3.（2024·盐城亭湖一模）如图，一个正多边形被树叶遮掩了一部分. 若直线a，b所夹锐角为36°，则这个正多边形的边数是______.

5

4.（2024·扬州）若用半径为10 cm的半圆形纸片围成一个圆锥的侧面，则这个圆锥底面圆的半径为______cm.

5

5.（2024·常州溧阳一模）如图，点A，B，C在半径为9的⊙O上，$\overset{\frown}{AB}$的长为2π，则∠ACB=______°.

20

6.（2024·南京建邺一模）如图，AE，DF是正八边形ABCDEFGH的两条对角线，则$\frac{AE}{DF}$=______.

$\sqrt{2}$

7.（2024·扬州邗江一模）如图，在△ABC中，AB=AC，以AB为直径的⊙O交BC于点D，DE⊥AC，垂足为E.
（1）求证：DE是⊙O的切线；
（2）若∠C=30°，CD=2$\sqrt{3}$，求涂色部分的面积.

(1) 如图，连接$OD$，$AD$.$\because AB$是$\odot O$的直径，$\therefore \angle ADB = 90^{\circ}$.$\therefore AD \perp BC$.$\because AB = AC$，$\therefore BD = CD$.$\because OB = OA$，$\therefore OD$是$\triangle ABC$的中位线.$\therefore OD// AC$.$\because DE \perp AC$，$\therefore OD \perp DE$.$\because OD$是$\odot O$的半径，$\therefore DE$是$\odot O$的切线 (2) $\because AB = AC$，$\angle C = 30^{\circ}$，$\therefore \angle B = \angle C = 30^{\circ}$.$\because OB = OD$，$\therefore \angle B = \angle ODB = 30^{\circ}$.$\therefore \angle AOD = 60^{\circ}$. 由(1)知，$AD \perp BC$，$\therefore \angle ADC = 90^{\circ}$.$\therefore \angle CAD = 60^{\circ}$.$\because DE \perp AC$，$\therefore \angle CED = \angle AED = 90^{\circ}$.$\therefore \angle ADE = 30^{\circ}$.$\because CD = 2\sqrt{3}$，$\therefore DE = \frac{1}{2}CD = \sqrt{3}$.$\therefore AE = \frac{\sqrt{3}}{3}DE = 1$.$\therefore AD = 2AE = 2$.$\because OA = OD$，$\angle AOD = 60^{\circ}$，$\therefore \triangle AOD$是等边三角形.$\therefore OA = AD = 2$.$\therefore S_{涂色}=S_{\triangle AOD}+S_{\triangle ADE}-S_{扇形AOD}=\frac{1}{2} \times 2 \times \sqrt{3}+\frac{1}{2} \times 1 \times \sqrt{3}-\frac{60\pi \times 2^{2}}{360}=\frac{3\sqrt{3}}{2}-\frac{2\pi}{3}$

8.（2023·泰州兴化期中）如图，正六边形ABCDEF内接于⊙O，⊙O的半径为4. 求：
（1）点O到CD的距离；
（2）正六边形ABCDEF的面积.

(1) 如图，连接$OC$，$OD$，过点$O$作$OH \perp CD$于点$H$.$\because$六边形$ABCDEF$是正六边形，$\therefore$易得$\angle COD = 60^{\circ}$.$\because OC = OD$，$OH \perp CD$，$\therefore \angle COH = \frac{1}{2}\angle COD = 30^{\circ}$.$\therefore$在$Rt\triangle OHC$中，$OH = OC\cdot \cos\angle COH = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3}$，即点$O$到$CD$的距离为$2\sqrt{3}$ (2) 正六边形$ABCDEF$的面积为$\frac{1}{2} \times 4 \times 2\sqrt{3} \times 6 = 24\sqrt{3}$