《2025年千里马单元测试卷八年级数学下册人教版》

1. 如图,在长方形纸片ABCD中,点E在BC边上,将△CDE沿DE翻折得到△FDE,点F落在AE上. 若CE = 3 cm,AF = 2EF,求AB的长.

答案： 解:依题意,得$FE = CE = 3\mathrm{cm},DC = DF$,$\angle DEC=\angle DEF$,$\angle DFE=\angle C=\angle DFA = 90^{\circ}$.
$\because AF = 2EF$,
$\therefore AF = 6\mathrm{cm}$,
$\therefore AE = AF + EF = 9\mathrm{cm}$.
$\because AD// BC$,
$\therefore \angle ADE=\angle DEC=\angle DEF$,
$\therefore AD = AE = 9\mathrm{cm}$.
在$Rt\triangle ADF$中,$AF^{2}+DF^{2}=AD^{2}$,
$\therefore 6^{2}+DF^{2}=9^{2}$,
$\therefore DF = 3\sqrt{5}\mathrm{cm}$,
$\therefore AB = DC = DF = 3\sqrt{5}\mathrm{cm}$.

2. (牡丹江爱民区期中)如图,把长方形纸片ABCD沿EF折叠,使点B落在边AD上的点B'处,点A落在点A'处.
(1)求证:B'E = BF;
(2)若AE = 3,AB = 4,求BF的长.

答案：
(1)证明:$\because$在长方形$ABCD$中,$AD// BC$,
$\therefore \angle B'EF=\angle EFB$.
由题意,得$\angle B'FE=\angle EFB$,
$\therefore \angle B'FE=\angle B'EF$,
$\therefore B'E = B'F$.
又易知$BF = B'F$,
$\therefore B'E = BF$.
(2)解:由题易知$\angle A'=\angle A = 90^{\circ}$.
在$Rt\triangle A'B'E$中,$A'B' = AB = 4,A'E = AE = 3$,
$\therefore B'E^{2}=A'B'^{2}+A'E^{2}=4^{2}+3^{2}=25$,
$\therefore B'E = 5$,
$\therefore BF = B'E = 5$.