12.(2025·重庆月考)设$x$,$y$为实数,且$y = 21-\sqrt{3x - 18}-\sqrt{18 - 3x}$,则$x + y$的立方根是______.

3

13. 如图,点$O在\triangle ABC$内且到三边的距离相等.若$\angle A = 58^{\circ}$,则$\angle BOC= $______$^{\circ}$.

119

14.(2024·泉州月考)如图,在$\triangle ABC$中,$\angle BAC = 90^{\circ}$,$AB = 15$,$AC = 20$,$AD\perp BC$,垂足为$D$,则$AD + BD= $______.

21

15.(2025·无锡期末)如图,$\angle B = 45^{\circ}$,$BC = 2$,在射线$BM上取一点A$,设$AC = d$,若对于$d$的一个数值,只能作出唯一个$\triangle ABC$,则$d$的取值范围是______.

$d = \sqrt{2}$或d≥2　解析:由题意可知,当CA⊥BA时,∵BC=2,∴AC=AB=$\sqrt{2}$,即$d = \sqrt{2}$时,能作出唯一一个△ABC;当CA=BC时,∵∠B=45°,BC=2,∴AC=BC=2,即d≥2时能作出唯一一个△ABC.综上所述,当$d = \sqrt{2}$或d≥2时能作出唯一一个△ABC,故$d = \sqrt{2}$或d≥2.

16.(2024·湛江月考)如图,已知$\angle MON = 30^{\circ}$,点$A_1$,$A_2$,$A_3$,…在射线$ON$上,点$B_1$,$B_2$,$B_3$,…在射线$OM$上,$\triangle A_1B_1A_2$,$\triangle A_2B_2A_3$,$\triangle A_3B_3A_4$,…均为等边三角形,若$OA_1 = 2$,则$\triangle A_{2024}B_{2024}A_{2025}$的边长为______.

$2^{2024}$　解析:由$OA_{1}=2,∠MON = 30°,△A_{1}B_{1}A_{2}$为等边三角形,可求得$△A_{1}B_{1}A_{2}$的边长$A_{1}B_{1}=OA_{1}=2$,同理,$△A_{2}B_{2}A_{3}$的边长$A_{2}B_{2}=OA_{2}=2×2 = 2^{2}$,$△A_{3}B_{3}A_{4}$的边长$A_{3}B_{3}=OA_{3}=2^{2}×2 = 2^{3}$,…,从而得$△A_{n}B_{n}A_{n + 1}$的边长为$2^{n}$,∴$△A_{2024}B_{2024}A_{2025}$的边长为$2^{2024}$.

17. 在$\triangle ABC$中,$DE垂直平分AB$,分别交$AB$,$BC于点D$,$E$,$MN垂直平分AC$,分别交$AC$,$BC于点M$,$N$.若$\angle BAC= \alpha(\alpha\neq90^{\circ})$,则$\angle EAN= $______(用含$\alpha的代数式表示\angle EAN$的大小).

$180° - 2α$或$2α - 180°$　解析:如图①,在△ABC中,∠B + ∠C =$180° - ∠BAC = 180° - α$.∵DE垂直平分AB,∴AE = BE,∴∠BAE = ∠B.同理可得∠CAN = ∠C,∴∠EAN = ∠BAC - ∠BAE - ∠CAN =$α - (180° - α) = 2α - 180°$;如图②,∵DE垂直平分AB,∴AE = BE,∴∠BAE = ∠B.同理可得∠CAN = ∠C,∴∠EAN = ∠BAE + ∠CAN - ∠BAC =$(180° - α) - α = 180° - 2α$.综上所述,∠EAN的度数为$180° - 2α$或$2α - 180°$.

18.(盐城中考改编)如图,在长方形$ABCD$中,$AB = 3$,$AD = 4$,$E$,$F分别是边BC$,$CD$上一点,$EF\perp AE$,将$\triangle ECF沿EF翻折得到\triangle EC'F$,连接$AC'$,当$BE= $______时,$\triangle AEC'是以AE$为腰的等腰三角形.

