《2025年学霸题中题九年级数学上册人教版》

8. (宜昌中考改编)如图,分别以正三角形的3个顶点为圆心,边长为半径画弧,三段弧围成的图形称为莱洛三角形.若正三角形边长为2cm,则该莱洛三角形的周长为$2π$cm,面积为$(2π - 2\sqrt{3})$$cm^{2}$.(圆周率用π表示)

答案： $2π$ $(2π - 2\sqrt{3})$

9. (自贡中考)如图,直线$y= -2x+2$与坐标轴交于A,B两点,点P是线段AB上的一个动点,过点P作y轴的平行线交直线$y= -x+3$于点Q,$△OPQ$绕点O顺时针旋转$45^{\circ}$,边PQ扫过区域(阴影部分)面积的最大值是______.

答案：
$\frac {2π}{3}$解析：设$\triangle OPQ$绕点$O$顺时针旋转$45^{\circ }$到$\triangle ODC$的位置，如图所示。设$P(m,-2m + 2)$，则$Q(m,-m + 3)$。

$\therefore OP^{2}=m^{2}+(-2m + 2)^{2}=5m^{2}-8m + 4$，$OQ^{2}=m^{2}+(-m + 3)^{2}=2m^{2}-6m + 9$。
$\because \triangle OPQ$绕点$O$顺时针旋转$45^{\circ }$，$\therefore \triangle OPQ\cong \triangle ODC$，$∠QOC = ∠POD = 45^{\circ }$，$\therefore PQ$扫过区域（阴影部分）面积$S = S_{扇形OQC}-S_{扇形OPD}=\frac {45}{360}×π\cdot OQ^{2}-\frac {45}{360}×π\cdot OP^{2}=\frac {π}{8}(-3m^{2}+2m + 5)=-\frac {3π}{8}(m - \frac {1}{3})^{2}+\frac {2π}{3}$。当$m = \frac {1}{3}$时，$S$的最大值为$\frac {2π}{3}$。

$10. ($河北中考$)$如图$,\odot O$的半径为$6,$将该圆周$12$等分后得到表盘模型$,$其中整钟点为$A_{n}(n$为$1~12$的整数$),$过点$A_{7}$作$\odot O$的切线交$A_{1}A_{11}$的延长线于点$P.$　　
$(1)$通过计算比较直径和劣弧$\overset{\large{\frown}}{A_{7}A_{11}}$的长度哪个更长$.$　　
$(2)$连接$A_{7}A_{11},$则$A_{7}A_{11}$和$PA_{1}$有什么特殊位置关系$?$请简要说明理由$.$　　
$(3)$求切线长$PA_{7}$的值$.$　　

答案：

(1)劣弧$\overset{\frown }{A_{7}A_{11}}=\frac {4}{12}×2π×6 = 4π$，直径$2r = 12$，$\because 4π＞12$，故劣弧$\overset{\frown }{A_{7}A_{11}}$更长。
(2)$A_{7}A_{11}⊥PA_{1}$。理由：如图所示，连接$A_{1}A_{7}$，由图可知$A_{1}A_{7}$是直径，$\therefore$对应的圆周角$∠A_{7}A_{11}A_{1}=90^{\circ }$，$\therefore A_{7}A_{11}$和$PA_{1}$互相垂直。

(3)如图所示，连接$A_{11}O$，$∠A_{11}A_{1}A_{7}=\frac {1}{2}∠A_{11}OA_{7}=\frac {1}{2}×\frac {4}{12}×360^{\circ }=60^{\circ }$。$\because PA_{7}$是$\odot O$的切线，$\therefore ∠PA_{7}A_{1}=90^{\circ }$，$\therefore ∠A_{1}PA_{7}=30^{\circ }$，$\therefore PA_{1}=2A_{1}A_{7}=24$，$\therefore PA_{7}=\sqrt {PA_{1}^{2}-A_{1}A_{7}^{2}}=12\sqrt {3}$。

11. (达州中考)如图,AB是$\odot O$的直径,C为$\odot O$上一点(C不与点A,B重合),连接AC,BC,过点C作$CD⊥AB$,垂足为点D.将$△ACD$沿AC翻折,点D落在点E处得$△ACE$,AE交$\odot O$于点F.
(1)求证:CE是$\odot O$的切线;
(2)若$∠BAC= 15^{\circ},OA= 2$,求阴影部分的面积.

