1. Given the right triangle $\triangle A B C$ with side lengths of 3, 4, and 5. If its incircle is removed, the remaining area is $\qquad$ .
$=, 1.6-\pi$
$S_{\triangle M B C}=\frac{1}{2} \times 3 \times 4=6$.
Let the radius of the inscribed circle be $r$. Then
$$
S_{\triangle A B C}=\frac{1}{2} r(3+4+5)=6 r \Rightarrow r=1 \text {. }
$$

Therefore, the area of the inscribed circle is $\pi$.
Thus, the area of the remaining part is $6-\pi$.