A geometry problem by Hana Wehbi

An easy way to solve this is to take numerical values. Suppose the side of the square is $2$ which is the same as the diameter of the circle. Thus by using Pythagoras's theorem, the diagonal of the square measures $\sqrt{2^2+2^2 = 8}$ . Thus, the percentage of the diagonal that is outside the circle is $\frac{\sqrt{8}-2}{\sqrt{8}}= 29.3\%$

$AF = AK = r$ (radius of the circle) & $\theta = 45$

$sin(\theta) =$ $\frac{1}{\sqrt{2}}$ $=$ $\frac{FC}{AC}$ $=$ $\frac{r}{r + KC}$

Solving for $KC$ we get $KC = (\sqrt{2} - 1)r$

To answer the question "Approximately what percentage of the diagonal is outside the circle?", we have to find $\frac{KC}{AC}$

$\frac{KC}{AC}$ $=$ $\frac{(\sqrt{2} - 1)r}{r + (\sqrt{2} - 1)r}$ $= 1 -$ $\frac{1}{\sqrt{2}}$ $=$ ~ $0.293$ So the answer is $29.3$ %