EXAMPLE 2. What is the area of a regular hexagon, each side being 8.75 in length? 8.752 × 2.598 199.009375 the area nearly. A Table of Multipliers for Polygons from Three to Twelve Sides. 1.-The breadth of a Polygon given, to find the Radius of a Circle to contain that polygon. RULE.-Multiply half the breadth of the polygon by the numbers in the first column opposite to its name, or number of sides, and the product will be the radius of a circle to contain that polygon nearly. And, if the polygon have an unequal number of sides, the half breadth is accounted from its centre to one of its sides. 2.--The radius of a circle given to find the length of side. RULE.-Multiply the radius of any circle by the numbers in the second column opposite the polygon required; and the product will be the length of side nearly that will divide that circle into the proposed number of sides. And, 3.-The length of side given to find the radius. RULE.-Multiply the given length of side by the numbers in the third column opposite the polygon required; and the product will be the radius of a circle to contain that polygon nearly. EXAMPLE 1.-Required the radius of a circle to contain an octagon, whose breadth A B = 18.5 inches. Half of 18.5 9.25 and, 9.25 × 1.08 = 9.99 or 10 inches D EXAMPLE 2.-Given the radius OD = 9.99 inches; required the length of side D C. 9.99 .7657.64235, the length of side. EXAMPLE 3.-Given the length of side DC = 7.64235; required the radius D O. 7.64235 x 1.3079.98855145, or 9.99 inches nearly. PROBLEM VI. Having the Diameter of a Circle given, to find the Circumference; or the Circumference given, to find the Diameter. RULE 1.-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. 2. As 1 is to 3.1416, so is the diameter to the circumference. Or, As 3.1416 is to 1, so is the circumference to the diameter. EXAMPLE 1.-Required the circumference of a circle, when the diameter is 23.5. 23.5 x 22 7 = 73, the circumference. EXAMPLE 2. The circumference of a circle is 739, required the diameter. EXAMPLE 3.-Required the circumference of a circle whose diameter is 30. 3.1416 x 30 94.248, the circumference. EXAMPLE 4.-What is the diameter of a circle when the circumference is 94.248? 94.248 3.1416 = 30, the diameter. PROBLEM VII. To find the Length of any Arc of a Circle. RULE. Subtract the chord of the whole arc from eight times the chord of half the arc; and of the remainder is the length of the arc nearly. EXAMPLE.-Required the length of the arc A.BC; the chord of half the arc A B 19.8, and chord of the whole arc A C 34.4. = To find the Diameter of a Circle, by having the chord and versed sine given. ROLE. Divide the square of half the chord by the versed sine, to the quotient of which add the versed sine; and the sum will be the diameter. Or, If the sum of the squares of the semichord and versed sine be divided by the versed sine, the quotient will be the diameter of the circle to which that segment corresponds. EXAMPLE. Given the chord A B = 24, and versed sine C D 8; required the diameter of the circle C E. A Table of the Relative Proportions of the Circle, its Equal and Inscribed Squares. Examples illustrative of the preceding Table. EXAMPLE 1.-The diameter of a circle is 12.5; required the side of a square equal in area to the given circle. 125 × .8862 = 11.07750, side of equal square. EXAMPLE 2.-The circumference of a circle being 53.4; required the side of a square equal in area. 53.4 x .2821 = 15.06414, side of equal square. EXAMPLE 3.-The diameter of a circle being 18; required the side of the greatest square that can be inscribed therein. 18 x .7071 = 12,7278, side of inscribed square. EXAMPLE 4.-The circumference of a circle is 86; required the side of inscribed square. 86 x 2251= 19.3586, side of inscribed square. EXAMPLE 5.-The area of a circle being 371.5; required the area of the greatest square that can be inscribed within the circle. 371.5 x .6366 = 236.49690, area of the required square. EXAMPLE 6. The side of a square being 19.375; required the diameter of its circumscribing circle. 19.375 x 1.4142 = 27.4001250, diameter. EXAMPLE 7.-Required the circumference of a circle to circumscribe a square, each side being 19.375. 19.375 x 4.443 86.083125, circumference of the circle required. EXAMPLE 8.-The side of a square being 13.5; réquired the diameter of a circle equal in area to the given square. 13.5 x 1.128 = 152.280, diameter of the circle required. EXAMPLE 9.-The side of a square being 13.5; rẻquired the circumference of a circle equal in area to the given square. 13.5 x 3.545 47.8575, circumference of the circle required. Some of the Properties of a Circle. 1. It is the most capacious of all plain figures, or contains the greatest area within the same perimeter or outline. 2. The areas of circles are to each other as the squares of their diameters, or of their radii. 3.-Any circle whose diameter is double that of another contains four times the area of the other. 4. The area of a circle is equal to the area of a triangle whose base is equal to the circumference, and perpendicular equal to the radius. 5.-The area of a circle is equal to the rectangle of its radius, and a right line equal to half its circumference. |