1-4 of 3.1416, being the area of a circle whose diameter is 1.

of the two circles; or equal to the difference of the squares of the diameters of the two circles multiplied by .7854

PROBLEM XI.-To find the area of an Ellipse.

RULE.-Multiply the tranverse diameter by the conjugate diameter, and the product by .7854, and the last product will be the area.

Ex. What is the area of an ellipse whose tranverse diameter is 88, and conjugate diameter 72?

Here 88x72x.7854-4976.2944, Ans. PROBLEM XII. To find the area of any long irregular figure.

RULE. Take or measure the breadth of the figure in several places, at equal distances: Then divide the sum of these breadths by the number of them, and the quotient will be the mean breadth, nearly; by which multiply the length of the figure, and the product will be the area, near the truth.

Ex. The breadths of an irregular field, at five equidistant places, are 10, 8.1, 9.4, 10.2, and 12.3 rods, and the length is 32 rods: Required the area?

Here

(10+8.1+9.4+10.2+12.3)+5=10 rods.

Then 32×10-320 sq. rods=2 acres, Ans.

MENSURATION OF SOLIDS.

DEFINITIONS.

1. A prism, is a solid of equal size from end to end; the ends being parallel, equal, and like plane figures, and the sides connecting those ends are parallelograms. A prism takes particular names, according to the figure of its base or ends, whether triangular, square, rectangular, or circular, &c.

2. A parallelopiped, or parallelopipedon, is a prism bounded by six parallelograms, every opposite two of which are equal, alike, and parallel; as a stick of hewn timber of equal bigness from end to end.

3. A cube, is a square prism, bounded by six equal square sides, every two adjoining sides being perpendicular to each other; as a square block of wood, whose length, breadth, and thickness, are equal.

4. A cylinder, is a round prism, having circles for its ends, or bases; as a round log of equal bigness from end to end.

5. Solids which decrease or taper gradually from the base till they come to a point, are generally called pyramids; and they are of different kinds, according to the figure of their bases: Thus; if it has a square base, it is called a square pyramid; if a triangular base, a triangular pyramid; if a circular base, a circular pyramid, or a cone. The point where the top of a pyramid ends, is called a vertix; a straight line drawn from the vertix to the centre of the base, is called the perpendicular height of the pyramid; and a line drawn from the vertix to the middle of one of the sides of the base, is called the slant height.

6. A frustum of a pyramid, is what remains after the top is cut off by a plane parallel to the base; and is in the form of a stick of timber greater at one end than the other, whether round, or hewn square, &c. If it be round, it is usually called the frustum of a cone.

7. A sphere or globe, is a round body, bounded by one curve surface, which is every where equally distant from a certain point within, called the centre.

8. A segment of a globe, is any part cut off by a plane. 9. A spheroid, is a solid resembling an egg in shape, only both its ends are alike.

10. The measure or quantity of a solid, is called its solidity, capacity, or content.

11. Solids are measured by cubes, whose sides are an inch, a foot, or a yard, &c.; and hence the solidity of a body is said to be so many cubic inches, feet, or yards, &c. as will fill its capacity, or another of an equal magnitude.

PROBLEM I. To find the solid content of any Prism.

RULE. Find the area of one of its ends or bases, by the rule for the figure, whether it be a triangle, a square, a polygon, or a circle, &c.; then multiply this area by the length or height of the prism, and the product will be the solid content.