Explicație pas cu pas:
înălțimea: DM ⊥ FG
[tex]Aria(DEFG) = \frac{(DE +FG) \times DM}{2} \\ 24 \sqrt{3} = \frac{(4 \sqrt{2} + 8 \sqrt{2}) \times DM}{2}[/tex]
[tex]DM = \frac{2 \times 24 \sqrt{3} }{12 \sqrt{2} } = \frac{4 \sqrt{3} }{ \sqrt{2} } = \frac{4 \sqrt{3} \times \sqrt{2} }{2} = 2 \sqrt{6} => DM = 2 \sqrt{6} \: cm [/tex]
[tex]GM = \frac{DE - FG}{2} = \frac{8 \sqrt{2} - 4 \sqrt{2} }{2} = \frac{4 \sqrt{2} }{2} = 2 \sqrt{2} => GM = 2 \sqrt{2}\: cm [/tex]
în ΔDMG dreptunghic:
DG² = DM² + GM² = 24 + 8 = 32
[tex]DG = \sqrt{32} => DG = 4 \sqrt{2} \: cm[/tex]
a) DG = 2×GM => ∢DGM = 30°
(DEFG) trapez isoscel => ∢F = 30°
b)
[tex]perimetrul(DEFG) = DE + EF + FG + DG = DE + FG + 2×DG =4 \sqrt{2} + 8 \sqrt{2} + 2 \times 4 \sqrt{2} = 12 \sqrt{2} + 8 \sqrt{2} = 20 \sqrt{2} \: cm [/tex]
c)
[tex]aria(DEG) = \frac{DM \times DE}{2} = \frac{2 \sqrt{6} \times 4 \sqrt{2} }{2} = 4 \sqrt{12} = 8 \sqrt{3} \: {cm}^{2} [/tex]
