Overlapping CirclesDate: 07/14/99 at 22:38:13 From: Hastings Subject: Geometry Each of two overlapping circles has a radius of 6 inches. How long is the darkened portion of the circumferences of these circles, in inches? a) 48pi b) 24pi c) 20pi d) 18pi e) 16pi The two circles overlap such that the perimeter of each circle touches the center of the other circle. The shaded region is the perimeter of the figure formed when the circles are placed as described above. The answer is e) 16pi, but I don't know why. I tried to subtract the length of the circle's perimeter inside the figure from the total perimeter of the two circles. The total perimeter is 2(2xpix6) or 24pi. Next I drew a line down the center of the figure, connecting the two points where the circles meet. Since the line is 180 degrees, or pi radians, and the radius from the origin (the point along this connecting line collinear to the centers) is 3, the length of the enclosed perimeter is 2(3pi), or 6pi. 24pi - 6pi = 18pi. I don't know what I'm doing wrong. Date: 07/15/99 at 13:03:37 From: Doctor Peterson Subject: Re: Geometry Hi, Hastings. You've done some good work, but you got sidetracked by a wrong assumption. Here's your figure: ******* ******* **** **** *** **** *** ..*.. ** * .. /A\ . ** * . / \ . * ** . / \ .. * * . / \ . * * . / \ . * * ./ \. * * ./ \. * * O +-------+-------+ P * * .\ X /. * * .\ /. * * . \ / . * * . \ / . * ** . \ / .. * * . \ / . * * .. \B/ . ** *** ..*.. ** **** **** *** **** ******* ******* It sounds as if you're imagining that O, A, P, and B are on a circle centered at X. They aren't. Since A, O, and B are already on a circle centered at P, they can't be the same distance from X as well. What you do know is that the arc AOB is part of a circle, and you can figure out the central angle APB by looking at the triangles APO and OPB. Think carefully about the lengths of the sides of these triangles. Once you know the angle, you can find the arc length as a fraction of the circumference. Then fit that number into your original calculation and you'll be back on track. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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