[tex]DB=1.8+ \frac{2}{3}DB \\ \frac{1}{3}DB =1,8;DB=5,4 cm\\ EB= \frac{2}{3}*5,4 =3,6cm \\ [/tex]
din ΔDAB-dreptunghic
[tex]AE= \sqrt{1,8*3,6}=1,8 \sqrt{2}cm [/tex]
dinΔDEA-dreptunghic
[tex]AD= \sqrt{DE^2+AE^2}= \sqrt{3,24+6,48}= \sqrt{9,72}=18 \sqrt{0,03}cm \\ [/tex]=1,8√3
dinΔAEB-dreptunghic
[tex]AB= \sqrt{AE^2+EB^2} = \sqrt{6,48+12,96}= \sqrt{19,44} =1,8 \sqrt{6}cm [/tex]
[tex]A=AB*AD=1,8 \sqrt{6}*1,8 \sqrt{3}=3,24 \sqrt{18}=9,72 \sqrt{2}cm^2 [/tex]
