On the Difference of Areas
The problem below that was originally proposed by Joseph Kennedy in School Science and Mathematics (52, 162, February 1952) appeared as a quickie in the Mathematics Magazine (Vol. 26, No. 5 (May - Jun., 1953), p. 287). It was also included in an eclectic collecton C. W. Trigg.
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A circle of radius 15 intersects another circle, radius 20, at right angles. What is the difference of areas of the non-overlapping portions?
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Solution
Reference
- C. W. Trigg, Mathematical Quickies, Dover, 1985, #12
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Solution
The fact that the circles meet at right angles is a red herring. The solution is trivial and does not take that fact into account:
Let X be the area of the intersection. Then the remaining portions of the two circles have the areas (π 20² - X) and (π 15² - X), with the difference
| | (π 20² - X) - (π 15² - X) = π 20² - π 15²,
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independent of X.
Indeed, it is not important that the two shapes be circles. The numerical answer will be the same for any two blobs of areas π 20² and π 15².
Red Herrings: a Sample List
- On the Difference of Areas
- Area of the Union of Two Squares
- Circle through the Incenter
- Circle through the Incenter And Antiparallels
- Circle through the Circumcenter
- Inequality with Logarithms
- Breaking Chocolate Bars
- Circles through the Orthocenter
- 100 Grasshoppers on a Triangular Board
- Simultaneous Diameters in Concurrent Circles
|Contact|
|Front page|
|Contents|
|Generalizations|
|Geometry|
|Store|
Copyright © 1996-2012 Alexander Bogomolny
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