# Snake eyes

Can you find the area of the yellow part of this snake's eye?

## Problem

The diameter CD of this snake's eye is 2 cm.

It is yellow with a black slit, made from two arcs of circles with centres at A and B.

Find the area of the yellow part.

Image

## Student Solutions

The diagram below shows the eye and also the two full circles whose arcs form the slit, and some lengths.

Notice that all of the angles in the triangles formed are $90^\text{o}$ or $45^\text{o}$. By applying Pythagoras' theorem to the larger triangle (or any of the right-angled triangles), $x^2+x^2=2^2\Rightarrow 2x^2=4\Rightarrow x^2=2$, so $x=\sqrt2$.

In the diagram below, the grey area and the green area together form a sector of the left hand circle. The angle at the middle of the sector is $90^\text{o}$, so the sector is a quarter of the whole circle.

The area of the green triangle is $\frac{1}{2}\times 2\times1=1$ cm$^2.$

The radius of the larger circle is $x=\sqrt2$ cm, so the area of the larger circle is $\pi\times\sqrt{2}^2=2\pi$ cm$^2.$ So the area of the sector is $\frac{1}{4}\times2\pi=\frac{\pi}{2}$ cm$^2.$

So the area of the grey segment is $\frac{\pi}{2}-1$ cm$^2.$

The yellow area is just the area of the eye without twice the grey area. The area of the eye is $\pi1^2=\pi$ cm$^2$. So the yellow area must be $\pi-2(\frac{\pi}{2}-1)=\pi-\frac{2\pi}{2}+2=2$ cm$^2.$

Image

Notice that all of the angles in the triangles formed are $90^\text{o}$ or $45^\text{o}$. By applying Pythagoras' theorem to the larger triangle (or any of the right-angled triangles), $x^2+x^2=2^2\Rightarrow 2x^2=4\Rightarrow x^2=2$, so $x=\sqrt2$.

In the diagram below, the grey area and the green area together form a sector of the left hand circle. The angle at the middle of the sector is $90^\text{o}$, so the sector is a quarter of the whole circle.

Image

The area of the green triangle is $\frac{1}{2}\times 2\times1=1$ cm$^2.$

The radius of the larger circle is $x=\sqrt2$ cm, so the area of the larger circle is $\pi\times\sqrt{2}^2=2\pi$ cm$^2.$ So the area of the sector is $\frac{1}{4}\times2\pi=\frac{\pi}{2}$ cm$^2.$

So the area of the grey segment is $\frac{\pi}{2}-1$ cm$^2.$

The yellow area is just the area of the eye without twice the grey area. The area of the eye is $\pi1^2=\pi$ cm$^2$. So the yellow area must be $\pi-2(\frac{\pi}{2}-1)=\pi-\frac{2\pi}{2}+2=2$ cm$^2.$