Here 321.51+214.24=535.75=sum of the parallel sides AB, DC. Whence 535.75 × 171.16 (the perp. DE)=91698.9700. And 91698.9700 =45849.485 the area required. 2. 2. The parallel sides of a trapezoid are 12.41 and 8.22 chains, and the perpendicular distance 5.15 chains; required the area. ac. ro. po. Ans. 5 1 9.956. 3. Required the area of a trapezoid whose parallel sides are 25 feet 6 inches and 18 feet 9 inches, and the perpendicular distance 10 feet 5 inches. fe. in. pa. Ans. 230 5 7 4. Required the area of a trapezoid whose parallel sides are 20.5 and 12.25, and perpendicular distance 10.75. Ans. 176.03125. PROBLEM VII. To find the area of a regular polygon. RULE.* Multiply half the perimeter of the figure by the perpendicular falling from its centre upon one of the sides, and the product will be the area. Note. The perimeter of any figure is the sum of all its sides. * Demon. Every regular polygon is composed of as many equal triangles as it has sides, consequently the area of one of those triangles being multiplied by the number of sides must give the area of the whole figure; but the area of either of the triangles is equal to the rectangle of the perpendicular and half the base, and therefore the rectangle of the perpendicular and half the sum of the sides is equal to the area of the whole polygon; thus, AB 5AB OPX-area of the A AOB, and OP X-=area of the 2 2 F polygon ABCDE. Q. E. D. 1. Required the area of the regular pentagon ABCDE, whose side AB, or BC, &c. is 25 feet, and perpendicular OP 17.2 feet. 62.5=half perimeter; and 62.5× 17.2= 2 1075 square feet area required. 2. Required the area of a hexagon whose side is 14.6 feet, and perpendicular 12.64. Ans. 553.632 square feet. 3. Required the area of a heptagon whose side is 19.38, and perpendicular from the centre 20. Ans. 1356.6. 4. Required the area of an octagon whose side is 9.941, and perpendicular 12. Ans. 477.168. PROBLEM VIII. To find the area of a regular polygon, when the side only is given. RULE.* Multiply the square of the side of the polygon by the number standing opposite to its name in the following table, and the product will be the area. * Demon. The multipliers in the table are the areas of the polygons to which they belong when the side is unity, or 1. Whence as all regular polygons, of the same number of sides, are similar to each other, and as similar figures are as the squares of their like sides, (Euc. VI. 20.) 12: multiplier in the table :: square of the side of any polygon: area of that polygon; or which is the same thing, the square of the side of any polygon X by its tabular number is area of the polygon. Q. E. D. The table is formed by trigonometry, thus: As radius =1:tang. BP X tan. OBP LOBP:: BP (1): PO= radius = tang. OBP: whence OP X BP tang. LOBP = area of the ▲ AOB; and tang. 4 OBP X number of sides-tabular number, or the area of the polygon. The angle OBP, together with its tangent, for any polygon of not more than 12 sides, is shown in the following table. 4 Tetragon 45° 1.00000+1x1 5 Pentagon 54° 1.37638+=√1+3√5 6 Hexagon 60° 1.73205+= √3 7 Heptagon 8 Octagon 64°2.07652+ 67° 2.41421+=1+√2 9 Nonagon 70° 2.74747+ 10 Decagon 72° 3.07768+=√5+2√5 12 Duodecagon 75° 3.73205+=2+√3 EXAMPLES. 1. Required the area of a pentagon whose side is 15. The number opposite pentagon in the table is 1.720477. Hence 1.720477 × 152=1.720477 × 225=387.107325= area required. 2. The side of a hexagon is 5 feet 4 inches; what is the area? Ans. 73.9. 3. Required the area of an octagon whose side is 16. Ans. 1236.0773. 4. Required the area of a decagon whose side is 20.5. Ans. 3233.4913. 5. Required the area of a nonagon whose side is 36. Ans. 8011.6439. 6. Required the area of a duodecagon whose side is 125. Ans. 174939.875. PROBLEM IX. The diameter of a circle being given, to find the circumference; or, the circumference being given, to find the di ameter. RULE.* Multiply the diameter by 3.1416, and the product will be the circumference, or * The proportion of the diameter of a circle to its circumference has never yet been exactly ascertained. Nor can a square or any other right lined figure, be found, that shall be equal to a given circle This is the celebrated problem called the squaring of the circle, which has exercised the abilities of the greatest mathematicians for ages, Divide the circumference by 3.1416, and the quotient will be the diameter. Note 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 113 is to 355, so is the diameter to the circumference; or, as 352 is to 115, so is the circumference to the diameter. and been the occasion of so many disputes. Several persons of considerable eminence, have, at different times, pretended that they had discovered the exact quadrature; but their error have soon been detected, and it is now generally looked upon as a thing impossible to be done. But though the relation between the diameter and circumference cannot be accurately expressed in known numbers, it may yet be approximated to any assigned degree of exactness. And in this manner was the problem solved by the great Archimedes, about two thousand years ago, who discovered the proportion to be nearly as 7 to 22, which is the same as our first note. This he effected by showing that the perimeter of a circumscribed regular polygon of 192 sides, is to the diameter in a less ratio than that of 340 to 1, and that the perimeter of an inscribed polygon of 96 sides is to the diameter in a greater ratio than that of 31 to 1, and from thence inferred the ratio above mentioned, as may be seen in his book De Dimensione Circuli. The same proportion was also discovered by Philo Gedarensis and Apollonius Pergeus at a still earlier period, as we are informed by Eutocius in his observations on a work called Ocyteboos. The proportion of Vieta and Metius is that of 113 to 355, which is something more exact than the former, and is the same as the second note. This is a very commodious proportion: for being reduced into decimals, it agrees with the truth as far as the sixth figure inclusively. It was derived from the pretended quadrature of a M. Van Eick, which first gave rise to the discovery. But the first who ascertained this ratio to any great degree of exactness was Van Ceulen, a Dutchman, in his book, De Circulo et Adscriptis. He found that if the diameter of a circle was 1, the circumference would be 3.141592653589793238462643383279502884 nearly; which is exactly true to 36 places of decimals, and was effected by the continual bisection of an arc of a circle, a method so extremely F 2 |