• 1. Strahlensatz umstellen
  • Deliah Herbstritt
  • 21.07.2020
  • Mathematik
  • Flächen
  • R
  • 9
  • Einzelarbeit
  • Arbeitsblatt
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1
Berechne die fehlende Seitenlänge X. Die Maßeinheit ist cm. Runde wenn nötig auf zwei Stellen nach dem Komma.

Beispiel:

SBSB=SASA\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{SB'}}{\overline{SB}}=\frac{\overline{SA'}}{\overline{SA}}

X2=96\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{X}{2}=\frac{9}{6} |2\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot2

X=962\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{9}{6}\cdot2

X=3\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=3
Die Strecke SB\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SB'} ist also 3cm lang.
a)
Lösung
SBSB=SASA\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{SB'}}{\overline{SB}}=\frac{\overline{SA'}}{\overline{SA}}

X22=3210\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{X}{22}=\frac{32}{10} |22\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot22

X=321022\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{32}{10}\cdot22

X=70,4\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=70{,}4
Die Strecke SB\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SB'} ist also 70, 4cm lang.
b)
Lösung
SASA=SBSB\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{SA'}{SA}=\frac{SB'}{SB}

x102=670,8215\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{x}{102}=\frac{670{,}8}{215} |102\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot102

X=670,8215102\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{670{,}8}{215}\cdot102

X=318,24\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=318{,}24
Die Strecke SA\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SA'} ist also 318,24cm lang.
c)
Lösung
SASA=SBSB\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{SA}}{\overline{SA'}}=\frac{\overline{SB}}{\overline{SB'}}

X440=55275\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{X}{440}=\frac{55}{275} |440\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot440

X=55275440\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{55}{275}\cdot440

X=88\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=88
Die Strecke SA\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SA} ist also 88cm lang.
2
Gegeben:
SB=4,7cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SB}=4{,}7cm
SC=2,2cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SC}=2{,}2cm
SD=6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SD}=6cm
Gesucht: Länge der Strecke SA\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SA}
Lösung2
SASB=SCSD\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{SA}}{\overline{SB}}=\frac{\overline{SC}}{\overline{SD}}

X4,7=2,26\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{X}{4{,}7}=\frac{2{,}2}{6} |4,7\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot4{,}7

X=2,264,7\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{2{,}2}{6}\cdot4{,}7

X=1,72cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=1{,}72cm
Die Länge der Strecke SA\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SA} beträgt 1,72cm.
3
Gegeben:
SA=750cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SA}=750cm
SC=825cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{SC}=825cm
CD=550cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{CD}=550cm
Gesucht: Länge der Strecke AB\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{AB}
Lösung3
ABSA=DCSC\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{AB}}{\overline{SA}}=\frac{\overline{DC}}{\overline{SC}}

X750=550825\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{X}{750}=\frac{550}{825} |750\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot750

X=550825750\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{550}{825}\cdot750

X=500m\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=500m
Die Länge der Strecke AB\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{AB} beträgt 500m.