aria tr ABC=144√3:4=36√3 cm²
36√3=l²√3/4
l²=144⇒l=12cm
Am=AN ca inaltimi in tr echilaterale congruente si=12√3/2=6√3
si avand baza MN=12/2=6, ca linie mijlocie in tr.CBD
fie AQ inaltimea acestui tr isoscel care va fi√((6√3)²-3²)=√(108-9)=√99=3√11
deci aria va fi (1/2) *6*3√11=9√11 cm²