《2025年高效精练九年级数学上册苏科版》

10. (2023·德州)如图,AC 为四边形 ABCD 的对角线,$∠CAD=60^{\circ },∠ACD=35^{\circ },∠ACB=90^{\circ },\triangle ABC$的外接圆交 CD 于点 E,AC 所对的圆心角的度数为$120^{\circ }$.
(1) 求证:AD 是$\triangle ABC$的外接圆的切线;
(2) 若$\triangle ABC$的外接圆的半径为 3,求$\overset{\frown }{CE}$的长为$\frac{5\pi}{6}$.

答案：
(1) 如图，设圆心为点 $O$，连接 $OC$。$ \because \overset{\frown}{AC} $ 所对圆心角的度数为 $ 120^{\circ} $，$ \therefore \angle AOC = 120^{\circ} $。$ \because OA = OC $，$ \therefore \angle OAC = \angle OCA = 30^{\circ} $。$ \because \angle CAD = 60^{\circ} $，$ \therefore \angle OAD = \angle OAC + \angle CAD = 90^{\circ} $，$ \therefore OA \perp AD $。$ \because \angle ACB = 90^{\circ} $，$ \therefore AB $ 是 $ \odot O $ 的直径，$ \therefore OA $ 是 $ \odot O $ 的半径，$ \therefore AD $ 是 $ \triangle ABC $ 外接圆的切线。
(2) 连接 $OE$。$ \because \angle OCA = 30^{\circ} $，$ \angle ACD = 35^{\circ} $，$ \therefore \angle OCD = \angle OCA + \angle ACD = 30^{\circ} + 35^{\circ} = 65^{\circ} $。$ \because OC = OE $，$ \therefore \angle OEC = \angle OCD = 65^{\circ} $，$ \therefore \angle COE = 180^{\circ} - \angle OCE - \angle OEC = 180^{\circ} - 65^{\circ} - 65^{\circ} = 50^{\circ} $，$ \therefore \overset{\frown}{CE} $ 的长 $ = \frac{50 \times \pi \times 3}{180} = \frac{5\pi}{6} $。

11. (2024·乐山)如图,$\odot O$是$\triangle ABC$的外接圆,AB 为直径,过点 C 作$\odot O$的切线 CD 交 BA 延长线于点 D,点 E 为$\overset{\frown }{CB}$上一点,且$\overset{\frown }{AC}=\overset{\frown }{CE}$.
(1) 求证:$DC// AE$;
证明：$\because CD$为$\odot O$的切线，点$C$在$\odot O$上，$\therefore \angle OCD=90^{\circ}$，$\therefore \angle DCA+\angle OCA=90^{\circ}$。$\because AB$为直径，$\therefore \angle ACB=90^{\circ}$，$\therefore \angle B+\angle OAC=90^{\circ}$。$\because OC=OA$，$\therefore \angle OAC=\angle OCA$，$\therefore \angle B=\angle DCA$。$\because \overset{\frown}{AC}=\overset{\frown}{CE}$，$\therefore \angle B=\angle CAE$，$\therefore \angle CAE=\angle DCA$，$\therefore CD//AE$；
(2) 若 EF 垂直平分 OB,$DA=3$,求阴影部分的面积.
解：连接$OE$，$BE$，$\because EF$垂直平分$OB$，$\therefore OE=BE$。$\because OE=OB$，$\therefore \triangle OEB$为等边三角形，$\therefore \angle BOE=60^{\circ}$，$\therefore \angle AOE=180^{\circ}-60^{\circ}=120^{\circ}$。$\because OA=OE$，$\therefore \angle OAE=\angle OEA=30^{\circ}$。$\because DC//AE$，$\therefore \angle D=\angle OAE=30^{\circ}$。$\because \angle OCD=90^{\circ}$，$\therefore OD=2OC=OA+AD$。$\because OA=OC$，$\therefore OC=AD=3$，$\therefore AO=OE=OC=3$，$\therefore EF=\frac{\sqrt{3}}{2}OE=\frac{3\sqrt{3}}{2}$，$\therefore \triangle OAE$的面积$=\frac{1}{2}AO\cdot FE=\frac{9\sqrt{3}}{4}$。$\because$扇形$OAE$的面积$=\frac{120\pi×3^{2}}{360}=3\pi$，$\therefore$阴影的面积$=$扇形$OAE$的面积$-\triangle OAE$的面积$=$$3\pi-\frac{9\sqrt{3}}{4}$。

