Note 1. If the radius of a circle be given, the side of any inscribed polygon may be found by dividing the given radius by the number standing opposite to the name of the polygon, in the fourth column of the preceding table.

2. By the assistance of this Problem, any regular polygon, whose side is given, may be easily constructed in the following manner: Find the radius of the circumscribing circle; and to the circumference of this circle apply the given side the proposed number of times, and you will have the polygon required. Or if the radius of a circle be given, find the side of the inscribed polygon, with which proceed as before.

3. The perpendicular let fall from the centre of a regular polygon upon one of its sides is equal to the radius of the inscribed circle; and the sum of the sides is equal to the product of one side multiplied by the number of sides.

EXAMPLES.

1. The side of a regular pentagon is 21.75; what are the radii of the inscribed and circumscribing circles?

Here .688191 x 21. 75 = 14.96815425 DC, the radius of the inscribed circle; and .8506508 × 21.75 18.5016549 AC, the radius of the circumscribing

circle.

2. If the side of a regular heptagon be 25.25; what is the radius of the circumscribing circle?

Ans. 29.097658. 3. If the side of an octagonal grass-plot, in a gentleman's pleasure-ground, measure 86 feet 10 inches; what will be the expense of making a gravel-walk from the middle of one of its sides to the middle of the opposite side, at 24d. per yard, lineal measure?

Ans. 14s. 6d. 4. If the radius of a circle be 65 feet; what is the sum of the sides of its inscribed nonagon?

PROBLEM XI.

Ans. 400.16356 feet.

To find the area of a regular polygon.

RULE.

Multiply the sum of the sides, or perimeter of the polygon, by the perpendicular demitted from its centre to one of the sides, and half the product will be the area.

Note. If double the area of a regular polygon be divided by the perpendicular, the quotient will be the sum of the sides.

EXAMPLES.

1. Required the area of the regular hexagon ABCD EF, whose side AB is 20 feet 6 inches, and perpendicular Po is 17 feet 9 inches.

E

D

F

AP B

2183.25

Here 20.5 × 6 × 17.75 2183.25; and

2

=1091.625 feet = 1091 ft. 7 in. 6 pa. the area required.

2. What is the area of a court-yard in the form of a regular pentagon, whose side measures 92 feet 6 inches, and perpendicular 63 feet 8 inches?

Ans. 14722.91666 feet. 3. Required the area of a heptagonal stone, whose side measures 8 feet 9 inches, and perpendicular 9 feet. Ans. 275 ft. 7 in. 6 pa.

4. What will the floor of an octagonal summer-house cost paving with black and white marble, at 4s. 6d. per square foot, the side of which measures 9 feet 6 inches, and the nearest distance from one of its sides, to the opposite side 22 feet 11 inches?

Ans. £97 19s. 4 d.

5. A hexagonal piece of ground, in a gentleman's park, cost £29 10s. 5 d. planting with trees, at £5 10s. per acre; and a gravel walk leading from the middle of one of its sides to the middle of the opposite side, cost £2 3s. 3 d. making, at 3d. per yard, lineal measure; what was the expense of fencing the perimeter of the polygon, at 6s. 6d. per rood?

Ans. £27 17s.

PROBLEM XII.

To find the area of a regular polygon, when the side only is given.

RULE.

Multiply the square of the given side by the number or area standing opposite to the name of the polygon, in the following Table, and the product will be the area. (This Rule is founded on Theorem XIII.)

SIDES.

No. of

II. TABLE OF POLYGONS, &c.

Multipliers, Angle | Angle

3 Trigon

4

5

Tetragon Pentagon 6 Hexagon

8

7 Heptagon
Octagon.

9 Nonagon

10

11

12

Decagon

Undecagon.

Duodecagon.

1.0000000 90° 450

1.7204774 72° 540

2.5980762 60o 60° 3.6339124

4.8284271 45°

5130 642°

671°

6.1818242 40°

700

Note 1. If the area of a polygon be divided by the number standing opposite to its name, in the foregoing table, the quotient will be the square of the polygon's side.

2. The multipliers in the table of polygons, Problem X, are the radii of the inscribed and circumscribing circles, when the side of the polygon is unity, or 1; and may be found by trigonometry, in the following manner: Divide 360 degrees by the number of sides, and the quotient will be the angle ACB at the centre of the polygon, half of which will be the angle ACD; then say, as the nat. sine of the angle ACD is to AD, so is the nat. co-sine of the angle ACD to CD, the radius of the inscribed circle, or the perpendicular of the polygon; and, as the nat. sine of ACD is to AD, so is radius (1) to AC, the radius of the circumscribing circle.

The multipliers in the last table, are the areas of their respective polygons, when the side is 1, and may be found thus: Multiply the perpendicular or number in column the third, table 1, by .5 (half of AB) and the product will be the area of the triangle ACB, which being multiplied by the number of sides, we obtain the area or multiplier in table 2.

Or if the nat, tangent of the angle DAC be inultiplied by the number of sides, one fourth of the product will be the multiplier. (See the figure, Problem X.)

EXAMPLES.

1. If the side of a pentagon be 8 feet 4 inches; what is its area?

2. What is the area of the base of a hexagonal stone pillar, whose side measures 1 foot 6 inches?

Ans. 5.84567 feet.

3. If the side of an octagonal brick pillar measure

1 foot 5 inches; what is the area of its base?

Ans. 9 ft. 8 in. 3 pa.

4. Required the area of a decagon, whose side mea

sures 25 feet 9 inches.

5. A gardener wishes to plot that shall contain 260 be the length of its side?

Ans. 5101 ft. 8 in. 10 pa. make a hexagonal grasssquare yards; what must Ans. 10 yards.

PROBLEM XIII.

The diameter of a circle being given, to find the circumference; or, the circumference being given, to find the diameter.

RULE I.

As 7 is to 22, so is the diameter to the circumference: or, as 22 is to 7, so is the circumference to the diameter.

RULE II.

As 113 is to 355, so is the diameter to the circumference: or, as 355 is to 113, so is the circumference to the diameter.

RULE III.

Multiply the diameter by 3.1416, and the product will be the circumference: or, divide the circumference by 3.1416, and the quotient will be the diameter.

Note 1. There is no figure that affords a greater variety of useful properties than the circle; nor is there any that contains so large an area within the same perimeter.

The ratio of the diameter of a circle to its circumference has never yet been exactly determined; although this celebrated Problem, called the squaring of the circle, has engaged the attention and exercised the abilities of the ablest mathematicians, both ancient and modern. But though the relation between the diameter and circumference cannot be exactly defined in known numbers; yet approximating ratios have been determined sufficiently correct for practical purposes.

Archimedes, a native of Syracuse, who flourished about 200 years before the Christian æra, after attempting in vain to determine the true ratio of the diameter to the circumference, found it to be nearly as 7 to 22..

The proportion given by Vieta, a Frenchman, and Metius, a Dutchman, about the end of the 16th century, is as 113 to 355, which is rather more accurate than the former; and is a very commodious ratio, for being reduced into decimals, it agrees with the truth to the sixth figure inclusively.