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2025年智慧学堂九年级数学下册华师大版
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年智慧学堂九年级数学下册华师大版 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
11.在公园的O处附近有E，F，G，H四棵树，位置如图所示（图中小正方形的边长均相等）.现计划修建一座以O为圆心，OA长为半径的圆形水池，要求池中不留树木，则E，F，G，H四棵树中需要被移除的为 （ ）
A.E，F，G
B.F，G，H
C.G，H，E
D.H，E，F
A.E，F，G
B.F，G，H
C.G，H，E
D.H，E，F
答案：
A
12.(2023广安中考)如图，△ABC内接于⊙O，圆的半径为7，∠BAC = 60°，则弦BC的长度为_________.
答案：
$7\sqrt{3}$
13.如图，在△ABC中，∠ACB = 90°，AB = 10，BC = 8，CD⊥AB于D，O为AB的中点.
(1)以C为圆心，6为半径作圆C，试判断A，D，B与⊙C的位置关系；
(2)⊙C的半径为多少时，点O在⊙C上？
(1)以C为圆心，6为半径作圆C，试判断A，D，B与⊙C的位置关系；
(2)⊙C的半径为多少时，点O在⊙C上？
答案：
解：$\because\angle ACB = 90^{\circ}$，$AB = 10$，$BC = 8$，$\therefore$易得$AC = 6$，$CD=\frac{24}{5}$.
(1)当以$6$为半径作$\odot C$时，$\because CA = 6$，$CD=\frac{24}{5}＜6$，$CB = 8＞6$，$\therefore$点$A$在$\odot C$上，点$D$在$\odot C$内，点$B$在$\odot C$外；
(2)$\because OC=\frac{1}{2}AB = 5$，$\therefore\odot C$的半径为$5$时，点$O$在$\odot C$上.
(1)当以$6$为半径作$\odot C$时，$\because CA = 6$，$CD=\frac{24}{5}＜6$，$CB = 8＞6$，$\therefore$点$A$在$\odot C$上，点$D$在$\odot C$内，点$B$在$\odot C$外；
(2)$\because OC=\frac{1}{2}AB = 5$，$\therefore\odot C$的半径为$5$时，点$O$在$\odot C$上.
14.(2023陕西中考)如图，△ABC内接于⊙O，∠BAC = 45°，过点B作BC的垂线，交⊙O于点D，并与CA的延长线交于点E，作BF⊥AC，垂足为M，交⊙O于点F.
(1)求证：BD = BC；
(2)若⊙O的半径r = 3，BE = 6，求线段BF的长.
(1)求证：BD = BC；
(2)若⊙O的半径r = 3，BE = 6，求线段BF的长.
答案：
(1)证明：连结$CD$，则$\angle BDC=\angle BAC = 45^{\circ}$，$\because BD\perp BC$，$\therefore\angle BCD = 90^{\circ}-\angle BDC = 45^{\circ}$，$\therefore\angle BCD=\angle BDC$，$\therefore BD = BC$；
(2)解：$\because\angle DBC = 90^{\circ}$，$\therefore CD$为$\odot O$的直径，$\therefore CD = 2r = 6$，$\therefore BC = CD\cdot\sin\angle BDC = 6\sin45^{\circ}=3\sqrt{2}$，$\therefore EC=\sqrt{BE^{2}+BC^{2}}=\sqrt{6^{2}+(3\sqrt{2})^{2}} = 3\sqrt{6}$.$\because\angle BMC=\angle EBC = 90^{\circ}$，$\angle BCM=\angle ECB$，$\therefore\triangle BCM\backsim\triangle ECB$，$\therefore\frac{EC}{EB}=\frac{BM}{CB}=\frac{CM}{EB}$，$\therefore BM=\frac{BC\cdot EB}{EC}=\frac{3\sqrt{2}\times6}{3\sqrt{6}} = 2\sqrt{3}$，$CM=\frac{BC^{2}}{EC}=\frac{(3\sqrt{2})^{2}}{3\sqrt{6}}=\sqrt{6}$，连结$CF$，则$\angle F=\angle BAC = 45^{\circ}$，$\therefore\angle MCF = 45^{\circ}$，$\therefore MF = MC=\sqrt{6}$，$\therefore BF = BM+MF = 2\sqrt{3}+\sqrt{6}$.