答案：
(1) $ \because CD $ 为 $ \odot O $ 的切线，点 $C$ 在 $ \odot O $ 上，$ \therefore \angle OCD = 90^{\circ} $，$ \therefore \angle DCA + \angle OCA = 90^{\circ} $。$ \because AB $ 为直径，$ \therefore \angle ACB = 90^{\circ} $，$ \therefore \angle B + \angle OAC = 90^{\circ} $。$ \because OC = OA $，$ \therefore \angle OAC = \angle OCA $，$ \therefore \angle B = \angle DCA $。$ \because \overset{\frown}{AC} = \overset{\frown}{CE} $，$ \therefore \angle B = \angle CAE $，$ \therefore \angle CAE = \angle DCA $，$ \therefore CD // AE $；
(2) 连接 $OE$，$BE$，$ \because EF $ 垂直平分 $OB$，$ \therefore OE = BE $。$ \because OE = OB $，$ \therefore \triangle OEB $ 为等边三角形，$ \therefore \angle BOE = 60^{\circ} $，$ \therefore \angle AOE = 180^{\circ} - 60^{\circ} = 120^{\circ} $。$ \because OA = OE $，$ \therefore \angle OAE = \angle OEA = 30^{\circ} $。$ \because DC // AE $，$ \therefore \angle D = \angle OAE = 30^{\circ} $。$ \because \angle OCD = 90^{\circ} $，$ \therefore OD = 2OC = OA + AD $。$ \because OA = OC $，$ \therefore OC = AD = 3 $，$ \therefore AO = OE = OC = 3 $，$ \therefore EF = \frac{\sqrt{3}}{2}OE = \frac{3\sqrt{3}}{2} $，$ \therefore \triangle OAE $ 的面积 $ = \frac{1}{2}AO \cdot FE = \frac{9\sqrt{3}}{4} $。$ \because $ 扇形 $OAE$ 的面积 $ = \frac{120\pi \times 3^{2}}{360} = 3\pi $，$ \therefore $ 阴影的面积 $ = $ 扇形 $OAE$ 的面积 $ - \triangle OAE $ 的面积 $ = 3\pi - \frac{9\sqrt{3}}{4} $。

12. 如图,$\odot O$的直径 AB 垂直于弦 DC 于点 F,点 P 在 AB 的延长线上,CP 与$\odot O$相切于点 C.
(1) 求证:$∠PCB=∠PAD$;
(2) 若$\odot O$的直径为 4,弦 DC 平分半径 OB,求图中阴影部分的面积.$\frac{2}{3}\pi$

答案：
(1) 如图，连接 $OC$，$ \because CP $ 与 $ \odot O $ 相切，$ \therefore OC \perp PC $，$ \therefore \angle PCB + \angle OCB = 90^{\circ} $。$ \because AB \perp DC $，$ \therefore \angle PAD + \angle ADF = 90^{\circ} $。$ \because OB = OC $，$ \therefore \angle OBC = \angle OCB $，由圆周角定理得：$ \angle ADF = \angle OBC = \angle OCB $，$ \therefore \angle PCB = \angle PAD $；
(2) 连接 $OD$，在 $Rt\triangle ODF$ 中，$ OF = \frac{1}{2}OD $，则 $ \angle ODF = 30^{\circ} $，$ \therefore \angle DOF = 60^{\circ} $。$ \because AB \perp DC $，$ \therefore DF = FC $。$ \because BF = OF $，$ AB \perp DC $，$ \therefore S_{\triangle CFB} = S_{\triangle DFO} $，$ \therefore S_{阴影部分} = S_{扇形 BOD} = \frac{60\pi \times 2^{2}}{360} = \frac{2}{3}\pi $。

13. 如图 1,半径为 R,圆心角为$n^{\circ }$的扇形面积是$S_{扇形}=\frac {nπR^{2}}{360}$.由弧长$l=\frac {nπR}{180}$,得:$S_{扇形}=\frac {nπR^{2}}{360}=\frac {1}{2}\cdot \frac {nπR}{180}\cdot R=\frac {1}{2}lR$.通过观察,我们发现$S_{扇形}=\frac {1}{2}lR$类似于$S_{三角形}=\frac {1}{2}×底×高$.类比扇形,我们探索扇环(如图 2,两个同心圆围成的圆环被扇形截得一部分叫作扇环)的面积公式及其应用.
设扇环的面积为$S_{扇环}$,$\overset{\frown }{AB}$的长为$l_{1}$,$\overset{\frown }{CD}$的长为$l_{2}$,线段 AD 的长为 h(即两个同心圆半径 R 与 r 的差),类比$S_{梯形}=\frac {1}{2}×$(上底+下底)×高,用含$l_{1},l_{2},h$的代数式表示$S_{扇环}$,并证明.


$S_{扇环}=$$\frac{1}{2}(l_{1}+l_{2})h$
证明：$ S_{扇环} = S_{扇形 AOB} - S_{扇形 COD} = \frac{n\pi R^{2}}{360} - \frac{n\pi r^{2}}{360} = \frac{n\pi}{360}(R^{2} - r^{2}) = \frac{n\pi}{360}(R + r)(R - r) = \frac{n\pi}{360}(R + r)h = \frac{1}{2} \cdot (\frac{n\pi R}{180} + \frac{n\pi r}{180})h = \frac{1}{2}(l_{1} + l_{2})h $。

答案： $ S_{扇环} = S_{扇形 AOB} - S_{扇形 COD} = \frac{n\pi R^{2}}{360} - \frac{n\pi r^{2}}{360} = \frac{n\pi}{360}(R^{2} - r^{2}) = \frac{n\pi}{360}(R + r)(R - r) = \frac{n\pi}{360}(R + r)h = \frac{1}{2} \cdot (\frac{n\pi R}{180} + \frac{n\pi r}{180})h = \frac{1}{2}(l_{1} + l_{2})h $。