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8. 如图,在扇形OBA中,$∠AOB=100°,OA=4$,分别以点A,B为圆心,4为半径画弧,交$\overset{\frown}{AB}$于点D,C,则图中涂色部分的面积为
$\frac{56}{9}π-8\sqrt{3}$
。答案
8. $\frac{56}{9}π-8\sqrt{3}$
解析:如图,连接OC,OD,AD,BC,过点C作$CH⊥OB$于点H.$\because OA=OB=OC=OD=AD=CB = 4$,$\therefore△ OBC$,$△ OAD$ 均为等边三角形.
$\therefore∠ AOD = ∠ BOC = ∠ OAD = ∠ OBC = 60°$.
$\because∠ AOB = 100°$,$\therefore∠ COD = ∠ AOD + ∠ BOC -∠ AOB= 2×60° - 100° = 20°$.$\because CH ⊥ OB$,$\therefore BH =\frac{1}{2}OB = 2$.$\therefore CH = \sqrt{BC^2-BH^2} = 2\sqrt{3}$.$\therefore S_{△ COB} =\frac{1}{2}OB· CH=4\sqrt{3}$.$\because$ 扇形OBC 和扇形 OAD 的面积相等,$△ OBC$ 和 $△ OAD$ 的面积相等,$\therefore$ 弓形 OC 与弓形 OD 的面积相等.$\therefore S_{涂色}=2(S_{扇形BCO}-S_{△ OBC})+S_{扇形ODC}=2×(\frac{60×π×4^2}{360}-4\sqrt{3})+\frac{20×π×4^2}{360}=\frac{56}{9}π-8\sqrt{3}$.
三、解答题(共44分)
9. (20分)某校开设了A,B,C三个食堂通道,某天中午,该校小明和小丽两名同学将随机通过这三个通道进入食堂.
(1) 小明从A通道通过的概率是
(2) 利用画树状图或列表的方法,求小明和小丽从同一个通道通过的概率.
9. (20分)某校开设了A,B,C三个食堂通道,某天中午,该校小明和小丽两名同学将随机通过这三个通道进入食堂.
(1) 小明从A通道通过的概率是
$\frac{1}{3}$
;(2) 利用画树状图或列表的方法,求小明和小丽从同一个通道通过的概率.
答案
9. (1) $\frac{1}{3}$
(2) 列表如下：
| 小 明 | 小 丽 | | |
| ---- | ---- | ---- | ---- |
| | A | B | C |
| A | (A,A) | (A,B) | (A,C) |
| B | (B,A) | (B,B) | (B,C) |
| C | (C,A) | (C,B) | (C,C) |
由上表可知,共有9种等可能的结果,其中这两人从同一个通道通过的结果有3种,$\therefore P$(小明和小丽从同一个通道通过)$=\frac{3}{9}=\frac{1}{3}$
(2) 列表如下：
| 小 明 | 小 丽 | | |
| ---- | ---- | ---- | ---- |
| | A | B | C |
| A | (A,A) | (A,B) | (A,C) |
| B | (B,A) | (B,B) | (B,C) |
| C | (C,A) | (C,B) | (C,C) |
由上表可知,共有9种等可能的结果,其中这两人从同一个通道通过的结果有3种,$\therefore P$(小明和小丽从同一个通道通过)$=\frac{3}{9}=\frac{1}{3}$
10. (24分)[烟台中考]如图，$△ ABC$内接于$\odot O$，$∠ ABC=2∠ C$，点D在线段CB的延长线上，且$BD=AB$，连接AD.
(1)求证:AD是$\odot O$的切线；
(2)当$AB=5$，$AC=8$时，求BC的长及$\odot O$的半径.
(1)求证:AD是$\odot O$的切线；
(2)当$AB=5$，$AC=8$时，求BC的长及$\odot O$的半径.
答案
10. (1) 连接 AO 并延长,交$\odot O$于点 E,连接 BE.
