Now let the right-angled triangles BCA and BCD, BED, and BEF, BGF, and BGH, etc., etc., be the given polygon, and let two circles be drawn around the center B; one with a radius of four or BC, and the other with a radius of five or BD; so that the first circle shall pass through the points which form the right angles, namely, CEGIK, and the second through the points at the other extremity of the bases, namely, ADFHJZ; then if a straight line starting at the point Z shall be made to pass around the polygon, changing its direction at every point where it intercepts the circle whose radius is five, so that it shall not at any time pass beyond it, nor at any time come within the circle whose radius is four, this straight line, if continued in this manner till it is applied exactly forty-nine times, will return to the point A. from whence it first started; thus forming a lap equal to the onefiftieth of the whole polygon, and the circle so formed by the intersection of these lines, which will have a radius of four, will be exactly forty-nine fiftieths of the entire polygon, or exactly one-fiftieth less than the polygon contained within the given circle. See Plate 5 Fig. 2. Therefore AZ th of entire polygon of the entire circle. Con sequently, If from the number of sides of the above polygon 4th be deducted, the square root of the remaining 48 can be extracted exactly; thus 50 — 1 49, and the 149 7; and if from the sum of the squares of the two sides of any square expressed in integers the one-fiftieth be deducted, the square root of the remaining forty-nine fiftieths can be extracted exactly. Demonstration. Let 5 be the side of the given square 5 × 525, which multiplied by two gives 50, then 50 1 49 and the √ 49— 7 ; Again, let the side of the given square be 1, then 1 × 1 = 1, which doubled, gives 2; the and the V1.96 th of 2 = or .04 and 2.00 -.04 1 1.96, CORALLARY 1. If the 30th part of the sums of the squares of the two sides of any square be added thereto, the square root of the sum can be extracted exactly. Demonstration. Let 1 be the side of the given square, then 1 × 1 = 1, which mul 00 2 tiplied by 22, the go of 2 = 9800 or 900; then 2400 = 9801 and the // 9801 99 = 48 QED. 4900 4900 PROPOSITION 3. THEOREM. ARGUMENT 1. If a circle be described with the square root of two for a radius, and the one-fiftieth of the square described upon the radius. be deducted therefrom, the square root of the remaining forty-nine fiftieths can be extracted exactly. 50 2. The square root of the so deducted will be the sine of the given arc. 3. The square root of the remaining will be the cosine of the given arc; For, let the straight line fh, fm, Plate 6, Fig. 1 and 2 be the given radius, and ADBC the given circle, whose arc is A, and let the straight lines gh, ge, lm, ln, be the sine of the given arc, then will the straight line fg, fl, be the cosine of the given are A; Substituting the numbers, we have (1/2)2= 2, and 3 of 2% (VŽ)2 1.4 25 .04. Then 2.00 — .04 1.96, and the V1.96 22 0 12 or 3, .04, and ARGUMENT 4. Now, if the cosine of the given are be multiplied by two, it will be 14, which is very near though not quite the true diameter; and if the sine of the given arc be multiplied by two it will be, which is very near though not quite the part of true circumference being the chord of twice the arc, whose sine is very near though not quite the part of the true circumference. If then we assume 14 to be the true diameter and 44 to be the true circumference of the given circle, and they are very near it, then dividing the cir44 14 cumference 44 by the diameter 14, we have, by cancellation, 55 |