Trapezoid 83

Trapezoid ABCD is composed of five triangles. Points E, and G divide segment AB in the ratio 2:4:3 (in this order) into three segments. Point F is the midpoint of segment AD. Triangle AEF is isosceles and right-angled. Triangles GBC and CDG are right-angled. The area of ​​triangle AEF is 8cm². What is the perimeter of the trapezoid?

Correct answer:

o =  54.9676 cm

Step-by-step explanation:

$$\begin{gathered} S_1=8{\text{cm}}^2 \\ \Delta AEF: \\ \angle AFE=90{}^{\circ } \\ |AF|=|FE| \\ |AE|=\sqrt{2}\cdot |AF|=\sqrt{2}\cdot |FE| \\ {S}_1=\frac{|AF|\cdot |FE|}{2}=\frac{|AF{|}^2}{2} \\ AF=\sqrt{2\cdot S_1}=\sqrt{2\cdot 8}=4\text{cm} \\ AD=2\cdot AF=2\cdot 4=8\text{cm} \\ AE=\sqrt{2}\cdot AF=\sqrt{2}\cdot 4=4\sqrt{2}\text{cm}\doteq 5.6569\text{cm} \\ AE:EG:GB=2:3:4=2x:4x:3x \\ AE=2x \\ x=AE/2=5.6569/2=2\sqrt{2}\text{cm}\doteq 2.8284\text{cm} \\ AB=(2+4+3)\cdot x=(2+4+3)\cdot 2.8284=18\sqrt{2}\text{cm}\doteq 25.4558\text{cm} \\ {S}_1=\frac{AE\cdot h_1}{2} \\ {h}_2=\frac{2\cdot S_1}{AE}=\frac{2\cdot 8}{5.6569}=2\sqrt{2}\text{cm}\doteq 2.8284\text{cm} \\ h=2\cdot h_2=2\cdot 2.8284=4\sqrt{2}\text{cm}\doteq 5.6569\text{cm} \\ h=|GC| \\ BG=3\cdot x=3\cdot 2.8284=6\sqrt{2}\text{cm}\doteq 8.4853\text{cm} \\ BC=\sqrt{h^2+B{G}^2}=\sqrt{5.6569^2+8.4853^2}=2\sqrt{26}\text{cm}\doteq 10.198\text{cm} \\ |EG|=|DC|=4\cdot x \\ CD=4\cdot x=4\cdot 2.8284=8\sqrt{2}\text{cm}\doteq 11.3137\text{cm} \\ o=AB+BC+CD+AD=25.4558+10.198+11.3137+8=54.9676\text{cm} \end{gathered}$$

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