$\frac{7}{8}$或$\frac{4}{3}$　解析:设BE = x,则CE = EC' = 4 - x.①当AE = EC'时,AE = 4 - x.∵四边形ABCD是长方形,∴∠B = 90°.在Rt△ABE中,$AB^{2}+BE^{2}=AE^{2}$,即$3^{2}+x^{2}=(4 - x)^{2}$,解得$x = \frac{7}{8}$,∴$BE = \frac{7}{8}$.②当AE = AC'时,如图①,过点A作AG⊥EC'于点G,则CE = EC' = 2EG.∵AE⊥EF,∴∠AEF = 90°,∴∠AEC' + ∠C'EF = 90°,∠AEB + ∠CEF = 90°.由折叠得∠C'EF = ∠CEF,∴∠AEC' = ∠AEB.在△ABE和△AGE中,$\left\{\begin{array}{l} ∠AEB = ∠AEG,\\ ∠B = ∠AGE,\\ AE = AE,\end{array}\right.$∴△ABE≌△AGE(AAS),∴BE = EG = x,∴CE = EC' = 2x,∴BC = BE + EC = x + 2x = 4,解得$x = \frac{4}{3}$,即$BE = \frac{4}{3}$.综上可知,BE的长为$\frac{7}{8}$或$\frac{4}{3}$.

一题多解
当AE = AC'时,也可利用下面的方法求解:如图②,延长EB到点H,使EH = EC'.∵AE⊥EF,∴∠AEF = 90°,∴∠AEC' + ∠C'EF = 90°,∠AEB + ∠CEF = 90°.由折叠得∠C'EF = ∠CEF,∴∠AEC' = ∠AEB.在△AEH和△AEC'中,$\left\{\begin{array}{l} AE = AE,\\ ∠AEB = ∠AEC',\\ EH = EC',\end{array}\right.$∴△AEH≌△AEC'(SAS),∴AH = AC'.又∵AE = AC',∴AH = AE.∵AB⊥HE,∴EH = 2BE = 2x.由折叠得EC' = EC = 4 - x,∴EH = 4 - x,∴2x = 4 - x,解得$x = \frac{4}{3}$,即$BE = \frac{4}{3}$.

19.(6分)(2025·泰州期末)
(1)计算：$\sqrt{9}+(-2)^2+\sqrt[3]{-64}$；
(2)求$(x - 1)^2 = 9中x$的值.

(1)原式=3+4+(−4)=3.
(2)∵$(x - 1)^{2}=9$,∴x - 1 = 3 或 x - 1 = -3,解得x = 4 或 x = -2.

20.(6分)(2023·苏州中考)如图,在$\triangle ABC$中,$AB = AC$,$AD为\triangle ABC$的角平分线.以点$A$为圆心,$AD$长为半径画弧,与$AB$,$AC分别交于点E$,$F$,连接$DE$,$DF$.
(1)求证：$\triangle ADE\cong\triangle ADF$；
(2)若$\angle BAC = 80^{\circ}$,求$\angle BDE$的度数.

(1)∵AD是△ABC的角平分线,∴∠BAD = ∠CAD.由作图知AE = AF.在△ADE和△ADF中,$\left\{\begin{array}{l} AE = AF,\\ ∠BAD = ∠CAD,\\ AD = AD,\end{array}\right.$∴△ADE≌△ADF(SAS).
(2)∵∠BAC = 80°,AD为△ABC的角平分线,∴∠EAD =$\frac{1}{2}∠BAC = 40°$.由作图知,AE = AD,∴∠AED = ∠ADE,∴∠ADE =$\frac{1}{2}×(180° - 40°)=70°$.∵AB = AC,∴AD⊥BC,∴∠BDE = 90° - ∠ADE = 20°.