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2. Find the area of a pentagon, whose side is 25. Ans. 1075.298375.

3. Find the area of a decagon, whose side is 20.

Ans. 3077.68352.

To find the circumference of a circle, when the diameter is

given.

102. From the principle demonstrated in Book V., Prop. XVI., we may write the following

RULE.

Multiply the given diameter by 3.1416; the product will be the circumference required.

EXAMPLES.

1. What is the circumference of a circle, whose diameter is 25 ? Ans. 78.54.

2. If the diameter of the earth is 7921 miles, what is the circumference? Ans. 24884.6136.

To find the diameter of a circle, when the circumference is

given.

103. From the preceding case, we may write the following

RULE.

Divide the given circumference by 3.1416; the quotient will be the diameter required.

EXAMPLES.

1. What is the diameter of a circle, whose circumference is 11652.1944 ?

Ans. 3709.

2. What is the diameter of a circle, whose circumference

is 6850 ?

Ans.

2180.41.

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 arc 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

is 3 feet?

diameter

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 are 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.

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 EA CB 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

denote the radii of the two circles,

R being greater than r. The area of the outer circle is R2 x 3.1416, and that of the inner circle is 2 × 3.1416; hence, the area of the ring is equal to (R2 — y2) × 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. I., 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.

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