• Grundaufgaben Pythagoras
  • Deliah Herbstritt
  • 21.07.2020
  • Mathematik
  • Flächen
  • R
  • 9
  • Einzelarbeit
  • Arbeitsblatt
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1
Be­schrif­te die Flä­chen­dia­go­na­le und die Raum­dia­go­na­le.
2
In jedem Qua­der lie­gen ver­schie­de­ne Drei­ecke. Zeich­ne bei den ent­stan­de­nen Drei­ecken rech­te Win­kel ein, die für die Be­rech­nung von Stre­cken nötig wären.
3
Be­rech­ne die Raum­dia­go­na­le die­ses Qua­ders.
Lösung3
Flä­chen­dia­go­na­le:
f2=(7cm)2+(4cm)2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=(7cm)^2+(4cm)^2
f2=49cm+16cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=49cm+16cm
f2=65cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=65cm
f2=(65cm)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=(\sqrt{65cm})
f8,06cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f\approx8{,}06cm
Raum­dia­go­na­le:
d2=(8,06cm)2+(3cm)2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=(8{,}06cm)^2+(3cm)^2
d2=65cm+9cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=65cm+9cm
d2=74cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=74cm
d2=(74cm)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=(\sqrt{74cm})
d8,60cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d\approx8{,}60cm

Ant­wort: Die Raum­dia­go­na­le be­trägt 8,60cm.
4
Be­rech­ne die Flächen-​/ und Raum­dia­go­na­le fol­gen­der Kör­per. Fer­ti­ge dazu eine Schräg­bild­skiz­ze an.
  • Qua­der mit den Maßen: I= 8cm, b= 4cm, h= 2cm
  • Wür­fel mit der Kan­ten­län­ge a= 3cm
Quader4
a)
Flä­chen­dia­go­na­le:
f2=(8cm)2+(4cm)2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=(8cm)^2+(4cm)^2
f2=64cm+16cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=64cm+16cm
f2=80cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=80cm
f2=(80cm)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=(\sqrt{80cm})
f8,94cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f\approx8{,}94cm

Raum­dia­go­na­le:
d2=(8,94cm)2+(2cm)2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=(8{,}94cm)^2+(2cm)^2
d2=80cm+4cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=80cm+4cm
d2=84cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=84cm
d2=(84cm)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=(\sqrt{84cm})
d9,17cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d\approx9{,}17cm
Ant­wort: Die Flä­chen­dia­go­na­le be­trägt 8,94cm, die Raum­dia­go­na­le be­trägt 9,17cm.
Würfel
b)
Flä­chen­dia­go­na­le:
f2=(3cm)2+(3cm)2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=(3cm)^2+(3cm)^2
f2=9cm+9cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=9cm+9cm
f2=18cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=18cm
f2=(18cm)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f^2=(\sqrt{18cm})
f4,24cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} f\approx4{,}24cm

Raum­dia­go­na­le:
d2=(4,24cm)2+(3cm)2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=(4{,}24cm)^2+(3cm)^2
d2=18cm+9cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=18cm+9cm
d2=27cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=27cm
d2=(27cm)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d^2=(\sqrt{27cm})
d5,20cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} d\approx5{,}20cm
Ant­wort: Die Flä­chen­dia­go­na­le be­trägt 4,24cm, die Raum­dia­go­na­le be­trägt 5,20cm.
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