Question

高二简单数学立体几何直观图求面积

已知△ABC的平面直观图△A'B'C'是边长为a的正三角形，求原△ABC的面积。

需要解答过程，谢谢！

Answer 1

设直观图为△A'B'C',
作A'D⊥B'C',交B'C'于D,在A'点作直线A'E,使与A'D成45度角,并交C'B'延长线于E,在E点作EA⊥EC',并取EA=2A'E,连结AB',AC',即为三角形的原图.
AE=2A'E=2√2A'D=2√2*(√3B'C'/2)=√6a,
S△ABC=AE*B'C'/2=√6a*a/2=√6a^2/2.设直观图为△A'B'C',
作A'D⊥B'C',交B'C'于D,在A'点作直线A'E,使与A'D成45度角,并交C'B'延长线于E,在E点作EA⊥EC',并取EA=2A'E,连结AB',AC',即为三角形的原图.
AE=2A'E=2√2A'D=2√2*(√3B'C'/2)=√6a,
S△ABC=AE*B'C'/2=√6a*a/2=√6a^2/2.

Answer 2

设直观图为△A'B'C',
作A'D⊥B'C',交B'C'于D,在A'点作直线A'E,使与A'D成45度角,并交C'B'延长线于E,在E点作EA⊥EC',并取EA=2A'E,连结AB',AC',即为三角形的原图.
AE=2A'E=2√2A'D=2√2*(√3B'C'/2)=√6a,
S△ABC=AE*B'C'/2=√6a*a/2=√6a^2/2.设直观图为△A'B'C',
作A'D⊥B'C',交B'C'于D,在A'点作直线A'E,使与A'D成45度角,并交C'B'延长线于E,在E点作EA⊥EC',并取EA=2A'E,连结AB',AC',即为三角形的原图.
AE=2A'E=2√2A'D=2√2*(√3B'C'/2)=√6a,
S△ABC=AE*B'C'/2=√6a*a/2=√6a^2/2.