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第68页

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$\pi$

$\frac{9}{4}$

 解：如图，连接$BE，$$DF，$$EF，$过点$P$作$PH\perp AB$于点$H，$则$\angle BHP = \angle AHP = 90^{\circ}。$

 因为四边形$ABCD$是边长为$3\sqrt{2}$的正方形，所以$AB = 3\sqrt{2}，$$\angle ABP = \angle ADP = 45^{\circ}，$所以$\angle BPH = 90^{\circ}-\angle ABP = 45^{\circ}，$所以$\angle ABP = \angle BPH，$所以$BH = PH。$

 设$BH = PH = x，$则$BP = \sqrt{BH^{2}+PH^{2}}=\sqrt{2}x。$又$BP = 2，$所以$\sqrt{2}x = 2，$解得$x = \sqrt{2}，$所以$BH = PH = \sqrt{2}，$所以$AH = AB - BH = 2\sqrt{2}，$所以$AP = \sqrt{AH^{2}+PH^{2}}=\sqrt{10}。$

 因为$E，$$F$分别是$\triangle APB$和$\triangle APD$的外心，所以$\angle AEP = 2\angle ABP = 90^{\circ}，$$\angle AFP = 2\angle ADP = 90^{\circ}，$$EA = EP，$$FA = FP，$所以$\angle EAP = \angle EPA = \frac{1}{2}(180^{\circ}-\angle AEP)=45^{\circ}，$$\angle FAP = \angle FPA = \frac{1}{2}(180^{\circ}-\angle AFP)=45^{\circ}，$所以$\angle EAF = \angle EAP + \angle FAP = 90^{\circ}，$所以四边形$AEPF$是正方形，所以$EF\perp AP，$$EF = AP = \sqrt{10}，$所以$S_{正方形AEPF}=\frac{1}{2}EF\cdot AP = 5。$故四边形$AEPF$的面积为$5。$

 解：如图，连接$OA，$$OM，$$ON。$

 因为半圆分别与$AB，$$AC$相切于点$M，$$N，$所以$AM = AN，$$OM\perp AB，$$ON\perp AC，$所以$\angle OMA = \angle ONA = 90^{\circ}。$

 因为$\angle BAC = 120^{\circ}，$所以$\angle MON = 360^{\circ}-\angle BAC-\angle OMA-\angle ONA = 60^{\circ}。$

 设半圆的半径为$r。$因为$\overset{\frown}{MN}$的长为$\pi，$所以$\frac{60\pi r}{180}=\pi，$解得$r = 3，$所以$OM = ON = 3。$

 在$Rt\triangle OAM$和$Rt\triangle OAN$中，$\begin{cases}OA = OA \\ OM = ON\end{cases}，$所以$Rt\triangle OAM\cong Rt\triangle OAN，$所以$\angle AOM = \angle AON = \frac{1}{2}\angle MON = 30^{\circ}，$所以$AM = AN = \frac{1}{2}OA。$

 设$AM = AN = x，$则$OA = 2x，$所以$OM = \sqrt{OA^{2}-AM^{2}}=\sqrt{3}x，$所以$\sqrt{3}x = 3，$解得$x = \sqrt{3}，$所以$AM = AN = \sqrt{3}。$

 因为$AB + AC = 16，$所以$BM + CN = AB + AC - AM - AN = 16 - 2\sqrt{3}，$所以$S_{\triangle OBM}+S_{\triangle OCN}=\frac{1}{2}BM\cdot OM+\frac{1}{2}CN\cdot ON=\frac{1}{2}(BM + CN)\cdot OM = 24 - 3\sqrt{3}。$

 因为$\angle MOE+\angle NOF = 180^{\circ}-\angle MON = 120^{\circ}，$所以$S_{扇形OEM}+S_{扇形OFN}=\frac{120\pi\times3^{2}}{360}=3\pi，$所以$S_{阴影}=S_{\triangle OBM}+S_{\triangle OCN}-(S_{扇形OEM}+S_{扇形OFN})=24 - 3\sqrt{3}-3\pi。$故阴影部分的面积为$24 - 3\sqrt{3}-3\pi。$

 (1) 解：如图①，连接$OB。$因为$OD\perp BC，$$BC = 4\sqrt{3}，$所以$\angle OEB = 90^{\circ}，$$BE = \frac{1}{2}BC = 2\sqrt{3}。$

 设$\odot O$的半径为$R，$则$OB = OD = R。$因为$DE = 2，$所以$OE = OD - DE = R - 2。$

 因为$BE^{2}+OE^{2}=OB^{2}，$所以$(2\sqrt{3})^{2}+(R - 2)^{2}=R^{2}，$

 展开得$12+R^{2}-4R + 4 = R^{2}，$

 移项得$R^{2}-R^{2}-4R=-12 - 4，$

 合并同类项得$-4R=-16，$

 解得$R = 4。$故$\odot O$的半径为$4。$

 (2) 解：如图②，连接$OB，$$BD，$过点$O$作$OF\perp BD$于点$F，$延长$FO$与$\odot O$交于点$A，$则此时阴影部分的面积最大。

 因为$OA = OB = OD = 4，$$DE = 2，$所以$OE = OD - DE = 2，$所以$OE = \frac{1}{2}OB。$

 因为$\angle OEB = 90^{\circ}，$所以$\angle OBE = 30^{\circ}，$所以$\angle BOE = 90^{\circ}-\angle OBE = 60^{\circ}，$所以$\triangle OBD$为等边三角形，所以$BD = OB = 4，$所以$BF = DF = \frac{1}{2}BD = 2，$所以$S_{\triangle OAB}=\frac{1}{2}OA\cdot BF = 4，$$S_{\triangle OAD}=\frac{1}{2}OA\cdot DF = 4。$

 因为$S_{扇形OBD}=\frac{60\pi\times4^{2}}{360}=\frac{8\pi}{3}，$所以$S_{阴影}=S_{扇形OBD}+S_{\triangle OAB}+S_{\triangle OAD}=\frac{8\pi}{3}+8。$故阴影部分面积的最大值为$\frac{8\pi}{3}+8。$