Răspuns :
[tex]\it a)\ \Delta ADC - pitagoreic\ (30,\ \ 40,\ \ 50),\ \ \mathcal{P}=30+40+50=120\ cm\\ \\ \ Ducem\ DF||AC,\ F\in AB\\ \\ BD\perp AC,\ \ AC||DF \Rightarrow BD\perp DF \Rightarrow \Delta BDF-dreptunghic, \hatD=90^o\\ \\ DA=\ h\ pentru\ \Delta BDF \stackrel{T.h}{\Longrightarrow}\ DA^2=FA\cdot AB \Rightarrow AB=\dfrac{DA^2}{FA}\ \ \ \ \ (1)[/tex]
ACDF-paralelogram ⇒ FA = CD = 30 cm (2)
[tex]\it (1),\ (2) \Rightarrow AB=\dfrac{40^2}{30}=\dfrac{1600}{30}=\dfrac{160}{3}\ cm\\ \\ \\ \mathcal{A}=\dfrac{\mathcal{B}+b}{2}\cdot h=\dfrac{\dfrac{160}{3}+30}{2}\cdot40=\\ \\ \\ =(\dfrac{160}{3}+\ ^{3)}30)\cdot20=\dfrac{250}{3}\cdot20=\dfrac{5000}{3}\approx 1667\ cm^2[/tex]