# 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 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

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 $\pi 20^{2} - X$ and $\pi 15^{2} - X$ with the difference

$(\pi 20^{2} - X) - (\pi 15^{2} - X) = \pi 20^{2} - \pi 15^{2},$

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 $\pi 20^2$ and $\pi 15^2.$

Let $A$ and $B$ be the areas of the two non-overlapping pieces such that the area of one shape is $A+X$ and the other $B+X.$ Then the difference in the areas of the shapes, $(A+X)-(B+X),$ is exactly the difference in the area of their non-overlapping parts:

$(A+X) - (B+X) = A-B.$ ### What Is Red Herring 