Für diese Aufgabe benötigst Du folgendes Grundwissen: Sinussatz und Cosinussatz
Lösung Teilaufgabe 1.1 Zur Berechnung der Strecke benötigst du den Cosinussatz .
B D ‾ \displaystyle \overline{BD} B D = = = A B ‾ 2 + A D ‾ 2 − 2 ⋅ A B ‾ ⋅ A D ‾ ⋅ cos ∢ B A D \displaystyle \sqrt {\overline{AB}^2+\overline{AD}^2-2\cdot\overline{AB}\cdot\overline{AD}\cdot\cos\sphericalangle BAD} A B 2 + A D 2 − 2 ⋅ A B ⋅ A D ⋅ cos ∢ B A D = = = 1 0 2 + 8 2 − 2 ⋅ 10 ⋅ 8 ⋅ cos 100 ° c m \displaystyle \sqrt{10^2+8^2-2\cdot10\cdot8\cdot\cos100°}cm 1 0 2 + 8 2 − 2 ⋅ 10 ⋅ 8 ⋅ cos 100° c m = = = 100 + 64 − 160 ⋅ ( − 0 , 1736 ) c m \displaystyle \sqrt{100+64-160\cdot(-0{,}1736)}\,cm 100 + 64 − 160 ⋅ ( − 0 , 1736 ) c m = = = 164 + 27 , 78 c m \displaystyle \sqrt {164+27{,}78}\,cm 164 + 27 , 78 c m = = = 191 , 78 c m \displaystyle \sqrt {191{,}78}\,cm 191 , 78 c m = = = 13 , 85 c m \displaystyle 13{,}85cm 13 , 85 c m
Zur Berechnung des Maßes des Winkels ∢ D B A \sphericalangle DBA ∢ D B A benötigst du den Sinussatz .
sin ∢ D B A A D ‾ \displaystyle \dfrac{\sin
\sphericalangle DBA}{\overline{AD}} A D sin ∢ D B A = = = sin 100 ° B D ‾ \displaystyle \dfrac{\sin 100°}{\overline{BD}} B D sin 100° sin ∢ D B A 8 c m \displaystyle \dfrac{\sin
\sphericalangle DBA}{8cm} 8 c m sin ∢ D B A = = = sin 100 ° 13 , 85 c m \displaystyle \dfrac{\sin 100°}{13{,}85cm} 13 , 85 c m sin 100° sin ∢ D B A \displaystyle \sin\sphericalangle DBA sin ∢ D B A = = = 0 , 9848 ⋅ 8 c m 13 , 85 c m \displaystyle \dfrac{0{,}9848\cdot8cm}{13{,}85cm} 13 , 85 c m 0 , 9848 ⋅ 8 c m sin ∢ D B A \displaystyle \sin\sphericalangle DBA sin ∢ D B A = = = 0 , 5688 \displaystyle 0{,}5688 0 , 5688 ∢ D B A \displaystyle \sphericalangle DBA ∢ D B A = = = 34 , 67 ° \displaystyle 34{,}67° 34 , 67°
Lösung Teilaufgabe 1.2 sin ∢ D C A 8 c m \displaystyle \dfrac{\sin
\sphericalangle DCA}{{8cm}} 8 c m sin ∢ D C A = = = sin ( 180 ° − 100 ° ) 10 c m \displaystyle \dfrac{\sin (180°-100°)}{{10cm}} 10 c m sin ( 180° − 100° ) sin ∢ D C A \displaystyle \sin\sphericalangle DCA sin ∢ D C A = = = sin ( 80 ° ) ⋅ 8 c m 10 c m \displaystyle \dfrac{\sin(80°)\cdot 8cm}{{10cm}} 10 c m sin ( 80° ) ⋅ 8 c m sin ∢ D C A \displaystyle \sin\sphericalangle DCA sin ∢ D C A = = = 0 , 9848 ⋅ 8 c m 10 c m \displaystyle \dfrac{0{,}9848\cdot 8cm}{10cm} 10 c m 0 , 9848 ⋅ 8 c m sin ∢ D C A \displaystyle \sin\sphericalangle DCA sin ∢ D C A = = = 0 , 7878 \displaystyle 0{,}7878 0 , 7878 ∢ D C A \displaystyle \sphericalangle DCA ∢ D C A = = = 51 , 98 ° \displaystyle 51{,}98° 51 , 98°
Die Winkel ∢ B A C \sphericalangle BAC ∢ B A C und ∢ D C A \sphericalangle DCA ∢ D C A sind Wechselwinkel an den zueinander parallelen Geraden AB und CD. Folglich gilt: ∢ B A C = ∢ D C A = 51 , 98 \sphericalangle BAC=\sphericalangle DCA=51{,}98 ∢ B A C = ∢ D C A = 51 , 98 °
Lösung Teilaufgabe 1.3 Das Viereck ABCD \text{Viereck}_\text{ABCD} Viereck ABCD setzt sich zusammen aus dem Dreieck ACD \text{Dreieck}_\text{ACD} Dreieck ACD und dem Dreieck ABC \text{Dreieck}_\text{ABC} Dreieck ABC
Berechne die Fläche der beiden Dreiecke und und addiere die Flächen.
