EXAMPLES.

1. What is the area of lune, the chord A B of which is 24 ft. and the heights of its two arcs 5 and 3 ft.? Ans. 25 sq. ft. 2. The chord of a lune is 40 feet, and the heights of its arcs 4 and 20 feet; required the area. Ans. 57 867 square yards.

PROBLEM XII.

To find the area of an ellipse.

RULE.-Multiply the product of the semiaxes TP, CP by 3.1416 for the area.

FORMULA.

A = abπ, in which a and b are the semiaxes.

EXAMPLES.

1. The axes of an elliptical shrubbery in a park are 300 and 200 feet; required the area.

Ans. 5236 square yards,

1 acre 396 square yards. 2. Required the area of an ellipse, the axes of which are 70 and 50 yards. Ans. 2748 square yards 8 feet.

PROBLEM XIII.

To find the area of an elliptical segment, the chord of which is parallel to one of the axes. (See last figure.)

RULE.-Divide the height of the segment by that axis of the ellipse of which it is a part; and find in the table of circular segments at the end of the book, a circular segment having the same versed sine as this quotient. Then multiply continually together, this segment, and the two axes, for the area required.

EXAMPLES.

1. What is the area of an elliptic segment m Rn, whose height Rr is 20; the tranverse T R being 70, and the conjugate CO 50 feet?

70) 20 ( 285 the tabular versed sine.

The corresponding segment

is

185166
70

12.961620

50

648 081000 square feet, the area required.

2. What is the area of an elliptic segment, cut off parallel to the shorter axis, the height being 10, and the axes 25 and 35 feet? Ans. 162.021 square feet.

3. What is the area of the elliptic segment, cut off parallel to the longer axis, the height being 5, and the axes 25 and 35 feet? Ans. 97.8458 square feet.

PROBLEM XIV.

To find the area of a parabola.

RULE.-Multiply the axis or height VE

by the base or double ordinate D F, and of the product will be the area.

A

FORMULA.

a d, in which a is the axis, and A d the double ordinate.

EXAMPLES.

1. Required the area of the parabola A V C, the axis V B being 2, and the D double ordinate A C 12 feet.

16 square feet, the area required.

2. The double ordinate of a parabola is 20 feet, and its axis or height 18; required the area of the parabola.

PROBLEM XV.

Ans. 240 square feet.

To find the area of a parabolic frustrum A C F D.

Cube each end of the frustrum, and subtract the one cube from the other; then multiply that difference by double the altitude, and divide the product by triple the difference of their squares, for the area.

A=3a.

parallel chords.

FORMULA.

in which a is the altitude, and C and c the

EXAMPLES.

1. Required the area of the parabolic frustrum AC FD, AC being 6, DF 10, and the altitude B E 4 feet.

2. What is the area of the parabolic frustrum, the two ends of which are 6 and 10, and its altitude 3 feet. Ans. 24 square feet.

NOTE. Those who wish for further information on the areas of the conic sections, are referred to the works of Emerson, Hamilton, &c., it being foreign to the object of this work to give more on this subject.

PROBLEM XVI.

To find the areas of irregular figures whether bounded by straight lines or curves.

CASE I.-When the figure is long and narrow.

RULE.-Take the perpendicular breadth at several places, at equal distances; to half the sum of the first and last two breadths, add the sum of all the intermediate breadths, and multiply the result by the common distance between the breadths for the

area.

CASE II.-When the breadths or perpendiculars are taken at unequal distances, the figure being long and narrow.

RULE I. Find the areas of all the trapezoids and triangles separately, and add them together for the area.

RULE II.—Add all the breadths together, and divide the sum by the whole number of them for the mean breadth, which multiply by the length for the area. This method is not very correct, but may do where great accuracy is not required.

EXAMPLES.

=

1. The perpendicular breadths, or offsets of an irregular figure at five equidistant places are A D = 8·2, m p = 7·4, n q 9.2, or= 10.2, BC= 8.6 feet; and the common distances A m — mn = &c. 50 feet; required the area.

=

2. The length of an irregular plank is 25 feet, and its perpendicular breadth at six equidistant places are 174, 20·6, 14·2, 16.5, 20·1, and 24.4 inches; required the area.

3. Take the dimensions and find the area of the annexed irregular figure, by Rule I. and II., Case II.

Ans. 30 square feet.

CASE III. When the breadth of the figure is large and its boundary curved or crooked.

RULE.-Divide the figure into trapeziums and triangles, in the most convenient manner, taking offsets to the curved or crooked portion of the boundary. Find the areas of the trapeziums, triangles, and the offset pieces separately, which, being added together, will give the required area of the whole figure.

ziums are found by letting fall perpendiculars on the diagonals AF, BE by Prob. IV., and the area of the triangle by Prob. II., the areas of the several offset pieces being found by one or other of the cases of this Problem.

PROMISCUOUS EXERCISES.

1. The sides of three squares are 6, 8, and 24 feet; required the side of a square that shall have an area equal to all the three. Ans. 26 feet.

2. In cutting a circle, the largest possible, out of a card-board 5 feet square, how much will be wasted.

Ans. 5.365 square feet.

3. The area of a square is 72 square feet; required the length of its diagonal. Ans. 12 feet.

4. A ditch 13 yards wide surrounds a circular fortress, the circumference of the fortress being 704 yards; required the area of the ditch. Ans. 2 acres nearly.

5. What is the area of a circular table the diameter of which is 59 inches. Ans. 19 square feet nearly.

6. What is the area of an isosceles triangle, the base of which is 5 feet 10 inches, and each side 8 feet?

Ans. 23 square feet 41 inches. 7. Required the side of a decagon the area of which is 9 square feet. Ans. 1 foot 1 inch nearly. 8. The side of a square is 50 yards, and its corners are cut off so as to form an octagon; required the area of the octagon. Ans. 2071 square yards.

PART IV.

MENSURATION OF SOLIDS.

DEFINITIONS.

1. A SOLID has three dimensions, length, breadth, and thickness.

2. A prism is a solid, or body, whose ends are any plane figures, which are parallel, equal, and similar; and its sides are parallelograms.

A prism is called a triangular one when its ends are triangles; a square prism, when its ends are squares; a pentagonal prism, when its ends are pentagons; and

so on.