《2025年5年中考3年模拟七年级数学下册人教版》

7.(2024四川德阳中考,3,★☆☆)如图所示的是某机械加工厂加工的一种零件的示意图,其中AB//CD,DE⊥BC,∠ABC = 70°,则∠EDC等于(M7207006)

( )
A.10°
B.20°
C.30°
D.40°

答案：
B 如图, 过$E$作$EF // AB$,
则$\angle 1 = \angle B = 70^{\circ}.$

$\because DE \perp BC, \therefore \angle BED = 90^{\circ}.$
$\therefore \angle 2 = \angle BED - \angle 1 = 90^{\circ} - 70^{\circ} = 20^{\circ}.$
$\because AB // CD, \therefore EF // CD.$
$\therefore \angle D = \angle 2 = 20^{\circ}.$ 故选B.

8.猪蹄模型(2023湖北鄂州中考,5,★☆☆)如图,直线AB//CD,GE⊥EF于点E.若∠BGE = 60°,则∠EFD的度数是

( )
A.60°
B.30°
C.40°
D.70°

答案：
B 如图, 过点$E$作$EM // AB$,

$\therefore \angle GEM = \angle BGE = 60^{\circ},$
$\because GE \perp EF, \therefore \angle GEF = 90^{\circ}, \therefore \angle FEM = \angle GEF - \angle GEM = 30^{\circ},$
$\because AB // CD, AB // EM, \therefore CD // EM,$
$\therefore \angle EFD = \angle FEM = 30^{\circ}.$ 故选B.
模型解读
如果两条平行线中间有向内的拐点, 如图, 就得到了猪蹄模型. 该模型常用结论: 若$AB // CD$, 则$\angle APC = \angle A + \angle C.$

9.靴子模型 情境题·中华优秀传统文化(2023江苏苏州吴江期中,15,★☆☆)某同学在研究传统文化“抖空竹”时有一个发现,他把它抽象成数学问题:如图所示,已知AB//CD,∠A = 88°,∠C = 121°,则∠E的度数是________.(M7207006)

答案：
答案 $33^{\circ}$
解析 如图, 过点$E$作$EF // CD$, 则$\angle 1 + \angle C = 180^{\circ}.$

$\because \angle C = 121^{\circ}, \therefore \angle 1 = 180^{\circ} - 121^{\circ} = 59^{\circ}.$
$\because AB // CD, \therefore EF // AB. \therefore \angle A + \angle AEF = 180^{\circ}.$
$\because \angle A = 88^{\circ}, \therefore \angle AEF = 180^{\circ} - 88^{\circ} = 92^{\circ}.$
$\therefore \angle AEC = \angle AEF - \angle 1 = 92^{\circ} - 59^{\circ} = 33^{\circ}.$
模型解读
靴子模型因两条平行线和三个角构成的图形像我们穿的靴子而得名. 靴子模型及其变形中的三个角之间存在如下的关系.
　　　
　　　　　　
$\angle 3 = \angle 1 + \angle 2 - 180^{\circ}.$ $\angle 3 = \angle 1 + \angle 2 - 180^{\circ}.$

10.(2022广东揭阳榕城期末,15,★☆☆)如图,已知长方形纸片ABCD,点E,F在AD边上,点G,H在BC边上,分别沿EG,FH折叠,使点D和点A都落在点M处,若α + β = 118°,则∠EMF的度数为________.
D　　　　　　E　　　　　F　　　　A
C　　　　G　　　　　−−H　　　B
　　　　　　　　　　　　!　　　B'

答案： 答案 $56^{\circ}$
解析 $\because$ 四边形$ABCD$是长方形,$\therefore \angle D = \angle C = 90^{\circ}, \therefore \angle D + \angle C = 180^{\circ}, \therefore AD // BC, \therefore \angle DEG = \alpha,$
$\angle AFH = \beta$, 由折叠得$\angle DEM = 2\angle DEG = 2\alpha, \angle AFM = 2\angle AFH = 2\beta, \therefore \angle FEM + \angle EFM = 360^{\circ} - (\angle DEM + \angle AFM) = 360^{\circ} - (2\alpha + 2\beta) = 360^{\circ} - 236^{\circ} = 124^{\circ},$
$\therefore \angle EMF = 180^{\circ} - (\angle FEM + \angle EFM) = 56^{\circ}.$

11.(2024广西百色田阳二模,21,★★☆)如图所示的是一种躺椅及其简化结构示意图,AB与CD都平行于EF,OE与OF分别与CD交于点G和点D,AB与DM交于点N,∠AOE = ∠BNM.(M7207006)
(1)求证:OE//DM.
(2)若OE平分∠AOF,∠ODC = 30°,求∠ANM的度数.

答案： 解析
(1) 证明:$\because \angle BNM = \angle AND, \angle AOE = \angle BNM,$
$\therefore \angle AOE = \angle AND, \therefore OE // DM.$
(2)$\because AB$与$CD$都平行于$EF$,
$\therefore AB // CD, \therefore \angle BOD = \angle ODC = 30^{\circ},$
$\because \angle AOF + \angle BOD = 180^{\circ}, \therefore \angle AOF = 150^{\circ},$
$\because OE$平分$\angle AOF,$
$\therefore \angle EOF = \frac{1}{2} \angle AOF = 75^{\circ},$
$\therefore \angle BOE = \angle BOD + \angle EOF = 105^{\circ},$
$\because OE // DM, \therefore \angle ANM = \angle BOE = 105^{\circ}.$