(1)证明：连结$CD$，则$\angle BDC=\angle BAC = 45^{\circ}$，$\because BD\perp BC$，$\therefore\angle BCD = 90^{\circ}-\angle BDC = 45^{\circ}$，$\therefore\angle BCD=\angle BDC$，$\therefore BD = BC$；
(2)解：$\because\angle DBC = 90^{\circ}$，$\therefore CD$为$\odot O$的直径，$\therefore CD = 2r = 6$，$\therefore BC = CD\cdot\sin\angle BDC = 6\sin45^{\circ}=3\sqrt{2}$，$\therefore EC=\sqrt{BE^{2}+BC^{2}}=\sqrt{6^{2}+(3\sqrt{2})^{2}} = 3\sqrt{6}$.$\because\angle BMC=\angle EBC = 90^{\circ}$，$\angle BCM=\angle ECB$，$\therefore\triangle BCM\backsim\triangle ECB$，$\therefore\frac{EC}{EB}=\frac{BM}{CB}=\frac{CM}{EB}$，$\therefore BM=\frac{BC\cdot EB}{EC}=\frac{3\sqrt{2}\times6}{3\sqrt{6}} = 2\sqrt{3}$，$CM=\frac{BC^{2}}{EC}=\frac{(3\sqrt{2})^{2}}{3\sqrt{6}}=\sqrt{6}$，连结$CF$，则$\angle F=\angle BAC = 45^{\circ}$，$\therefore\angle MCF = 45^{\circ}$，$\therefore MF = MC=\sqrt{6}$，$\therefore BF = BM+MF = 2\sqrt{3}+\sqrt{6}$.
15.(2023贵州中考)如图，已知⊙O是等边三角形ABC的外接圆，连结CO并延长交AB于点D，交⊙O于点E，连结EA，EB.
(1)写出图中一个度数为30°的角：_________，图中与△ACD全等的三角形是_________；
(2)求证：△AED∽△CEB；
(3)连结OA，OB，判断四边形OAEB的形状，并说明理由.
(1)写出图中一个度数为30°的角：_________，图中与△ACD全等的三角形是_________；
(2)求证：△AED∽△CEB；
(3)连结OA，OB，判断四边形OAEB的形状，并说明理由.
答案：
(1)$\angle 1$(或$\angle 2$，$\angle 3$，$\angle 4$)；$\triangle BCD$；
(2)证明：$\because\triangle ABC$是等边三角形，$\therefore AC = BC$，$\therefore\overset{\frown}{AC}=\overset{\frown}{BC}$，$\therefore\angle 5=\angle 6$，$\because\angle 2=\angle 3$，$\therefore\triangle AED\backsim\triangle CEB$；
(3)解：四边形$OAEB$是菱形.理由：$\because\angle 5=\angle ABC = 60^{\circ}$，$OA = OE = OB$，$\therefore\triangle AOE$和$\triangle BOE$是边长相等的等边三角形，$\therefore OA = AE = BE = OB$，$\therefore$四边形$OAEB$是菱形.
(1)$\angle 1$(或$\angle 2$，$\angle 3$，$\angle 4$)；$\triangle BCD$；
(2)证明：$\because\triangle ABC$是等边三角形，$\therefore AC = BC$，$\therefore\overset{\frown}{AC}=\overset{\frown}{BC}$，$\therefore\angle 5=\angle 6$，$\because\angle 2=\angle 3$，$\therefore\triangle AED\backsim\triangle CEB$；
(3)解：四边形$OAEB$是菱形.理由：$\because\angle 5=\angle ABC = 60^{\circ}$，$OA = OE = OB$，$\therefore\triangle AOE$和$\triangle BOE$是边长相等的等边三角形，$\therefore OA = AE = BE = OB$，$\therefore$四边形$OAEB$是菱形.
