《2025年通城学典全程测评卷九年级数学全一册苏科版》

19. (8 分)如图,$\odot O的直径AB\perp弦CD$,垂足为$E$,$AE = 2$,$CD = 8$.
(1) 求$\odot O$的半径;
(2) 连接$BC$,过点$O作OF\perp BC于点F$,求$OF$的长.

答案：
(1)连接OD.设⊙O的半径为r.
∵直径AB⊥弦CD,
∴∠OED = 90°,DE = CE = $\frac{1}{2}$CD = $\frac{1}{2}$×8 = 4.在Rt△ODE中,
∵OE = r - 2,OD = r,DE = 4,
∴(r - 2)² + 4² = r²,解得r = 5,即⊙O的半径为5
(2)在Rt△BCE中,
∵CE = 4,BE = AB - AE = 8,
∴BC = $\sqrt{4² + 8²}$ = 4$\sqrt{5}$.
∵OF⊥BC,
∴BF = CF = $\frac{1}{2}$BC = 2$\sqrt{5}$,∠OFB = 90°.在Rt△OBF中,OF = $\sqrt{OB² - BF²}$ = $\sqrt{5² - (2\sqrt{5})²}$ = $\sqrt{5}$,即OF的长为$\sqrt{5}$.

20. (8 分)(2024·扬州邗江期末)如图,在$\triangle ABC$中,$AB = AC$,以$AB为直径的\odot O与BC$、$AC分别交于点D$、$E$,过点$D作DF\perp AC$,垂足为$F$.
(1) 求证:$DF为\odot O$的切线;
(2) 若$\odot O$的半径为 2,$\angle CDF = 22.5^{\circ}$,求涂色部分的面积.
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答案：

(1)如图,连接AD、OD.
∵AB是⊙O的直径,
∴∠ADB = 90°.
∴AD⊥BC.又
∵AB = AC,
∴D是BC的中点.
∴BD = DC.
∵BO = OA,
∴易得DO//AC.又
∵DF⊥AC,
∴DF⊥OD.
∵OD是⊙O的半径,
∴DF为⊙O的切线
(2)如图,连接OE.
∵DF⊥AC,∠CDF = 22.5°,
∴∠ABC = ∠ACB = 67.5°.
∴∠BAC = 45°.
∵OA = OE,
∴∠AOE = 90°.
∵⊙O的半径为2,
∴S扇形AOE = π,S△AOE = 2.
∴S涂色 = S扇形AOE - S△AOE = π - 2