A A C D = 1 2 ⋅ s i n ∢ C A D ⋅ A D ‾ ⋅ A C ‾ ∢ C A D = 100 ° − 51 , 98 ° = 48 , 02 ° \text{A}_{ACD}\ =\ \dfrac{1}{2}\cdot sin\sphericalangle CAD\cdot\overline{AD}\cdot\overline{AC}\hspace{28 mm} \sphericalangle CAD\ =\ 100°-\ 51{,}98°\ =\ 48{,}02° A A C D = 2 1 ⋅ s in ∢ C A D ⋅ A D ⋅ A C ∢ C A D = 100° − 51 , 98° = 48 , 02°
A A C D = 1 2 ⋅ s i n 48 , 02 ° ⋅ 8 c m ⋅ 10 c m \text{A}_{ACD}\ =\ \dfrac{1}{2}\cdot sin\ 48{,}02°\cdot 8\ cm\cdot10\ cm A A C D = 2 1 ⋅ s in 48 , 02° ⋅ 8 c m ⋅ 10 c m
A A C D = 1 2 ⋅ 0 , 7434 ⋅ 80 c m 2 \text{A}_{ACD}\ =\dfrac{1}{2}\cdot 0{,}7434\cdot80\ cm^2 A A C D = 2 1 ⋅ 0 , 7434 ⋅ 80 c m 2
A A C D = 29 , 74 c m 2 \text{A}_{ACD}\ =\ 29{,}74\ cm^2 A A C D = 29 , 74 c m 2
A A B C = 1 2 ⋅ s i n ∢ B A C ⋅ A B ‾ ⋅ A C ‾ \text{A}_{ABC}\ =\ \dfrac{1}{2}\cdot sin\sphericalangle BAC\cdot\overline{AB}\cdot\overline{AC} A A BC = 2 1 ⋅ s in ∢ B A C ⋅ A B ⋅ A C
A A B C = 1 2 ⋅ s i n 51 , 98 ° ⋅ 10 c m ⋅ 10 c m \text{A}_{ABC}\ =\ \dfrac{1}{2}\cdot sin\ 51{,}98°\cdot 10\ cm\cdot10\ cm A A BC = 2 1 ⋅ s in 51 , 98° ⋅ 10 c m ⋅ 10 c m
A A B C = 1 2 ⋅ 0 , 7878 ⋅ 100 c m 2 \text{A}_{ABC}\ =\dfrac{1}{2}\cdot 0{,}7878\cdot100\ cm^2 A A BC = 2 1 ⋅ 0 , 7878 ⋅ 100 c m 2
A A B C = 39 , 39 c m 2 \text{A}_{ABC}\ =\ 39{,}39\ cm^2 A A BC = 39 , 39 c m 2
Viereck ABCD = 29 , 74 c m 2 + 39 , 39 c m 2 \text{Viereck}_\text{ABCD}\ =\ 29{,}74 cm^2\ +\ 39{,}39\ cm^2 Viereck ABCD = 29 , 74 c m 2 + 39 , 39 c m 2
Viereck ABCD = 69 , 13 c m 2 \text{Viereck}_\text{ABCD}\ =\ 69{,}13\ cm^2 Viereck ABCD = 69 , 13 c m 2
Lösung Teilaufgabe 1.4
Lösung Teilaufgabe 1.5 Berechne den Flächeninhalt der Figur F B E \text{Figur}_{FBE} Figur FBE
Die Figur F B E \text{Figur}_{FBE} Figur FBE setzt sich zusammen aus einem dem Dreieck MBE \text{Dreieck}_\text{MBE} Dreieck MBE und dem Kreissektor FME \text{Kreissektor}_\text{FME} Kreissektor FME
Berechne die Fläche des Kreissektors F M E \text{Kreissektors}_{FME} Kreissektors FME
A FME = ∢ F M E 36 0 ∘ ⋅ M E ‾ 2 ⋅ π M E ‾ = M B ‾ ⋅ sin ∢ A B D M E ‾ = 5 c m ⋅ sin 34 , 6 7 ∘ M E ‾ = 5 c m ⋅ 0 , 5688 M E ‾ = 2 , 84 c m \text{A}_\text{FME}\ =\ \dfrac{\sphericalangle FME}{360^\circ}\cdot\overline{ME}^2\cdot\pi\hspace{28mm}\overline{ME}\ =\ \overline{MB}\cdot\sin\sphericalangle ABD\\\hspace{66mm}\overline{ME}\ =\ 5\ cm\cdot\sin 34{,}67^\circ\\\hspace{66mm}\overline{ME}\ =\ 5\ cm\cdot 0{,}5688\\\hspace{66mm}\overline{ME}\ =\ 2{,}84\ cm A FME = 36 0 ∘ ∢ FME ⋅ ME 2 ⋅ π ME = MB ⋅ sin ∢ A B D ME = 5 c m ⋅ sin 34 , 6 7 ∘ ME = 5 c m ⋅ 0 , 5688 ME = 2 , 84 c m
