12. [2026 南通海门中学附属学校月考] 如图, 四边形ABCD内接于$\odot O$, AE是$\odot O$的直径. 若$\odot O$的半径为6, $∠ ADC - ∠ ABC = 40°$, 则$\overset{\frown}{CE}$的长度为 ()

A.$\frac{2}{3}π$
B.$\frac{4}{3}π$
C.$3π$
D.$8π$

如图,连接$OC$. $\because$ 四边形$ABCD$是$\odot O$的内接四边形,$\therefore ∠ ADC + ∠ ABC = 180°$. 又$\because ∠ ADC - ∠ ABC = 40°, \therefore ∠ ADC = 110°, ∠ ABC = 70°$. $\because \overset{\frown}{AC} = \overset{\frown}{AC}, \therefore ∠ AOC = 2∠ ABC = 140°, \therefore ∠ COE = 180° - ∠ AOC = 40°, \therefore l_{\overset{\frown}{CE}} = \frac{40 × π × 6}{180} = \frac{4}{3}π$.

13. 一题多解 [2026 苏州中学园区校期中] 如图,正方形ABCD的边长为1,分别以A,D为圆心,1为半径作$\overset{\frown}{BD}$和$\overset{\frown}{AC}$,两处阴影部分的面积分别记为$S_1,S_2$,则$S_1 - S_2 =$（）

A.$\frac{π}{2} -1$
B.$1 - \frac{π}{4}$
C.$\frac{π}{3} -1$
D.$1 - \frac{π}{6}$

通解 如图1,设$AB,\overset{\frown}{AO}$和$\overset{\frown}{BO}$围成的图形面积为$S_3$,则$S_1 - S_2 = (S_1 + S_3) - (S_2 + S_3) = S_{\mathrm{扇形}ABD} - (S_{\mathrm{正方形}ABCD} - S_{\mathrm{扇形}DAC}) = 2S_{\mathrm{扇形}ABD} - S_{\mathrm{正方形}ABCD} = 2 × \frac{1}{4}π × 1^2 - 1^2 = \frac{π}{2} - 1$.

另解 如图2,过点$O$作$EF//DC$,分别交$AD,BC$于点$E,F$,连接$DO,AO$. $\because$ 四边形$ABCD$为正方形,$\therefore DC//AB,AB = BC = CD = DA = 1, ∠ ABC = ∠ BCD = ∠ CDA = ∠ DAB = 90°$. $\because EF//DC, \therefore DC//EF//AB$,易得四边形$DEFC$是矩形. $\because DO = DA = AO = 1, \therefore △ ADO$是等边三角形,$\therefore ∠ ADO = 60°, ∠ CDO = 30°$. $\because S_1 = 2(S_{\mathrm{扇形}DAO} - S_{△ DAO}) + S_{△ DAO} = 2(S_{\mathrm{扇形}DAO} - S_{△ DEO}), S_2 = 2(S_{\mathrm{矩形}DEFC} - S_{△ DEO} - S_{\mathrm{扇形}DCO}), \therefore S_1 - S_2 = 2(S_{\mathrm{扇形}DAO} - S_{△ DEO}) - 2(S_{\mathrm{矩形}DEFC} - S_{△ DEO} - S_{\mathrm{扇形}DCO}) = 2(S_{\mathrm{扇形}DAO} + S_{\mathrm{扇形}DCO} - S_{\mathrm{矩形}DEFC}) = 2S_{\mathrm{扇形}DCA} - S_{\mathrm{正方形}ABCD} = 2 × \frac{90 × π × 1^2}{360} - 1^2 = \frac{π}{2} - 1$.

14. ▶一题多解 如图,在每个小正方形的边长均为1的网格中,一段圆弧经过格点A,B,C,格点C,D的连线交$\overset{\frown}{BC}$于点E,则$\overset{\frown}{EC}$的长为.(结果保留π)

解题思路:求$\overset{\frown}{EC}$的长,要先求$\overset{\frown}{EC}$所对的圆心角的度数及其所在圆的半径,故需先找到$\overset{\frown}{EC}$所在圆的圆心,根据勾股定理求出半径,再构造等腰直角三角形求出圆周角$∠ ACD$的度数.
通解 如图1,作$AB,BC$的垂直平分线,两条直线交于点$O$,则$O$为圆弧所在圆的圆心,记$F$为$BC$的中点,连接$OA,OC,OE,AD$.易知$O$为$AC$的中点.由垂径定理,得$BF = CF = \frac{3}{2}, \therefore OC = \sqrt{OF^2 + CF^2} = \sqrt{1^2 + (\frac{3}{2})^2} = \frac{\sqrt{13}}{2}$. 根据网格图形,知$AC = AD = \sqrt{13}, CD = \sqrt{26}, \therefore AC^2 + AD^2 = 26 = CD^2, \therefore △ ACD$是等腰直角三角形,$\therefore ∠ CAD = 90°, ∠ ACD = 45°, \therefore ∠ EOA = 2∠ ACD = 90°, \therefore ∠ EOC = 90°, \therefore l_{\overset{\frown}{EC}} = \frac{90π × \frac{\sqrt{13}}{2}}{180} = \frac{\sqrt{13}π}{4}$.

