Table relative to the Construction and Estimation of Polygons. POLYGONS. sides, and half the product is the area. cular height to the centre, and by the number of gons, multiply the length of a side by the perpendicourse be similarly measured: hence in regular polyPolygons being composed of triangles, may of Application of the Table. 1. The radius of a circle being 6 feet, required the side of the greatest heptagon that may be inscribed therein. 867 x 6.5=5.6355, or 5 feet 7 inches nearly. 2. Each side of a pentagon is required to be 9 feet; required the radius of circumscribing circle. 852 x 9=7.668, or 7 feet 8 inches. 3. A perpendicular from the centre to either side of an octagon is required to be 12 feet; what must be the radius of circumscribing circle? 1·08 × 12=12.96, or 12 feet 11 inches. 4. Each side of a hexagon is 4 yards; required its superficial contents. 42×2-598-52.6095 square yards. THE CIRCLE AND ITS SECTIONS. Observations and Definitions. 1. The circle contains a greater area than any other plane figure bounded by the same perimeter or outline. 2. The areas of circles are to each other as the squares of their diameters; any circle twice the diameter of another contains four times the area of the other. 3. The radius of a circle is a straight line drawn from the centre to the circumference, as o B. 4. The diameter of a circle is a straight line drawn through the centre, and terminated both ways at the circumference, as co A. C B G D F E A 5. A chord is a straight line joining any two points of the circumference, as D F. 6. The versed sine is a straight line joining the chord and circumference, as E G. 7. An arc is any part of the circumference, as C D E. 8. A semicircle is half the circumference cut off by a diameter, as C E A. 9. A segment is a chord, as De F. any portion of a circle cut off by 10. A sector is a part of a circle cut off by two radii, as A O B. General rules in relation to the circle. 1. Multiply the diameter by 3.1416, the product is the circumference. 2. Multiply the circumference by 31831, the product is the diameter. 3. Multiply the square of the diameter by 7854, the product is the area. 4. Multiply the square root of the area by 1.12837, the product is the diameter. 5. Multiply the diameter by 8862, the product is the side of a square of equal area. 6. Multiply the side of a square by 1.128, the produet is the diameter of a circle of equal area. Application of the rules as to purposes of practice. 1. The diameter of a circle being 7 quired its circumference. 16 inches, re 7.1875 x 3.1416-22-58025 inches, the circumference. Or, the diameter being 30 feet, required the circumference. 3.1416 x 30.5=95.8188 feet, the circumference. 2. A straight line, or the circumference of a circle, being 274.89 inches, required the circle's diameter corresponding thereto. •274.89 × 31831-87.5 inches diameter. Or, what is the diameter of a circle when the circumference is 39 feet? •31831 × 39=12.41409 feet, and 41409 × 12=4.96908 inches, or 12 feet 5 inches, very nearly the diameter. 3. The diameter of a circle is 3 inches; what is its area in square inches? 3.752 14.0625 × 7854=11·044, &c. inches area. Or, suppose the diameter of a circle 25 feet 6 inches, required the area. 25.52 650.25 x 7854-510.706, &c. feet, the area. 4. What must the diameter of a circle be, to contain an area equal to 706.86 square inches? 706.86=26.586 × 1·12837=29.998 or 30 inches, the diameter required. 5. The diameter of a circle is 144 inches; what must I make each side of a square, to be equal in area to the given circle? 14.25 x 8862=12.62835 inches, length of side required. Any chord and versed sine of a circle being given to find the diameter. D Rule.-Divide the sum of the squares of the chord and versed sine by the versed sine; the quotient is the diameter of corresponding circle. 1. The chord of a circle A B equal 6 feet, and the versed sine CD equal 2 feet; required the circle's diameter. Α 6.5%+22=46.25÷2=23·125 feet, the diameter. B 2. In a curve of a railway, I stretched a line 72 feet in length, and the distance from the line to the curve I found to be 14 ft.; required the radius of the curve. 722+1.252=5185.5625, and 5185.5625 =2074 225 ft.the radius. To find the length of any given arc of a circle. Rule. From eight times the chord of half the arc, subtract the chord of the whole arc, and onethird of the remainder is equal the length of the arc. Required the length of the arc A B C, the chord A B of half the arc B and chord A c of the whole arc 8 feet 4 inches. 4.25 x 8=34, and 34-8.333= length of the arc. 25.667 =8.555 feet, the To find the area of the sector of a circle. Rule.-Multiply the length of the arc by its radius, and half the product is the area. The length of the arc A C B equal 9 feet, and the radii f a, F B, equal each 7 feet; required the area. F A e B C 9.5x7=65.5÷÷2=32.75, the area. Note. The most simple means whereby to find the area of the segment of a circle, is, to first find the area of a sector whose arc is equal to that of the given segment; and if it be less than a semicircle, subtract the area of the triangle formed by the chord of the segment and radii of its extremities; but if more than a semicircle, add the area of the triangle to the area of the sector, and the remainder or sum is the area of the segment. |
