Area of Union of Two Circles
Date: 6/10/96 at 10:26:43 From: West Craig Subject: Area of union of two circles Hi, I understand from Doctor Bob's Accessing The Internet By E-Mail that you guys can help with maths problems. Well here goes... (1) Take a circular field of radius R (2) Plant a tree at some point on the circumference of the circular field (3) Tie a rope to the tree and attach the other end to a passing goat (4) If the effective length of the rope is L, and the goat can eat exactly half of the grass in the field, express L in terms of R Any solution would be greatly appreciated Cheers, Craig West
Date: 6/10/96 at 11:42:51 From: Doctor Anthony Subject: Re: Area of union of two circles I will slightly change the notation, to r = radius of given circle, and R = radius of circle defined by L the length of rope. Since one cannot draw a decent diagram with ascii, I suggest you draw the following diagram on a piece of paper and refer to it while I go through the working. Draw a circle with suitable radius r. Now take a point C on the circumference and with a slightly larger radius R draw an arc of a circle to cut the first circle in points A and B. Join AC and BC. Let O be the centre of the first circle of radius r. Let angle OCA = x (radians). This will also be equal to angle OCB. The area we require is made up of a sector of a circle radius R with angle 2x at the centre, C, of this circle, plus two small segments of the first circle of radius r cut off by the chords AC and BC. The area of the sector of circle R is (1/2)R^2*2x = R^2*x Area of two segments = 2[(1/2)r^2(pi-2x) - (1/2)r^2sin(pi-2x)] = r^2[pi - 2x - sin(2x)] We also have R = 2rcos(x) so R^2*x = 4r^2*x*cos^2(x) We add the two elements of area and equate to (1/2)pi*r^2 4r^2*x*cos^2(x) + r^2[pi-2x-sin(2x)] = (1/2)pi*r^2 divide out r^2 4x*cos^2(x) + pi - 2x - sin(2x) = (1/2)pi 4x*cos^2(x) + (1/2)pi - 2x - sin(2x) = 0 We must solve this for x and we can then find R/r from R/r = 2cos(x) Newton-Raphson is a suitable method for solving this equation, using a starting value for x at about 0.7 radians The solution I get is x = 0.95284786466 and from this cos(x) = 0.579364236509 and so finally R/r = 2cos(x) = 1.15872847 In terms of the notation you used this is L = 1.1587*radius of given circle. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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