《2025年综合应用创新题典中点九年级数学全一册浙教版浙江专版》

10. 扇文化是中华优秀传统文化的组成部分,在我国有着深厚的底蕴.如图,某折扇张开的角度为$120^{\circ}$时,扇面面积为S,该折扇张开的角度为$n^{\circ}$时,扇面面积为$S_{n}$,若$m = \frac{S_{n}}{S}$,则m与n关系的图象大致是(C)

答案： C

11. 如图,点C为扇形AOB的半径OB上一点,将$\triangle AOC$沿AC折叠,点O恰好落在$\overset{\frown}{AB}$上的点D处,且$\overset{\frown}{AD}:\overset{\frown}{DB} = 3:1$,若此扇形AOB的面积为$\frac{32}{9}\pi$,则$\overset{\frown}{AB}$的长为(C)

A.$\frac{2}{9}\pi$
B.$\frac{8}{9}\pi$
C.$\frac{16}{9}\pi$
D.$\frac{32}{9}\pi$

答案： C 【点拨】连结OD，则OD = OA，由折叠的性质可得OA = DA，$\therefore \triangle OAD$是等边三角形，$\therefore \angle AOD = 60^{\circ }$．由$\overset{\frown }{AD}:\overset{\frown }{DB}=3:1$，可得$\angle AOD:\angle BOD = 3:1$，$\therefore \angle BOD = 20^{\circ }$，$\therefore \angle AOB = 80^{\circ }$，又$\because S_{扇形AOB}=\frac{80}{360}\pi OA^{2}=\frac{32}{9}\pi$，$\therefore OA = 4$，$\therefore \overset{\frown }{AB}=\frac{80}{180}\pi OA=\frac{16}{9}\pi$．

12. 如图,在$\triangle ABC$中,已知$\angle ABC = 90^{\circ}$,$\angle BAC = 30^{\circ}$,$BC = 1$.将$\triangle ABC$绕点A按逆时针方向旋转$90^{\circ}后得到\triangle AB'C'$,则图中阴影部分的面积为(B)

A.$\frac{\pi}{4}$
B.$\frac{\pi - \sqrt{3}}{2}$
C.$\frac{\pi - \sqrt{3}}{4}$
D.$\frac{\sqrt{3}}{2}\pi$

答案： B 【点拨】因为$\angle ABC = 90^{\circ }$，$\angle BAC = 30^{\circ }$，BC = 1，所以$AB = \sqrt{3}BC = \sqrt{3}$，AC = 2BC = 2．由旋转易得$\angle CAB' = 60^{\circ }$，所以题图中阴影部分的面积$=\frac{90\cdot \pi × 2^{2}}{360}-\frac{1}{2}× 1× \sqrt{3}-\frac{60\cdot \pi × (\sqrt{3})^{2}}{360}=\frac{\pi - \sqrt{3}}{2}$．

13. [2025杭州期末]如图,在$\triangle ABC$中,$AB = AC$,点E在AC上,经过A,B,E三点的圆O交BC于点D,且D是$\overset{\frown}{BE}$的中点.
(1)求证：AB是圆的直径.
(2)连结BE,若$AB = 10$,$\angle C = 75^{\circ}$,求阴影部分的面积.

答案： （1）【证明】连结AD．
∵D是$\overset{\frown }{BE}$的中点，$\therefore \overset{\frown }{BD}=\overset{\frown }{DE}$，$\therefore \angle BAD = \angle CAD$．
∵AB = AC，$\therefore \angle ADB = 90^{\circ }$，$\therefore$AB是圆的直径．（2）【解】连结OE．
∵AB = AC，$\therefore \angle ABC = \angle C = 75^{\circ }$，$\therefore \angle BAC = 180^{\circ } - \angle ABC - \angle C = 30^{\circ }$，$\therefore \angle ABE = 90^{\circ } - 30^{\circ } = 60^{\circ }$，$BE = \frac{1}{2}AB = 5$．$\therefore \angle AOE = 2\angle ABE = 120^{\circ }$，$AE = 5\sqrt{3}$，$\therefore$阴影部分的面积 = 扇形AOE的面积 + $\triangle BOE$的面积$=\frac{120\pi × 5^{2}}{360}+\frac{1}{2}\triangle ABE$的面积$=\frac{25\pi }{3}+\frac{1}{2}× \frac{1}{2}AE\cdot BE$$=\frac{25\pi }{3}+\frac{1}{4}× 5\sqrt{3}× 5$$=\frac{25\pi }{3}+\frac{25\sqrt{3}}{4}$．

14. 如图,在扇形AOB中,$\angle AOB = 120^{\circ}$,半径$OA = 2$,点C是OA上一点,连结BC,沿BC将扇形折叠,使得点A落在BO的延长线上的点D处,连结CD,求图中阴影部分的面积(结果保留$\pi$).

答案： 【解】如图，过点B作$BE\perp AO$，垂足为E，设点O关于BC的对称点为$O'$．
∵$\angle AOB = 120^{\circ }$，$\therefore \angle BO'D = 120^{\circ }$，$\therefore S_{扇形BO'D}=\frac{120\pi × 2^{2}}{360}=\frac{4}{3}\pi$．
∵$O'B = O'D$，$\therefore \angle O'BD = \angle O'DB = 30^{\circ }$，$\therefore \angle OBC = \angle O'BC = \frac{1}{2}\angle O'BD = 15^{\circ }$．
∵$\angle AOB = 120^{\circ }$，$\therefore \angle BOE = 60^{\circ }$，$\therefore \angle EBO = 30^{\circ }$，$\therefore$易得OE = 1，$BE = \sqrt{3}$，$\angle CBE = 45^{\circ }$，$\therefore \triangle BEC$是等腰直角三角形，$\therefore CE = BE = \sqrt{3}$，$\therefore OC = CE - OE = \sqrt{3} - 1$，$\therefore S_{\triangle BOC}=\frac{1}{2}OC\cdot BE=\frac{1}{2}× (\sqrt{3} - 1)× \sqrt{3}=\frac{3 - \sqrt{3}}{2}$，$\therefore S_{阴影部分}=S_{扇形BO'D}-2S_{\triangle BOC}=\frac{4}{3}\pi - (3 - \sqrt{3})=\frac{4}{3}\pi - 3 + \sqrt{3}$．