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 21d. 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 1. If double the area of a regular polygon be divided by the perpendicular, the quotient will be the sum of the sides.

2. In any figure whatever, the sum of all the inward angles is equal to twice as many right angles, wanting four, as the figure has sides.

3. When every side of any figure is produced out, the sum of all the outward angles is equal to four right angles.

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. 4d.

5. A hexagonal piece of ground, in a gentleman's park, cost £29. 10s. 5d. 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. 38. 32d. 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.)

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 the radius (I) 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 multiplied 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?

Here 8 feet 4 inches = 81 feet =253; and (25 ÷ 3)2=6259, the square of the side; then 1.7204774 1075.2983759 119.4775972 feet, the

× 625÷9 area required.

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 measures 25 feet 9 inches. Ans. 5101 ft. 8 in. 10 pa. 5. A gardener wishes to make a hexagonal grass-plot that shall contain 260 square yards; what must be the length of its side? 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.

The first, however, who ascertained this ratio to any great degree of exactness, was Ludolph Van Ceulen, a Dutchman. He found that if the diameter of a circle be 1, the circumference will be 3.141592653589793238462643383279502884 nearly, which is true to 36 places of decimals. This was thought so extraordinary a performance, that the numbers were cut on his tomb-stone, in St. Peter's church-yard, at Leyden.

Since the invention of fluxions, by the illustrious Sir Isaac Newton, the squaring of the circle has become more easy; and the late ingenious Mr. Abraham Sharp, of Little Horton, near Bradford, in Yorkshire, has not only confirmed Ceulen's ratio, but extended it to 72 places of decimals.

Mr. John Machin, professor of astronomy in Gresham College, London, has also given us a quadrature of the circle, which is true to 100 places of figures; and even this has been extended, by the French mathematicians, to 128.

2. The first Rule is the proportion of Archimedes; the second that of Vieta and Metius; and the third is an abridgment of Van Ceulen's ratio. This Rule is not quite so accurate as the second; but is most commonly used, as being most convenient, and, in most cases, correct enough for practice.

As 113 355 12: 37.699115, the circumference required.

By Rule III.

Here 3.1416 x 12 37.6992, the circumference required.

2. If the circumference of a circle be 45; what is the diameter ?

By Rule I.

As 22 : 7 :: 45: 14.318181, the diameter required.

By Rule II.

As 355: 113 :: 45: 14.323943, the diameter required.

By Rule III.

Here 45 3.1416 14.323911, the diameter required.

3. If the diameter of a well be 3 feet 9 inches; what is its circumference. Ans. 11 ft. 9 in. 4 pa. 4. The diameter of a circular plantation is 100 yards; what did it cost fencing round, at 6s. 9d. per rood? Ans. £15. 2s. 11 d. 5. What is the diameter of a stone column whose circumference measures 9 feet 6 inches?

6. The circumference of the what is its diameter, supposing it

;

Ans. 3 ft. 0 in. 3 pa. earth is 25000 miles a perfect sphere? Ans. 7957.72854 miles.

7. The diameter of the sun is 883220 miles; what is

his circumference?

Ans. 2774723.952 miles.

8. The circumference of the moon is 6850 miles; what is her diameter ?

Ans. 2180.41762 miles.

9. The diameter of Venus is 7680 miles; what is her circumference? Ans. 24127.488 miles.

Note.-Those who are desirous of making themselves acquainted with the method of finding the distances of the sun, moon, and planets from the earth, and also their diameters, are referred to Martin's Trigonometry, vol. i. page 208.; Ferguson's Astronomy, page 100.; and Bonnycastle's Astronomy, page 277.

PROBLEM XIV.

To find the length of any arc of a circle.

RULE I.

From 8 times the chord of half the arc subtract the chord of the whole arc, and 3 of the remainder will be the length of the arc, nearly.

Note 1. Half the chord of the whole arc, the chord of half the arc, and the versed sine are sides of a right-angle triangle; any two of which being given, the third may be found by Problem VI.

2. The difference of the diameters of any two circles, multiplied by 3.1416, will give the difference of their circumference; and vice versa.

3. If a line drawn through or from the centre of a circle, bisect a chord, it will be perpendicular to it; or, if it be perpendicular to the chord, it will bisect both the chord and the arc of the chord.

4. If more than two equal lines can be drawn from any point within a circle to the circumference, that point will be the centre.

5. Any chords in a circle which are equally distant from the centre are equal to each other.

6. A line perpendicular to the extremity of the radius, is a tangent to the circle.