1. As shown in Figure 1, in the right triangle $\triangle ABC$, $AB = 35$, and a square CDEF with side length 12 is inscribed in $\triangle ABC$. Then the perimeter of $\triangle ABC$ is ( ).
(A) 35
(B) 40
(C) 81
(D) 84
-、1.D.
Let $B C=a, A C=b$. Then
$$
a^{2}+b^{2}=35^{2}=1225.
$$

Also, since Rt $\triangle A F E \backsim$ Rt $\triangle A C B$, we have $\frac{F E}{C B}=\frac{A F}{A C}$, which means $\frac{12}{a}=\frac{b-12}{b}$.
Thus, $12(a+b)=a b$.
From equations (1) and (2), we get
$$
(a+b)^{2}=1225+24(a+b).
$$

Solving this, we get $a+b=49(a+b=-25$ is discarded).
Therefore, the perimeter is 84.