• Übungen Pythagoras
  • P. Ruppaner
  • 21.07.2020
  • Mathematik
  • Flächen
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  • 9
  • Einzelarbeit
  • Arbeitsblatt
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1
Berechne die die fehlende Länge des rechtwinkligen Dreiecks wie im Beispiel.












  • a=13cmb=5cmc\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad a=13cm\qquad b=5cm\qquad c\approx13,93cm













Beispiel: gegeben: a=6cmb=3cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} a = 6 cm \qquad b=3cm\qquadgesucht: c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c


 a2+b2=c2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad\qquad \qquad \nobreakspace a^2+b^2=c^2
(6cm)2+(3cm)2=c2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad(6 cm)^2 + (3 cm)^2=c^2
36cm2+9cm2=c2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad\quad36cm^2+9cm^2=c^2
  45cm2=c2 \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad\qquad\qquad\nobreakspace\nobreakspace45cm^2=c^2\qquad|\sqrt{\nobreakspace}
6,71cmc\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad\qquad\qquad\underline{6{,}71cm\approx c}

Lösung1
a) Berechnung der Hypotenuse c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c:
 c2=a2+b2  \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=a^2+b^2\qquad \qquad \qquad\nobreakspace |\sqrt{\nobreakspace}

 c2=13cm2+5cm2 \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=13cm^2+5cm^2\qquad\quad\nobreak|\sqrt{\nobreakspace}

  13,93cm13cm2+5cm2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \nobreakspace \nobreakspace13{,}93 cm \approx \sqrt{13cm^2+5cm^2}

b) Berechnung der Hypotenuse c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c:
 c2=a2+b2  \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=a^2+b^2\qquad \qquad \qquad\nobreakspace |\sqrt{\nobreakspace}

 c2=64cm2+48cm2   \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=64cm^2+48cm^2\qquad\nobreakspace\nobreakspace|\sqrt{\nobreakspace}

   80cm=64cm2+48cm2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \nobreakspace \nobreakspace\quad\nobreakspace80 cm = \sqrt{64cm^2+48cm^2}

c) Berechnung der Hypotenuse c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c:
 c2=a2+b2  \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=a^2+b^2\qquad \qquad \qquad\nobreakspace |\sqrt{\nobreakspace}

 c2=12cm2+5cm2 \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=12cm^2+5cm^2\qquad\quad\nobreak|\sqrt{\nobreakspace}

   13cm=12cm2+5cm2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \nobreakspace \nobreakspace\quad\nobreakspace13 cm = \sqrt{12cm^2+5cm^2}

b)

c)

2
Berechne die fehlende Länge des rechtwinkligen Dreiecks. Dafür musst du die Formel des Satz des Pythagoras umstellen wie im Beispiel.












  • a=13cmb\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad a=13cm\qquad b\approx 9,8cmc=10cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad c=10cm

Beispiel: gegeben: a=5cmc=9cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} a = 5 cm \qquad c=9cm\qquadgesucht: b\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} b


c2a2=b2(9cm)2(5cm)2=b281cm225cm2=b256cm2=b2  67,48cmc\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} c^2-a^2&=b^2\\ (9 cm)^2 - (5 cm)^2&=b^2\\ 81cm^2-25cm^2&=b^2\\ 56cm^2&=b^2\qquad|\sqrt{\nobreakspace}\\\ \underline{67{,}48cm\approx c} \end{aligned}

Lösung2
a) Berechnung der Hypotenuse c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c:
 c2=a2+b2  \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=a^2+b^2\qquad \qquad \qquad\nobreakspace |\sqrt{\nobreakspace}

 c2=13cm2+5cm2 \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=13cm^2+5cm^2\qquad\quad\nobreak|\sqrt{\nobreakspace}

  13,93cm13cm2+5cm2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \nobreakspace \nobreakspace13{,}93 cm \approx \sqrt{13cm^2+5cm^2}

b) Berechnung der Hypotenuse c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c:
 c2=a2+b2  \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=a^2+b^2\qquad \qquad \qquad\nobreakspace |\sqrt{\nobreakspace}

 c2=64cm2+48cm2   \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=64cm^2+48cm^2\qquad\nobreakspace\nobreakspace|\sqrt{\nobreakspace}

   80cm=64cm2+48cm2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \nobreakspace \nobreakspace\quad\nobreakspace80 cm = \sqrt{64cm^2+48cm^2}

c) Berechnung der Hypotenuse c\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} c:
 c2=a2+b2  \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=a^2+b^2\qquad \qquad \qquad\nobreakspace |\sqrt{\nobreakspace}

 c2=12cm2+5cm2 \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \qquad \quad\nobreakspace c^2=12cm^2+5cm^2\qquad\quad\nobreak|\sqrt{\nobreakspace}

   13cm=12cm2+5cm2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \nobreakspace \nobreakspace\quad\nobreakspace13 cm = \sqrt{12cm^2+5cm^2}
Hinweis

Die Variable c in der Formel steht für die Hypotenuse im Dreieck. Die Benennung in den Dreiecken kann jedoch nochmal anders sein.

b)

c)

3
Berechne die Seitenlänge a, b und c des rechtwinkligen Dreiecks

a)a2=25cm2a=25cmb)a2156,42cm2a=156,42cmc)a2=81dm2a81dmd)a2=202,98mm2a=202,98mm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} a)\qquad a^2&=25cm^2\qquad\quad \\a&= \cloze{25cm}\\\\\\b)\qquad a^2&\approx\cloze{156{,}42cm^2}\\a&=\cloze{156{,}42cm}\\\\\\c)\qquad a^2&=81dm^2\\a&\approx\cloze{81dm}\\\\\\d)\qquad a^2&=\cloze{202{,}98mm^2}\\a&=\cloze{202{,}98mm} \end{aligned}

b2=25cm2b=25cmb2=64cm2b=64cmb2=391,71cm2b=391,71cmb2=156mm2b=156cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} b^2&=25cm^2\qquad\quad \\ b&=\cloze{25cm}\\\\\\ b^2&=64cm^2\\b&=\cloze{64cm}\\\\\\b^2&=\cloze{391{,}71cm^2}\\b&=\cloze{391{,}71cm}\\\\\\ b^2&=156mm^2\\b&=\cloze{156cm} \end{aligned}

c2=35,36cm2c=156,42cmc2=169m2c=169cmc2=400dm2c=400dmc2=256mm2c=256mm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} c^2&=\cloze{35{,}36cm^2}\qquad\quad \\ c&=\cloze{156{,}42cm}\\\\\\ c^2&=169m^2\\c&=\cloze{169cm}\\\\\\c^2&=400dm^2\\c&=\cloze{400dm}\\\\\\ c^2&=256mm^2\\c&=\cloze{256mm} \end{aligned}