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Rombo di base:
diagonale incognita $\small \dfrac{2×\cancel{96}^6}{\cancel{16}_1} = 2×6 = 12\,cm;$
lato $\small l=\sqrt{\left(\dfrac{D}{2}\right)^2+ \left(\dfrac{d}{2}\right)^2} = \sqrt{\left(\dfrac{16}{2}\right)^2+ \left(\dfrac{12}{2}\right)^2} = \sqrt{8^2+6^2} = 10\,cm$ (teorema di Pitagora);
perimetro $\small 2p= 4×l =4×10 = 40\,cm.$
Prisma:
area di base $\small Ab= 96\,cm^2;$
perimetro di base $\small = 40\,cm;$
altezza = lato del rombo $\small h= l= 10\,cm;$
area laterale $\small Al= 2p×h = 40×10 = 400\,cm^2;$
area totale $\small At= Al+2×Ab = 400+2×96 = 592\,cm^2.$