The area of the outer square is indeed $16$. The area of the two right triangles on either side of a shaded bar are each $(1/2) (3)(4)=6$, so that the area of one shaded bar is $16-12=4$. The area of the shaded region is thus $4+4-$ the area of the square center. The side of that square is the width of a shaded bar, which is the sine of the right triangle $4/5$, so that the shaded area is
Best Answer
Hint:
The shaded region consists of two segments. Each segments comes from a sector with subtended angle of $120^{\circ}$.
\begin{align*} \text{The required area} &= 2\left( \pi \times 7^2 \times \frac{120^{\circ}}{360^{\circ}}- \frac{7^2}{2} \sin 120^{\circ} \right) \end{align*}