The perimeter is equal to 120: hence, denoting the area by Q,

2. What is the area of an octagon, one of whose sides is 20? Ans. 1931.36886.

The areas of some of the most important of the regular polygons have been computed by the preceding method, on the supposition that each side is equal to 1, and the results are given in the following

The areas of similar polygons are to each other as the squares of their homologous sides (Book IV., Prop. XXVII.).

Denoting the area of a regular polygon whose side is 8, by Q, and that of a similar polygon whose side is 1, by T, the tabular area, we have,

Q : T:: s2 : 12;

hence, the following

RULE.

Q = Ts2;

Multiply the corresponding tabular area by the square of the given side; the product will be the area required.

EXAMPLES.

1. What is the area of a regular hexagon, each of whose sides is 20 ?

We have, T = 2.5980762,

and s2 = 400: hence,

Q = 2.5980762 × 400 = 1039.23048 Ans.

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.

3700.

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

is 6850 ?

Ans. 2180.41

To find the length of an

arc containing any number of

degrees.

arc

of 1°,

104. The length of an 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? 4.712364 ft.

Ans.

2. What is the length of an arc of 12° 10', or 121°, the diameter being 20 feet?

Ans.

To find the area of a circle.

2.123472 ft.

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

I.

RULE.

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.

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.

AEB is

equal to

Solving the triangle AEB, we find the angle equal to 73° 44', the area of the sector EA CB 31.35, and the area of the triangle AEB equal to 48; ence, 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 r. R2 3.1416, and that

denote the radii of the two circles,

The area of the outer circle is of the inner circle is 2 × 3.1416; hence, the area of the ring is equal to (R2 — p2) × 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