$\because BD = AB$,$\therefore∠ D = ∠ BAD$.$\therefore∠ ABC = ∠ D +∠ BAD=2∠ BAD$.$\because∠ ABC=2∠ C$,$\therefore∠ C = ∠ BAD$.
$\because∠ E = ∠ C$,$\therefore∠ E = ∠ BAD$.$\because AE$ 为$\odot O$ 的直径,
$\therefore∠ ABE = 90°$.$\therefore∠ E + ∠ BAE = 90°$.$\therefore∠ BAD +∠ BAE = 90°$,即$∠ DAE = 90°$.$\therefore AE ⊥ AD$.$\because AE$ 为$\odot O$的直径,$\therefore AD$ 是$\odot O$ 的切线
(2) 过点 A 作$AH ⊥ BC$ 于点 H.$\therefore∠ AHC = 90°$.$\because∠ D = ∠ C$,
$\therefore AD = AC = 8$.$\because∠ BAD = ∠ C$,$∠ ADB = ∠ CDA$,
$\therefore△ DAB ∽ △ DCA$.$\therefore DB:DA = DA:DC$.$\because BD =AB=5$,$\therefore 5:8=8:DC$,解得$DC=\frac{64}{5}$.$\therefore BC=\frac{64}{5}-5=\frac{39}{5}$.$\because AD = AC$,$AH ⊥ CD$,$\therefore CH=\frac{1}{2}CD=\frac{32}{5}$.在$Rt△ ACH$ 中,$\because AC = 8$,$CH = \frac{32}{5}$,$\therefore AH =\sqrt{8^2-(\frac{32}{5})^2}=\frac{24}{5}$.$\because∠ E = ∠ C$,$∠ ABE = ∠ AHC = 90°$,$\therefore△ ABE ∽ △ AHC$.$\therefore AE:AC=AB:AH$,即$AE:8=5:\frac{24}{5}$,解得$AE=\frac{25}{3}$.$\therefore\odot O$ 的半径为$\frac{25}{6}$
$\because BD = AB$,$\therefore∠ D = ∠ BAD$.$\therefore∠ ABC = ∠ D +∠ BAD=2∠ BAD$.$\because∠ ABC=2∠ C$,$\therefore∠ C = ∠ BAD$.
$\because∠ E = ∠ C$,$\therefore∠ E = ∠ BAD$.$\because AE$ 为$\odot O$ 的直径,
$\therefore∠ ABE = 90°$.$\therefore∠ E + ∠ BAE = 90°$.$\therefore∠ BAD +∠ BAE = 90°$,即$∠ DAE = 90°$.$\therefore AE ⊥ AD$.$\because AE$ 为$\odot O$的直径,$\therefore AD$ 是$\odot O$ 的切线
(2) 过点 A 作$AH ⊥ BC$ 于点 H.$\therefore∠ AHC = 90°$.$\because∠ D = ∠ C$,
$\therefore AD = AC = 8$.$\because∠ BAD = ∠ C$,$∠ ADB = ∠ CDA$,
$\therefore△ DAB ∽ △ DCA$.$\therefore DB:DA = DA:DC$.$\because BD =AB=5$,$\therefore 5:8=8:DC$,解得$DC=\frac{64}{5}$.$\therefore BC=\frac{64}{5}-5=\frac{39}{5}$.$\because AD = AC$,$AH ⊥ CD$,$\therefore CH=\frac{1}{2}CD=\frac{32}{5}$.在$Rt△ ACH$ 中,$\because AC = 8$,$CH = \frac{32}{5}$,$\therefore AH =\sqrt{8^2-(\frac{32}{5})^2}=\frac{24}{5}$.$\because∠ E = ∠ C$,$∠ ABE = ∠ AHC = 90°$,$\therefore△ ABE ∽ △ AHC$.$\therefore AE:AC=AB:AH$,即$AE:8=5:\frac{24}{5}$,解得$AE=\frac{25}{3}$.$\therefore\odot O$ 的半径为$\frac{25}{6}$
