《2025年零障碍导教导学案九年级数学全一册人教版》

17. (2024·新会区校级期中)如图,有一圆弧形拱桥,它的跨度(所对弦长)为60 m,拱高18 m,当水面涨至其跨度只有30 m时,就要采取紧急措施. 某次洪水来到时,拱顶离水面只有4 m,问是否需要采取紧急措施？
解:设圆的半径为34m.
∵ AB = 60 m，MP = 18 m，
OP ⊥ AB，
∴ AM = $\frac{1}{2}$AB = 30(m)，
OM = OP - MP = (x - 18)(m).
在Rt△OAM中，
OA² = AM² + OM²，
∴ x² = 30² + (x - 18)²，
解得x = 34.
∴ OA' = OP = 34 m.
当PN = 4 m时，
ON = 34 - 4 = 30(m)，
设A'N = y m.
在Rt△OA'N中，
OA'² = A'N² + ON²，
∴ 34² = y² + 30²，
解得y = 16或y = -16(舍去).
∴ A'N = 16 m.
∴ A'B' = 2A'N = 2 × 16 = 32(m).
∵ 32 ＞ 30，
∴ 不需要采取紧急措施.

答案： 解:设圆的半径为$x$m.
$\because AB = 60$m，$MP = 18$m，
$OP \perp AB$，
$\therefore AM = \frac{1}{2}AB = 30(m)$，
$OM = OP - MP = (x - 18)(m)$.
在$Rt\triangle OAM$中，
$OA^{2} = AM^{2} + OM^{2}$，
$\therefore x^{2} = 30^{2} + (x - 18)^{2}$，
解得$x = 34$.
$\therefore OA' = OP = 34$m.
当$PN = 4$m时，
$ON = 34 - 4 = 30(m)$，
设$A'N = y$m.
在$Rt\triangle OA'N$中，
$OA'^{2} = A'N^{2} + ON^{2}$，
$\therefore 34^{2} = y^{2} + 30^{2}$，
解得$y = 16$或$y = -16$(舍去).
$\therefore A'N = 16$m.
$\therefore A'B' = 2A'N = 2 \times 16 = 32(m)$.
$\because 32 ＞ 30$，
$\therefore$不需要采取紧急措施.

18. 如图,在四边形ABCD中,AB//CD,点O在BD上,以点O为圆心的圆恰好经过A,B,C三点,⊙O交BD于点E,交AD于点F,且$\overset{\frown}{AE}= \overset{\frown}{CE}$,连接OA,OF.
(1)求证：四边形ABCD是菱形；
(2)若$\angle AOF = 3\angle FOE$,求$\angle ABC$的度数.

(1)证明:$\because \overset{\frown}{AE} = \overset{\frown}{CE}$，

$\therefore \angle CBD = \angle ABD$.
$\because CD // AB$，
$\therefore \angle ABD = \angle CDB$.
$\therefore \angle CBD = \angle CDB$.
$\therefore CB = CD$.
$\because BE$是$\odot O$的直径，
$\therefore \overset{\frown}{BAE} - \overset{\frown}{AE} = \overset{\frown}{BCE} - \overset{\frown}{CE}$，
$\therefore \overset{\frown}{AB} = \overset{\frown}{BC}$.
$\therefore AB = BC = CD$.
$\because CD // AB$，
$\therefore$四边形ABCD是菱形.
(2)解:$\because \angle AOF = 3\angle FOE$，
设$\angle FOE = x$，
则$\angle AOF = 3x$.
$\therefore \angle AOD = \angle FOE + \angle AOF = 4x$.
$\because OA = OF$，
$\therefore \angle OAF = \angle OFA = \frac{1}{2}(180^{\circ} - 3x)$.
$\because OA = OB$，
$\therefore \angle OAB = \angle OBA$.
$\because \angle AOD = \angle OAB + \angle OBA = 2\angle OAB = 2\angle OBA = 4x$，
$\therefore \angle OAB = \angle OBA = 2x$.
$\because$四边形ABCD是菱形，
$\therefore \angle ABC = 2\angle ABO = 4x$，
$BC // AD$.
$\therefore \angle ABC + \angle BAD = 180^{\circ}$.
$\therefore 4x + 2x + \frac{1}{2}(180^{\circ} - 3x) = 180^{\circ}$.
解得$x = 20^{\circ}$.
$\therefore \angle ABC = 4x =$$80^{\circ}$.

答案：
(1)证明:$\because \overset{\frown}{AE} = \overset{\frown}{CE}$，
$\therefore \angle CBD = \angle ABD$.
$\because CD // AB$，
$\therefore \angle ABD = \angle CDB$.
$\therefore \angle CBD = \angle CDB$.
$\therefore CB = CD$.
$\because BE$是$\odot O$的直径，
$\therefore \overset{\frown}{BAE} - \overset{\frown}{AE} = \overset{\frown}{BCE} - \overset{\frown}{CE}$，
$\therefore \overset{\frown}{AB} = \overset{\frown}{BC}$.
$\therefore AB = BC = CD$.
$\because CD // AB$，
$\therefore$四边形ABCD是菱形.
(2)解:$\because \angle AOF = 3\angle FOE$，
设$\angle FOE = x$，
则$\angle AOF = 3x$.
$\therefore \angle AOD = \angle FOE + \angle AOF = 4x$.
$\because OA = OF$，
$\therefore \angle OAF = \angle OFA = \frac{1}{2}(180^{\circ} - 3x)$.
$\because OA = OB$，
$\therefore \angle OAB = \angle OBA$.
$\because \angle AOD = \angle OAB + \angle OBA = 2\angle OAB = 2\angle OBA = 4x$，
$\therefore \angle OAB = \angle OBA = 2x$.
$\because$四边形ABCD是菱形，
$\therefore \angle ABC = 2\angle ABO = 4x$，
$BC // AD$.
$\therefore \angle ABC + \angle BAD = 180^{\circ}$.
$\therefore 4x + 2x + \frac{1}{2}(180^{\circ} - 3x) = 180^{\circ}$.
解得$x = 20^{\circ}$.
$\therefore \angle ABC = 4x = 80^{\circ}$.