答案：

(1)如图，连接$OC$。$\because CD⊥AB$，$\therefore ∠ADC = 90^{\circ }$。$\because \triangle ACD$沿$AC$翻折得到$\triangle ACE$，$\therefore ∠EAC = ∠BAC$，$∠AEC = ∠ADC = 90^{\circ }$。$\because OA = OC$，$\therefore ∠ACO = ∠BAC$，$\therefore ∠ACO = ∠EAC$，$\therefore OC// AE$，$\therefore ∠AEC + ∠ECO = 180^{\circ }$，$\therefore ∠ECO = 90^{\circ }$，即$OC⊥CE$，$\therefore CE$是$\odot O$的切线。

(2)如图，连接$OF$，过点$O$作$OG⊥AE$于点$G$。$\because ∠BAC = 15^{\circ }$，$\therefore ∠BAE = 2∠BAC = 30^{\circ }$，$∠COF = 2∠EAC = 2∠BAC = 30^{\circ }$。$\because OA = 2$，$\therefore OG = \frac {1}{2}OA = 1$，$AG = \sqrt{3}$。$\because OA = OF$，$\therefore AF = 2AG = 2\sqrt{3}$。$\because ∠BOC = 2∠BAC = 30^{\circ }$，$CD⊥AB$，$OC = OA = 2$，$\therefore CD = \frac {1}{2}OC = 1$，$OD = \sqrt{3}$，$\therefore AE = AD = AO + OD = 2 + \sqrt{3}$，$\therefore EF = AE - AF = 2 - \sqrt{3}$，$CE = CD = 1$，$\therefore S_{阴影}=S_{梯形OCEF}-S_{扇形OCF}=\frac {1}{2}×(2 - \sqrt{3}+2)×1-\frac {30}{360}×π×2^{2}=2-\frac {\sqrt{3}}{2}-\frac {1}{3}π$。

12. 教材P123复习题T6变式(聊城中考)如图,线段$AB= 2$,以AB为直径画半圆,圆心为$A_{1}$,以$AA_{1}$为直径画半圆①;取$A_{1}B的中点A_{2}$,以$A_{1}A_{2}$为直径画半圆②;取$A_{2}B的中点A_{3}$,以$A_{2}A_{3}$为直径画半圆③……按照这样的规律画下去,大半圆内部依次画出的8个小半圆的弧长之和为______$\frac {255}{256}π$.

答案： $\frac {255}{256}π$解析：$\because AB = 2$，$\therefore AA_{1}=1$，半圆①的弧长为$\frac {π×1}{2}=\frac {1}{2}π$，同理$A_{1}A_{2}=\frac {1}{2}$，半圆②的弧长为$\frac {π×\frac {1}{2}}{2}=(\frac {1}{2})^{2}π$，$A_{2}A_{3}=\frac {1}{4}$，半圆③的弧长为$\frac {π×\frac {1}{4}}{2}=(\frac {1}{2})^{3}π$……半圆⑧的弧长为$\frac {π×(\frac {1}{2})^{7}}{2}=(\frac {1}{2})^{8}π$，$\therefore 8$个小半圆的弧长之和为$\frac {1}{2}π+(\frac {1}{2})^{2}π+(\frac {1}{2})^{3}π+... +(\frac {1}{2})^{8}π=\frac {255}{256}π$。

13. 如图,将半径为2,圆心角为$120^{\circ}$的扇形OAB绕点A逆时针旋转$60^{\circ}$,点O,B的对应点分别为$O',B'$,连接$BB'$,则图中阴影部分的面积是______$2\sqrt{3}-\frac {2π}{3}$.

答案： $2\sqrt{3}-\frac {2π}{3}$解析：连接$OO'$，$BO'$，$\because$将半径为$2$，圆心角为$120^{\circ }$的扇形$OAB$绕点$A$逆时针旋转$60^{\circ }$，$\therefore ∠OAO' = 60^{\circ }$，$\therefore \triangle OAO'$是等边三角形，$\therefore ∠AOO' = ∠AO'O = 60^{\circ }$。$\because ∠AOB = 120^{\circ }$，$\therefore ∠O'OB = 60^{\circ }$，$\therefore \triangle OO'B$是等边三角形，$\therefore ∠AO'B = 120^{\circ }$。$\because ∠AO'B' = 120^{\circ }$，$\therefore ∠AO'O + ∠AO'B' = 180^{\circ }$，$\therefore O$，$O'$，$B'$三点共线，$\therefore ∠B'O'B = 120^{\circ }$。又$\because O'B' = O'B$，$\therefore ∠O'B'B = ∠O'BB' = 30^{\circ }$。过点$O'$作$O'C⊥BB'$，垂足为$C$，易得$O'C = 1$，$BB' = 2\sqrt{3}$，$\therefore$题图中阴影部分的面积$=S_{\triangle B'O'B}-(S_{扇形OO'B}-S_{\triangle OO'B})=\frac {1}{2}×1×2\sqrt{3}-(\frac {60π\cdot 2^{2}}{360}-\frac {1}{2}×2×\sqrt{3})=2\sqrt{3}-\frac {2π}{3}$。