第109页 - 通城学典课时作业本九年级数学人教版南通专版

C

$\frac{16\pi}{9}$

 解：由题意，得$\frac{90\pi R}{180}=2\pi r，$

 $\therefore R = 4r.$

 解：设圆锥的高为$h\ cm，$底面圆的半径为$r\ cm，$母线$AC$长为$l\ cm.$

 （1）由题意，得$\frac{180\pi l}{180}=2\pi r，$$\therefore \frac{l}{r}=2，$即圆锥的母线长与底面圆的半径的比值为$2.$

 （2）$\because \frac{l}{r}=2，$$\therefore$易得圆锥的高与母线的夹角为$30^{\circ}.$

 $\because AB = AC，$$AO\perp BC，$$\therefore \angle BAC = 2\times30^{\circ}=60^{\circ}.$

 （3）由题意，可知$l^{2}=h^{2}+r^{2}.$ 又$\because \frac{l}{r}=2，$$h = 3\sqrt{3}，$

 $\therefore (2r)^{2}=(3\sqrt{3})^{2}+r^{2}，$解得$r = 3$（负值舍去）.$\therefore l = 2r = 6.$

 $\therefore$圆锥的侧面积为$\frac{180\pi l^{2}}{360}=18\pi(cm^{2}).$

 解：（1）设$\angle BAC=\alpha.$ 根据题意，得$\overset{\frown}{EF}$的长就是圆锥底面圆的周长，

 $\therefore \frac{\alpha}{180^{\circ}}\times\pi\times AD = ED\times\pi.$ 又$\because ED:AD = 1:2，$$\therefore AD = 2ED.$

 $\therefore \alpha = 90^{\circ}，$即$\angle BAC = 90^{\circ}.$

 （2）$\because$圆锥底面圆的直径$ED$为$5\ cm，$$\therefore AD = 2ED = 10\ cm.$

 $\because \angle BAC = 90^{\circ}，$$AB = AC，$$\therefore \triangle ABC$是等腰直角三角形.

 $\because AD\perp BC，$$\therefore$易得$BC = 2AD = 20\ cm.$

 $\therefore S_{涂色部分}=S_{\triangle ABC}-S_{扇形AEF}=\frac{1}{2}BC\cdot AD-\frac{90\pi\times AD^{2}}{360}=\frac{1}{2}\times20\times10-\frac{90\pi\times10^{2}}{360}=(100 - 25\pi)cm^{2}.$