$21 \cdot 10 \triangle ABC$ has three sides $a=2, \quad b=2 \sqrt{2}, \quad c=\sqrt{2}+\sqrt{6}$, then the area of the smallest circular paper piece that can cover this triangle is 
(A) $2 \pi$.
(B) $4 \pi$.
(C) $2 \sqrt{3} \pi$.
(D) $(2+\sqrt{3}) \pi$.
(China Hubei Province Huanggang Junior High School Mathematics Competition, 1990)
［Solution］From the given and the cosine theorem, we have
$$
\cos \angle C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{4+8-8-4 \sqrt{3}}{2 \cdot 2 \cdot 2 \sqrt{2}}=\frac{1-\sqrt{3}}{2 \sqrt{2}}<0,
$$

thus $\angle C$ is an obtuse angle, and the circle with radius $c$ is the smallest circle that can cover the triangle. Its area is
$$
\left(\frac{\sqrt{2}+\sqrt{6}}{2}\right)^{2} \pi=(2+\sqrt{3}) \pi
$$

Therefore, the answer is $(D)$.