N, the square described on GH will be to the square described
on KL as MM the square of the
number M to NN the square of the U

number N. COR. 2.-If the square on GH be to the square on KL as a

number to a number, the line GH will be to the line KL as the square root of the first number to the square root of the second number; for, if GH were to KL in any other ratio, the square described on GH would not be to the square described on KL in the ratio supposed.

[blocks in formation]

The rectangle under the radius of a circle, and the sum of the diameter and supplemental chord of an arc, is equal to the square on the supplemental chord of half that are.

Given ABE a circle, AB any arc, AD its half; AB and AD their chords, and OF and OC perpendiculars upon them from the centre; to prove that then 2EC. EO = AE?.

(Const.) For, produce EC to D. (Dem.) And since the angle EAD is a right angle (III. 31), therefore (I. 28) EA is parallel to OF; therefore (VÌ. 4) the triangles OFD, EAD are similar, and therefore DE: DO=EA:OF. But DE=2DO, therefore EA = 20F, or the perpen- B dicular on AD from the centre is half of the supplemental chord EA. Similarly it is shewn that OC is half the supplemental chord of the arc AB. The triangles CEA and EAD are similar, the angles at A and C being right angles,

and that at E common; therefore CE : EA = EA : ED, and therefore (VI. 16) CE · ED = EA’, or 2CE. EO = EA?, but 2CE = 2EO + 200 = DE + 20C, and 20C is the supplemental chord of AB; therefore 2CE · EO = AE”.

[ocr errors]

Cor. 2.--Let C denote the chord on which P is the perpen

dicular, then 4C = 1-P. For if AD = C, ADP = 4AF?, and AF? = 0A’ - OF? : do COR. 3.—Let R= the side of an inscribed regular polygon,

and S that of the corresponding circumscribed one,
P the perpendicular on the former; then S= 1


[ocr errors][ocr errors]

Cor. 4.-The triangles DCA, DAE are similar, and there

fore ED : DA = DA : DC, or AD2 = ED DC = ED(OD – OC) = 2(1 - OC). Schol.-In these four corollaries and the next proposition, the radius is supposed = 1, and numbers proportional to the other lines are taken instead of the lines. When the square of a number is found, the number itself can of course be found by extracting the square root of the former.


To find the approximate ratio of the diameter of a circle to its circumference as nearly as may be required.

This may be done by calculating first the apothem of an inscribed regular polygon, by taking some polygon, the length of the side of which is accurately known, as that of a square or hexagon. Then calculate (10, Cor. 1) the apothems of the polygons found by doubling successively the number of sides, till at last the apothem of a polygon of a sufficient number of sides be found; its side may then be found (10, Cor. 2); and then the side of the corresponding circumscribed polygon may be found (10, Cor. 3). The perimeters of the two last polygons will be found by multiplying one of their sides by the number of sides, and it will be found that the numbers expressing their values are the same for a certain number of places of figures ; and since the value of the circumference of the circle is intermediate between these, it will be accurately expressed to that number of places by the figures common to both these former numbers ; and this approximate value may, by the same method, be carried to any degree of accuracy required., . riser

[ocr errors][ocr errors][ocr errors][ocr errors][ocr errors]

Now (10, Cor. 4) if AB be the side of the polygon of 768 sides, then OC = H, and AD2 = 2(1 – H), and AD = .004090612, which is the side of the inscribed polygon of 1536 sides. But (10, Cor. 3) if AD = R, and GH = S, then S= = *004090618 = the side of the corresponding circumscribing polygon. And the perimeters of these two polygons, found by multiplying these two sides by 1536, are respectively 6.283180032 and 6.283189248; and consequently the approximate value of the circumference of the circle, carried to 6 places in the decimal part, is 6•283185.

If the first inscribed regular polygon be a square, instead of a hexagon, its apothem is the square root of or of •5, and the apothems and sides of the successive inscribed and circumscribed polygons of 8, 16, 32, &c., may be calculated in the same manner as above. If the sides of these polygons be found, and then their areas be calculated by multiplying their perimeter by half

their apothem, these areas carried to 7 places in the decimal, will be found to be the same as in the subjoined table:

[merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors]

Since the areas of the last inscribed and circumscribed polygons agree as far as the seventh decimal place, this must be also the area of the circle, which is always of an intermediate value; and since the area of a circle is equal to the rectangle under its circumference and half its radius, therefore the radius being 1, the value of the semicircumference is 3.1415926, which is also the ratio of the circumference of any circle to its diameter, carried to the seventh decimal place.

The value of this ratio may be carried in the same manner to any required degree of approximation ; but much more expedi. tious methods of effecting this are afforded by analytical principles. This approximate value was found by Archimedes to be 37, or 22 to 7; by Peter Metius 355 to 113; it was carried by Vieta to 11 figures; by Adrianus Romanus to 17; by Ludolph Van Ceulen to 36; by Abraham Sharp to 74; by Machin to 100; and by De Lagny to 127 figures. This ratio, carried to 36 figures, is 3.14159,26535,89793,23846,26433,83279,50288.

Dr Rutherford has since carried it to 208 figures.



1. Of all magnitudes that fulfil the same conditions, the greatest is called a maximum, and the least a minimum.

2. Figures that have equal perimeters are said to be isoperimetrical.

AXIOM. Of all isoperimetrical figures, there must be at least one such that its area is not exceeded by that of any other, and if there be only one such, it is the maximum.

PROPOSITION I. THEOREM. Of all straight lines that can be drawn from a given point to a given straight line, the perpendicular is the least.

Given the point P, and the line AB; draw PC perpendicular to AB, and any other line PD meeting AB; to prove that

PC PD. (Dem.) For since the angles at Care right angles, angle PDC is less than a right angle (I. 17);

therefore PC PD. CoR.-A line nearer to the perpendicular is less than one more

remote. For angle PDE is greater than a right angle (I. 16), and PEC is less (I. 17), therefore PD <PE.

PROPOSITION II. THEOREM. Of all triangles having the same base and equal perimeters, that is a maximum whose undetermined sides are equal. ,

« ForrigeFortsett »