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.

  A circle of radius 15 intersects another circle, radius 20, at right angles. What is the difference of areas of the non-overlapping portions?

Solution

Reference

  1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #12

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Copyright © 1996-2012 Alexander Bogomolny

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,

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

  1. On the Difference of Areas
  2. Area of the Union of Two Squares
  3. Circle through the Incenter
  4. Circle through the Incenter And Antiparallels
  5. Circle through the Circumcenter
  6. Inequality with Logarithms
  7. Breaking Chocolate Bars
  8. Circles through the Orthocenter
  9. 100 Grasshoppers on a Triangular Board
  10. Simultaneous Diameters in Concurrent Circles

|Contact| |Front page| |Contents| |Generalizations| |Geometry| |Store|

Copyright © 1996-2012 Alexander Bogomolny

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