第9页 - 亮点给力提优课时作业本九年级数学江苏版

2$\pi$

6$\sqrt{3}+\pi$

$\frac{\pi}{8}$

$\sqrt{13}-1$

证明:(1) 因为$\angle AOB = 2\angle ACB，$$\angle BOC = 2\angle BAC，$$\angle ACB = 2\angle BAC，$所以$\angle AOB = 2\angle BOC。$

 (2) 过点$O$作$OD\perp AB$于点$D，$延长$OD$交$\odot O$于点$E，$连接$BE，$则$\angle BDO=\angle BDE = 90^{\circ}，$$BD=\frac{1}{2}AB，$$\overset{\frown}{AE}=\overset{\frown}{BE}，$所以$\angle AOE=\angle BOE，$所以$\angle AOB = 2\angle BOE。$因为$\angle AOB = 2\angle BOC，$所以$\angle BOE=\angle BOC，$所以$BE = BC=\sqrt{5}。$因为$AB = 4，$所以$BD = 2，$所以$DE=\sqrt{BE^{2}-BD^{2}} = 1。$设$\odot O$的半径为$r，$则$OB = OE = r，$所以$OD = OE - DE = r - 1。$因为$OD^{2}+BD^{2}=OB^{2}，$所以$(r - 1)^{2}+2^{2}=r^{2}，$解得$r=\frac{5}{2}。$故$\odot O$的半径为$\frac{5}{2}。$

证明: (1) 连接$OC。$因为$C$是$\overset{\frown}{BE}$的中点，所以$\overset{\frown}{BC}=\overset{\frown}{CE}，$所以$\angle OAC=\angle CAD。$因为$OA = OC，$所以$\angle OAC=\angle OCA，$所以$\angle CAD=\angle OCA，$所以$AE// OC。$因为$AE\perp CD，$所以$CD\perp OC。$因为$OC$是$\odot O$的半径，所以$CD$是$\odot O$的切线。

 (2) 因为$AB$是$\odot O$的直径，所以$\angle ACB = 90^{\circ}。$因为$\angle ABC = 60^{\circ}，$所以$\angle CAD=\angle OAC = 90^{\circ}-\angle ABC = 30^{\circ}，$所以$\angle DAF=\angle OAC+\angle CAD = 60^{\circ}。$因为$AE\perp CD，$所以$\angle D = 90^{\circ}，$所以$\angle F = 90^{\circ}-\angle DAF = 30^{\circ}，$$AC = 2CD。$因为$CD=\sqrt{3}，$所以$AC = 2\sqrt{3}，$所以$AD=\sqrt{AC^{2}-CD^{2}} = 3，$所以$AF = 2AD = 6。$