《2025年5年中考3年模拟七年级数学下册人教版》

1.(2024甘肃陇南武都二模)如图,直线a、b被直线c、d所截,若∠1 = 100°,∠2 = 80°,∠3 = 125°,则∠4的度数是(M7207006)

( )
A.55°
B.75°
C.100°
D.125°

答案： D $\because \angle 1 = 100^{\circ}, \angle 2 = 80^{\circ},$
$\therefore \angle 1 + \angle 2 = 180^{\circ}, \therefore a // b,$
$\therefore \angle 4 = \angle 3 = 125^{\circ}.$
故选D.

2.(2024广东中考)如图,一把直尺、两个含30°角的三角尺拼接在一起,则∠ACE的度数为 ( )

A.120°
B.90°
C.60°
D.30°

答案： C $\because \angle ABC = \angle ECD = 30^{\circ}, \therefore AB // CE,$
$\therefore \angle ACE = \angle A = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}.$
故选C.

3.(2024湖南益阳二模)如图,∠B + ∠DCB = 180°,AC平分∠DAB,且∠D：∠DAC = 4：1,则∠D的度数是(M7207006)

( )
A.120°
B.110°
C.105°
D.100°

答案： A $\because \angle B + \angle DCB = 180^{\circ}, \therefore AB // CD,$
$\therefore \angle D + \angle DAB = 180^{\circ},$
设$\angle D = 4x^{\circ}$, 则$\angle DAC = x^{\circ},$
$\because AC$平分$\angle DAB,$
$\therefore \angle DAB = 2\angle DAC = 2x^{\circ},$
$\therefore 4x^{\circ} + 2x^{\circ} = 180^{\circ}, \therefore x = 30,$
$\therefore \angle D = 4x^{\circ} = 4 \times 30^{\circ} = 120^{\circ}.$ 故选A.

4.(2024云南曲靖二模)如图,a⊥c,b⊥c于点B,过点B的直线d交直线a于点A,若∠1 = 40°,则∠2的度数是(M7207006)

( )
A.50°
B.40°
C.30°
D.20°

答案：
A 如图,

$\because \angle 1 = 40^{\circ}, \therefore \angle 3 = \angle 1 = 40^{\circ}.$
$\because a \perp c, b \perp c,$
$\therefore a // b, \angle 2 + \angle 4 = 90^{\circ},$
$\therefore \angle 4 = \angle 3 = 40^{\circ}.$
$\therefore \angle 2 = 90^{\circ} - \angle 4 = 90^{\circ} - 40^{\circ} = 50^{\circ}.$
故选A.

5.铅笔模型(2024浙江杭州上城期中)图1是某景区电动道闸门,将其抽象为几何图形,如图2所示,BA垂直于地面AE于A,当CD平行于地面AE时,∠ABC + ∠BCD =(M7207006)



( )
A.180°
B.210°
C.250°
D.270°

答案：
D 过点$B$作$BF // AE$, 如图,

$\because CD // AE, \therefore BF // CD,$
$\therefore \angle BCD + \angle CBF = 180^{\circ},$
$\because AB \perp AE, BF // AE, \therefore AB \perp BF, \therefore \angle ABF = 90^{\circ},$
$\therefore \angle ABC + \angle BCD = \angle ABF + \angle CBF + \angle BCD = 90^{\circ} + 180^{\circ} = 270^{\circ}.$
故选D.
模型解读
如果两条平行线中间有向外的拐点, 如图, 就得到了铅笔模型. 该模型常用结论: 若$AB // CD$, 则$\angle A + \angle AEC + \angle C = 360^{\circ}.$

6.教材变式 如图,MN//BC,BD⊥DC,∠1 = ∠2 = 60°,DC是∠NDE的平分线.(M7207006)
(1)AB与DE平行吗?请说明理由.
(2)试说明∠ABC = ∠C.
(3)求∠ABD的度数.

答案： 解析
(1)$AB // DE$. 理由:$\because MN // BC, \angle 1 = 60^{\circ},$
$\therefore \angle ABC = \angle 1 = 60^{\circ}. \because \angle 1 = \angle 2, \therefore \angle ABC = \angle 2.$
$\therefore AB // DE.$
(2)证明:$\because MN // BC, \therefore \angle NDE + \angle 2 = 180^{\circ}.$
$\therefore \angle NDE = 180^{\circ} - \angle 2 = 180^{\circ} - 60^{\circ} = 120^{\circ}.$
$\because DC$是$\angle NDE$的平分线,$\therefore \angle EDC = \angle NDC = \frac{1}{2} \angle NDE = 60^{\circ}. \because MN // BC, \therefore \angle C = \angle NDC = 60^{\circ}.$
$\because \angle ABC = 60^{\circ}, \therefore \angle ABC = \angle C.$
(3)$\because \angle ADC + \angle NDC = 180^{\circ}, \angle NDC = 60^{\circ},$
$\therefore \angle ADC = 180^{\circ} - \angle NDC = 180^{\circ} - 60^{\circ} = 120^{\circ}.$
$\because BD \perp DC, \therefore \angle BDC = 90^{\circ}.$
$\therefore \angle ADB = \angle ADC - \angle BDC = 120^{\circ} - 90^{\circ} = 30^{\circ}.$
$\because MN // BC, \therefore \angle DBC = \angle ADB = 30^{\circ}.$
$\because \angle ABC = 60^{\circ}, \therefore \angle ABD = 60^{\circ} - 30^{\circ} = 30^{\circ}.$