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Relevant wiki: Length and Area - Composite Figures

$\text{ Area of triangle }ABM + \text{ Area of triangle }BNC + \text{ Area of quarter circle }DMN + \text{Area of blue region} = \text{ Area of square }ABCD$

$\text{ Area of triangle }ABM = \dfrac12 \times AM \times AB = \dfrac 12 \times \dfrac{AD}2 \times 16 = \dfrac12 \times 8\times 16 = 64 \qquad\qquad (1)$ .

Similarly, $\text{ Area of triangle }BNC = \dfrac12 \times NC \times BC = \dfrac 12 \times \dfrac{DC}2 \times 16 = \dfrac12 \times 8\times 16 = 64\qquad\qquad (2)$ .

$\text{ Area of quarter circle }DMN = \dfrac14 \pi r^2 = \dfrac14 \pi \cdot 8^2 = 16\pi \qquad\qquad (3)$ .

$\text{ Area of the square }ABCD = 16\times16 = 256 \qquad\qquad (4)$ .

Substitute $(1), (2), (3), (4)$ into the very first equation gives:

$64 + 64 + 16\pi + \text{Area of blue region} = 256 \qquad \Rightarrow \qquad \text{Area of blue region}= 256 - 64 - 64 - 16\pi = 128 - 16\pi \approx \boxed{77.73 } \; .$

The area of the blue region is equal to the area of the square minus the area of the quarter circle and the two triangles. Considering my diagram, we have

$A_{BLUE}=16^2-A_1-A_2-A_3= 256-\dfrac{1}{4}(\pi)(8^2)-\dfrac{1}{2}(8)(16)-\dfrac{1}{2}(8)(16) \approx$ $\boxed{77.7345}$