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2025年千里马单元测试卷八年级数学下册人教版
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年千里马单元测试卷八年级数学下册人教版 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
1. 下列说法正确的是（ ）
A. 有一个角是直角的平行四边形是正方形
B. 对角线互相垂直的矩形是正方形
C. 有一组邻边相等的菱形是正方形
D. 各边都相等的四边形是正方形
A. 有一个角是直角的平行四边形是正方形
B. 对角线互相垂直的矩形是正方形
C. 有一组邻边相等的菱形是正方形
D. 各边都相等的四边形是正方形
答案：
B
2. 如图,边长为3的正方形OBCD两边与坐标轴正半轴重合,点C的坐标是（ ）
A. (3,-3) B. (-3,3)
C. (3,3) D. (-3,-3)
A. (3,-3) B. (-3,3)
C. (3,3) D. (-3,-3)
答案：
C
3. 如图,在正方形ABCD中,AE平分∠BAC交BC于点E,点F是边AB上一点,连接DF,若BE = AF,则∠CDF的度数为（ ）
A. 45° B. 60° C. 67.5° D. 77.5°
A. 45° B. 60° C. 67.5° D. 77.5°
答案：
C
4. (龙东地区中考)如图,在矩形ABCD中,对角线AC,BD相交于点O,试添加一个条件:____________,使得矩形ABCD为正方形.
答案：
$AC\perp BD$（或$AB = AD$等）
5. 如图,在正方形ABCD中,对角线AC与BD相交于点O,E为BC上一点,CE = 7,F为DE的中点,若△CEF的周长为32,则OF的长为______.
答案：
$\frac{17}{2}$ [解析] $\because CE = 7,\triangle CEF$的周长为$32,\therefore CF + EF=32 - 7 = 25.\because F$为$DE$的中点，$\angle BCD = 90^{\circ},\therefore DF =EF = CF=\frac{1}{2}DE,\therefore DE = 25,\therefore CD=\sqrt{DE^{2}-CE^{2}} = 24.$
$\because$四边形$ABCD$是正方形，$\therefore BC = CD = 24.\because O$为$BD$的中点，$F$为$DE$的中点，$\therefore OF$是$\triangle BDE$的中位线
$\therefore OF=\frac{1}{2}BE=\frac{1}{2}(BC - CE)=\frac{1}{2}\times(24 - 7)=\frac{17}{2}.$
$\because$四边形$ABCD$是正方形，$\therefore BC = CD = 24.\because O$为$BD$的中点，$F$为$DE$的中点，$\therefore OF$是$\triangle BDE$的中位线
$\therefore OF=\frac{1}{2}BE=\frac{1}{2}(BC - CE)=\frac{1}{2}\times(24 - 7)=\frac{17}{2}.$
6. 如图,在矩形ABCD中,∠BAD的平分线交BC于E,∠ABC的平分线交AD于F,求证:四边形ABEF是正方形.
答案：
证明：如答图. $\because$四边形$ABCD$是矩形，
$\therefore \angle BAF = 90^{\circ},AD// BC,$
$\therefore \angle 2=\angle 3.$
$\because AE$平分$\angle BAF,$
$\therefore \angle 1=\angle 2,\therefore \angle 1=\angle 3,$
$\therefore AB = BE$，同理可得$AB = AF,$
$\therefore BE = AF.$
$\because AD// BC,\therefore AF// BE,$
$\therefore$四边形$ABEF$是平行四边形.
$\because \angle BAF = 90^{\circ},\therefore$四边形$ABEF$是矩形.
$\because AB = AF,\therefore$四边形$ABEF$是正方形.
证明：如答图. $\because$四边形$ABCD$是矩形，
$\therefore \angle BAF = 90^{\circ},AD// BC,$
$\therefore \angle 2=\angle 3.$
$\because AE$平分$\angle BAF,$
$\therefore \angle 1=\angle 2,\therefore \angle 1=\angle 3,$
$\therefore AB = BE$，同理可得$AB = AF,$
$\therefore BE = AF.$
$\because AD// BC,\therefore AF// BE,$
$\therefore$四边形$ABEF$是平行四边形.
$\because \angle BAF = 90^{\circ},\therefore$四边形$ABEF$是矩形.
$\because AB = AF,\therefore$四边形$ABEF$是正方形.
7. 如图,E,F,M,N分别是正方形ABCD四条边上的点,且AE = BF = CM = DN.
(1)求证:四边形EFMN是正方形;
(2)若AB = 7,AE = 3,求四边形EFMN的周长.
(1)求证:四边形EFMN是正方形;
(2)若AB = 7,AE = 3,求四边形EFMN的周长.
答案：
(1) 证明：$\because AE = BF = CM = DN,\therefore AN = DM = CF = BE.$
$\because \angle A=\angle B=\angle C=\angle D = 90^{\circ},$
$\therefore \triangle ANE\cong \triangle DMN\cong \triangle CFM\cong \triangle BEF,$
$\therefore FE = EN = NM = MF,\angle ENA=\angle NMD,$
$\therefore$四边形$EFMN$是菱形.
$\because \angle ENA=\angle NMD,\angle NMD+\angle DNM = 90^{\circ},$
$\therefore \angle ENA+\angle DNM = 90^{\circ},$
$\therefore \angle ENM = 90^{\circ},$
$\therefore$四边形$EFMN$是正方形.
(2) 解：$\because AB = 7,AE = 3,$
$\therefore AN = BE = AB - AE = 4,\therefore EN=\sqrt{AE^{2}+AN^{2}} = 5,$
$\therefore$正方形$EFMN$的周长$=4\times5 = 20.$
(1) 证明：$\because AE = BF = CM = DN,\therefore AN = DM = CF = BE.$
$\because \angle A=\angle B=\angle C=\angle D = 90^{\circ},$
$\therefore \triangle ANE\cong \triangle DMN\cong \triangle CFM\cong \triangle BEF,$
$\therefore FE = EN = NM = MF,\angle ENA=\angle NMD,$
$\therefore$四边形$EFMN$是菱形.
$\because \angle ENA=\angle NMD,\angle NMD+\angle DNM = 90^{\circ},$
$\therefore \angle ENA+\angle DNM = 90^{\circ},$
$\therefore \angle ENM = 90^{\circ},$
$\therefore$四边形$EFMN$是正方形.
(2) 解：$\because AB = 7,AE = 3,$
$\therefore AN = BE = AB - AE = 4,\therefore EN=\sqrt{AE^{2}+AN^{2}} = 5,$
$\therefore$正方形$EFMN$的周长$=4\times5 = 20.$
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