1. [2026 洛阳二模] 如图,在边长为1的正方形网格中,O,A,B,C,D是网格线的交点.若扇形OAB与扇形OCD所在圆的圆心都为点O,则阴影部分的面积为 ()

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

1. 由题意得∠COD = 90°,∠COB = 45°,OB = BD = 2. 由勾股定理得OC = $2\sqrt{2}$,$\therefore S_{阴影}=S_{扇形OCD}-S_{扇形OEB}-S_{△ OBD} = \frac{90 × π × (2\sqrt{2})^2}{360} - \frac{45 × π × 2^2}{360} - \frac{1}{2} × 2 × 2 = \frac{3}{2}π -2$.

2. [2026苏州高新实验中学考]如图,在扇形AOB中,∠AOB=30°,OA=2√3,点C在OB上,且OC=AC,延长CB到D,使CD=CA.以CA,CD为邻边作平行四边形ACDE,则图中阴影部分的面积为(结果保留π).

2. $3\sqrt{3} - π$ 如图,过点A作AH⊥OD于点H,则∠AHC = 90°. $\because ∠ AOB = 30°, OA = 2\sqrt{3}, \therefore AH = \frac{1}{2}OA = \sqrt{3}. \because OC = AC, \therefore ∠ OAC = ∠ AOB = 30°, \therefore ∠ ACB = ∠ OAC + ∠ AOB = 30° + 30° = 60°. \because ∠ AHC = 90°, \therefore ∠ CAH = 30°, \therefore AC = 2CH.$
设$CH = x$,则$AC = 2x$. 在$Rt△ ACH$中,由勾股定理得$x^2 + (\sqrt{3})^2 = (2x)^2$,解得$x_1 = 1, x_2 = -1$(不符合题意,舍去),$\therefore CH = 1, AC = 2, \therefore CD = CA = OC = 2, \therefore S_{阴影} = S_{△ AOC} + S_{□ ACDE} - S_{扇形OAB} = \frac{1}{2} × 2 × \sqrt{3} + 2 × \sqrt{3} - \frac{30 × π × (2\sqrt{3})^2}{360} = 3\sqrt{3} - π$.

3. [2025盐城东台期中]如图，一个扇形纸片的圆心角为$90°$，半径为4. 将这张扇形纸片折叠，使点A与点O恰好重合，折痕为CD，图中阴影为重合部分，则阴影部分的面积为.(结果保留π)

3. $\frac{8}{3}π - 2\sqrt{3}$ 连接OD. 在$Rt△ OCD$中,$OC = \frac{1}{2}OA = 2$,$OD = 4, \therefore ∠ ODC = 30°, CD = \sqrt{OD^2 - OC^2} = 2\sqrt{3}, \therefore ∠ COD = 60°, \therefore S_{阴影} = S_{扇形OAD} - S_{△ COD} = \frac{60 × π × 4^2}{360} - \frac{1}{2} × 2 × 2\sqrt{3} = \frac{8}{3}π - 2\sqrt{3}$.

4. [2025徐州铜山区清华中学月考]如图,正方形ABCD的边长为2,O为对角线的交点,点E,F分别为BC,AD的中点.以点C为圆心,2为半径作$\overset{\frown}{BD}$,再分别以点E,F为圆心,1为半径作$\overset{\frown}{BO}$,$\overset{\frown}{OD}$,则图中阴影部分的面积为.(结果保留π)

4. $π-2$ 如图,连接BD,则$S_{阴影} = S_{扇形CBD} - S_{△ BCD} = \frac{90}{360} × π × 2^2 - \frac{1}{2} × 2 × 2 = π - 2$.

5. [2026无锡二泉中学月考]如图,已知点C是半圆O上一点,将$\overset{\frown}{BC}$沿弦BC折叠后恰好经过点O.若半圆O的半径是2,则图中阴影部分的面积是.

5. $\frac{2}{3}π$ 如图,过点O作$OD ⊥ BC$于点D,延长OD交$\overset{\frown}{BC}$于点E,连接OC,OE. $\because OD ⊥ BC, \therefore$ 点E是$\overset{\frown}{BC}$的中点. 由折叠得点O为$\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{OC} = \overset{\frown}{OB}, \therefore S_{阴影} = S_{扇形OAC}$. 由折叠得CB垂直平分OE,$\therefore OC = CE, \therefore OC = CE = OE, \therefore △ OCE$是等边三角形,$\therefore ∠ COE = 60°$. 同理可得$∠ EOB = 60°, \therefore ∠ AOC = 180° - ∠ COE - ∠ EOB = 60°, \therefore S_{阴影} = \frac{60 × π × 2^2}{360} = \frac{2}{3}π$.

6. [2025 连云港灌南一模] 如图,在扇形 OAB 中,∠AOB = 90°,以 OA 为直径在扇形 OAB 内部作半圆,圆心为点 E,C 为$\overset{\frown}{AB}$的中点,连接 OC 交半圆 E 于点 D.若 OA = 2,则阴影部分的面积为.

6. $\frac{π-2}{2}$ 如图,连接AD. $\because ∠ AOB = 90°$,点C是$\overset{\frown}{AB}$的中点,$\therefore ∠ BOC = ∠ AOC = 45°. \because OA$是半圆E的直径,$\therefore ∠ ADO = 90°, \therefore ∠ DAO = ∠ DOA = 45°, \therefore AD = OD = \frac{OA}{\sqrt{2}} = \sqrt{2}, \therefore \overset{\frown}{AD} = \overset{\frown}{OD}, \therefore S_{阴影} = S_{扇形OAC} - S_{△ AOD} = \frac{45 × π × 2^2}{360} - \frac{1}{2} × \sqrt{2} × \sqrt{2} = \frac{π-2}{2}$.

7. 如图,在半径为10的扇形OAB中,∠AOB = 90°,C为$\overset{\frown}{AB}$上一点,CD⊥OA,CE⊥OB,垂足分别为D,E.若∠CDE为36°,则图中阴影部分的面积为 ()

A.$10π$
B.$9π$
C.$8π$
D.$6π$

7. A 如图,连接OC. $\because ∠ AOB = 90°, CD ⊥ OA, CE ⊥ OB, \therefore$ 四边形CDOE是矩形,$\therefore CD // OE, \therefore ∠ DEO = ∠ CDE = 36°$. 易知$S_{△ DOE} = S_{△ CEO}, ∠ COB = ∠ DEO = 36°, \therefore S_{阴影} = S_{扇形OBC} = \frac{36 × π × 10^2}{360} = 10π$.

8.如图,半圆O的直径AB=40,C,D是半圆上的三等分点,E是OA的中点,则阴影部分的面积等于.(结果保留π)

8. $\frac{200π}{3}$ 如图,连接OC,OD,CD.
$\because C,D$是半圆上的三等分点,
$\therefore ∠ AOC = ∠ COD = ∠ BOD = 60°$.
$\because OC = OD, \therefore △ OCD$为等边三角形,$\therefore ∠ OCD = 60°$,
$\therefore ∠ OCD = ∠ AOC, \therefore CD // AB, \therefore S_{△ ECD} = S_{△ OCD}, \therefore S_{阴影} = S_{扇形OCD} = \frac{60π × 20^2}{360} = \frac{200π}{3}$.