Für diese Aufgabe benötigst Du folgendes Grundwissen: Kreisumfang und Kreisfläche
Im Kreis gilt:
U = 2 r ⋅ π ; A = r 2 ⋅ π ; d = 2 r \ U=2r\cdot\pi;\quad\ A=r^2\cdot\pi;\quad d=2r U = 2 r ⋅ π ; A = r 2 ⋅ π ; d = 2 r
Berechnungen Spalte a.)
d = 2 r ⇒ d = 2 ⋅ 1 , 5 cm \ d=2r\ \Rightarrow\ d=2\cdot1{,}5\ \text{cm}\ d = 2 r ⇒ d = 2 ⋅ 1 , 5 cm
d = 3 cm \ \boxed{d=3\ \text{cm}} d = 3 cm
U = 2 r ⋅ π ⇒ U = 2 ⋅ 1 , 5 cm ⋅ π \ U=2r\cdot\pi\ \Rightarrow\ U=2\cdot1{,}5\ \text{cm}\cdot\pi U = 2 r ⋅ π ⇒ U = 2 ⋅ 1 , 5 cm ⋅ π
U = 9 , 4 cm \ \boxed{U=9{,}4\ \text{cm}} U = 9 , 4 cm
A = r 2 ⋅ π ⇒ A = 2 , 25 cm 2 ⋅ π \ A=r^2\cdot\pi\ \Rightarrow\ A=2{,}25\ \text{cm}^2\cdot\pi A = r 2 ⋅ π ⇒ A = 2 , 25 cm 2 ⋅ π
A = 7 , 1 cm 2 \ \boxed{A=7{,}1\ \text{cm}^2} A = 7 , 1 cm 2
Berechnungen Spalte b.) wie a.) jedoch r = 33 , 0 cm r=33{,}0\ \text{cm} r = 33 , 0 cm
Berechnungen Spalte c.)
r = d 2 ⇒ r = 2 , 4 m 2 \ r=\dfrac{d}{2}\ \Rightarrow\ r=\dfrac{2{,}4\ \text{m}}{2} r = 2 d ⇒ r = 2 2 , 4 m
r = 1 , 2 m \ \boxed{r=1{,}2\ \text{m}} r = 1 , 2 m
U = 2 r ⋅ π ⇒ U = 2 ⋅ 1 , 2 m ⋅ π \ U=2r\cdot\pi\ \Rightarrow\ U=2\cdot1{,}2\ \text{m}\cdot\pi U = 2 r ⋅ π ⇒ U = 2 ⋅ 1 , 2 m ⋅ π
U = 7 , 5 m \ \boxed{U=7{,}5\ \text{m}} U = 7 , 5 m
A = r 2 ⋅ π ⇒ A = 1 , 44 cm 2 ⋅ π \ A=r^2\cdot\pi\ \Rightarrow\ A=1{,}44\ \text{cm}^2\cdot\pi A = r 2 ⋅ π ⇒ A = 1 , 44 cm 2 ⋅ π
A = 4 , 5 m 2 \ \boxed{A=4{,}5\ \text{m}^2} A = 4 , 5 m 2
Berechnungen Spalte d.)
U = 2 r ⋅ π ⇒ 2 r = U π ⇒ 2 r = 71 , 6 m π \ U=2r\cdot\pi\ \Rightarrow\ 2r=\dfrac{U}{\pi}\ \Rightarrow\ 2r=\dfrac{71{,}6\text{\ m}}{\pi} U = 2 r ⋅ π ⇒ 2 r = π U ⇒ 2 r = π 71 , 6 m
2 r = d = 22 , 8 m \ 2r=\boxed{d=22{,}8\text{\ m}} 2 r = d = 22 , 8 m
r = 11 , 4 m \hspace{9mm}\boxed{r=11{,}4\text {\ m}} r = 11 , 4 m
A = r 2 ⋅ π ⇒ A = 129 , 96 m 2 ⋅ π \ A=r^2\cdot\pi\ \Rightarrow\ A=129{,}96\ \text{m}^2\cdot\pi A = r 2 ⋅ π ⇒ A = 129 , 96 m 2 ⋅ π
A = 408 , 3 m 2 \ \boxed{A=408{,}3\text{\ m}^2} A = 408 , 3 m 2
Berechnungen Spalte e.)
A = r 2 ⋅ π ⇒ r = A π ⇒ r = 12 , 6 cm 2 π \ A=r^2\cdot\pi\ \Rightarrow\ r=\sqrt{\dfrac{A}{\pi}}\ \Rightarrow\ r=\sqrt{\dfrac{12{,}6\text{\ cm}^2}{\pi}} A = r 2 ⋅ π ⇒ r = π A ⇒ r = π 12 , 6 cm 2
r = 2 , 0 cm ⇒ d = 2 ⋅ 2 , 0 cm \ \boxed{r=2{,}0\ \text{cm}}\ \Rightarrow\ d=\ 2\cdot2{,}0\text{\ cm} r = 2 , 0 cm ⇒ d = 2 ⋅ 2 , 0 cm
d = 4 , 0 cm \ \boxed{d=4{,}0\text{\ cm}} d = 4 , 0 cm
U = 2 r ⋅ π ⇒ U = 2 ⋅ 2 , 0 cm ⋅ π \ U=2r\cdot\pi\ \Rightarrow\ U=2\cdot2{,}0 \text{\ cm}\cdot\pi U = 2 r ⋅ π ⇒ U = 2 ⋅ 2 , 0 cm ⋅ π
U = 12 , 6 cm \ \boxed{U=12{,}6\text{\ cm}} U = 12 , 6 cm
Ergänze die Tabelle entsprechend.