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Diagonale $AC= \sqrt{14^2+14^2} = 14\sqrt{2} ≅ 19,8~cm$ (teorema di Pitagora) (è in pratica la diagonale del quadrato AHCD);
base maggiore $AB= 14+10,5 = 24,5~cm$;
lato obliquo $BC= \sqrt{CH^2+HB^2} = \sqrt{14^2+10,5^2} = 17,5~cm$ (teorema di Pitagora);
perimetro $2p= AB+BC+AD+DC = 24,5+17,5+14+14 = 70~cm$;
area $A= \dfrac{(AB+DC)·CH}{2} = \dfrac{(24,5+14)×14}{2} = \dfrac{38,5×14}{2}=269,5~cm^2$.
AC = 14 √2 cm (19,80)
CH = AD = 14 cm
AB = BH+CD = 24,5 cm
BC = √CH^2+BH^2 = 3,5√4^2+3^2 = 3,5*5 = 17,5 cm
perimetro 2p = 14+14+17,5+24,5 = 70 cm
area A = (24,5+14)*14/2 = 269,5 cm^2