A = r x b/2n =r/2bn or, the area equals half the radius by a side into the number of sides; or, half the radius into the periphery of the polygon. Now, if a polygon have very small sides and many of them, its periphery will approximate the circumference of the circle inscribed within it; indeed when the number of sides becomes infinite, and consequently infinitely small, the periphery and circumference become equal. Consequently, for the area of the circle, we have -
A = r/2c, (164.)
where c represents the circumference.
By computing the area of a polygon inscribed within a given circle, and that of one circumscribed about the circle, the area of one will approximate the area of the other in proportion as the number of the sides of the polygon are increased.
For example: if polygons of 4 sides be inscribed within and circumscribed about a circle, the radius of which is 1, the areas will be respectively 2 and 4. If the polygons have 16 sides, the areas are each 3 and a fraction, the fractions being unlike; when they have 128 sides the areas are each 3.14 and with unlike fractions; when the sides are increased to 2048, the areas each equal 3.1415 and unlike fractions, and when the sides reach 32768 in number the areas are equal each to 3.1415926, having like decimals to seven places. The computations have been continued to 127 places (Gregory's "Math, for Practical Men"), but for all possible uses in building operations seven places will be found to be sufficient. From this result we have the diameter in proportion to the circumference as 1: 3.1415926, or as -
1: 3.14159 1/4,
1: 3.14159, 1:3-1416.
Of these proportions, that one may be used which will give a result most nearly approximating the degree of accuracy required. For many purposes the last proportion will be sufficiently near the truth.
For ordinary purposes the proportion 7: 22 is very useful, and is correct for two places of decimals; it fails in the third place.
The proportion 113: 355 is correct to six places of decimals.
For the quantity 3.1415926 putting the Greek letter π (calledpy), and 2r = d for the diameter, we have -
c = π d. (165.)
To apply this: in a circle of 50 feet diameter, what is the circumference?
c = 3.1416 x 50 c = 157.08 ft.
If the more accurate value of π be used, we have -
c = 3.1415926 x 50, c = 157.07963.
The difference between the two results is 0.00037, which for all ordinary purposes, would be inappreciable. By the rule of 7: 22, we have -
C = 50 X 22/7 = 157.1428571,
an excess over the more accurate result above, of 0.0632271, which is about 3/4 of an inch.
By the rule of 113: 355, we have -
C = 50 X 355/113 = 157.079646.
This result gives an excess of only 0.000016; it is sufficiently near for any use required in building.
From these results we have these rules, namely: To obtain the circumference of a circle, multiply its diameter by
22, and divide the product by 7; or, more accurately, multiply the diameter by 355 and divide the product by 113; or, by multiplication only, multiply the diameter by 3.1416; or, by 3.14159 1/4; or, by 3.1415926; according to the degree of accuracy required.
And conversely: To obtain the diameter from the circumference, multiply the circumference by 7 and divide the product by 22; or, multiply by 113 and divide by 355; or, divide the circumference by 3.1416; or, by 3.14159 1/4; or, by 3.1415926.
447. - Circle: Length of an Arc. - Considering the circle divided into 3600, the length of an arc of one degree in a circle the diameter of which is unity may be thus found.
The circumference for 3600 is 3.14159265;
3.14159265/360 = 0.00872664625
which equals an arc of one degree in a circle having unity as its diameter; or, for ordinary use the decimal 0.008727 or 0.0087 1/4 may be taken; or putting a for the arc and g for the number of degrees, we have -
a - 0.00872665 dg. (166.)
Wherefore: To obtain the length of an arc of a circle, multiply the diameter of the circle by the number of degrees in the arc, and by the decimal 0.0087 1/4, or, instead thereof, by 0.008727.
448. - Circle: Area. - The area of a circle may be obtained in a manner similar to that for the area of polygons
(Art. 439), in which A=Bn; B= r b/2, or -
A = 1/2 b nr,
where b equals a side of the polygon and n the number of sides; so that b n equals the perimeter of the polygon.
Now, if for the perimeter of the polygon there be sub-
4/6 THE CIRCLE.
stituted the circumference of the circle, we shall have, putting for the circumference 3.1416 d, or, π d (Art. 446) -
A = 1/2 π dr.
in which r is the radius. Since 2 r = d, the diameter, and
r = d/2, we
A = 1/2π d d/2
A = 1/4 π d2
And since -
π = 3.14159265,
1/2 π = 0.78539816, or -
1/4 π = 0.7854, nearly. Therefore -
A = 0.7854 d2. (167.)
Or: The area of a circle equals the square of the diameter multiplied by 0.7854.
