(1) Länge $\overline{DE}\backslash overline\{DE\}$ :

$\sin {60}^{\circ }={\displaystyle \frac{\overline{DE}}{\overline{AE}}}={\displaystyle \frac{\overline{DE}}{8}}\backslash sin\; 60^\backslash circ=\backslash dfrac\{\backslash overline\{DE\}\}\{\backslash overline\{AE\}\}=\backslash dfrac\{\backslash overline\{DE\}\}\{8\}$ $\Rightarrow \backslash Rightarrow$ $\overline{DE}\approx \mathrm{0,866}\cdot 8\approx \mathrm{6,9}\backslash overline\{DE\}\backslash approx0\{,\}866\backslash cdot\; 8\backslash approx6\{,\}9$

(2) Länge $\overline{AD}\backslash overline\{AD\}$:

$\cos {50}^{\circ }={\displaystyle \frac{\overline{AD}}{\overline{AE}}}={\displaystyle \frac{\overline{AD}}{8}}\backslash cos\; 50^\backslash circ=\backslash dfrac\{\backslash overline\{AD\}\}\{\backslash overline\{AE\}\}=\backslash dfrac\{\backslash overline\{AD\}\}\{\{8\}\}$ $\Rightarrow \backslash Rightarrow$ $\overline{AD}=\mathrm{0,5}\cdot 8=4\backslash overline\{AD\}=0\{,\}5\backslash cdot\; 8=4$

(3) Länge $\overline{GF}\backslash overline\{GF\}$:

${\overline{GF}}^2=1\cdot (11+4)=15\backslash overline\{GF\}^2=1\backslash cdot(11+4)=15$ $\Rightarrow \backslash Rightarrow$ $\overline{GF}\approx \mathrm{3,9}\backslash overline\{GF\}\backslash approx3\{,\}9$

Flächeninhalt Trapez $DGFEDGFE$:

$A_{DGFE}={\displaystyle \frac{(\mathrm{6,9}+\mathrm{3,9})\cdot 11}{2}}\approx \mathrm{59,4}A\_\{DGFE\}=\backslash dfrac\{(6\{,\}9+3\{,\}9)\backslash cdot\; 11\}\{2\}\backslash approx59\{,\}4$

Der Flächeninhalt des Trapezes $DGFEDGFE$ beträgt $\approx \mathrm{59,4}{\text{ cm}}^2\mathrm{.}\backslash approx\; 59\{,\}4\backslash cm^2.$$$

Länge $\overline{EF}\backslash overline\; \{EF\}$:

$\overline{EF}=\sqrt{[\overline{DE}-\overline{GF}{]}^2+{\overline{DG}}^2}=\sqrt{[\mathrm{6,9}-\mathrm{3,9}{]}^2+{11}^2}\approx \backslash overline\{EF\}=\backslash sqrt\{[\backslash overline\{DE\}-\backslash overline\; \{GF\}]^2+\backslash overline\{DG\}^2\}=\backslash sqrt\{[6\{,\}9-3\{,\}9]^2+11^2\}\backslash approx$ 11,4

$U_{DGFE}=11+\mathrm{3,9}+\mathrm{11,4}+\mathrm{6,9}=\mathrm{33,2}U\_\{DGFE\}=11+3\{,\}9+11\{,\}4+6\{,\}9=33\{,\}2$

Der Umfang des Trapezes beträgt $\mathrm{33,2}\text{ cm}33\{,\}2\backslash cm$.