$2p=4*l=4*2=8~cm$
$S_l=2p*h=8*2,5=20~cm^2$
$d=\sqrt{l^2-(\frac{D}{2})^2}=\sqrt{2^2-1,6^2}=\sqrt{4-2,56}=\sqrt{1,44}=1,2~cm$
$S_b=\frac{D*d}{2}=\frac{3,2*1,2}{2}=1,92~cm^2$
$S_t=S_l+2*S_b=20+2*1,92=23,84~cm^2$
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Diagonale minore del rombo di base $d= \sqrt{2^2-\big(\frac{3.2}{2}\big)^2}=1,2~cm$ (teorema di Pitagora);
perimetro di base $2p_b= 4l= 4×2 = 8~cm$;
area di base $Ab= \frac{D×d}{2}=\frac{3.2×1.2}{2}= 1,92~cm^2$;
area laterale $Al= 2p_b×h = 8×2,5 = 20~cm^2$;
area totale $At= Al+2Ab = 20+2×1,92=23,84~cm^2$.