17 sides, and of 257 sides; but the construction of the latter polygon is so lengthy, it is not likely that it has ever been attempted. Proposition 15. Problem. 457. Given the radius and the side of a regular inscribed polygon, to compute the side of a similar circumscribed polygon. Given, AB a side of the regular in- C scribed polygon, and OFR, the radius of the ABH. Required, to compute CD, a side of the similar circumscribed polygon. Cons. Join OC, OD. They will cut the Oce in A and B. The similar AS OCF, OAE give F E H (402) If we know the sides of the regular polygons of n, 2n, 4n, . . . . sides, inscribed in the circle of radius 1, we have, by this formula, the sides, and therefore the perimeters, of the regular circumscribed similar polygons, Proposition 16. Problem. 459. Given the radius and the side of a regular inscribed polygon, to compute the side of the regular inscribed polygon of double the number of sides. Given, AB, a side of the regular inscribed polygon, and OC R, the radius of the ABD. = Required, to compute AC, a side of the regular inscribed polygon of double the number of sides. Cons. C is the mid. pt. of the arc AB. E D (403) CD is to AB at its mid. pt. (204) AC is a mean proportional between CD and CE. (325) ... AC2 = CD × CE = R (2R — 20E). But OE = √4RAB'. ... AC = √/R (2R - √4R2 – ABo) 460. SCH. When R 1, this gives AC = √ 2 = (457) Q.E.F. By repeated applications of this formula, we may compute successively the sides, and therefore the perimeters, of the regular inscribed polygons of 2n, 4n, 8n, 16n, . sides, n being the number of sides of the first polygon. EXERCISES. .... 1. If the radius of a circle is 4, find its circumference and area. 2. If the circumference of a circle is 28, find its diameter and area. Proposition 17. Problem. 461. To compute the ratio of the circumference of a circle to its diameter. Given, the Oce C, and the radius R. Required, to find the number 7. Cons. C = 2πR. Let R = 1, then π = C. (436) That is, the number = semi Oce of radius 1. Therefore, the semi-perimeter of each polygon inscribed in this Oce is an approximate value of : and as the number of sides of the polygons increases indefinitely, the lengths of the perimeters approach to that of the Oce as the limit. (430) Hence, by the process of (460), we may obtain a succession of nearer and nearer approximations to the length of the semice. It is convenient to begin the computation with the inscribed hexagon. ... making AB = 1, we have from (460), the following: The last two results show that the first four decimals do not change as the number of sides is increased. Hence the approximate value of π is 3.1415, correct to the fourth decimal place. In practice we generally take π = 3.14159. Q.E.F. 462. SCH. The above is called the method of perimeters. For the method of isoperimeters see Rouché et Comberousse, Edition of 1883, p. 194. NOTE.-The number is of such fundamental importance in geometry that mathematicians have sought for its value in a great variety of ways, all of which agree in the conclusion that it cannot be expressed exactly in decimals, but only approximately. Archimedes (born 287 B.C.) was the first to assign an approximate value of π. He proved that it is included between the numbers 24 and 34, or, in decimals, between 3.1428 and 3.1408; he therefore assigned its value correctly within a unit of the third decimal place. The number 34, or 4 is often used in rough computations. Adrian Metius, a Dutch geometer of the 16th century, has given us the much more accurate value of, which is correct to within a half-millionth, and which is remarkable for the manner in which it is formed by the first three odd numbers 1, 3, 5, each written twice. More recently, the value has been found to a great number of decimals by the aid of series. Dase and Clausen, German computers, carried the calculation to 200 decimal places, independently of each other, and their results agreed to the last figure. The first 20 figures of their result are as follows: 3.14159 26535 89793 23846. (437) This result is far beyond all the wants of mathematics. Ten decimals are sufficient to give the circumference of the earth to the fraction of an inch, and thirty decimals would give the circumference of the whole visible universe to a quantity imperceptible with the most powerful microscope.* EXERCISES. 1. The area of the regular inscribed hexagon is equal to twice the area of the inscribed equilateral triangle. 2. The square of a side of the inscribed equilateral triangle is three times the square of a side of the regular inscribed hexagon. 3. The area of a regular inscribed hexagon is half the area of the circumscribed equilateral triangle. 4. If the diameter of a circle be produced to Cuntil the produced part is equal to the radius, the two tangents from C and their chord of contact form an equilateral triangle. * Newcomb's Geom., p. 235. 5. Divide an angle of an equilateral triangle into five equal parts. Describe a about the A, then use (453). 6. The square inscribed in a semicircle is equal to twofifths the square inscribed in the whole circle. 7. The area of a given circle is 314.16; if this circle be circumscribed by a square, find the area of the part between the circumference and the perimeter of the square. 8. The area of a circle is 40 feet; find the side of the inscribed square. 9. Find the angle subtended at the centre of a circle by an arc 6 feet long, if the radius is 8 feet long. 10. Find the length of the arc subtended by one side of a regular octagon inscribed in a circle whose radius is 20 feet. 11. Find the area of a circular sector whose arc contains 18°, the radius of the circle being 4 feet. 12. Find the area of a circular sector, the chord of half the arc being 20 inches, and the radius 45 inches. 13. The radius of a circle is 5 feet: find the area of a circle 7 times as large. 14. The radius of a circle is 7 feet: find the radius of a circle 16 times as large. 15. Find the height of an arc, the chord of half the arc being 10 feet, and the radius 16 feet. 16. Find the area of a segment whose height is 4 inches, and chord 30 inches. 17. Find the area of a segment whose height is 16 inches, the radius of the circle being 20 inches. 18. Find the area of a segment whose arc is 100°, the radius being 12 feet. 19. Find the area of a sector, the radius of the circle being 24 feet, and the angle at the centre 30°. 20. A circular park 500 feet in diameter has a carriageway around it 25 feet wide: find the area of the carriageway. |