At=2Aria bazei +Aria laterala=2l²+4lh
2l²+4lh=576
l²+2l*12=288
l²+24l-288=0
ec de grad 2
l1,2=(-24+/-√(576+1152))/2=(-24+/-√1728)/2 ∉Q putin probabil..
are sens doar valoarea pozitiva
(-24+√1728)/2
dar
√1728=24√3
atunci latura este
(24√3-24)/2=12(√3-1)
verificare
aria totala
2*144(4-2√3)+4*12*12(√3-1)=2*144*4-288*2√3+144*4√3-4*144=4*144(2-1)=576*1=576
in mod neasteptat, se verifica
