RULE.

Join the vertices of two opposite angles by a diagonal; from each of the other vertices let fall perpendiculars upon this diagonal; multiply the diagonal by half of the sum of the perpendiculars, and the product will be the area required.

2. How many square yards of paving are there in the quadrilateral, whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 33 feet? Ans. 222.

To find the area of any polygon.

100. From what precedes, we have the following

RULE.

Draw diagonals dividing the proposed polygon into tra pezoids and triangles: then find the areas of these figures separately, and add them together for the area of the whole polygon.

EXAMPLE.

1. Let it be required to determine the area of the polygon ABCDE, having five sides.

E

Let us suppose that we have meaured the diagonals and perpendiculars, and found AC 36.21, EC 7.26, Aa = 4.18: required the area.

Dd=

A

39.11,

Bb = 4

Ans. 296.1292.

To find the area of a regular polygon.

101. Let AB, denoted by s, represent one side of a regular polygon, whose centre is C. Draw CA and CB, and from C draw CD perpendicular to AB. Then will CD be the

apothem, and we shall have AD = BD.

A

D

Denote the number of sides of the polygon by n; then

360°

will the angle ACB, at the centre, be equal to n (B. V., Page 138, D. 2), and the angle ACD, which is half

180°

of ACB, will be equal to n

In the right-angled triangle ADC, we shall have, For mula (3), Art. 37, Trig.,

a formula by means of which the apothem may be computed. But the area is equal to the perimeter multiplied by half the apothem (Book V., Prop. VIII.): hence the following

RULE

Find the apothem, by the preceding formula; multiply the perimeter by half the apothem; 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,

CD 10 x cot 30°; or, log CD = log 10+ log cot 30°-10

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,

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 = 2.598)762, 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 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 121°, 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.