XK is parallel to AB, ZX : ZA :: ZK: ZB (VI. 2.); but ZX is to ZA as the triangle ZXK is to the triangle ZAK (V. 25. cor. 2.), and, for the same reason, ZK is to ZB as the triangle ZAK is to the triangle ZAB; whence ZXK: ZAK :: ZAK: ZAB, and consequently the derivative inscribed polygon AKCNEQGT is a mean proportional between the inscribed and circumscribed figures TKNQ and HBDF.

Again, because ZI bisects

L

K

X

M

the angle AZB, ZA is to ZB, B or ZX is to ZK, as AI to IB (VI. 10.), and consequently (V. 25. cor. 2.) the triangle XZK is to the triangle AZK, as the triangle AZI to the triangle IZB. Hence the inscribed figure TKNQ is to its derivative incribed figure AKCNEQGT as the trian- H gle AZI to the triangle IZB; wherefore (V. 11. and 13.) TKNQ and AKCNEQGT together are to twice TKNQ, as the triangles AZI and IZB, or AZB, to twice the triangle AZI, or the space AIKZ,—that is, as HBDF to VILMOPRS. And thus the two inscribed polygons are to twice the simple inscribed polygon, as the surface of the circumscribing polygon to the surface of the derivative circumscribing polygon with double the number of sides.

S G R

F

Cor. Hence the area of a circle is equivalent to the rectangle under its radius and a straight line equal to half its circumference. For the surface of any regular circumscribing polygon, such as VILMOPRS, being composed of a number of triangles AZI, which have all the same altitude ZA, is equivalent (II. 6.) to the rectangle under ZA and half the sum of their bases, or the semiperimeter of the polygon. But the circle itself, as it forms the ultimate limit of the polygon, must have its area, therefore, equivalent to the rectangle under the radius ZA, and the semicircumference ACE.

Scholium. This solution, it was observed, affords one of the

1

best elementary methods of approximating to the numerical expression for the area of a circle. Supposing the radius of a circle to be denoted by unit; the surface of the circumscribing square will be expressed by 4, and consequently (IV. 15. cor.) that of its inscribed square by 2. Wherefore the surface of the inscribed octagon is 2x4=2,8284271247; and the surface of the circumscribing octagon is found by the analogy, 2+2.8284271247: 2 x 24 : 3.3137084990. Again, (2.8284271247 × 3.3137064990) = 3.0614674589, which expresses the area of the inscribed polygon of 16 sides; and 2.8284271247+ 3.0614674589: 2×2,8284271247, or 5,8898945836 : 5.6568542494 : : 3.313708499: 3.1825979781, which denotes the area of the circumscribing polygon of 16 sides. Pursuing this mode of calculation, by alternately extracting a square root and finding a fourth proportional, the following Table will be formed, in which the numbers expressing the surfaces of the inscribed and circumscribed polygons continually approach to each other, and consequently to the measure of their intermediate circle.

The computation of this table might be greatly abridged, by attending to the successive formation of the numbers. Let a and b denote the area of an inscribed and circumscribing polygon of the same number of sides, and a' and b' the areas of corresponding polygons having double the number of sides. Since a' ab, when a and b approach to equality, it is ob

a+b
2

b-a 2

vious that a'= nearly, or a'—a— : Wherefore, af

ter the sides of the polygon are multiplied, the numbers of the first column will be formed, by constantly adding half their difference from those of the second column. Again, because

b

the fraction

; but, since a and b come to differ little,

may be reckoned to 4, or b—b'= 3a+6 to,

b a 4

very nearly. Consequently the higher numbers in the second column may be filled up, by subtracting one-fourth of the common difference. It follows likewise, from combining this result with what has been shown before, that a number in the second column, diminished by the third part of the common difference, must give very nearly the final result. Thus, the areas of the inscribed and circumscribing polygon of 2048 sides, being 3.1415877253 and 3.1415951177, their difference is 73924, and the third of this, or 24641, taken away from the greater, leaves 3.1415926536, for the ultimate value, or the area of the circle itself.

Of the two modes of approximating to the mensuration of the circle, the one contained in the text, though not so direct, is on the whole simpler than the other. In the course of my geometrical lectures, I generally mentioned, that the first proposition of the fourth book, by enabling us to discover a series of regular polygons with the same sides continually doubled, admitted of an easy application. But not having pursued the calculation to any length, I neglected the obvious advantage which results from reducing the perimeter at each step to the same extent, till I was led to reconsider the subject, in consequence of meeting with the small work of Schwab, before. quoted. It somehow had escaped my notice, that M. Legendre, in the additions to his Geometry, has cursorily treated the subject in the same way.

The numbers contained in the last table were copied and interpolated from the tract of James Gregory, entitled Vera Circuli et Hyperbola Quadratura, as reprinted in the Opera Varia of Huygens. For the calculation of the table contained in the text, and of other two tables which will be annexed to this note, accompanied by several acute remarks concerning the formation of the successive numbers, I am indebted to the very obliging assiduity of a young friend, Mr G. A. Walker Arnott, whose solid talents and unwearied application promise the happiest fruits.

Let the same mode of computation be applied to the successive polygons derived from the hexagon. The radius of the circle being unit, the perpendicular from the centre to the base of each component triangle of the inscribed hexagon will be = √, and consequently the area of the figure = 3√3 = 2.5980762114. Again, each side of the circumscribing hexagon is 2/3, and therefore its area, or that of the six contained triangles, is 6/2/3/12 = 3.4641016151, or one-third more than the former. Hence the following table is constructed:

If the method employed in the text for discovering the radius of the circle, which has twice the number of sides under the same extent of perimeter, be applied to the hexagon or its elementary equilateral triangle, the numbers will stand as below.

Wherefore, .95492965855:1::3:3.1415926536; and hence 3.1415926536 is the nearest expression, consisting of ten decimal places, for the area of the circle whose radius is 1. But the semicircumference in this case denoting also the surface, the same number must represent the circumference of a circle whose diameter is 1. Consequently, if D denote the diameter of any circle, the circumference will be expressed approximately, by 3.1415926536 × D; whence the area will be D1 x 3.1415926536, or DX.7853981634.

By help of the note to Prop. 27. Book V. lower numbers may be found, approximating to the same results. For in this case