16.(凉山州中考)如图，⊙O的半径为R，其内接锐角三角形ABC中，∠A，∠B，∠C所对的边分别是a，b，c.
(1)求证：$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$；
(2)若∠A = 60°，∠C = 45°，BC = 4$\sqrt{3}$，利用(1)的结论求AB的长和sinB的值.
(1)求证：$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$；
(2)若∠A = 60°，∠C = 45°，BC = 4$\sqrt{3}$，利用(1)的结论求AB的长和sinB的值.
答案：
(1)证明：作直径$BE$，连结$CE$，则$\angle BCE = 90^{\circ}$，$\angle E=\angle A$，$\therefore\sin A=\sin E=\frac{BC}{BE}=\frac{a}{2R}$，$\therefore\frac{a}{\sin A}=2R$，同理：$\frac{b}{\sin B}=2R$，$\frac{c}{\sin C}=2R$，$\therefore\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$；
(2)解：由
(1)得：$\frac{AB}{\sin C}=\frac{BC}{\sin A}$，即$\frac{AB}{\sin45^{\circ}}=\frac{4\sqrt{3}}{\sin60^{\circ}} = 2R$，$\therefore AB=\frac{4\sqrt{3}\times\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 4\sqrt{2}$，$2R=\frac{4\sqrt{3}}{\frac{\sqrt{3}}{2}} = 8$，过$B$作$BH\perp AC$于$H$，$\because\angle AHB=\angle BHC = 90^{\circ}$，$\therefore AH = AB\cdot\cos60^{\circ}=4\sqrt{2}\times\frac{1}{2}=2\sqrt{2}$，$CH=\frac{\sqrt{2}}{2}BC = 2\sqrt{6}$，$\therefore AC = AH+CH = 2(\sqrt{2}+\sqrt{6})$，$\therefore\sin B=\frac{AC}{2R}=\frac{2(\sqrt{2}+\sqrt{6})}{8}=\frac{\sqrt{2}+\sqrt{6}}{4}$.则$AB$的长为$4\sqrt{2}$，$\sin B$的值为$\frac{\sqrt{2}+\sqrt{6}}{4}$.
(1)证明：作直径$BE$，连结$CE$，则$\angle BCE = 90^{\circ}$，$\angle E=\angle A$，$\therefore\sin A=\sin E=\frac{BC}{BE}=\frac{a}{2R}$，$\therefore\frac{a}{\sin A}=2R$，同理：$\frac{b}{\sin B}=2R$，$\frac{c}{\sin C}=2R$，$\therefore\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$；
(2)解：由
(1)得：$\frac{AB}{\sin C}=\frac{BC}{\sin A}$，即$\frac{AB}{\sin45^{\circ}}=\frac{4\sqrt{3}}{\sin60^{\circ}} = 2R$，$\therefore AB=\frac{4\sqrt{3}\times\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 4\sqrt{2}$，$2R=\frac{4\sqrt{3}}{\frac{\sqrt{3}}{2}} = 8$，过$B$作$BH\perp AC$于$H$，$\because\angle AHB=\angle BHC = 90^{\circ}$，$\therefore AH = AB\cdot\cos60^{\circ}=4\sqrt{2}\times\frac{1}{2}=2\sqrt{2}$，$CH=\frac{\sqrt{2}}{2}BC = 2\sqrt{6}$，$\therefore AC = AH+CH = 2(\sqrt{2}+\sqrt{6})$，$\therefore\sin B=\frac{AC}{2R}=\frac{2(\sqrt{2}+\sqrt{6})}{8}=\frac{\sqrt{2}+\sqrt{6}}{4}$.则$AB$的长为$4\sqrt{2}$，$\sin B$的值为$\frac{\sqrt{2}+\sqrt{6}}{4}$.
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