• X-Figur Übungen
  • Deliah Herbstritt
  • 21.07.2020
  • Mathematik
  • Flächen
  • R
  • 9
  • Einzelarbeit
  • Arbeitsblatt
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1
Berechne die fehlende Strecke x.
Lösung1
SDSB=SASC\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{SD}}{\overline{SB}}=\frac{\overline{SA}}{\overline{SC}}

X5,6cm=4,1cm3,7cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{X}{5{,}6cm}=\frac{4{,}1cm}{3{,}7cm} 5,6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cdot5{,}6cm

X=4,1cm3,7cm5,6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X=\frac{4{,}1cm}{3{,}7cm}\cdot5{,}6cm

X6,21cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} X\approx6{,}21cm

Antwort: Die Strecke x ist 6,21cm lang.
2
Ergänze zu der Strahlensatzfigur die Verhältnisgleichungen so oft wie vorgegeben.

c+dc\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \dfrac{c+d}{c} = f+gf\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cloze{\frac{f+g}{f}}

hc+d\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \dfrac{h}{c+d} = ic\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cloze{\frac{i}{c}}

ih\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \dfrac{i}{h} = fg+f\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cloze{\frac{f}{g+f}}

cf\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \dfrac{c}{f} = c+df+g\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \cloze{\frac{c+d}{f+g}}

3
Berechne z mithilfe der Strahlensatzgleichung. Markiere die gegebenen Seiten in der Figur farbig.

Gegeben:

AB=8,1cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{AB}=8{,}1cm
DB=6,6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{DB}=6{,}6cm
DE=1,9cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{DE}=1{,}9cm
EB=4,8cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{EB}=4{,}8cm
EF=1,3cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{EF}=1{,}3cm
CB=5,9cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{CB}=5{,}9cm
DC=4,6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{DC}=4{,}6cm
AD=2,6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \overline{AD}=2{,}6cm
Lösung3
AEEB=ADDC\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{\overline{AE}}{\overline{EB}}=\frac{\overline{AD}}{\overline{DC}}

z4.8cm=2,6cm4,6cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \frac{z}{4.8cm}=\frac{2{,}6cm}{4{,}6cm} 4,8cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} |\cdot4{,}8cm

z=2,6cm4,6cm4,8cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} z=\frac{2{,}6cm}{4{,}6cm}\cdot4{,}8cm

z2,71cm\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} z\approx2{,}71cm

Antwort: Die Strecke z ist 2,71cm lang.