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.