Berechnung der Strecke $\mathrm{\mid }\overline{DE}\mathrm{\mid }|\backslash overline\; \{DE\}|$

Berechnung der Strecke $\mathrm{\mid }\overline{DC}\mathrm{\mid }|\backslash overline\; \{DC\}|$

Hierfür nutzen wir den Tangens von $\mathrm{\sphericalangle }BAD=60\mathrm{°}\backslash sphericalangle\{BAD\}=\; 60°$

$\tan 60\mathrm{°}={\displaystyle \frac{GK}{AK}}={\displaystyle \frac{\mathrm{\mid }\overline{DE}\mathrm{\mid }}{4\text{ cm}}}\backslash quad\; \backslash tan60°=\backslash dfrac\{GK\}\{AK\}=\backslash dfrac\{|\backslash overline\; \{DE\}|\}\{\backslash color\{green\}4\backslash text\{\; cm\}\}$

$\iff \mathrm{\mid }\overline{DE}\mathrm{\mid }=4\text{ cm}\cdot \tan 60\mathrm{°}=4\cdot \sqrt{3}\text{ cm}\backslash Leftrightarrow\; |\backslash overline\; \{DE\}|=\; \{4\backslash text\{\; cm\}\backslash cdot\; \backslash tan60°=\backslash color\{red\}4\backslash cdot\backslash sqrt3\}\backslash text\{\backslash color\{red\}\; cm\}$

Flächeninhalt des gleichschenkligen Trapezes

$A_{\text{Trapez}}={\displaystyle \frac{(a+c)}{2}}\cdot hA\_\{\backslash text\{Trapez\}\}=\backslash dfrac\{\{(a+c)\}\}\{2\}\backslash cdot\; h$

• $a=AB=11\text{cm}a=AB=11\backslash :\; \backslash text\{cm\}$

• $c=DC=3\text{cm}c=DC=\; \backslash color\{blue\}3\backslash :\backslash text\{cm\}$

• $h=DE=4\cdot \sqrt{3}cmh=DE=\backslash color\{red\}4\backslash cdot\backslash sqrt\{3\}\backslash :cm$

$\Rightarrow A_{\text{Trapez}\text{ABCD}}={\displaystyle \frac{(AB+CD)}{2}}\cdot DE={\displaystyle \frac{(11+3)}{2}}\cdot 4\cdot \sqrt{3}{\text{cm}}^2=28\cdot \sqrt{3}\text{ cm}\backslash Rightarrow\; A\_\{\backslash text\{Trapez\backslash :ABCD\}\}=\backslash dfrac\{\{(AB+CD)\}\}\{2\}\backslash cdot\; DE=\backslash dfrac\{\{(11+3)\}\}\{2\}\backslash cdot\; \{4\backslash cdot\backslash sqrt\{3\}\backslash :\backslash text\{cm\}^2\}=28\backslash cdot\; \backslash sqrt\{3\}\backslash text\{\; cm\}$ ${}^2^2$