| rGeoffrey on Tue, 7 May 2002 14:27:30 -0400 |
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Last night we were discussing a problem of finding the area around a merry-go-round. Below is the problem and the solution with lots of white space to give you time to solve it yourself. The problem is you have a round building with a round merry-go-round in the center. Your company is hired to put down a new floor in the area around the merry-go-round and inside the building. You find you cannot take a measurement through the center to get the diameter of the whole building because there is stuff there. So you take a measurement that is a chord of the larger circle and touches the inner circle exactly once, at the midpoint of the chord. This measurement is 70 feet. Your task is to determine the area that must be surfaced by the company and prove you have the right answer knowing only the length of the chord. For this problem consider these measurements... R = radius of the building (the larger circle) r = radius of the merry-go-round (the smaller circle) n = half the length of the chord, from the point on the inner circle to the point the line meets the outer circle (35 feet) A = area needing flooring ^2 means squared in all my equations Draw the two circles and the chord and create the triangle made from r, R, and n where r and R connect the center of the circles to the ends of n. The area is then... A = (PI * R^2) - (PI * r^2) Another hint Notice that r and n are perpendicular and form a right angle so r^2 + n^2 = R^2
A = (PI * R^2) - (PI * r^2) r^2 + n^2 = R^2 A = (PI * (r^2 + n^2) ) - (PI * r^2) A = (PI * r^2) + (PI * n^2) - (PI * r^2) A = ((PI * r^2) - (PI * r^2)) + (PI * n^2) A = (PI * n^2) And the actual area for a chord of 70 feet A = PI * 35^2 A = PI * 1225 A = 3848.4510006474967171167381445174
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