《2025年北大绿卡九年级数学下册人教版》

16. 如图，一副三角板拼成如图所示的图形. 若AB = 2\sqrt{2}，则BC的长为( )．

A. 2\sqrt{2}
B. 3
C. 2 + \frac{2\sqrt{3}}{3}
D. 2\sqrt{3} + 2

答案：
C 【解析】如图，过$A$作$AD\perp BC$于$D$.

$\because\angle ABD = 45^{\circ}$，$AB = 2\sqrt{2}$，
$\therefore AD = BD = \frac{\sqrt{2}}{2}\times2\sqrt{2} = 2$.
$\because\angle ACB = 60^{\circ}$，
$\therefore CD = \frac{AD}{\tan 60^{\circ}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$，
$\therefore BC = BD + CD = 2 + \frac{2\sqrt{3}}{3}$.
故选 C.

17. 如图所示，在半径为3的⊙O中，直径AB与弦CD相交于点E，连接AC，BD，若AC = 2，则tan D = ________．

答案： $2\sqrt{2}$ 【解析】连接$BC$.$\because AB$是$\odot O$的直径，
$\therefore\angle ACB = 90^{\circ}$.$\because AB = 6$，$AC = 2$，
$\therefore BC = \sqrt{AB^{2} - AC^{2}} = \sqrt{6^{2} - 2^{2}} = 4\sqrt{2}$.
又$\because\angle D = \angle A$，
$\therefore\tan D = \tan A = \frac{BC}{AC} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

18. 如图，在矩形ABCD中，BC = \sqrt{2}AB，O为BC中点，OE = AB = 4，则扇形EOF的面积为________．

答案： $4\pi$ 【解析】$\because BC = \sqrt{2}AB$，$AB = 4$，$\therefore BC = 4\sqrt{2}$.
$\because O$为$BC$中点，
$\therefore OB = OC = \frac{1}{2}BC = 2\sqrt{2}$.
$\because OE = 4$，
在$Rt\triangle OBE$中，$\cos\angle BOE = \frac{OB}{OE} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$，
$\therefore\angle BOE = 45^{\circ}$.
同理$\angle COF = 45^{\circ}$.
$\therefore\angle EOF = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$，
$\therefore$扇形$EOF$的面积为$\frac{90\pi\cdot4^{2}}{360} = 4\pi$.
故答案为$4\pi$.

19. 在△ABC中，AC = 4\sqrt{2}，BC = 6，∠C为锐角且tan C = 1.
(1)求△ABC的面积；
(2)求AB的值；
(3)求cos ∠ABC的值．

答案：
【解】
(1)过点$A$作$AD\perp BC$，垂足为$D$，

则$\angle ADC = \angle ADB = 90^{\circ}$.
$\because\angle C$为锐角且$\tan C = 1$，$\therefore\angle C = 45^{\circ}$，
$\therefore\angle DAC = 90^{\circ} - \angle C = 45^{\circ}$，
$\therefore\angle DAC = \angle C = 45^{\circ}$，$\therefore AD = DC$.
在$Rt\triangle ACD$中，
$\because\sin C = \frac{AD}{AC}$，$AC = 4\sqrt{2}$，
$\therefore DC = AD = AC\cdot\sin C = 4\sqrt{2}\times\frac{\sqrt{2}}{2} = 4$.
$\because BC = 6$，
$\therefore S_{\triangle ABC} = \frac{1}{2}BC\cdot AD = \frac{1}{2}\times6\times4 = 12$.
(2)$\because DC = AD = 4$，$BC = 6$，
$\therefore BD = BC - DC = 6 - 4 = 2$.
在$Rt\triangle ABD$中，
$AB = \sqrt{AD^{2} + BD^{2}} = \sqrt{4^{2} + 2^{2}} = 2\sqrt{5}$.
(3)$\because$在$Rt\triangle ABD$中，$AB = 2\sqrt{5}$，$BD = 2$，
$\therefore\cos\angle ABC = \frac{BD}{AB} = \frac{2}{2\sqrt{5}} = \frac{\sqrt{5}}{5}$.