2026年学霸题中题七年级数学下册苏科版第168页解析答案

1. (2025·苏州期末)有一副三角尺,其中△ABC中,∠A=90°,∠B=45°;△DEF中,∠D=90°,∠F=60°.将这副直角三角尺按如图①放置.此时边BC与EF在同一直线上,且三角尺DEF的顶点E落在边BC的中点处.若将三角尺DEF绕点E按逆时针方向旋转,旋转角为θ(0°＜θ＜90°).
(1)当θ=

15°

时,DE//AB;当θ=

45°

时,EF//AB;
(2)如图②,设边EF所在直线与边AB所在直线交于点M,边DE所在直线与边AC所在直线交于点N,记∠AME=α,∠CNE=β.在整个旋转过程中,请探究α与β的数量关系,并说明理由;
(3)在(2)的条件下,若4α+β＜150°,直接写出θ的取值范围.

答案：
(1)15°　45°　解析:在△ABC中,∠A = 90°,∠ABC = 45°,
∴当∠DEC = ∠ABC = 45°时,DE //AB,
∴θ = ∠FEC = 45° - 30° = 15°;在△DEF中,∠D = 90°,∠F = 60°,
∴∠DEF = 30°,当∠FEC = ∠ABC = 45°时,EF //AB,
∴θ = ∠FEC = 45°.
(2)当0°＜θ＜45°时,β - α = 60°,理由如下:如图①,△ABC中,∠A = 90°,∠ABC = 45°,
∴∠MBE = 180° - 45° = 135°,
∴∠BEM = ∠CEF = 180° - (∠BME + 135°) = 45° - ∠BME.
∵∠DEF = 30°,∠NCE = 180° - 90° - 45° = 45°,
∴∠CNE = 180° - (∠NEC + ∠NCE) = 180° - (30° + 45° - ∠BME + 45°) = 60° + ∠BME,即β - α = 60°.
　　　　　

当45°＜θ＜90°时,β + α = 60°,理由如下:如图②,在△ABC中,∠A = 90°,∠ABC = 45°,
∴∠BEM = 180° - (∠BME + 45°) = 135° - ∠BME,
∴∠CEM = 45° + ∠BME.
∵∠DEF = 30°,∠NCE = 180° - 90° - 45° = 45°,
∴∠CNE = 180° - (∠NEC + ∠NCE) = 180° - (30° + 45° + ∠BME + 45°) = 60° - ∠BME,即β + α = 60°.
当θ = 45°时,EF//AB,点M不存在,故舍去该情况.
综上所述,当0°＜θ＜45°时,β - α = 60°;当45°＜θ＜90°时,β + α = 60°.
(3)当0°＜θ＜45°时,由(2)得β - α = 60°,θ + α = 180° - 135° = 45°,
∴β = 60° + α,α = 45° - θ.
∵4α + β＜150°,
∴4(45° - θ) + 60° + (45° - θ)＜150°,
∴θ＞27°,
∴27°＜θ＜45°.
当45°＜θ＜90°时,由(2)得β + α = 60°,θ = 180° - ∠BEM = 180° - (180° - 45° - α) = 45° + α,
∴β = 60° - α,α = θ - 45°.
∵4α + β＜150°,
∴4(θ - 45°) + 60° - (θ - 45°)＜150°,
∴θ＜75°,
∴45°＜θ＜75°.综上所述,θ的取值范围是27°＜θ＜45°或45°＜θ＜75°.

2. (2025·扬州期末)综合与实践——折纸中的数学
折纸是同学们喜欢的手工活动之一,通过折纸我们可以得到许多美丽的图形,折纸的过程还蕴含着丰富的数学知识.现有等腰三角形ABC,如图①所示,其中AB=AC,∠B=∠C=70°.
【初步感知】
问题1:如图②所示,直线AP与线段BC相交于点D.将△ABD沿着直线AP翻折,使B点与C点重合,则BD与BC的数量关系是

BD = $\frac{1}{2}$BC

;直线AP与BC的位置关系是

AP⊥BC

.
【深入探究】
问题2:如图③,直线AP与线段BC相交于点D,将△ABD沿直线AP翻折至△AB'D处,点B的对应点为B'.
(1)当AB'⊥BC时,∠BAD=

10°

;
(2)探究∠BAD与∠CDB'之间的数量关系.
【拓展延伸】
问题3:如图④所示,点E在AB上,过点E作EF//BC交AC于点F,直线AP与射线BC相交于点D,将△AEF沿直线AP翻折得到△AE'F',点E的对应点为E',点F的对应点为F',当△E'AF'的某一边与BC垂直时,∠BAD的度数为

