《2025年5年中考3年模拟九年级数学下册沪科版》

14. 鸡爪模型 (2024湖南岳阳期中,18,★★☆)如图,P是等边△ABC内部的一个点,PA = 2,PB = 2√3,PC = 4,则△ABC的边长是________.

答案：
答案：$2\sqrt{7}$ 解析：已知$\triangle ABC$是等边三角形，如图，将$\triangle APC$绕点$A$逆时针旋转$60^{\circ}$到$\triangle AQB$的位置，连接$PQ$，过点$A$作$AM\perp BP$，交$BP$的延长线于点$M$. 由旋转的性质得$AQ = AP = 2$，$\angle PAQ = 60^{\circ}$，$BQ = PC = 4$，$\therefore \triangle APQ$为等边三角形，$\therefore \angle APQ = 60^{\circ}$，$PQ = PA = 2$，$\because PQ^{2}+BP^{2}=2^{2}+(2\sqrt{3})^{2}=4^{2}=BQ^{2}$，$\therefore \angle BPQ = 90^{\circ}$，$\therefore \angle APB = 90^{\circ}+60^{\circ}=150^{\circ}$，$\therefore \angle APM = 30^{\circ}$，$\therefore AM=\frac{1}{2}PA = 1$，$PM=\frac{\sqrt{3}}{2}PA=\sqrt{3}$，$\therefore BM = BP + PM = 3\sqrt{3}$，由勾股定理得，$AB^{2}=BM^{2}+AM^{2}=27 + 1 = 28$，$\therefore AB = 2\sqrt{7}$.

15. 新考向·新定义试题 (2024四川泸州中考,16,★★☆)定义:在平面直角坐标系中,将一个图形先向上平移a(a ＞ 0)个单位,再绕原点按逆时针方向旋转θ角度,这样的图形运动叫做图形的ρ(a,θ)变换.如:点A(2,0)按照ρ(1,90°)变换后得到的点A'的坐标为(-1,2),则点B(√3,-1)按照ρ(2,105°)变换后得到的点B'的坐标为________. (M9224001)

答案：
答案：$(-\sqrt{2},\sqrt{2})$ 解析：根据题意，先将点$B(\sqrt{3},-1)$向上平移$2$个单位，得到点$C(\sqrt{3},1)$，如图，$\therefore CE = 1$，$OE=\sqrt{3}$，$\therefore OC=\sqrt{1^{2}+(\sqrt{3})^{2}} = 2$，$\sin\angle COE=\frac{CE}{OC}=\frac{1}{2}$，$\therefore \angle COE = 30^{\circ}$，再将点$C(\sqrt{3},1)$绕原点按逆时针方向旋转$105^{\circ}$，$\therefore OB' = OC = 2$，$\angle B'OE = 105^{\circ}+30^{\circ}=135^{\circ}$，$\therefore \angle B'OD = 180^{\circ}-135^{\circ}=45^{\circ}$，作$B'D\perp x$轴于点$D$，$\therefore B'D = OD = OB'\cdot\sin45^{\circ}=\sqrt{2}$，$\therefore$点$B'$的坐标为$(-\sqrt{2},\sqrt{2})$.

16. (2022安徽中考,16,★★☆)如图,在由边长为1个单位长度的小正方形组成的网格中,△ABC的顶点均为格点(网格线的交点).
(1)将△ABC向上平移6个单位,再向右平移2个单位,得到△A₁B₁C₁,请画出△A₁B₁C₁.
(2)以边AC的中点O为旋转中心,将△ABC按逆时针方向旋转180°,得到△A₂B₂C₂,请画出△A₂B₂C₂.

答案：
解析： （1）如图，$\triangle A_1B_1C_1$即为所求. （2）如图，$\triangle A_2B_2C_2$即为所求.

17. 推理能力 数形结合思想 如图,在Rt△ABC中,∠ACB = 90°,BC = 8,E为BC上一点,且CE = 3,D为直线AC上一动点,将DE绕点E逆时针旋转90°得到FE,连接BF,CF,则BF + CF的最小值为________.

