Krawędź podstawy
[tex]a \sqrt{2} = 4 \sqrt{2} \\ a = 4[/tex]
Obliczmy krawędź boczną
[tex]x - kr.boczna[/tex]
[tex] \frac{1}{2} a \sqrt{2} = \frac{1}{2} \times 4 \sqrt{2} = 2 \sqrt{2} [/tex]
Czyli
[tex]y = 2 \sqrt{2} \\ y \sqrt{2} = 2 \sqrt{2} \times \sqrt{2} = 2 \sqrt{4} = 2 \times 2 = 4 \\ \\ x = 4[/tex]
Krawędź boczna wynosi 4
Zatem wysokość ostrosłupa wynosi
[tex] {4}^{2} = (2 \sqrt{2} {)}^{2} + {y}^{2} \\ 16 = 4 \times 2 + {y}^{2} \\ 16 = 8 + {y}^{2} \\ {y}^{2} = 8 \\ y = \sqrt{8} \\ y = \sqrt{4 \times 2} \\ y = 2 \sqrt{2} [/tex]
Pole podstawy
[tex]4 \times 4 = 16[/tex]
Objętość
[tex] \frac{16 \times 2 \sqrt{2} }{3} = \frac{32 \sqrt{2} }{3} = 10 \frac{2}{3} \sqrt{2} [/tex]
