# 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.