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201)
Base minore $DC= \dfrac{3}{7}AB = \dfrac{3}{7}×35 = 15~cm\,$;
proiezione lato obliquo $HB = \dfrac{AB-DC}{2} = \dfrac{35-15}{2} = 10~cm\,$;
altezza $CH= \sqrt{BC^2-HB^2}=\sqrt{14,5^2-10^2} = 10,5~cm$ (teorema di Pitagora);
area $A= \dfrac{(AB+DC)·CH}{2} = \dfrac{(35+15)×10,5}{2} = 262,5~cm^2\,$.
DC = 35*3/7 = 15 cm
CH =√14,5^2-((35-15)/2)^2 = 10,5 cm
area A = (35+15)*10,5/2 = 25*10,5 = 262,5 cm^2