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1. [2025绥化中考]在$\odot O$中,如果$75°$的圆心角所对的弧长是$2.5π$ cm,那么$\odot O$的半径是（
A.6 cm
B.8 cm
C.10 cm
D.12 cm
A
）A.6 cm
B.8 cm
C.10 cm
D.12 cm
答案
设$\odot O$的半径是$r$ cm,则$\frac{75 × π r}{180}=2.5π$,解得$r=6$.
2. [2025西藏中考]如图,在$\odot O$中,直径$AB=6$,BC是$\odot O$的弦,若$∠ B=60°$,则$\overset{\frown}{AC}$的长为 (
A.$6π$
B.$4π$
C.$2π$
D.$π$
C
)A.$6π$
B.$4π$
C.$2π$
D.$π$
答案
连接$OC. \because \overset{\frown}{AC} = \overset{\frown}{AC}, \therefore ∠ AOC = 2∠ B. \because ∠ B = 60°, \therefore ∠ AOC = 120°. \because$ 直径$AB = 6, \therefore$ 半径$r = 3, \therefore l_{\overset{\frown}{AC}} = \frac{120 × π × 3}{180} = 2π$.
3. 跨学科·物理 [2026无锡宜兴一模] 如图，一个半径为6 cm的定滑轮带动重物上升，假设绳索与滑轮之间没有相对滑动，若滑轮上某一点P旋转了100°，则重物上升的高度为
$\boldsymbol{\frac{10}{3}π}$
cm.答案
由题意得重物上升的高度为$\frac{100 × π × 6}{180} = \frac{10}{3}π(\mathrm{cm})$.
4.［2026常州溧阳燕山中学模拟］如图，点A，B，C均在$\odot O$上，$\odot O$的半径为2 cm，$∠ C = 130°$，则$\overset{\frown}{AB}$的长为
$\boldsymbol{\frac{10}{9}π}$
cm（结果保留$π$）。答案
如图,在优弧$AB$上取点$D$,连接$OB,OA,BD$和$AD$.
$\because$ 四边形$ACBD$是$\odot O$的内接四边形,$\therefore ∠ D + ∠ C = 180°$.
$\because ∠ C = 130°, \therefore ∠ D = 50°, \therefore ∠ BOA = 2∠ D = 2 × 50° = 100°, \therefore l_{\overset{\frown}{AB}} = \frac{100 × π × 2}{180} = \frac{10}{9}π(\mathrm{cm})$.
5. [2025 盐城盐都区模拟] 如图,在扇形AOB中,∠AOB=30°,点C为半径OA上一点,现以点O为圆心,OC长为半径作弧,该弧交半径OB于点D,记$\overset{\frown}{AB}$的长为$m$,BD的长为$d$,则$\overset{\frown}{CD}$的长为
$\boldsymbol{\frac{6m - π d}{6}}$
(用含$m,d$的式子表示).答案
$\because \overset{\frown}{AB}$的长为$m, \therefore \frac{30 × π × OB}{180} = m, \therefore OB = \frac{6m}{π}$.
$\because BD = d, \therefore OD = \frac{6m}{π} - d, \therefore l_{\overset{\frown}{CD}} = \frac{30 × π × (\frac{6m}{π} - d)}{180} = \frac{6m - π d}{6}$.
$\because BD = d, \therefore OD = \frac{6m}{π} - d, \therefore l_{\overset{\frown}{CD}} = \frac{30 × π × (\frac{6m}{π} - d)}{180} = \frac{6m - π d}{6}$.
6.如图,阴影部分是一广告标志,已知两圆弧所在圆的半径分别为20 cm,10 cm,∠AOB = 120°,求这个广告标志面的周长.(结果保留π)
答案
解:由题意,得 $l_{\mathrm{外边的弧}} = \frac{(360 - 120) · π · 20}{180} = \frac{80π}{3} (\mathrm{cm})$,
$l_{\mathrm{里边的弧}} = \frac{(360 - 120) · π · 10}{180} = \frac{40π}{3} (\mathrm{cm})$,
则 $C_{\mathrm{广告标志面}} = l_{\mathrm{外边的弧}} + l_{\mathrm{里边的弧}} + 2 × (20 - 10) = \frac{80π}{3} + \frac{40π}{3} + 20 = (40π + 20)(\mathrm{cm})$.
答:这个广告标志面的周长为$(40π + 20)\mathrm{cm}$.
$l_{\mathrm{里边的弧}} = \frac{(360 - 120) · π · 10}{180} = \frac{40π}{3} (\mathrm{cm})$,
则 $C_{\mathrm{广告标志面}} = l_{\mathrm{外边的弧}} + l_{\mathrm{里边的弧}} + 2 × (20 - 10) = \frac{80π}{3} + \frac{40π}{3} + 20 = (40π + 20)(\mathrm{cm})$.
答:这个广告标志面的周长为$(40π + 20)\mathrm{cm}$.
7. [2025盐城大丰区一模] 在一个直径为6 cm的圆中，小明画了一个圆心角为$120°$的扇形，则这个扇形的面积为（
A.$π \ \mathrm{cm}^2$
B.$2π \ \mathrm{cm}^2$
C.$3π \ \mathrm{cm}^2$
D.$6π \ \mathrm{cm}^2$
C
）A.$π \ \mathrm{cm}^2$
B.$2π \ \mathrm{cm}^2$
C.$3π \ \mathrm{cm}^2$
D.$6π \ \mathrm{cm}^2$
答案
$\because$ 直径为6 cm,$\therefore$ 半径为3 cm,$\therefore S_{\mathrm{扇形}} = \frac{120 × π × 3^2}{360} = 3π(\mathrm{cm}^2)$.
