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Overlapping Circles
Date: 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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