过A点作平面ADE‖平面A'B'C',交BB'于D,交CC'于E,则BD=5-3=2,CE=4-3=1
则△ADE≌△A'B'C',设正三角形边长=a
由AB²=AC²+BC²===>(a²+2²)=(a²+1²)+[a²+(2-1)²]===>a=√2
∴S△A'B'C'=a²sin60º/2=(√2)²(√3/2)/2=√3/2
过A点作平面ADE‖平面A'B'C',交BB'于D,交CC'于E,则BD=5-3=2,CE=4-3=1
则△ADE≌△A'B'C',设正三角形边长=a
由AB²=AC²+BC²===>(a²+2²)=(a²+1²)+[a²+(2-1)²]===>a=√2
∴S△A'B'C'=a²sin60º/2=(√2)²(√3/2)/2=√3/2