Fie AD⊥BC D∈BC
B=60° ⇒BD=8√3cos60° =4√3
AD=ABsin60°=8√3*√3/2=8*3/2=12
facem AD=12
AC=12√2
AD⊥DC si rezulta ca DC=12 ⇒(12√2²-12²) ( TOT SUB RADICAL)
atunci ne da ca BC=BD+DC=4√3+12
Aria Triughiului ABC= BC*AD/2=(12+4√3)*12/2
=6(12+4√3)=24(3+√3) cm²
aflam diagonala de la C la dreapta AB
d(C,AB)*AB=BC*AD
d(C,AB)*8√3=14√3+12*12 simplifici
d(C,AB)=(4√3+12)*3/2√3 =(2√3+6)*√3 =6+6√3
