• LÖSUNG: Flächeninhalt von Figuren (2)
  • MNWeG
  • 14.01.2022
  • Mathematik
  • Messen
  • E (Expertenstandard)
  • 6
  • Arbeitsblatt
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1
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Dein Lösungsweg kann sich hiervon unterscheiden (z.B. wenn du die Teilflächen anders eingeteilt hast). Solange aber das Ergebnis stimmt, hast du alles richtig gemacht!
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A1\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{1}

A2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{2}

A3\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{3}

A5\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{5}

A4\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{4}

A6\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{6}

A9\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{9}

A11\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{11}

A8\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{8}

A7\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{7}

A10\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{10}

A12\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} A_{12}

Aufgabe 1

Schritt 1: Figur in sinnvolle Teilflächen unterteilen und benennen (siehe vorherige Seite).


Schritt 2: Teilflächen berechnen.

A2=12aha=123cm1cm=123cm²=1,5cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{2}&=\frac{1}{2}\cdot a \cdot h_{a}\\ &=\frac{1}{2}\cdot 3cm \cdot 1cm\\ &=\frac{1}{2}\cdot 3cm²\\ &={\underline{\underline{1{,}5cm²}}} \end{aligned}
A1=ab=4cm1cm=4cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{1}&=a \cdot b\\ &=4cm \cdot 1cm\\ &={\underline{\underline{4cm²}}} \end{aligned}
A1=ab=15cm2cm=30cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{1}&=a \cdot b\\ &=15cm \cdot 2cm\\ &={\underline{\underline{30cm²}}} \end{aligned}
A4=ab=5cm3cm=15cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{4}&=a \cdot b\\ &=5cm \cdot 3cm\\ &={\underline{\underline{15cm²}}} \end{aligned}
A5=aha=5cm3cm=15cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{5}&=a \cdot h_{a}\\ &=5cm \cdot 3cm\\ &={\underline{\underline{15cm²}}} \end{aligned}
A6=ab=3cm2cm=6cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{6}&=a \cdot b\\ &=3cm \cdot 2cm\\ &={\underline{\underline{6cm²}}} \end{aligned}
A7=aha=6cm3cm=18cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{7}&=a \cdot h_{a}\\ &=6cm \cdot 3cm\\ &={\underline{\underline{18cm²}}} \end{aligned}
A8=ab=3cm11cm=33cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{8}&=a \cdot b\\ &=3cm \cdot 11cm\\ &={\underline{\underline{33cm²}}} \end{aligned}
A9=12aha=129cm5cm=1245cm²=22,5cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{9}&=\frac{1}{2}\cdot a \cdot h_{a}\\ &=\frac{1}{2}\cdot 9cm \cdot 5cm\\ &=\frac{1}{2}\cdot 45cm²\\ &={\underline{\underline{22{,}5cm²}}} \end{aligned}
A11=(a+c)ha2=(5cm+1cm)2cm2=(6cm)2cm2=12cm²2=6cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{11}&=\frac{(a+c) \cdot h_{a}}{2}\\ &=\frac{(5cm+1cm) \cdot 2cm}{2}\\ &=\frac{(6cm) \cdot 2cm}{2}\\ &=\frac{12cm² }{2}\\ &={\underline{\underline{6cm²}}} \end{aligned}
A12=12aha=121cm5cm=125cm²=2,5cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{12}&=\frac{1}{2}\cdot a \cdot h_{a}\\ &=\frac{1}{2}\cdot 1cm \cdot 5cm\\ &=\frac{1}{2}\cdot 5cm²\\ &={\underline{\underline{2{,}5cm²}}} \end{aligned}
A10=ab=8cm5cm=40cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{10}&=a \cdot b\\ &=8cm \cdot 5cm\\ &={\underline{\underline{40cm²}}} \end{aligned}

Schritt 3: Teilflächen addieren.

Agesamt=A1+A2+A3+A4+A5+A6+A7+A8+A9+A10+A11+A12=4cm²+1,5cm²+30cm²+15cm²+15cm²+6cm²+18cm²+33cm²+22,5cm²    +40cm²+6cm²+2,5cm2=193,5cm²\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} A_{gesamt}&=A_{1}+A_{2}+A_{3}+A_{4}+A_{5}+A_{6}+A_{7}+A_{8}+A_{9}+A_{10}+A_{11}+A_{12}\\ &=4cm²+1{,}5cm²+30cm²+15cm²+15cm²+6cm²+18cm²+33cm²+22{,}5cm²\\ &\ \ \ \ +40cm²+6cm²+2{,}5cm^2\\ &=\textbf{\underline{\underline{193{,}5cm²}}} \end{aligned}