To find the length of an

arc cortaining any number of degrees.

104. The length of an arc of 1°, in a circle whose diameter is 1, is equal to the circumference, or 3.1416 divided by 360; that is, it is equal to 0.0087266: hence, the length of an arc of n degrees, will be, n × 0.0087266. To find the length of an are containing n degrees, when the diameter is d, we employ the principle demonstrated in‹ Book V., Prop. XIII., C. 2: hence, we may write the following

RULE.

Multiply the number of degrees in the arc by .0087266, and the product by the diameter of the circle; the result will be the length required.

EXAMPLES.

1. What is the length of an arc of 30 degrees, the diameter being 18 feet? Ans. 4.712364 ft. 2. What is the length of an arc of 12° 10', or 121o, the diameter being 20 feet? Ans. 2.123472 ft.

To find the area of a circle.

105. From the principle demonstrated in Book V., Prop. XV., we may write the following

RULE.

Multiply the square of the radius by 3.1416; the product will be the area required.

EXAMPLES.

1. Find the area of a circle, whose diameter is 10, and circumference 31.416. Ans. 78.54.

2. How many square yards in a circle whose diameter is 3 feet? Ans. 1.069016.

3. What is the area of a circle whose circumference is

12 feet?

Ans. 11.4595.

To find the area of a circular sector.

106. From the principle demonstrated in Book V., Prop. XIV., C. 1 and 2, we may write the following

RULE.

I. Multiply half the arc by the radius; or,

II. Find the area of the whole circle, by the last rule; then write the proportion, as 360 is to the number of degrees in the sector, so is the area of the circle to the area of the sector.

EXAMPLES.

1. Find the area of a circular sector, whose arc contains 18°, the diameter of the circle being 3 feet. 0.35343 sq. ft.

2. Find the area of a sector, whose arc is 20 feet, the radius being 10.

Ans. 100.

3. Required the area of a sector, whose arc is 147° 29', and radius 25 feet. Ans. 804.3986 sq. ft.

To find the area of a circular segment.

107. Let AB represent the chord corresponding to the two segments ACB and AFB. Draw AE and BE The segment ACB is equal to the sector EA CB, minus the triangle AEB. The segment AFB is equal to the sector EAFB, plus the triangle AEB. Hence, we have the following

RULE.

D

Find the area of the corresponding sector, and also of the triangle formed by the chord of the segment and the two extreme radii of the sector; subtract the latter from the former when the segment is less than a semicircle, and take their sum when the segment is greater than a semicircle ; the result will be the area required.

EXAMPLES.

1. Find the area of a segment, whose chord is 12 and the radius 10.

Solving the triangle AEB, we find the angle AEB is equal to 73° 44', the area of the sector EACB equal to 64.35, and the area of the triangle AEB equal to 48; hence, the segment ACB is equal to 16.35 Ans.

2. Find the area of a segment, whose height is 18, the diameter of the circle being 50. Ans. 636.4834.

3. Required the area of a segment, whose chord is 16, the diameter being 20. Ans. 44.764.

To find the area of a circular ring contained between the circumferences of two concentric circles.

108. Let R and R being greater than R2 x 3.1416, and that

r.

denote the radii of the two circles,

The area of the outer circle is of the inner circle is p2 × 3.1416; hence, the area of the ring is equal to (R2 — r2) × 3.1416. Hence, the following

RULE.

Find the difference of the squares of the radii of the two circles, and multiply it by 3.1416; the product will be the area required.

EXAMPLES.

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.

MENSURATION

OF BROKEN AND CURVED

SURFACES.

To find the area of the entire surface of a right prism.

109. From the principle demonstrated in Book VII., Prop, L, we may write the following

RULE.

Multiply the perimeter of the base by the altitude, the product will be the area of the convex surface; to this add the areas of the two bases; the result will be the area required.

EXAMPLES.

1. Find the surface of a cube, the length of each side being 20 feet. Ans. 2400 sq. ft.

2. 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 sq. ft.

To find the area of the entire surface of a right pyramid.

110. From the principle demonstrated in Book VII., Prop. IV., we may write the following

RULE.

Multiply the perimeter of the base by half the slant height; the product will be the area of the convex surface; to this add the area of the base; the result will be the area required.

EXAMPLES.

1. Find the convex surface of a right triangular pyramid, 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 right pyramid, whose slant height is 15 feet, and the base a pentagon, of which each side is 25 feet? Ans. 2012.798 sq. ft.

To find the area of the convex surface of a frustum of a right pyramid.

111. From the principle demonstrated in Book XII., Prop. IV., C., we may write the following

RULE.

Multiply the half sum of the perimeters of the two bases by the slant height; the product will be the area required.

EXAMPLES.

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.

2. What is the convex surface of the heptagonal pyramid, whose slant height is 55

ft. frustum of a

feet, each side of the lower base 8 feet, and each side of the upper base 4 feet? Ans. 2310 sq. ft.

112. Since a cylinder may be regarded as a prism whose base has an infinite number of sides, and a cone as a pyra mid whose base has an infinite number of sides, the rules just given, may be applied to find the areas of the surfaces of right cylinders, cones, and frustums of cones, by simply changing the term perimeter, to circumference.

EXAMPLES.

1. What is the convex surface of a cylinder, the diameter of whose base is 20, and whose altitude 50? Ans. 3141.6.

2. What is the entire surface of a cylinder, the altitude

being 20, and diameter of the base 2 feet?

131.9472 sq.

ft.

3. Required the convex surface of a cone, whose slant height is 50 feet, and the diameter of its base 8 feet. Ans. 667.59 sq. ft.