$17 \cdot 172$ As shown in the figure, points $P, Q, R$ are on the sides $AB, BC, CA$ of $\triangle ABC$, respectively, and $BP=PQ=QR=RC=1$. Then, the maximum value of the area of $\triangle ABC$ is
(A) $\sqrt{3}$.
(B) 2 .
(C) $\sqrt{5}$.
(D) 3 .
(China Junior High School Mathematics League, 1990)
[Solution] First, if we use the numbers $\mathrm{I}, \mathrm{II}, \mathrm{III}, \mathrm{I}$ to represent $\triangle A P R, \triangle B Q P, \triangle C R Q, \triangle P Q R$, then $S_{\mathbb{I}}, S_{\mathbb{U}}, S_{\mathbb{R}}$ are all no greater than $\frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.
Since $\angle P Q R = 180^{\circ} - (\angle B + \angle C) = \angle A$,
it is easy to prove: $h_{2} \leqslant h_{1}$ (where $h_{1}, h_{2}$ are the heights from the common side $P R$ of $\triangle Q R P$ and $\triangle A P R$. If we construct the symmetric figure $\triangle P Q^{\prime} R$ of $\triangle P Q R$ about $P R$, and $A$ lies on the arc containing $\angle A$ with $P R$ as the chord, and because $P Q^{\prime} = Q^{\prime} R$, $Q^{\prime}$ is the midpoint of this arc, hence $h_{2} \leqslant h_{1}$).
Thus, $S_{I} \leqslant S_{\Pi} \leqslant \frac{1}{2}$,
and $S_{\triangle A B C} = S_{I} + S_{\mathbb{I I}} + S_{\mathbb{W}} + S_{\Pi} \leqslant 4 \cdot \frac{1}{2} = 2$.
Finally, when $A B = A C = 2, \angle A = 90^{\circ}$, $S_{\triangle A B C} = 2$ can achieve the maximum value.
Therefore, the answer is $(B)$.