Ya its Geometry!

Let the portion used for the square have length $4x$ and that used for the circle have length $36 - 4x = 4(9 - x)$ . The circle then has a radius $r$ such that

$2\pi r = 4(9 - x) \Longrightarrow r = \dfrac{2}{\pi}(9 - x)$ .

With $A(x)$ being the sum of the respective areas of the square and circle, we see that

$A(x) = \left(\dfrac{4x}{4}\right)^{2} + \pi r^{2} = x^{2} + \pi*\dfrac{4}{\pi^{2}}(9 - x)^{2} = x^{2} + \dfrac{4}{\pi}(9 - x)^{2}$ .

Any critical points will then occur when

$\dfrac{dA}{dx} = 2x - \dfrac{8}{\pi}(9 - x) = 0 \Longrightarrow 2x\left(1 + \dfrac{4}{\pi}\right) = \dfrac{72}{\pi}$

$\Longrightarrow x = \dfrac{36}{\pi + 4} \Longrightarrow 4x = \dfrac{144}{\pi + 4}.$

Now since $\dfrac{d^{2}A}{dx^{2}} = 2 + \dfrac{8}{\pi} \gt 0$ we know by the second derivative test that $A(x)$ will achieve a minimum at the above critical point. Thus the desired length of wire used for the square has length $4x = \dfrac{144}{\pi + 4}$ , and that used for the circle has length $36 - \dfrac{144}{\pi + 4} = \dfrac{36\pi}{\pi + 4}.$

This results in the sum $a + b + c = 144 + 4 + 36 = \boxed{184}$ .