答案：9.解:(1)$S_{△ ABC}=4×4-\frac{1}{2}×1×2-\frac{1}{2}×2×4-\frac{1}{2}×4×$
$3=5$.
(2)$△ ABC$是直角三角形.理由:由勾股定理,得$AB=$
$\sqrt{1^{2}+2^{2}}=\sqrt{5}$,$AC=\sqrt{2^{2}+4^{2}}=2\sqrt{5}$,$BC=\sqrt{3^{2}+4^{2}}=5$,
$\therefore AB^{2}+AC^{2}=BC^{2}$,$\therefore △ ABC$是直角三角形.

答案：
10.解:如答图,作点A关于x轴的对称点$A'$,连接$A'B$,交
x轴于点C,作$DB// A'A$,$A'D// OC$,两线交于点D,
$\therefore A'(0,-1)$,$AC=A'C$,
故这束光从点A到点B所经过的路程$=A'B=$
$\sqrt{A'D^{2}+DB^{2}}=\sqrt{(0+3)^{2}+(3+1)^{2}}=5$.

答案：11.解:(1)$CE=OD$.
理由:$\because OB⊥ OC$,$BD⊥ OA$,$CE⊥ OA$,
$\therefore ∠ BOC=∠ BDO=∠ CEO=90°$,$∠ BOD+∠ COE=$
$90°$,$∠ OCE+∠ COE=90°$,
$\therefore ∠ BOD=∠ OCE$.
又$\because OB=OC$,$\therefore △ CEO≌ △ ODB(\mathrm{AAS})$.$\therefore CE=OD$.
(2)由题可知:$CE=24\ \mathrm{cm}$,$BD=18\ \mathrm{cm}$,
由(1)知$△ CEO≌ △ ODB$,$\therefore OE=BD=18\ \mathrm{cm}$.
在$\mathrm{Rt}△ CEO$中,$OC=\sqrt{18^{2}+24^{2}}=30(\mathrm{cm})$.
$\therefore OA=OC=30\ \mathrm{cm}$.
答:钟摆OA的长为30 cm.

答案：
12.解:(1)$\because AC=5$,$CE=1$,$AE=\sqrt{26}$,
$\therefore AC^{2}+CE^{2}=26$,$AE^{2}=26$,
$\therefore AC^{2}+CE^{2}=AE^{2}$,$\therefore ∠ ACE=90°$.
$\because BC=CE+BE=5$,$AC=5$,
$\therefore AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{5^{2}+5^{2}}=5\sqrt{2}$.
(2)①当$BF=BE=4$时,$AF=AB-BF=5\sqrt{2}-4$.
②如答图①,当$BF=EF$时,由(1)知$AC=BC$,$∠ C=$
$90°$,$\therefore ∠ B=45°$,$\therefore ∠ FEB=∠ B=45°$,
$\therefore ∠ BFE=90°$.
设$BF=EF=x$,
$\because BF^{2}+EF^{2}=BE^{2}$,$\therefore x^{2}+x^{2}=4^{2}$,
$\therefore x=2\sqrt{2}$(负值舍去),
$\therefore AF=AB-BF=5\sqrt{2}-2\sqrt{2}=3\sqrt{2}$.

③如答图②,当$BE=EF$时,有$∠ EFB=∠ B=45°$,
$\therefore ∠ BEF=90°$,$EF=BE=4$,
$\therefore BF=\sqrt{BE^{2}+EF^{2}}=4\sqrt{2}$,
$\therefore AF=AB-BF=5\sqrt{2}-4\sqrt{2}=\sqrt{2}$.
综上所述,$AF$的长为$5\sqrt{2}-4$或$3\sqrt{2}$或$\sqrt{2}$.