Trapez ABCD

Trapez ABCD

Strecke AD e mal Wurzel 6

Strecke CD e mal wurzel 2

Winkel bei C 120° Winkel bei D 135°

Umfang und Flächeninhalt nur mit e und Wurzeln berechnen

$$\\ \small{\text{$ \begin{array}{rcl} a=\overline{AD}=e\sqrt{6}\quad b=\overline{CD}=e\sqrt{2}\quad c=\overline{CB}= \?\quad d=\overline{AB}=p+b+q\quad \alpha=ADC = 135 \ensurement{^{\circ}}\quad \beta=DCB = 120 \ensurement{^{\circ}} \end{array} $}}\\ \small{\text{$ \begin{array}{rcl} \cos{(\alpha-90\ensurement{^{\circ}})}&=&\dfrac{h}{a}\\ h &=& a\cdot \cos{(\alpha-90\ensurement{^{\circ}})}\\ h &=& e\sqrt{6} \cdot \cos{(135-90\ensurement{^{\circ}})}\\ h &=& e\sqrt{6} \cdot \cos{(45\ensurement{^{\circ}})} \quad | \quad \cos{(45\ensurement{^{\circ}})} = \dfrac{ \sqrt{2} } {2}\\ h &=& e\sqrt{6} \cdot \dfrac{ \sqrt{2} } {2}\\ h &=& e \cdot \dfrac{ \sqrt{4\cdot 3} } {2}\\ h &=& e \cdot 2 \cdot \dfrac{ \sqrt{3} } {2}\\ h &=& e \sqrt{3} \\ \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} \sin{(\alpha-90\ensurement{^{\circ}})}&=&\dfrac{p}{a}\\ p &=& a\cdot \sin{(\alpha-90\ensurement{^{\circ}})}\\ p &=& e\sqrt{6} \cdot \sin{(135-90\ensurement{^{\circ}})}\\ p &=& e\sqrt{6} \cdot \sin{(45\ensurement{^{\circ}})} \quad | \quad \sin{(45\ensurement{^{\circ}})} = \dfrac{ \sqrt{2} } {2}\\ p &=& e\sqrt{6} \cdot \dfrac{ \sqrt{2} } {2}\\ p &=& e \cdot \dfrac{ \sqrt{4\cdot 3} } {2}\\ p &=& e \cdot 2 \cdot \dfrac{ \sqrt{3} } {2}\\ p &=& e \sqrt{3} \\ \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcl} \tan{( 180\ensurement{^{\circ}} - \beta)}&=&\dfrac{h}{q}\\ q &=& \dfrac{ h } { \tan{( 180\ensurement{^{\circ}} - \beta)} } \\\\ q &=& \dfrac{ e\sqrt{3} } { \tan{( 180\ensurement{^{\circ}} - 120\ensurement{^{\circ}} )} } \\\\ q &=& \dfrac{ e\sqrt{3} } { \tan{( 60\ensurement{^{\circ}} )} } \quad | \quad \tan{( 60\ensurement{^{\circ}}) } = \sqrt{3} \\\\ q &=&\dfrac{ e\sqrt{3} } { \sqrt{3} }\\\\ q &=& e \end{array} $}}$$

$$\begin{array}{rcl} \sin{( 180\ensurement{^{\circ}} - \beta)}&=&\dfrac{h}{c}\\ c &=& \dfrac{ h } { \sin{( 180\ensurement{^{\circ}} - \beta)} } \\\\ c &=& \dfrac{ e\sqrt{3} } { \sin{( 180\ensurement{^{\circ}} - 120\ensurement{^{\circ}} )} } \\\\ c &=& \dfrac{ e\sqrt{3} } { \sin{( 60\ensurement{^{\circ}} )} } \quad | \quad \sin{( 60\ensurement{^{\circ}}) } = \dfrac{1}{2}\sqrt{3} \\\\ c &=&\dfrac{ e\sqrt{3} } {\dfrac{1}{2}\sqrt{3} }\\\\ c &=& 2e \end{array} $}}$$

$$\\d = p + b + q \qquad d = e\sqrt{3}+ e\sqrt{2} + e\\\\ \begin{array}{rcl} U&=&a+b+c+d \\ U&=& e\sqrt{6} +e\sqrt{2} + 2e+e\sqrt{3}+ e\sqrt{2} + e\\ U&=& 3e +2e\sqrt{2}+ e\sqrt{3} + e\sqrt{6}\\ U&=& e \left( 3 +2\sqrt{2}+ \sqrt{3} + \sqrt{6} \right) \end{array}$$

$$\boxed{ U&=& e \left( 3 +2\sqrt{2}+ \sqrt{3} + \sqrt{6} \right) }$$

$$\begin{array}{rcl} A&=& \left( \dfrac{d+b}{2}\ \right) \cdot h\\\\ A&=& \left( \dfrac{ e\sqrt{3}+ e\sqrt{2} + e+ e\sqrt{2} }{2}\ \right) \cdot e\sqrt{3}\\\\ A&=& \left( \dfrac{ e + 2e\sqrt{2} + e\sqrt{3} }{2}\ \right) \cdot e\sqrt{3}\\\\ A&=& \left( e + 2e\sqrt{2} + e\sqrt{3} \right) \cdot \dfrac{ e\sqrt{3} } {2}\\\\ A&=& \left( 1 + 2\sqrt{2} + \sqrt{3} \right) \cdot \dfrac{ e^2\sqrt{3} } {2} \end{array}$$

$$\boxed{ A= \left( 1 + 2\sqrt{2} + \sqrt{3} \right) \cdot \dfrac{ e^2\sqrt{3} } {2} }$$

Soll die Zeichnung so aussehen?

Ja, genau so.