2025年畅行课堂七年级数学下册华师大版


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《2025年畅行课堂七年级数学下册华师大版》

1. 如图,AB,CD相交于点O,连结AD,BC. 若$\angle A = 43^{\circ}$,$\angle D = 57^{\circ}$,$\angle C = 37^{\circ}$,则$\angle B$的度数为______.

答案: $63^{\circ}$
2. 如图,$\angle A+\angle B+\angle C+\angle D+\angle E=$______$^{\circ}$.

答案: $180$
3. 如图,$\angle A = 65^{\circ}$,$\angle B = 40^{\circ}$,$\angle C = 25^{\circ}$,则$\angle D+\angle E=$______.

答案: $50^{\circ}$
4. 如图,AP,CP分别平分$\angle BAD$,$\angle BCD$,若$\angle ABC = 36^{\circ}$,$\angle ADC = 16^{\circ}$,求$\angle P$的度数.
答案: 解:$\because AP,CP$分别平分$\angle BAD,\angle BCD$,
$\therefore\angle BAP=\angle PAD$,$\angle BCP=\angle PCD$.
$\because\angle P+\angle BCP=\angle ABC+\angle BAP$,
$\angle P+\angle PAD=\angle ADC+\angle PCD$,
$\therefore 2\angle P+\angle BCP+\angle PAD=\angle BAP+\angle PCD+\angle ABC+\angle ADC$.
$\therefore 2\angle P=\angle ABC+\angle ADC$.
$\because\angle ABC = 36^{\circ},\angle ADC = 16^{\circ}$,
$\therefore\angle P=\frac{1}{2}(\angle ABC+\angle ADC)$
$=\frac{1}{2}(36^{\circ}+16^{\circ}) = 26^{\circ}$.
5. 如图,已知$\angle A = 60^{\circ}$,$\angle BDC = 120^{\circ}$,$\angle C = 37^{\circ}$,则$\angle B=$______$^{\circ}$.

答案: $23$
6. 如图,若$\angle CGE = 125^{\circ}$,则$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$______.

答案: $250^{\circ}$
7. 如图,在四边形ABCD中,AM,CM分别平分$\angle DAB$和$\angle DCB$,AM与CM相交于点M,探究$\angle AMC$与$\angle ABC$,$\angle D$之间的数量关系.
答案:
解:如图,由模型可知,$\angle AMC=\angle 1+\angle D+\angle 4$,①
$\angle ABC=\angle 2+\angle AMC+\angle 3$,②
①$-$②,得$\angle AMC-\angle ABC=\angle 1+\angle D+\angle 4-\angle 2-\angle AMC-\angle 3$.
$\because AM$平分$\angle DAB$,$CM$平分$\angle DCB$,
$\therefore\angle 1=\angle 2,\angle 3=\angle 4$.
$\therefore\angle AMC-\angle ABC=\angle D-\angle AMC$,
$\therefore 2\angle AMC=\angle ABC+\angle D$.
3B

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