《2025年名师面对面学科素养评价九年级数学全一册浙教版》

8. 如图，$AB$是$\odot O$的直径，$CD$是弦，点$C$，$D$在直径$AB$的两侧.若$\angle AOC:\angle AOD:\angle DOB = 2:7:11$，$CD = 4$，则$\overset{\frown}{CD}$的长为(D)


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

答案： 8.D

9. 如图，已知$\triangle ABC$是$\odot O$的内接三角形，$AD$是$\odot O$的直径，连结$BD$，$BC$平分$\angle ABD$.
(1) 求证：$\angle CAD = \angle ABC$.
(2) 若$AD = 6$，求$\overset{\frown}{CD}$的长.

答案： 9.解：
(1)
∵$BC$平分$\angle ABD$，$\therefore\angle DBC = \angle ABC$，
∵$\angle CAD = \angle DBC$，$\therefore\angle CAD = \angle ABC$.
(2)
∵$\angle CAD = \angle ABC$，$\therefore\widehat{CD}=\widehat{AC}$，
∵$AD$是$\odot O$的直径，
$AD = 6$，$\therefore\widehat{CD}$的长$=\frac{1}{2}×\frac{1}{2}×\pi×6=\frac{3}{2}\pi$.

10. 如图，在矩形$ABCD$中，点$F$在边$BC$上，且$AF = AD$，过点$D$作$DE\perp AF$，垂足为$E$.
(1) 求证：$DE = AB$.
(2) 以$D$为圆心、$DE$为半径作圆弧交$AD$于点$G$.若$BF = FC = 1$，试求$\overset{\frown}{EG}$的长.

答案： 10.解：
(1)证明：
∵$DE\perp AF$，$\therefore\angle AED = 90^{\circ}$.
又
∵四边形$ABCD$是矩形，$\therefore AD// BC$，$\angle B = 90^{\circ}$.
$\therefore\angle DAE = \angle AFB$，$\angle AED = \angle B = 90^{\circ}$.
又
∵$AF = AD$，$\therefore\triangle ADE\cong\triangle FAB(AAS)$，$\therefore DE = AB$.
(2)
∵$BF = FC = 1$，$\therefore AD = BC = BF + FC = 2$.
又
∵$\triangle ADE\cong\triangle FAB$，$\therefore AE = BF = 1$.
$\therefore$在$Rt\triangle ADE$中，$AE=\frac{1}{2}AD$，$\therefore\angle ADE = 30^{\circ}$.
又
∵$DE = \sqrt{AD^{2}-AE^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}$，
$\therefore\widehat{EG}=\frac{n\pi R}{180}=\frac{30\cdot\pi\cdot\sqrt{3}}{180}=\frac{\sqrt{3}}{6}\pi$.