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  • UeT Nr. 2 - Winkel bei Vierecken
  • Christian Leeser
  • 09.02.2022
  • Mathematik
  • 8
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UeT Nr. 2 - Winkel bei Vierecken
08.12.2021
1
Berechne jeweils die fehlenden drei Winkeln (Parallelogramm / Raute)
12 / 12
  • α=33°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha= 33°
  • β=133°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta= 133°
  • γ=56°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma= 56°
  • δ=99°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta= 99°
Lösung1
a) γ=33°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 33^°; β=147°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 147^°; δ=147°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 147^°

b) δ=133°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 133^°; α=47°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 47^°; γ=47°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 47^°

c) α=56°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 56^°; β=124°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 124^°; δ=124°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 124^°

c) β=99°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 99^°; γ=81°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 81^°; α=81°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 81^°
2
Berechne jeweils die fehlenden zwei Winkeln (Trapez)
8 / 8
  • α=33°,β=122°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha= 33°, \beta= 122°
  • γ=45°,δ=78°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma= 45°, \delta= 78°
  • α=46°,β=100°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha= 46°, \beta= 100°
  • γ=77°,δ=121°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma= 77°, \delta= 121°
Lösung2
a) δ=147°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 147^°; γ=58°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 58^°

b) β=135°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 135^°; α=102°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 102^°

c) δ=134°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 134^°; γ=80°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 80^°

a) β=103°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 103^°; α=59°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 59^°
3
Berechne jeweils die fehlenden zwei Winkeln (Drachenviereck)
8 / 8
  • α=33°,β=122°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha= 33°, \beta= 122°
  • β=45°,γ=78°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta= 45°, \gamma= 78°
  • γ=46°,δ=100°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma= 46°, \delta= 100°
  • α=77°,δ=121°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha= 77°, \delta= 121°
Lösung3
a) δ=122°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 122^°; γ=83°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 83^°

b) δ=45°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \delta = 45^°; α=192°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 192^°

c) β=100°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 100^°; α=114°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \alpha = 114^°

d) β=121°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \beta = 121^°; γ=41°\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \gamma = 41^°
Note
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Unterschrift
Mathematik
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