# Problem of the Week Problem D and Solution Coloured Areas

## Problem

Chandra wishes to paint a sign. The sign is composed of three concentric semi-circles, creating an outer band, a middle band, and an inner semi-circle.

The outer band is divided into five regions of equal area and will be painted blue. Each of these regions is labelled with a $$B$$. The middle band is divided into three regions of equal area and will be painted red. Each of these regions is labelled with an $$R$$. The inner semi-circle will be painted white.

The diameter of the largest semi-circle is $$10$$ m and the diameter of the middle semi-circle is $$6$$ m.

If the ratio of the area of one region marked with an $$R$$ to the area of one region marked with a $$B$$ is $$5:6$$, what is the diameter of the inner semi-circle?

## Solution

Since the area of a circle with radius $$r$$ is $$\pi r^2$$, the area of a semi-circle with radius $$r$$ is $$\frac{\pi r^2}{2}$$.
The large semi-circle has diameter $$10$$ m and therefore has a radius of $$5$$ m. Thus, the area of the large semi-circle is $$\frac{\pi (5)^2}{2}= \frac{25\pi }{2}$$ m$$^2$$. The middle semi-circle has diameter $$6$$ m and therefore a radius of $$3$$ m. Thus, the area of the middle semi-circle is $$\frac{\pi (3)^2}{2}= \frac{9\pi}{2}$$ m$$^2$$.

The area of the large semi-circle is made up of the areas of $$5$$ regions marked with a $$B$$ plus the area of the middle semi-circle. Therefore, \begin{aligned} 5B+\frac{9\pi}{2}&=\frac{25\pi }{2}\\ 5B&=8\pi\\ B&=\frac{8\pi}{5}\end{aligned} Since ratio of the area of one red region marked with an $$R$$ to the area of one blue region marked with a $$B$$ is $$5:6$$, we have $$\frac{R}{B} = \frac{5}{6}$$. And so, \begin{aligned} R &= \frac{5B}{6}\\ & = \frac{5}{6}\left(\frac{8\pi}{5}\right)\\ &= \frac{4\pi}{3}\end{aligned}

Let the radius of the smallest semi-circle be $$r$$.
The area of the middle semi-circle is made up of the areas of $$3$$ regions marked with an $$R$$ plus the area of the smallest semi-circle. Therefore, $\frac{9\pi}{2} = 3R + \frac{\pi r^2}{2}$ Since $$R = \dfrac{4\pi}{3}$$, we have $\frac{9\pi}{2} = 4\pi + \frac{\pi r^2}{2}$ Therefore, $$\dfrac{\pi}{2} = \dfrac{\pi r^2}{2}$$ or $$r^2=1$$. Thus $$r=1$$, since $$r>0$$.

Therefore, the diameter of the smallest semi-circle is $$2$$ m.