There is no escape...believe it or not. Mathematics is everywhere.
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Steve, it looks like you have added up four areas that total to the unit area. There is an easier way. Add up two lengths that total to the unit length, or 1. The width of the square, which is 1, has two components. Left of the origin, in the 9 o'clock or negative-x direction, the distance to the edge is r. Right of the origin in the 3 o'clock or x direction the distance is r/sqrt(2), by pythagoras' theorem on the 45 degree triangle. This expression can be factorised as follows to calculate r, the radius of the circle. We both get the same answer for r, in different form. Now, we just need a volunteer to calculate the area of the semicircle, which was the original question.
The LaTex code for the evaluation is:
r+\frac{r}{\sqrt[]{2}} = 1\\\\
r\:\sqrt[]{2}+r = \sqrt[]{2}\\\\
r(1+\sqrt[]{2}) = \sqrt[]{2}\\\\
r = \frac{\sqrt[]{2}}{1+\sqrt[]{2}}\\\\
r = \frac{(\sqrt[]{2}-1)\:\sqrt[]{2}}{(\sqrt[]{2}-1)(\sqrt[]{2}+1)}\\\\
r = 2 -\sqrt[]{2}\\\\
A_{semicircle}=\frac{\pi r^{2}}{2}\\\\
This code was entered at http://thornahawk.unitedti.org/equationeditor/equationeditor.php to generate the equation.
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