# Half an Annulus

Can you locate the point on an annulus that splits it into two areas?

Image

The shaded region in the diagram, bounded by two concentric circles (circles whose centres are at the same point), is called an annulus.

The circles have radii 2 cm and 14 cm.

The dashed circle has the same centre as the other two and divides the area of this annulus into two equal areas.

What is its radius?

This problem is taken from the UKMT Mathematical Challenges.

Image

The radii of the three circles are shown on the diagram, where we are trying to find $r.$

**Using the whole annulus and the darker grey annulus**

The area of the whole annulus is the area of the full circle (raduis $14$ cm) without the area of the empty circle (radius $2$ cm), so it is given by $\pi14^2-\pi2^2=196\pi-4\pi=192\pi$ cm$^2.$

The area of the darker grey annulus is the area within the dotted circle without the area in the empty circle - so $\pi r^2-\pi2^2=\pi r^2-4\pi$ cm$^2.$

This area is equal to half of the whole annulus, so $\pi r^2-4\pi=\frac{1}{2}\times192\pi\Rightarrow \pi r^2-4\pi=96\pi.$

So $\pi r^2=100\pi$, so $r^2=100$, so $r=10.$

**Using the whole annulus and the lighter grey annulus**

The area of the whole annulus is the area of the full circle (raduis $14$ cm) without the area of the empty circle (radius $2$ cm), so it is given by $\pi14^2-\pi2^2=196\pi-4\pi=192\pi$ cm$^2.$

The area of the lighter grey annulus is the area within the full circle without the area in the dotted circle - so $\pi14^2-\pi r^2=196\pi-\pi r^2$ cm$^2.$

This area is equal to half of the whole annulus, so $196\pi-\pi r^2=\frac{1}{2}\times192\pi\Rightarrow 196\pi-\pi r^2=96\pi.$

So $196\pi-96\pi=\pi r^2\Rightarrow\pi r^2=100\pi$, so $r^2=100$, so $r=10.$

**Using the lighter grey annulus and the darker grey annulus**

The area of the lighter grey annulus is the area within the full circle without the area in the dotted circle - so $\pi14^2-\pi r^2=196\pi-\pi r^2$ cm$^2.$

The area of the darker grey annulus is the area within the dotted circle without the area in the empty circle - so $\pi r^2-\pi2^2=\pi r^2-4\pi$ cm$^2.$

These two areas are equal, so $196\pi-\pi r^2=\pi r^2-4\pi.$ Solving this equation for $r$, $$\begin{align}196\pi-\pi r^2&=\pi r^2-4\pi\\

\Rightarrow 196\pi +4\pi&=\pi r^2+\pi r^2\\

\Rightarrow 200\pi &=2\pi r^2\\

\Rightarrow 100&=r^2\\

\Rightarrow 10&=r\end{align}$$