(14) Given that the inradius of $\triangle A B C$ is 2, and $\tan A=-\frac{4}{3}$, find the minimum value of the area of $\triangle A B C$.
(14) Let $A B=c, B C=a, A C=b, D$ be the point of tangency, then $2 A D+2 a=a+b+c$, so $A D=\frac{1}{2}(b+c-a)$.

From $\tan A=-\frac{4}{3}$ we get $\tan \angle D A O=2$, so $D O=2 A D=b+c-a=2$, and also $\sin A=\frac{4}{5}$, hence $S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{2}{5} b c$.
$$
\text { Also, } S_{\triangle A B C}=\frac{1}{2}(a+b+c) \cdot 2 \text {, }
$$

Therefore,
$$
\frac{2}{5} b c=a+b+c=2(b+c)-2,
$$

Thus,
$$
b c=5(b+c)-5 \geqslant 10 \sqrt{b c}-5,
$$

Therefore,
$$
b c-10 \sqrt{b c}+5 \geqslant 0,
$$

Hence,
$$
\sqrt{b c} \geqslant 5+2 \sqrt{5}, b c \geqslant 45+20 \sqrt{5} \text {. }
$$

Therefore,
$$
S_{\triangle A B C}=\frac{2}{5} b c \geqslant 18+8 \sqrt{5}
$$
(when $b=c=5+2 \sqrt{5}$, equality holds), so the minimum area of $\triangle A B C$ is $18+$ $8 \sqrt{5}$. (Provided by Yang Hanlin, Urumqi Senior High School)