8. [2026 泰州姜堰区月考] 一个扇形的弧长是$20π \ \mathrm{cm}$，面积是$240π \ \mathrm{cm}^2$，则这个扇形的圆心角等于（
A.$60°$
B.$90°$
C.$120°$
D.$150°$
D
）A.$60°$
B.$90°$
C.$120°$
D.$150°$
答案
设这个扇形的半径为$r$,圆心角是$n°$. $\because$ 弧长是$20π$ cm,面积是$240π \mathrm{ cm}^2, \therefore \frac{1}{2} × 20π × r = 240π$,解得$r = 24$. $\because$ 弧长是$20π$ cm,$\therefore \frac{nπ × 24}{180} = 20π$,解得$n = 150$,即这个扇形的圆心角等于$150°$.
9. [2026泰州二附中月考]如图,在$△ ABC$中,$AB=AC=4$,$∠ C=67.5°$,以$AB$为直径作$\odot O$,交$AC$于点$E$,连接$BE$,则图中阴影部分的面积为
$\boldsymbol{2+π}$
.答案
如图,连接$OE$,则$OE = \frac{1}{2}AB = 2$. $\because AB = AC, ∠ C = 67.5°, \therefore ∠ ABC = ∠ C = 67.5°, \therefore ∠ BAC = 180° - ∠ ABC - ∠ C = 180° - 67.5° - 67.5° = 45°$. $\because AB$是$\odot O$的直径,$\therefore ∠ AEB = 90°$. 又$\because ∠ BAC = 45°, \therefore ∠ ABE = 45°, \therefore ∠ BAC = ∠ ABE, \therefore AE = BE$. $\because \overset{\frown}{AE} = \overset{\frown}{AE}, \therefore ∠ AOE = 2∠ ABE = 90°$.
$\because ∠ AEB = 90°, \therefore AB^2 = AE^2 + BE^2 = 2BE^2, \therefore 4^2 = 2BE^2, \therefore AE = BE = 2\sqrt{2}, \therefore S_{\mathrm{阴影}} = S_{△ BOE} + S_{\mathrm{扇形}OAE} = \frac{1}{2}S_{△ ABE} + S_{\mathrm{扇形}OAE} = \frac{1}{2} × \frac{1}{2} × 2\sqrt{2} × 2\sqrt{2} + \frac{90 × π × 2^2}{360} = 2 + π$,即图中阴影部分的面积为$2 + π$.
10. 如图,在$△ ABC$中,$∠ A=72°,BC=10$,若以$BC$为直径作$\odot O$分别交$AB,AC$于点$M,N$,连接$OM,ON$,则图中阴影部分的面积为________(结果保留$π$).
答案
$\because ∠ A = 72°, ∠ A + ∠ B + ∠ C = 180°, \therefore ∠ B + ∠ C = 108°$. $\because OB = OM = OC = ON = \frac{1}{2}BC = \frac{1}{2} × 10 = 5, \therefore ∠ B = ∠ OMB, ∠ C = ∠ ONC, \therefore ∠ OMB + ∠ ONC = ∠ B + ∠ C = 108°, \therefore ∠ BOM + ∠ CON = 180° - (∠ B + ∠ OMB) + 180° - (∠ C + ∠ ONC) = 360° - (∠ B + ∠ C) - (∠ OMB + ∠ ONC) = 144°, \therefore S_{\mathrm{阴影}} = S_{\mathrm{扇形}OBM} + S_{\mathrm{扇形}OCN} = \frac{144 × π × 5^2}{360} = 10π$.
11.如图,四边形ABCD是⊙O的内接四边形,AB是⊙O的直径,∠B=72°,连接AC.若AB=8,∠DCA=27°,求图中阴影部分的面积.(结果保留π)
答案
解:连接$OD,OC$.
$\because ∠ B = 72°, \therefore ∠ ADC = 108°$.
$\because ∠ DCA = 27°$,
$\therefore ∠ DAC = 180° - 108° - 27° = 45°$,
$\therefore ∠ DOC = 90°$.
又$\because OD = OC, \therefore △ COD$是等腰直角三角形.
$\because AB = 8, \therefore OC = OD = 4$,
$\therefore S_{\mathrm{阴影部分}} = S_{\mathrm{扇形}COD} - S_{△ COD} = \frac{90 × π × 4^2}{360} - \frac{1}{2} × 4^2 = 4π - 8$.
$\because ∠ B = 72°, \therefore ∠ ADC = 108°$.
$\because ∠ DCA = 27°$,
$\therefore ∠ DAC = 180° - 108° - 27° = 45°$,
$\therefore ∠ DOC = 90°$.
又$\because OD = OC, \therefore △ COD$是等腰直角三角形.
$\because AB = 8, \therefore OC = OD = 4$,
$\therefore S_{\mathrm{阴影部分}} = S_{\mathrm{扇形}COD} - S_{△ COD} = \frac{90 × π × 4^2}{360} - \frac{1}{2} × 4^2 = 4π - 8$.
