12. As shown in Figure 4, given that the two diagonals of rhombus $A B C D$ are $a$ and $b$, respectively, and semicircles are drawn inside the rhombus with each side as the diameter. Then the area of the petal-shaped region (the shaded area) formed by the 4 semicircular arcs is $\qquad$
12. $\frac{1}{2} \pi\left(a^{2}+b^{2}\right)-\frac{a b}{2}$.

From the problem, we can find the side length of the rhombus to be
$$
\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}}=\frac{\sqrt{a^{2}+b^{2}}}{2}
$$

Therefore, $S_{\text {field }}=4 S_{\text {half circle }}-S_{\text {rhombus }}$
$$
\begin{array}{l}
=4 \times \frac{1}{2} \pi\left(\frac{\sqrt{a^{2}+b^{2}}}{2}\right)^{2}-\frac{1}{2} a b \\
=\frac{1}{2} \pi\left(a^{2}+b^{2}\right)-\frac{a b}{2} .
\end{array}
$$