Area of Annulus as Area of Rectangle

Theorem

Let $A$ be an annulus whose inner radius is $r$ and whose outer radius is $R$.

The area of $A$ is given by:

$\map \Area A = 2 \pi \paren {r + \dfrac w 2} \times w$

where $w$ denotes the width of $A$.

That is, it is the area of the rectangle contained by:

the width of $A$

the circle midway in radius between the inner radius and outer radius of $A$.

Proof

\(\ds \map \Area A\)

\(=\)

\(\ds \pi \paren {R^2 - r^2}\)

Area of Annulus

\(\ds \)

\(=\)

\(\ds \pi \paren {\paren {r + w}^2 - r^2}\)

Definition of Width of Annulus

\(\ds \)

\(=\)

\(\ds \pi \paren {r^2 + 2 r w + w^2 - r^2}\)

Square of Sum

\(\ds \)

\(=\)

\(\ds \pi \paren {2 r + w} w\)

simplifying

\(\ds \)

\(=\)

\(\ds 2 \pi \paren {r + \frac w 2} \times w\)

extracting $2$

The result follows from Area of Rectangle.

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): annulus

• Weisstein, Eric W. "Annulus." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Annulus.html