Construim inaltimea AM
IN Δ AMC, m< M=90, m< C=60⇒m< MAC=30⇒Th 30-60-90 MC=AC:2
MC=6√6 cm
In ΔAMC, Th Pitagora AC²=AM²+MC²
AM²=144*6-36*6=864-216=648
AM=18√2
In ΔAMB, m<M=90, m<ABM=45, AM=18√2⇒Δ AMB- isoscel dreptunghic⇒BM=18√2
Aplicam Th Pitagora AB²=AM²+BM²
AB=18√2*√2=36
AB=36Perimetrul ΔABC=36+12√6+6√6+18√2=36+18√6+18√2=18(2+√6+√2)
Aria=[tex] \frac{(18 \sqrt{2}+6 \sqrt{6})* 18\sqrt{2} }{2} = {54(3+ \sqrt{3} )[/tex]
