Zeige, dass die beiden Kreisbögen $\backslash stackrel\{\backslash frown\}\{PR\}$ und $\backslash stackrel\{\backslash frown\}\{QR\}$ gleich lang sind

Für den Kreisbogen gilt allgemein:

$b={\displaystyle \frac{\alpha }{{360}^{\circ }}}\cdot 2\pi rb\; =\backslash dfrac\{\backslash alpha\}\{360^\{\backslash circ\}\}\; \backslash cdot\; 2\backslash pi\; r$

Berechne die Kreisbogenlänge von $\backslash stackrel\{\backslash frown\}\{PR\}$:

Hier ist $\alpha ={180}^{\circ }\backslash alpha=180^\{\backslash circ\}$ und $r=ar=a$.

$\Rightarrow ={\displaystyle \frac{\alpha }{{360}^{\circ }}}\cdot 2\cdot \pi \cdot r={\displaystyle \frac{{180}^{\circ }}{{360}^{\circ }}}\cdot 2\cdot \pi \cdot a={\displaystyle \frac{1}{2}}\cdot 2\cdot \pi \cdot a=\pi \cdot a\backslash Rightarrow\backslash ;\backslash stackrel\{\backslash frown\}\{PR\}=\backslash dfrac\{\backslash alpha\}\{360^\{\backslash circ\}\}\; \backslash cdot\; 2\backslash cdot\backslash pi\; \backslash cdot\; r=\backslash dfrac\{180^\{\backslash circ\}\}\{360^\{\backslash circ\}\}\; \backslash cdot\; 2\backslash cdot\backslash pi\backslash cdot\; a=\backslash dfrac\{1\}\{2\}\; \backslash cdot\; 2\backslash cdot\backslash pi\backslash cdot\; a=\backslash pi\backslash cdot\; a$

Berechne die Kreisbogenlänge von $\backslash stackrel\{\backslash frown\}\{QR\}$:

Hier ist $\alpha ={90}^{\circ }\backslash alpha=90^\{\backslash circ\}$ und $r=2ar=2\; a$.

$\Rightarrow ={\displaystyle \frac{\alpha }{{360}^{\circ }}}\cdot 2\cdot \pi \cdot r={\displaystyle \frac{{90}^{\circ }}{{360}^{\circ }}}\cdot 2\cdot \pi \cdot 2a={\displaystyle \frac{1}{4}}\cdot 2\cdot \pi \cdot 2a=\pi \cdot a\backslash Rightarrow\backslash ;\backslash stackrel\{\backslash frown\}\{QR\}=\backslash dfrac\{\backslash alpha\}\{360^\{\backslash circ\}\}\; \backslash cdot\; 2\backslash cdot\backslash pi\backslash cdot\; r=\backslash dfrac\{90^\{\backslash circ\}\}\{360^\{\backslash circ\}\}\; \backslash cdot\; 2\backslash cdot\backslash pi\backslash cdot\; 2a=\backslash dfrac\{1\}\{4\}\; \backslash cdot\; 2\backslash cdot\backslash pi\backslash cdot\; 2a=\backslash pi\backslash cdot\; a$

Die beiden Kreisbögen $\backslash stackrel\{\backslash frown\}\{PR\}$ und $\backslash stackrel\{\backslash frown\}\{QR\}$ sind gleich lang.