[tex] A_{T}=2 A_{B}+ A_{L} \\ 72 \sqrt{3} =2* \frac{a^2 \sqrt{3} }{4}+54 \sqrt{3} \\ 18 \sqrt{3}= \frac{a^2 \sqrt{3} }{2} \\ 36=a^2 \\ a=6 \\ [/tex]
deci,lungimea laturii bazei este a=6cm
[tex] A_{L}= P_{B}*h \\ 54 \sqrt{3}=3*3*h \\ h= \frac{54 \sqrt{3} }{9} \\ h=6 \sqrt{3}cm [/tex]