8. As shown in the figure, points $P, Q, R$ are on the sides $AB$, $BC$, and $CA$ of $\triangle ABC$, respectively, and $BP=PQ=QR=RC=1$. The maximum area of $\triangle ABC$ is
(A) $\sqrt{3}$;
(B) 2;
(C) $\sqrt{5}$;
(D) 3.
8. (B)

Solution First, if we denote $\triangle A P R, \triangle B Q P, \triangle C R Q, \triangle P Q R$ as I, II, III, N respectively, then $S_{1}, S_{1}, S_{\mathrm{N}}$ are all no greater than $\frac{1}{2} \times 1 \times 1=\frac{1}{2}$. Also, since $\angle P Q R=180^{\circ}-(\angle B+\angle C)$
$$
=\angle A \text {, }
$$

it is easy to prove: $h_{2} \leqslant h_{1} \quad\left(h_{1}, h_{2}\right.$ are the heights from the common side $P R$ of $\triangle Q R P, \triangle A P R$. If we construct the symmetric figure $P Q^{\prime} R$ of $\triangle P Q R$ about $P R$, at this time $Q^{\prime}, A$ are both 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 we can conclude $h_{2} \leqslant h_{1}$ ).
Thus $S_{1} \leqslant S_{N} \leqslant \frac{1}{2}$, so
$$
S_{\triangle A B C}=S_{\mathrm{I}}+S_{\mathrm{I}}+S_{\mathbf{I}}+S_{\mathrm{N}} \leqslant 4 \times \frac{1}{2}=2 .
$$

Finally, when $A B=A C=2, \angle A=90^{\circ}$, $S_{\triangle A B C}=2$ can achieve the maximum value of 2.