答案：
答案：$10$ 解析：过$F$作$FM\perp BC$于$M$，如图. $\because$将$DE$绕点$E$逆时针旋转$90^{\circ}$得到$FE$，$\therefore DE = FE$，$\angle DEF = 90^{\circ}$，$\therefore \angle CED = 90^{\circ}-\angle FEM=\angle EFM$，$\because \angle DCE = 90^{\circ}=\angle EMF$，$\therefore \triangle DCE\cong\triangle EMF(AAS)$，$\therefore DC = EM$，$MF = CE = 3$，设$DC = EM = x$，则$CM = x + 3$，$\therefore BM = BC - CM = 8-(x + 3)=5 - x$，$\therefore CF + BF=\sqrt{CM^{2}+MF^{2}}+\sqrt{BM^{2}+MF^{2}}=\sqrt{(x + 3)^{2}+3^{2}}+\sqrt{(5 - x)^{2}+3^{2}}=\sqrt{[x-(-3)]^{2}+[0-(-3)]^{2}}+\sqrt{(x - 5)^{2}+(0 - 3)^{2}}$，而$\sqrt{[x-(-3)]^{2}+[0-(-3)]^{2}}+\sqrt{(x - 5)^{2}+(0 - 3)^{2}}$相当于$x$轴上的点$(x,0)$到点$(-3,-3)$和到点$(5,3)$的距离之和，$\therefore$当点$(x,0)$与$(-3,-3)$，$(5,3)$共线时，$CF + BF$的值最小，最小值即为点$(-3,-3)$和$(5,3)$之间的距离，$\because \sqrt{(-3 - 5)^{2}+(-3 - 3)^{2}} = 10$，$\therefore BF + CF$的最小值为$10$.

1. (2024广西柳州期末)如图,将△ABC绕点A顺时针旋转60°得到△AED,若线段AB = 5,则BE的长度为________.

答案： 答案：$5$ 解析：$\because \triangle ABC$绕点$A$顺时针旋转$60^{\circ}$得到$\triangle AED$，$\therefore AB = AE$，$\angle BAE = 60^{\circ}$，$\therefore \triangle AEB$是等边三角形，$\therefore BE = AB$，$\because AB = 5$，$\therefore BE = 5$.

2. (2024安徽淮南大通期末)如图,把Rt△ABC绕点A逆时针旋转40°,得到Rt△AB'C',点C'恰好落在边AB上,连接BB',则∠BB'C' = ________度.

答案： 答案：$20$ 解析：$\because Rt\triangle ABC$绕点$A$逆时针旋转$40^{\circ}$得到$Rt\triangle AB'C'$，点$C'$恰好落在边$AB$上，$\therefore \angle AC'B'=\angle C = 90^{\circ}$，$\angle BAB' = 40^{\circ}$，$AB = AB'$，$\therefore \angle AB'B=\angle AB'B=\frac{1}{2}\times(180^{\circ}-40^{\circ}) = 70^{\circ}$，$\therefore \angle BB'C' = 90^{\circ}-\angle C'BB' = 20^{\circ}$.

3. 如图,在△ABC中,已知∠ACB = 45°,BC = 1,AB = √5,将△ABC绕点A逆时针旋转得到△ADE,点B与点D对应,点C与点E对应,且C,D,E三点恰好在同一条直线上,则CE的长度为________.

答案：
答案：$4$ 解析：如图，连接$BD$，$\because \triangle ABC$绕点$A$逆时针旋转得到$\triangle ADE$，$\therefore \triangle ABC\cong\triangle ADE$，$\therefore AB = AD$，$AC = AE$，$\angle E=\angle ACB = 45^{\circ}$，$DE = BC = 1$，$\because C$，$D$，$E$三点恰好在同一条直线上，$AC = AE$，$\therefore \angle ACE=\angle E = 45^{\circ}$，$\therefore \angle CAE=\angle BAD = 90^{\circ}$，$\therefore \triangle BAD$是直角三角形三角形，$\therefore BD=\sqrt{AB^{2}+AD^{2}}=\sqrt{10}$，$\because \angle ACB=\angle ACE = 45^{\circ}$，$\therefore \angle BCE = 90^{\circ}$，$\therefore CD=\sqrt{BD^{2}-BC^{2}} = 3$，$\therefore CE = CD + DE = 4$.