A FME = 124 , 6 7 ∘ 36 0 ∘ ⋅ ( 2 , 84 c m ) 2 ⋅ π \text{A}_\text{FME}\ =\ \dfrac{124{,}67^\circ}{360^\circ}\cdot{(2{,}84\ cm)}^2\cdot\pi A FME = 36 0 ∘ 124 , 6 7 ∘ ⋅ ( 2 , 84 c m ) 2 ⋅ π
A FME = 8 , 77 c m 2 \text{A}_\text{FME}\ =\ 8{,}77\ cm^2 A FME = 8 , 77 c m 2
Berechne die Fläche des Dreiecks MBE \text{Dreiecks}_\text{MBE} Dreiecks MBE
A MBE = 1 2 ⋅ sin ∢ E M B ⋅ M E ‾ ⋅ M B ‾ \text{A}_\text{MBE}\ =\ \dfrac{1}{2}\cdot\sin\sphericalangle{EMB}\cdot\overline{ME}\cdot\overline{MB} A MBE = 2 1 ⋅ sin ∢ EMB ⋅ ME ⋅ MB
A MBE = 1 2 ⋅ sin ∢ 55 , 33 ° ⋅ 2 , 84 c m ⋅ 5 c m \text{A}_\text{MBE}\ =\ \dfrac{1}{2}\cdot\sin\sphericalangle{55{,}33°}\cdot{2{,}84\ cm}\cdot{5\ cm} A MBE = 2 1 ⋅ sin ∢ 55 , 33° ⋅ 2 , 84 c m ⋅ 5 c m
A MBE = 5 , 84 c m 2 \text{A}_\text{MBE}\ =\ 5{,}84\ cm^2 A MBE = 5 , 84 c m 2
Addiere die beiden Flächen
A FBE = A FME + A MBE \text{A}_\text{FBE}\ =\ \text{A}_\text{FME}\ +\ \text{A}_\text{MBE} A FBE = A FME + A MBE
A FBE = 8 , 77 c m 2 + 5 , 84 c m 2 \text{A}_{\text{FBE}}\ =\ 8{,}77\ cm^2\ +\ 5{,}84\ cm^2 A FBE = 8 , 77 c m 2 + 5 , 84 c m 2
A FBE = 14 , 61 c m 2 \text{A}_\text{FBE}\ =\ 14{,}61\ cm^2 A FBE = 14 , 61 c m 2
Berechne den prozentualen Anteil des Flächeninhalts A F B E \text{A}_{FBE} A FBE der Figur F B E FBE FBE am Flächeninhalt A ABCD \text{A}_\text{ABCD} A ABCD des Vierecks A B C D . ABCD. A BC D .
Anteil P r o z e n t = A FBE A ABCD ⋅ 100 % \text{Anteil}_{Prozent}\ =\ \dfrac{\text{A}_\text{FBE}}{\text{A}_\text{ABCD}}\cdot100\% Anteil P roze n t = A ABCD A FBE ⋅ 100%
Anteil P r o z e n t = 14 , 61 c m 2 69 , 12 c m 2 ⋅ 100 % \text{Anteil}_{Prozent}\ =\ \dfrac{14{,}61\ cm^2}{69{,}12\ cm^2}\cdot100\% Anteil P roze n t = 69 , 12 c m 2 14 , 61 c m 2 ⋅ 100%
Anteil P r o z e n t = 21 , 14 % \text{Anteil}_{Prozent}\ =\ 21{,}14\% Anteil P roze n t = 21 , 14%
Lösung Teilaufgabe 1.6 ∢ C G D = 180 ° − 51 , 98 ° − 34 , 67 ° ⇒ ∢ C G D = 93 , 35 ° \sphericalangle CGD \ =\ 180°-51{,}98°-34{,}67°\qquad\Rightarrow\qquad \sphericalangle CGD \ =\ \ 93{,}35° ∢ CG D = 180° − 51 , 98° − 34 , 67° ⇒ ∢ CG D = 93 , 35°
Wegen ∢ C G D ≠ 90 ° \sphericalangle CGD \neq\ 90° ∢ CG D = 90° gilt: D G ‾ > d ( D ; [ A C ] ) \ \ \overline{DG} > d(D;[AC]) D G > d ( D ; [ A C ]) .