check the picture below.
since, we know that for the circle, the perimeter is s+s+s+s or 4s, then 4s = x and so on.
now, for the circle, the perimeter is just the circumference, circumference of a circle is 2πr, thus 2πr = 5-x and so on.
anyway... so... the equation of the area made by those two in x-terms, is just their sum... so let's do that then
[tex]\bf \pi \left( \cfrac{5-x}{2\pi } \right)^2\quad +\quad \cfrac{x^2}{4^2}\\\\
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\pi\cdot \cfrac{(5-x)^2}{(2\pi )^2}+\cfrac{x^2}{4^2}\implies \cfrac{\pi (5-x)^2}{2^2\pi^2}+\cfrac{x^2}{4^2}\implies \cfrac{(5-x)^2}{4\pi}+\cfrac{x^2}{4^2}
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\textit{let's use the LCD of }4^2\pi \textit{ to add them up}
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\cfrac{4(5-x)^2+\pi x^2}{4^2\pi }\implies \cfrac{4(5-x)^2+\pi x^2}{16\pi }[/tex]
now, you can expand the squared binomial on the numerator, but there will not be any like-terms to simplify, so, it won't make much difference, so.... you can expand it or not.
