《2025年5年中考3年模拟九年级数学下册湘教版》

9.教材变式·P60T4如图,在以点O为圆心的两个同心圆中,大圆的弦AB交小圆于点C,D.
(1)求证:AC = BD;
(2)若大圆的半径R = 10,小圆的半径r = 8,且圆心O到直线AB的距离为6,求AC的长.

答案：
解析 \n（1）证明：如图，过$O$点作$OE \perp AB$于点$E$，则$CE = DE$，$AE = BE$，$\therefore AE - CE = BE - DE$，即$AC = BD$.\n（2）由（1）可知$OE \perp AB$且$OE \perp CD$，连接$OC$，$OA$，如图，$\because OE = 6$，$\therefore CE=\sqrt{OC^{2}-OE^{2}}=\sqrt{8^{2}-6^{2}} = 2\sqrt{7}$，$AE=\sqrt{OA^{2}-OE^{2}}=\sqrt{10^{2}-6^{2}} = 8$，$\therefore AC = AE - CE = 8 - 2\sqrt{7}$.

10.新考向·尺规作图(2023山西忻州五台期中)如图,AB是⊙O的直径,点P是AB上一点,且点P是弦CD的中点.(M9202003)
(1)依题意画出弦CD;(尺规作图不写作法,保留作图痕迹)
(2)若AP = 4,CD = 16,求⊙O的半径.

答案：
解析 \n（1）弦$CD$如图所示.\n（2）如图，连接$OD$，$\because P$是弦$CD$的中点，点$P$在直径$AB$上，$\therefore AB \perp CD$，$PD=\frac{1}{2}CD = 8$. 设$\odot O$的半径为$r$，则$OD = r$，$OP = OA - AP = r - 4$，在$\odot Rt \triangle ODP$中，$OD^{2}=OP^{2}+PD^{2}$，$\therefore r^{2}=(r - 4)^{2}+8^{2}$，解得$r = 10$，$\therefore \odot O$的半径为$10$.

11.情境题·中华优秀传统文化(2023广西中考,10,★★☆)赵州桥是当今世界上建造最早、保存最完整的中国古代单孔敞肩石拱桥.如图,主桥拱呈圆弧形,跨度约为37 m,拱高约为7 m,则赵州桥主桥拱半径约为(M9202003) ( )

A.20 m
B.28 m
C.35 m
D.40 m

答案：
B 如图，由题意可知，$AB = 37\ \text{m}$，$CD = 7\ \text{m}$，设主桥拱半径为$r\ \text{m}$，$\therefore OD = OC - CD=(r - 7)\ \text{m}$，$\because OC$是半径，$OC \perp AB$，$\therefore AD = BD=\frac{1}{2}AB=\frac{37}{2}\ \text{m}$，在$\text{Rt} \triangle ADO$中，$AD^{2}+OD^{2}=OA^{2}$，$\therefore (\frac{37}{2})^{2}+(r - 7)^{2}=r^{2}$，解得$r=\frac{1565}{56}\approx28$，$\therefore$赵州桥主桥拱半径约为$28\ \text{m}$. 故选 B.

12.(2024湖南邵阳新宁模拟,10,★★☆)如图,已知AB是⊙O的直径,弦CD⊥AB,垂足为E,∠ACD = 22.5°,AB = 4,则CD的长为 ( )

A.$\sqrt{2}$
B.2
C.2$\sqrt{2}$
D.4$\sqrt{2}$

答案： C 连接$OD$，$\because AB$是$\odot O$的直径，弦$CD \perp AB$，$AB = 4$，$\therefore OD = 2$，$CE = DE=\frac{1}{2}CD$.$\because \angle ACD = 22.5^{\circ}$，$\therefore \angle AOD = 2\angle ACD = 45^{\circ}$，$\therefore \triangle DOE$为等腰直角三角形，$\therefore DE=\frac{\sqrt{2}}{2}OD=\sqrt{2}$，$\therefore CD = 2DE = 2\sqrt{2}$，故选 C.

13.方程思想(2024湖南岳阳汨罗弼时片区月考,9,★★☆)如图,在⊙O中,半径OD⊥弦AB于点C,连接AO并延长交⊙O于点E,连接EC,若AB = 8,CD = 2,则EC的长度为 ( )

A.2$\sqrt{15}$
B.8
C.2$\sqrt{10}$
D.2$\sqrt{13}$

答案： D 连接$BE$，$\because OD \perp AB$，$\therefore AC = BC=\frac{1}{2}AB=\frac{1}{2}\times8 = 4$. 设$\odot O$的半径为$r$，则$OA = r$，$OC = r - 2$，$\because OC^{2}+AC^{2}=OA^{2}$，$\therefore (r - 2)^{2}+4^{2}=r^{2}$，解得$r = 5$，$\therefore OC = 5 - 2 = 3$，$\because O$是$AE$的中点，$C$是$AB$的中点，$\therefore OC$是$\triangle ABE$的中位线，$\therefore BE = 2OC = 6$.$\because AE$为$\odot O$的直径，$\therefore \angle ABE = 90^{\circ}$，在$\text{Rt} \triangle BCE$中，$CE=\sqrt{BC^{2}+BE^{2}} = 2\sqrt{13}$. 故选 D.

14.(2023湖南永州中考,18,★★☆)如图,⊙O是一个盛有水的容器的横截面,⊙O的半径为10 cm,水的最深处到水面AB的距离为4 cm,则水面AB的宽度为________cm.

答案：
 $16$
解析 如图，过点$O$作$OD \perp AB$于点$C$，交$\odot O$于点$D$，连接$OA$，则$AC = BC=\frac{1}{2}AB$，由题意知，$OD = OA = 10\ \text{cm}$，$CD = 4\ \text{cm}$，$\therefore OC = 6\ \text{cm}$，在$\text{Rt} \triangle AOC$中，$AC=\sqrt{OA^{2}-OC^{2}}=\sqrt{10^{2}-6^{2}} = 8(\text{cm})$，$\therefore AB = 2AC = 16\ \text{cm}$.

15.(2022湖南长沙中考,13,★★☆)如图,A、B、C是⊙O上的点,OC⊥AB,垂足为D,且D为OC的中点,若OA = 7,则BC的长为________.

答案：   $7$
解析 $\because D$为$OC$的中点，$\therefore OD = CD$.$\because OC \perp AB$，$\therefore \angle ODA=\angle CDB = 90^{\circ}$，$AD = BD$. 在$\triangle AOD$和$\triangle BCD$中，$\begin{cases}OD = CD,\\\angle ADO=\angle BDC,\\AD = BD,\end{cases}$ $\therefore \triangle AOD\cong\triangle BCD(\text{SAS})$，$\therefore BC = OA = 7$.