PROBLEM XI. To find the Area of a Circle, or an Ellipsis. RULE. Multiply the square of the circle's diameter; or the product of the longest and shortest diameters of the Ellipsis, by .7854, for the area. Or, subtract 0.10491 from the double logarithm of the circle's diameter, or from the sum of the logarithms of those elliptic diameters, and the remainder will be the logarithm of the area. Note. In any Circle, the produces the circumference. Diameter multiplied Circumference divided by 3.14159 EXAMXPE I. How many acres are there in a circle of one mile di an elliptical fish pond 10 perches long, and 5 wide. Required the length of the wall, content of the inclosure, and area of the pond? Diameter 100 Multiplied by 3.14159 ་ The circumfer. 314.15900 perches. Or take 314.16, as in the former example, the content of the circumference 49A. OR. 14P. PROBLEM XII. The Area of a Circle given, to find the Diameter. RULE. To the logarithm of the area, add 0.10491, and half the sum will be the logarithm of the diameter, Or, divide the area by .7854, and the square root of the quotient will be the diameter. EXAMPLE. Required the radius of a Circle, to inclose one acre of ground? A. P. PROBLEM XIII. To lay out an Oblong Piece of Ground, so that the Length shall bear a given Proportion to the Breadth. As the less number, RULE. Is to the given area in square perches ; To the square of the longest side; (the square root of which will be the longest side in perches.) And, ‘ As the greater number, Is to the given area, in square perches So is the less number To the square of the shortest side; (the square root of which, will be that side in perches.) EXAMPLE. Let it be required to lay out an oblong piece of ground to contain 864 acres, and the breadth to bear the same proportion to the length, as 3 does to 5. |