解：当

0 < t \leq \frac{2}{3}

时

过点C做CG⊥BO于点G

∴ $\sin ∠ A B O = \frac{\sqrt{5}}{5} = \frac{C G}{B C}$

∴ $G C = \frac{4}{5} t$

∴ $S = \frac{1}{2} \times 2 t \times \frac{4}{5} t$ = $\frac{4}{5} t^{2}$

当 $\frac{2}{3} < t \leq \frac{5}{4}$ 时

设OA交CE于点F

∵CD=BD=t， $G C = \frac{4}{5} t$

∴由勾股定理得 $G D = \frac{3}{5} t$

∴ $G E = 2 t - \frac{3}{5} t = \frac{7}{5} t$ ， $G O = 2 - t - \frac{3}{5} t = 2 - \frac{8}{5} t$

∴ $O E = E G - O G = \frac{7}{5} t - 2 + \frac{8}{5} t = 3 t - 2$

∵OF∥CG

∴△EOF∽△CGE

∴ $\frac{O F}{C G} = \frac{O E}{O G}$

∴ $O F = \frac{4}{7} (3 t - 2)$

∴ $S_{Δ O F E} = \frac{1}{2} O E \cdot O F$ = $= \frac{1}{2} \cdot (3 t - 2) \cdot \frac{4}{7} (3 t - 2)$ = $\frac{2}{7} {(3 t - 2)}^{2}$

∴ $S = S_{Δ D C E} - S_{Δ O F E}$

= $- \frac{62}{35} t^{2} + \frac{24}{7} t - \frac{8}{7}$ ，

设CD交OA于点P

∵OP∥CG

∴△DOP∽DGC

∴ $\frac{O P}{C G} = \frac{O D}{D G}$

∵OD=2-t-

∴OP= $O P = \frac{4}{3} (2 - t)$

∴ $S = \frac{1}{2} \cdot O D \cdot O P$ = $\frac{2}{3} t^{2} - \frac{8}{3} t + \frac{8}{3}$

∴综上所述 $S = {\begin{array}{l} \frac{4}{5} t^{2} (0 < t \leq \frac{2}{3}) \\ - \frac{62}{35} t^{2} + \frac{24}{7} t - \frac{8}{7} (\frac{2}{3} < t \leq \frac{5}{4}) \\ \frac{2}{3} t^{2} - \frac{8}{3} t + \frac{8}{3} (\frac{5}{4} < t \leq 2) \end{array}$ .

辽宁省大连市中山区2020年数学中考一模试卷