第39页 - 亮点给力提优课时作业本九年级数学江苏版

2

$\sqrt{15}$

$9\pi$

B

24

 解：延长$AO$交$BC$于点$D，$过点$O$作$OE\perp BC$于点$E，$则$\angle OED = 90^{\circ}，$$BE=\frac{1}{2}BC。$

 因为$\angle A=\angle B = 60^{\circ}，$所以$\angle ADB = 180^{\circ}-\angle A-\angle B = 60^{\circ}，$所以$\angle A=\angle B=\angle ADB，$所以$\triangle ADB$为等边三角形，所以$BD = AD = AB = 12。$

 因为$OA = 8，$所以$OD = AD - OA = 4。$

 因为$\angle DOE = 90^{\circ}-\angle ADB = 30^{\circ}，$所以$DE=\frac{1}{2}OD = 2，$所以$BE = BD - DE = 10，$所以$BC = 2BE = 20。$

 解：如图，延长$AE$交$\odot O$于点$H，$连接$OC，$$OH，$$BH，$过点$O$作$ON\perp BH$于点$N，$交$CD$于点$M，$则$\angle MNH = 90^{\circ}，$$HN = BN。$

 因为$AE\perp CD，$$BF\perp CD，$所以$\angle HEF = \angle BFE = 90^{\circ}。$

 因为$OA = OH，$$OB = OH，$所以$\angle OHA = \angle OAH，$$\angle OHB = \angle OBH。$

 因为$\angle OAH+\angle OHA+\angle OHB+\angle OBH = 180^{\circ}，$所以$2(\angle OHA+\angle OH B)=180^{\circ}，$所以$\angle OHA+\angle OH B = 90^{\circ}，$所以$\angle AHB = 90^{\circ}，$所以四边形$HEMN$和四边形$HEFB$都是矩形，所以$MN = EH = BF，$$EF// HB，$所以$OM\perp CD，$所以$\angle OMC = 90^{\circ}，$$CM=\frac{1}{2}CD。$

 因为$CD = 16，$所以$CM = 8。$

 因为$\odot O$的半径为$10，$所以$OC = 10，$所以$OM=\sqrt{OC^{2}-CM^{2}} = 6。$

 因为$OA = OB，$所以$ON$为$\triangle ABH$的中位线，所以$AH = 2ON，$所以$AE + EH = 2(OM + MN)，$所以$AE + BF = 2(OM + BF)，$所以$AE - BF = 2OM = 12。$