《2026年启东中学作业本九年级数学下册苏科版》

8. (2024·苏州期中)如图，在$\mathrm{Rt}△ ABC$中，$∠ ACB = 90^{\circ}$，$CD$是$AB$边上的高，$CE$是$AB$边上的中线，$AD = 3$，$CE = 5$，则$\tan ∠ BCE$的值为(B)

A.$\dfrac{1}{2}$
B.$\dfrac{\sqrt{21}}{7}$
C.$\dfrac{\sqrt{5}}{5}$
D.$\dfrac{\sqrt{30}}{10}$

答案： 8. B

9. 如图，在$△ ABC$中，$AD$是$BC$边上的高，$DC = 1$，$BD = 2$，$\tan B = \cos ∠ DAC$，则$AB$的长为√7.

答案： 9. √7

10. 在$△ ABC$中，$AB = 3\sqrt{6}$，$AC = 6$，$∠ B = 45^{\circ}$，则$BC =$3√3 + 3或3√3 - 3.

答案： 10. 3√3 + 3或3√3 - 3

11. 如图，在四边形$ABCD$中，$∠ ABC = 90^{\circ}$，$∠ ADC = 90^{\circ}$，$AB = 6$，$CD = 4$，$BC$的延长线与$AD$的延长线交于点$E$.
(1)若$∠ A = 60^{\circ}$，求$BC$的长；（结果保留根号）
(2)若$\sin A = \dfrac{4}{5}$，求$AD$的长.

答案： 11. 解:
(1)
∵∠A = 60°, ∠ABE = 90°, AB = 6, tanA = BE/AB,
∴∠E = 30°, BE = 6·tan60° = 6√3.
又
∵∠CDE = 90°, CD = 4, sinE = CD/CE, ∠E = 30°,
∴CE = 4/(1/2) = 8,
∴BC = BE - CE = 6√3 - 8.
(2)
∵∠ABE = 90°, AB = 6, sinA = 4/5 = BE/AE,
∴设BE = 4x, 则AE = 5x, 由勾股定理, 得AB = 3x,
∴3x = 6, 解得x = 2,
∴BE = 8, AE = 10,
∴tanE = AB/BE = 6/8 = CD/DE = 4/DE, 解得DE = 16/3,
∴AD = AE - DE = 10 - 16/3 = 14/3.

12. (2023·雅安)如图，在四边形$ABCD$中，$AB = AD$，$BC = DC$，$∠ C = 60^{\circ}$，$AE // CD$交$BC$于点$E$，$BC = 8$，$AE = 6$，求$AB$的长.

答案：
12. 解: 如答图, 连接AC, BD交于点O, 过点E作EF⊥AC, 交AC于点F.
∵BC = DC, ∠BCD = 60°,
∴△BCD是等边三角形,
∴BD = BC = CD = 8.
∵AB = AD, BC = DC,
∴AC⊥BD, BO = DO = 1/2BD = 4,
∴∠ACD = ∠ACB = 1/2∠BCD = 30°.
∵AE//CD,
∴∠EAC = ∠ACD = ∠ACB = 30°,
∴AE = EC = 6,
∴CF = CE·cos30° = 6×√3/2 = 3√3,
AF = AE·cos30° = 6×√3/2 = 3√3, CO = BC·cos30° = 8×√3/2 = 4√3,
∴AC = CF + AF = 6√3,
∴AO = AC - CO = 6√3 - 4√3 = 2√3.
在Rt△BOA中,
AB = √(BO² + AO²) = √(4² + (2√3)²) = 2√7.