另解 如图2,连接$AE,AC,AD$. $\because ∠ ABC = 90°, \therefore AC$是直径,$\therefore ∠ AEC = 90°$. 根据网格图形,知$AC = AD = \sqrt{13}, CD = \sqrt{26}, \therefore AC^2 + AD^2 = 26 = CD^2, \therefore △ ACD$是等腰直角三角形,$\therefore ∠ CAD = 90°, ∠ ACD = 45°, \therefore ∠ EAC = 45°, \therefore \overset{\frown}{EC}$所对的圆心角是$90°, \therefore \overset{\frown}{EC}$的长为以$AC$为直径的圆周长的$\frac{1}{4}$,即$l_{\overset{\frown}{EC}} = \frac{1}{4} × π × \sqrt{13} = \frac{\sqrt{13}π}{4}$.

15. 教材复习题变式 如图，AB为$\odot O$的直径，CD，EF是$\odot O$的两条弦，且$AB// CD// EF$，连接DB，CB，AE，AF，若$AB=8$，$CD=5$，图中阴影部分的面积为$8π$，则$EF=$.

如图1,连接$OC,OD,OE,OF$. $\because AB//CD//EF, \therefore S_{△ BCD} = S_{△ OCD}, S_{△ AEF} = S_{△ OEF}$,即图1中阴影部分①的面积与扇形$OCD$的面积相等,图1中阴影部分②的面积与扇形$OEF$的面积相等. 在图1中,$\because \odot O$的面积为$π × (\frac{8}{2})^2 = 16π$,而图1中阴影部分的面积为$8π, \therefore$ 图1中阴影部分的面积占圆面积的一半. 如图2,画一个与图1中相同的圆,则扇形$OBC$的面积与图1中阴影部分①的面积相等,即图2中的$BC$与图1中的$CD$相等,扇形$OAC$的面积与图1中阴影部分②的面积相等,即图2中的$AC$与图1中的$EF$相等. 在图2中,$\because AB$是直径,$\therefore ∠ ACB = 90°, \therefore AC = \sqrt{AB^2 - BC^2} = \sqrt{39}$,即图1中的$EF = \sqrt{39}$.

16. 推理能力 [2025 盐城东台期中] 学习下面方框内的内容,并解答下列问题.
小明在反思学习时,发现解决下列2个问题时都用到了同一种数学思想方法.
问题1:若 $a - 2b = 3$,求 $2a - 4b + 1$ 的值. 解决思路:$2a - 4b +1 = 2(a - 2b) +1 = 2 × 3 +1 =7$.
问题2:如图 ,分别以△ABC的3个顶点为圆心,2为半径画圆,求图中3块阴影面积之和. 解决思路:图中3个扇形的半径都是2,可以将3块阴影扇形拼成一个半径为2的……求出这个图形的面积即可.
(1)问题1,2的解决都用到了思想方法.
A.分类讨论 B.数形结合
C.整体 D.从特殊到一般
(2)问题2中阴影部分的面积为.
(结果保留 π)
(3)如图 ,已知⊙O的半径为5,AB,CD是⊙O的弦,且 $AB = 8$, $CD = 6$,求$\overset{\frown}{AB}$与$\overset{\frown}{CD}$的长度之和.(结果保留 π)

16. 解:(1)C
(2)$2π$
$\because ∠ A + ∠ B + ∠ C = 180°, \therefore$ 阴影部分的面积和为$\frac{180 · π · 2^2}{360} = 2π$.
(3)如图,作直径$BE$,连接$AE$.

$\because BE$是直径,$\therefore ∠ EAB = 90°$.
$\because \odot O$的半径为5,$\therefore BE = 10$.
在$\mathrm{Rt}△ ABE$中,$AB = 8$,
$\therefore AE = \sqrt{BE^2 - AB^2} = 6$.
$\because CD = 6, \therefore AE = CD, \therefore \overset{\frown}{AE} = \overset{\frown}{CD}$,
$\therefore \overset{\frown}{AB} + \overset{\frown}{CD} = \overset{\frown}{AB} + \overset{\frown}{AE} = \overset{\frown}{EAB}$.
$\because l_{\overset{\frown}{EAB}} = \frac{180 · π · 5}{180} = 5π, \therefore \overset{\frown}{AB}$与$\overset{\frown}{CD}$的长度之和为$5π$.