RATIO OF THE CIRCUMFERENCE TO THE DIAMETER. = FIRST METHOD. We know that the side of a regular inscribed hexagon is equal to the radius (T. IX.). If we suppose, then, the radius R = 1, we can find the side of a regular inscribed polygon of twelve sides, by making a R=1, in formula (2) of Theorem VII., that which gives ✓2 - √3 for a side of a regular inscribed dodecagon. Its perimeter is 12√2-√3; consequently the ratio of this perimeter to the diameter is 6 √2 — √3. Were we to substitute anew, in the same formula, 1 for R, and √2−√3 for a, we should find a side of an inscribed regular polygon of twenty-four sides. Thus we might continue to find the sides of the successive polygons, each having double as many sides as the preceding one, to any extent we please. As the number of sides of the polygons increase, their perimeters will approximate more nearly to the circumference of the circle; we are thus enabled, by this method, to obtain the value of to as close a degree of approximation as we please. SECOND METHOD. We have demonstrated (T. XIV., S.) that the area of a circle which has R for its radius, is equal to R2. Now, if we make R = 1 in this expression, it will become, which shows that The ratio of the circumference to the diameter is equal to the area of a circle whose radius is unity. This being supposed, let us start with an inscribed and circumscribed square, whose areas are respectively 2 and 4 (T. X., S.). And the formulas (T. VIII.), if we suppose A =2, B = 4, will become which, converted into decimals, will give 2.8284271, etc., 3.3137085, etc., for the areas, respectively, of the inscribed and circumscribed octagons. Using these values in the same formulas, we shall obtain the areas of the inscribed and circumscribed polygons of sixteen sides, which in turn will make known the areas of polygons of thirty-two sides, and so on, till we arrive at an inscribed and circumscribed polygon, differing from each other, and consequently from the circle, so little that either may be considered as equivalent to it. The subjoined table exhibits the area, or numerical expression for the surface of each succeeding polygon, carried to seven places of decimals. It appears, then, that the inscribed and circumscribed polygons of 8192 sides differ so little from each other, that the numerical value of each, as far as six places of decimals, is absolutely the same; and as the circle is between the two, it cannot, strictly speaking, differ from either so much as they do from each other; so that the number 3.141592 expresses the area of a circle whose radius is 1: that is, the value of to six places of decimals is 3.141592. THIRD METHOD. Instead of seeking the approximate value of the circumference, or of the area of a circle, of which the radius is 1, we shall seek the value of the radius corresponding to a given circumference. For this purpose, we will premise the following LEMMA. Having given the radius R and the apothem r of a regular polygon, it is required to find the radius R' and the apothem r' of a second regular polygon having the same perimeter as the first and double the number of sides. Suppose the given polygon to be circumscribed, and that AB is one of the sides, AOB the angle at the centre, OAR the radius, and OP == the apothem. = This being supposed, the angle at the centre of the second polygon is half the angle AOB. If we produce PO until it P B meets the circumference at C, and draw the chords CA, CB, the angle ACB, half of AOB is the angle at the centre of the second polygon. Also, if we draw OA' perpendicular to CA, and draw A'B' parallel to AB, we shall have A'B' =}AB (B. I., T. XXXIII., Cor.) for the side of the second polygon, and CA' will be its radius, and CP' its apothem. The similar triangles CAP, CA'P', give CP' = CP = (CO+OP) = {(OA+OP), consequently, r' = }(R + r). Again, the right-angled triangle OA'C gives CA”= CO x CP=0AxCP; that is, R' = √R ×r'; and as r' has already been found, the problem is resolved. These formulas are much more simple than those already used in the first and second methods. Now, to make an application of these formulas, we will take for our given polygon a square whose side is 1, and consequently its perimeter is 4. Its radius is √2, being obviously half its diagonal, and its apothem is . = If, then, in the formula r' (R+r), we make R= }√2= 0.7071068, etc., r == 0.5, we shall find r'=0.6035534, etc. Formula R' = √Rx' will now give R' = 0.6532815, etc. We thus find the radius and apothem of a regular polygon of eight sides, and whose perimeter is 4. Using these values in the same formulas, they will in turn make known the radius and apothem of a regular polygon of sixteen sides, having the same perimeter, 4. We give the results of these successive operations, to seven places of decimals, in the following table: From this table, we see that a circle whose circumference is 4 has for its radius 0.6366196, etc., and consequently for its diameter 1.2732392, etc. Hence, the ratio of this circumference to its diameter, or the value of «, is This value of differs less than a unit in the sixth decimal place from the true value. There are other methods, depending upon a knowledge of the higher branches of mathematics, by which this value of has been extended to more than two hundred decimal places. The value to thirty-one decimals is T= 3.1415926535897932384626433832795. The following original geometrical construction is very simple, and gives this ratio sufficiently accurate for all practical purposes: Let ACB be the diameter of the given circle: produce it to wards N; take BD and DE each equal to AB; through E draw EG perpendicular to AE, and take EF and FG each equal to AB; join AG, AF, DG, and DF. Set off on the line EN, from E, the distances EH and HK, each equal to AG; then set off, in the opposite direction, the distance KL equal to AF, and from L set off LM equal to DG; also set off MN equal to DF. Then bisect EN at the point P; bisect EP at the point R, and, finally, trisect ER at the point T; then will CT be the circumference of the circle, nearly. For, by construction, we have, if we call the diameter a unit, CE=21; EL=2EH-KL=2√13 - √10; LM = √5; MN √2. Therefore, CT=2}+(2 √13 − √10 + √5+ √2) = 3.1415922, etc., which is the ratio true to six decimals. For simplicity and accuracy, a better graphic method of finding this ratio can hardly be expected, or even desired. |