《2025年千里马单元测试卷八年级数学下册人教版》

3. 如图,长方形ABCD中,AB = 8,BC = 6,P为AD上一点,将△ABP沿BP翻折至△EBP,PE与CD相交于点O,且OE = OD,BE与CD相交于点F.
(1)求证:OP = OF;
(2)求AP的长.

答案：
(1)证明:$\because$四边形$ABCD$是长方形,
$\therefore \angle D=\angle A=\angle C = 90^{\circ},AD = BC = 6,CD = AB = 8$.
由翻折的性质可知$EP = AP,\angle E=\angle A = 90^{\circ},BE = AB = 8$.
在$\triangle ODP$和$\triangle OEF$中,$\begin{cases}\angle D=\angle E = 90^{\circ},\\OD = OE,\\\angle DOP=\angle EOF,\end{cases}$
$\therefore \triangle ODP\cong\triangle OEF(ASA)$,
$\therefore OP = OF$.
(2)解:$\because \triangle ODP\cong\triangle OEF$,
$\therefore OP = OF,PD = EF$.
$\because OE = OD$,
$\therefore DF = EP$.
设$AP = EP = DF = x$,
则$PD = EF = 6 - x,CF = 8 - x$,
$\therefore BF = 8-(6 - x)=2 + x$.
在$Rt\triangle FCB$中,根据勾股定理,得$BC^{2}+CF^{2}=BF^{2}$,
$\therefore 6^{2}+(8 - x)^{2}=(x + 2)^{2}$,
解得$x = 4.8$,
$\therefore AP = 4.8$.

4. 在△ABC中,∠C = 90°,AC = 6,BC = 8,D,E分别是斜边AB和直角边CB上的点,把△ABC沿着直线DE折叠,顶点B的对应点是B'.
(1)如图①,如果点B'和顶点A重合,求CE的长;
(2)如图②,如果B'是AC的中点,求CE的长.

答案： 解:
(1)设$CE = x$,则$BE = 8 - x$.
由题意,得$AE = BE = 8 - x$.
由勾股定理,得$x^{2}+6^{2}=(8 - x)^{2}$,
解得$x=\frac{7}{4}$,即$CE$的长为$\frac{7}{4}$.
(2)$\because B'$是$AC$的中点,
$\therefore CB'=\frac{1}{2}AC = 3$.
设$CE = y$,类比
(1)中的解法,可列出方程
$y^{2}+3^{2}=(8 - y)^{2}$,解得$y=\frac{55}{16}$,
即$CE$的长为$\frac{55}{16}$.