TitelX-Figur Übungen
AutorDeliah Herbstritt
veröffentlicht21.07.2020
FachMathematik
KompetenzbereichFlächen
NiveaustufeR
Phase9
SozialformEinzelarbeit
MaterialformArbeitsblatt

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Berechne die fehlende Strecke x.

Lösung1

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{\overline{SD}}{\overline{SB}}=\frac{\overline{SA}}{\overline{SC}}

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

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \cdot5{,}6cm

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

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

Antwort: Die Strecke x ist 6,21cm lang.

Ergänze zu der Strahlensatzfigur die Verhältnisgleichungen so oft wie vorgegeben.

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

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

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

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

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

Gegeben:

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{AB}=8{,}1cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{DB}=6{,}6cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{DE}=1{,}9cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{EB}=4{,}8cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{EF}=1{,}3cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{CB}=5{,}9cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{DC}=4{,}6cm

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \overline{AD}=2{,}6cm

Lösung3

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{\overline{AE}}{\overline{EB}}=\frac{\overline{AD}}{\overline{DC}}

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} \frac{z}{4.8cm}=\frac{2{,}6cm}{4{,}6cm}

\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}}}} |\cdot4{,}8cm

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

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

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