Explicație pas cu pas:
a)
[tex]{AC}^{2} = AB^{2} + BC^{2} \\ AC^{2} = {4}^{2} + {4}^{2} = 32 = > AC = 4 \sqrt{2}[/tex]
[tex]AA'^{2} = A'C^{2} - AC^{2} = {(4 \sqrt{3})}^{2} - 32 = 48 - 32 = 16 = > AA' = 4 \: cm[/tex]
[tex]AB = BC = AA' = 4 \: cm =>ABCDA'B'C'D \: este \: cub[/tex]
b)
[tex]BA' = AC = 4 \sqrt{2} \: cm[/tex]
[tex]BM⊥A'C = > [/tex]
[tex]d(B;A'C) = BM = \frac{BA' \times BC}{A'C'} = \frac{4 \sqrt{2} \times 4}{4 \sqrt{3} } =\frac{4 \sqrt{6} }{3} = > BM = \frac{4 \sqrt{6} }{3} \: cm [/tex]
