The diagonal of the square is $\(7 \sqrt{2}\)$, hence, $\(\sqrt{2}$ side $=7 \sqrt{2} \Rightarrow$ side $=7\)$.
Therefore, the area of the square $\(\left(S^2\right)=49\)$ sq. units and the area of two semi-circles

$$
\(\begin{aligned}
& =2\left(\frac{1}{2} \times \pi \times r^2\right) \\
& =2 \times \frac{1}{2} \times \pi \times\left(\frac{7}{2}\right)^2=\frac{22}{7} \times \frac{49}{4}=\frac{77}{2} \text { sq. units. }
\end{aligned}\)
$$


The area of two equilateral $\(\Delta^{l e s}=2 \times \sqrt{3} \times \frac{7^2}{4} \Rightarrow 2 \times\left(\frac{\sqrt{3}}{4} a^2\right)=\frac{49}{2} \times \sqrt{3}=42.434\)$ sq. units
Therefore, the total area of the design $\(=49+38.5+42.434=129.93\)$ sq. units.
Hence, the approximate painting cost $\(=129.93 \times 10=\$ 1299.3\)$.
Thus, the correct answer is $D$.