Sometimes we are presented with a geometry problem that requires us to find the area of an irregular shape which can’t easily be partitioned into simple shapes. For example, the one in the following drawing, formed by intersecting curves. Here’s a good strategy to employ for such problems.

## Problem

Two circles, with radii 2 and 1 respectively, are externally tangent (that is, they intersect at exactly one point). An outer circle is tangent to both of these circles. Find the area of the shaded region.

## Strategy

Our usual strategy when presented with complex shapes is to partition them into simpler shapes whose areas are given by formulas we know.

But in this case, and in many similar geometry problems where the shape is formed by intersecting curves rather than straight lines, it is very hard to do so. For such cases, it is often possible to calculate the area of the desired shape by calculating the area of the outer shape, and then subtracting the areas of the inner shapes.

Tip: When asked to find the “shaded” region, that is often a hint to use this approach – as in another problem we’ve solved.

In this problem, it is easy to find the area of the two inner circles, since their radii are given. We can also find the area of the outer circle when we realize that its diameter is equal to the sum of the diameters of the two inner circles.

## Solution

(1) r_{smallInner}= 1 //Given

(2) A_{smallInner}=π*r_{smallInner}^{2} = π //(1), Area of a circle

(3) r_{largeInner}= 2 //Given

(4) A_{largeInner}=π*r_{largeInner}^{2} = 4π //(3), Area of a circle

(5) D_{outer} = 2*(r_{smallInner} +r_{largeInner}) //The outer diameter is equal to the sum of the diameters of the two inner circles.

(6) D_{outer}=2*(1+2) //(1), (3), (5)

(7) R_{outer}= 3 //(6), the radius is half the diameter

(8) A_{outer}=π*R_{outer}^{2} = 9π //(7), Area of a circle

(9) A_{shaded}=A_{outer}-(A_{smallInner}+A_{largeInner}) = 9π-( 4π+π )=4π