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Snake Eyes
Age 14 to 16
ShortChallenge Level 
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$^2.$
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$^2.$
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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