Dynamic Geometry: P6

Let the point $A$ that represents the bottom-left angle of the square be $(k,0)$ . Then the coordinates of the cyan point is $(k+\frac{\sqrt{k}}{2} ,\frac{\sqrt{k}}{2}).$ When the cyan point touches the green segment, $S_{\square}=\frac{9}{4}$ , implying $(\sqrt{k} )^2 =\frac{9}{4}$ . This solves to $k=\frac{9}{4}$ . The integral is therefore $S_{I} =\int ^{k=\frac{9}{4}} _{k=0} y~dx =\int ^{\frac{9}{4}} _0 y \frac{dx}{dk} dk= \int ^{\frac{9}{4}} _0 \frac{\sqrt{k}}{2} (1+\frac{1}{4\sqrt{k}})=\int ^{\frac{9}{4}} _0 \frac{\sqrt{k}}{2}+\frac{1}{8} =\left. \frac{1}{3} k^\frac{3}{2} +\frac{1}{8} k \right |^\frac{9}{4} _0 =\frac{45}{32}.$
Hence we input $45+32=77$ as the solution.