As an example, the area of a circle 10 feet in diameter is found thus -
10 x 10 = 100.
100 x 0.7854 = 78.54 feet.
449. - Circle: Area of a Sector. - The area of A B CD (Fig. 297), a sector of a circle, is proportionate to that of the whole circle. For, as the circumference of the whole circle is to its area, so is the arc A B C to the area of A B C D.
The circumference of a circle is (165.) C = n d. The area of a circle is (167.) A = .7854 d2. For the arc ABC put a, and for the area of A B C D put s. Then we have from the above-named proportion -
π d: .7854 d2 :: a : s,
s =.7854d2/πd a.
The coefficient 0.7854 is π (Art. 448).
Therefore, multiplying the fraction by 4, we have -
or - S = 1/4da = 1/2ra. (168.)
Wherefore: To obtain the area of a sector of a circle, multiply a quarter of the diameter by the length of the arc.
Thus: let A D equal 10; also let A B C = a, equal 12. Then the area of A B C D is -
S = 1/2x 10 x 12, S = 60. The length of the arc may be had by the rule in Art. 447.
450. - Circle: Area of a Segment. - In the last article, A BCD (Fig. 297) is called the sector of a circle. Of this the portion included within A B C B is a. segment of a circle. The area of this equals the area of the sector minus the area of the triangle ADC; or, putting M for the area of the segment, S for the area of the sector, and T for the area of the triangle, then -
M = S - T
Putting c for A C (Fig. 297) and h for D E, then T = c/2 h.
In the last article, s = 1/2 ra, in which a = the length of the arc ABC. Substituting this value of s in the above, we have -
M=a/2r- c/2h =ar-ch/2; (169.)
Or: When the length of the arc is known, also that of the chord and the perpendicular from the centre of the circle, then the area of the segment equals the difference between the product of half the arc into the radius, aud half the chord into its perpendicular to the centre of the circle.
But ordinarily the length of the arc and of the chord are unknown. If in this case the number of degrees contained between the two radii, DA, DC, are known, then the area of the segment may be found by a rule which will now be developed.
In Fig. 298 (a repetition of Fig. 297) upon D as a centre, and with D F = unity for a radius, describe the arc HF. Then GF is the sine of the angle C D B, and D G is the cosine; and we have -
DF: GF:: DC: EC, or -
1 : sin:: r: c/2 = r sin.
DF: D G:: D C: D E, or -
1 : cos:: r: h = r cos.
By equation (166.) we have -
a - 0.00872665 dg,
in which a is the length of the arc; g the number of degrees contained in the arc; and d is the diameter of the circle. Since d=2r, therefore -
a = 0.0174533 rg.
Putting B for the decimal coefficient, we have -
a = Br g.
The expression (169.), by substitution of values as above, becomes -
M = a/2r - c/2h,
M =Brg/2 r - r sin. x r cos.
M = 1/2 Bgr2 - sin. cos. r2
M = r2 (1/2.Bg - sin. cos.)
M = r2 (0.00872665 g - sin. cos.) (170.)
Or: The area of a segment of a circle equals the square of the radius into the difference between 0.00872665 times the number of degrees contained in the arc of the circle, and the product of the sine and cosine of half the arc.
When the number of degrees subtended by the arc is unknown, or tables of sines and cosines are not accessible, then the area may be obtained by equation (169.), provided the chord and versed sine are known; but before this equation can be used,for this purpose, expressions giving their values in terms of the chord and versed sine must be obtained, for a, the arc, r, the radius, and h, the perpendicular to the chord from the centre of the circle.
For the value of the arc we have (from "Penny Cycl.," Art. Segment) as a close approximation -
a = 1/3 (8 f - c)
By equation (162.) we have -
r = f2/2v;
h = r - v, or -
h = f2/2v - v
Substituting these values in equation (169.) we have -
M=1/2[1/3-(8 f-c) f2/2v - c (f2/2v - v)] (171.)
This rule is the rule (169.) expanded.
The written rule for equation (169.) may be used, substituting for "half the arc," one sixth of the difference between eight times the chord of half the arc and the chord (or 1/6 of 8 times A B, Fig. 298, minus A C, the chord). Also substitute for "the radius" the square of the chord of half the arc divided by twice the versed sine. Also, for." its perpendicular to the centre of the circle" substitute, the quotient of the square of the chord of half the arc divided by twice the versed sine, minus the versed sine.
When the arc is small the curve approximates that of a parabola. In this case the equation for the area of the parabola, which is quite simple, may be used. It is this -
Or, in segments of circles where the versed sine is small in comparison with the chord, the area equals approximately two thirds of the chord into the versed sine.