Area of a Segment from Arc and Chord Length

Date: 11/27/2000 at 11:07:32
From: Donald Castricone
Subject: Area

Dear Dr. Math,

Is it possible to find the area of a segment (minor segment) of a 
circle given ONLY the chord length and the arc length? Can it be done 
using geometry, trigonometry, and algebra, or only with calculus?

I am also interested in seeing if I can derive the radius of the 
circle. (I realize that calculus may be able to give me the area using 
integration, but I am not sure if it would also let me derive the 
radius.) It's been a long time since I graduated from college (15 
years ago) and I have not used this math in a long time. Your help 
would be much appreciated. I have been working on this problem (on and 
off of course) for two years. 

Thank you for your consideration in this matter.

Date: 11/27/2000 at 12:59:20
From: Doctor Rob
Subject: Re: Area

Thanks for writing to Ask Dr. Math, Donald.

Suppose you have a segment of a circle bounded by an arc of the circle 
and the chord subtending it. Let the length of the arc be s, the 
length of the chord be c, the radius of the circle be r, the distance 
from the midpoint of the chord to the midpoint of the arc be h (the 
height), the measure in radians of the central angle subtending the 
arc be theta, and the distance from the midpoint of the chord to the 
center of the circle (the apothem) be d. (Recall that 1 degree = 
pi/180 radians.)

You know s and c. Then solve

     c/s = sin(x)/x

for x, which must be done numerically (see below). Then

     theta = 2*x
         r = c/(2*sin[x])
         h = r*(1-cos[x])
         d = r - h

Once you know these quantities, finding the area is standard:

     K = r^2*[theta-sin(theta)]/2


To solve sin(x)/x = k for some constant k > 0 is the same as finding a 
root of the equation

     f(x) = sin(x) - k*x = 0

This can be done using Newton's Method as follows. Guess a starting 
value of

     x(0) = sqrt(6-6*k)

Then for each n = 0, 1, 2, ..., compute

     x(n+1) = x(n) - (sin[x(n)]-k*x[n])/(cos[x(n)]-k)

Continue this until |x(n+1)-x(n)| is smaller than the accuracy sought. 
Then x(n+1) agrees with the actual root to that level of accuracy.

Example: Solve sin(x)/x = 3/4 to five decimal places of accuracy. We 
carry seven places of accuracy in our calculations:

     n     x(n)      sin(x[n])   cos(x[n])   sin(x[n])/x(n)
     0   1.2247449   0.9407193   0.3391860   0.7680941
     1   1.2786882   0.9576389   0.2879717   0.7489229
     2   1.2757074   0.9567763   0.2908250   0.7499967
     3   1.2756981   0.9567736   0.2908338   0.7500000
     4   1.2756981

Since |x(4)-x(3)| < .000005, we have the answer correct to five 
decimal places, x = 1.27570.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/