《2026年启东中学作业本九年级数学下册苏科版宿迁专版》

8. 如图,边长分别为10,6,4的三个正方形拼接在一起,它们的底边在同一直线上,则阴影部分的面积为15.

答案： 8.15

9. (10分)如图,在$\odot O$中,$AB$是$\odot O$的直径,弦$CD$交$AB$于点$E$,$\overset{\frown}{AD}=\overset{\frown}{BD}$.

(1) 求证:$\triangle ACD \sim \triangle ECB$;
(2) 若$AC = 3$,$BC = 1$,求$CE$的长.

答案：
9.
(1)证明:
∵$\overset{\frown}{AD}=\overset{\frown}{BD}$,
∴∠ACD = ∠BCE;
∵∠ADC = ∠EBC,
∴△ACD∽△ECB.
(2)解:过点B作BH⊥CD于点H,如答图.

　由
(1)知∠ACD = ∠BCD.
∵$\overset{\frown}{AD}=\overset{\frown}{BD}$,
∴AD = BD.
∵AB是⊙O的直径，
∴∠ACB = ∠ADB = 90°,
∴∠BCD = $\frac{1}{2}$∠ACB = 45°.
　在Rt△ACB中,AB = $\sqrt{BC^2 + AC^2}$ = $\sqrt{1^2 + 3^2}$ = $\sqrt{10}$,
∴在Rt△ABD中,BD = $\frac{\sqrt{2}}{2}$AB = $\frac{\sqrt{2}}{2}$×$\sqrt{10}$ = $\sqrt{5}$.
　在Rt△BCH中，
∵∠BCH = 45°,
∴CH = BH = $\frac{\sqrt{2}}{2}$BC = $\frac{\sqrt{2}}{2}$.
　在Rt△BDH中,
DH = $\sqrt{BD^2 - BH^2}$ = $\sqrt{(\sqrt{5})^2 - (\frac{\sqrt{2}}{2})^2}$ = $\frac{3\sqrt{2}}{2}$,
∴CD = CH + DH = $\frac{\sqrt{2}}{2}$ + $\frac{3\sqrt{2}}{2}$ = 2$\sqrt{2}$,
　由
(1)知△ACD∽△ECB,
∴CA:CE = CD:CB,
　即3:CE = 2$\sqrt{2}$:1,解得CE = $\frac{3\sqrt{2}}{4}$.

10. (14分)如图,$\triangle ABC$是等边三角形,点$D$,$E$分别在边$BC$,$AC$上,且$BD = CE$,$AD$与$BE$相交于点$F$.

(1) $\triangle AEF$与$\triangle ABE$相似吗? 说明理由.
(2) 求证:$BD^2 = AD · DF$.

答案：
10.
(1)解:△AEF与△ABE相似,理由如下:
　如答图，
∵△ABC为等边三角形，
∴AB = BC,∠ABC = ∠C = ∠BAC = 60°.
　在△ABD和△BCE中，$\begin{cases} AB = BC, \\ \angle ABD = \angle C, \\ BD = CE, \end{cases}$
∴△ABD≌△BCE(SAS),
∴∠1 = ∠2.
∵∠AFE = ∠2 + ∠3,
∴∠AFE = ∠1 + ∠3 = 60°,
∴在△AEF和△BEA中,∠AEF = ∠BEA,∠AFE = ∠BAE = 60°,
∴△AEF∽△BEA.
(2)证明:
∵在△ABD和△BFD中,∠BDF = ∠ADB,∠1 = ∠2,
∴△ABD∽△BFD,
∴$\frac{BD}{AD}$ = $\frac{FD}{BD}$,
∴$BD^2$ = AD·DF.
　　　　　　　

11. (20分)(2024·高邮期末)如图,在矩形$ABCD$中,$AB = 6$,$BC = 8$,点$P$在射线$AD$上运动,以$BP$为直角边向右作$Rt\triangle PBQ$,使得$\angle BPQ = 90°$,$BP = 2PQ$,连接$CQ$.
(1) 若$\triangle ABP$与$\triangle BPQ$相似,则$AP =$3或12;
(2) 当$AP = 2$时,求$CQ$的长.

答案：
11.
(1)3或12
(2)解:如答图,过点Q作MN⊥AD于点M,与BC交于点N,
　则∠A = ∠PMQ = ∠CNQ = 90°,AB = MN = 6.
∵∠BPQ = 90°,
∴∠APB + ∠MPQ = ∠MPQ + ∠PQM = 90°,
∴∠APB = ∠MQP,
∴△APB∽△MQP,
∴$\frac{AP}{MQ}$ = $\frac{AB}{MP}$ = $\frac{BP}{PQ}$.
　设MQ = x,则NQ = 6 - x,
∵BP = 2PQ,AP = 2,
∴$\frac{2}{x}$ = $\frac{6}{PM}$ = 2,
∴x = 1,MP = 3,
∴CN = DM = AD - MP - AP = 8 - 3 - 2 = 3,
∴$CQ^2$ = $QN^2$ + $CN^2$ = $5^2$ + $3^2$ = 34,
∴CQ = $\sqrt{34}$.