10°或30°或65°或100°

答案：
[初步感知]BD = $\frac{1}{2}$BC　AP⊥BC　解析:折叠后△ABD和△ACD关于直线AD对称,所以BD = DC,AP⊥BC,所以BD = $\frac{1}{2}$BC.
[深入探究](1)10°　解析:
∵AB = AC,直线AP与线段BC相交于点D,将△ABD沿直线AP翻折至△AB'D处，点B的对应点为B',AB'⊥BC,
∴AB'是等腰三角形ABC的对称轴.
∵∠B = ∠C = 70°,
∴∠BAB' = ∠CAB' = $\frac{1}{2}$∠BAC = $\frac{1}{2}$(180° - 70° - 70°) = 20°,由折叠性质可得∠BAD = ∠B'AD = $\frac{1}{2}$∠BAB' = $\frac{1}{2}$×20° = 10°.
(2)设∠BAD = ∠B'AD = α,
∵∠B = ∠C = 70°,由折叠的性质可得∠BDA = ∠B'DA = 180° - 70° - α = 110° - α.
又
∵∠BDA + ∠ADC = 180°,
∴∠ADC = 180° - (110° - α) = 70° + α,
∴∠CDB' = ∠ADB' - ∠ADC = 110° - α - (70° + α) = 40° - 2α,
∴∠CDB' = 40° - 2∠BAD,
则∠CDB' + 2∠BAD = 40°.
[拓展延伸]10°或30°或65°或100°　解析:如图①,当△E'AF'的AE'的边与BC垂直时,AE'在等腰三角形ABC的对称轴上，
∵∠B = ∠C = 70°,AE'⊥BC,
∴∠BAC = 40°,∠BAE' = ∠CAE' = $\frac{1}{2}$∠BAC = 20°.
∵将△AEF沿直线AP翻折得到△AE'F',
∴∠BAD = ∠E'AD = $\frac{1}{2}$∠BAE’ = $\frac{1}{2}$×20° = 10°.
　　　　　　　

如图②,当△E'AF'的AF'的边与BC垂直时,AF'在等腰三角形ABC的对称轴上，
∵∠B = ∠C = 70°,AF'⊥BC,
∴∠BAC = 40°,∠BAF' = ∠CAF' = $\frac{1}{2}$∠BAC = 20°.
∵将△AEF沿直线AP翻折得到△AE'F',
∴∠CAD = ∠F'AD = $\frac{1}{2}$∠CAF' = $\frac{1}{2}$×20° = 10°,
∴∠BAD = ∠BAF' + ∠F'AD = 20° + 10° = 30°.
如图③,当△E'AF'的E'F'的边与射线BC垂直时,过点A作HG⊥BC,HG为等腰三角形ABC的对称轴且HG//E'F',
∵EF//BC,
∴∠B = ∠C = ∠AEF = ∠AFE = ∠E' = ∠F' = 70°,
∴∠GAF' = ∠F' = 70°.
∵∠B = ∠C = 70°,HG⊥BC,
∴∠BAC = 40°,∠BAG = ∠CAG = $\frac{1}{2}$∠BAC = 20°,
∴∠CAF' = 70° - 20° = 50°.
∵将△AEF沿直线AP翻折得到△AE'F',
∴∠BAD = ∠E'AD,即∠CAD = ∠F'AD = $\frac{1}{2}$∠CAF' = $\frac{1}{2}$×50° = 25°,
∴∠BAD = ∠BAC + ∠CAD = 40° + 25° = 65°.
　　　　　　　

如图④,当△E'AF'的AE'的边与BC垂直时,过点A作HG⊥BC,HG为等腰三角形ABC的对称轴,
∵EF//BC,
∴∠B = ∠C = ∠AEF = ∠AFE = ∠E' = ∠F' = 70°,
∴∠E'AF' = 40°.
∵∠B = ∠C = 70°,HG⊥BC,
∴∠BAC = 40°,∠BAG = ∠CAG = $\frac{1}{2}$∠BAC = 20°,
∴∠CAF' = 180° - 40° - 20 = 120°.
∵将△AEF沿直线AP翻折得到△AE'F',
∴∠BAD = ∠E'AD,即∠CAD = ∠F'AD = $\frac{1}{2}$∠CAF' = $\frac{1}{2}$×120° = 60°,
∴∠BAD = ∠BAC + ∠CAD = 40° + 60° = 100°.综上所述,∠BAD的度数为10°,30°,65°,100°.
　　　　