Pole podstawy
[tex]6 \times \frac{ {8}^{2} \sqrt{3} }{4} = 6 \times \frac{64 \sqrt{3} }{4} = 6 \times 16 \sqrt{3} = 96 \sqrt{3} {cm}^{2} [/tex]
Wysokość ściany bocznej
[tex] {4}^{2} + {x}^{2} = {5}^{2} \\ 16 + {x}^{2} = 25 \\ {x}^{2} = 9 \\ x = \sqrt{9} \\ x = 3cm[/tex]
Pole ściany bocznej
[tex] \frac{8 \times 3}{2} = \frac{24}{2} = 12 {cm}^{2} [/tex]
Pole całkowite
[tex]96 \sqrt{3} + 6 \times 12 = \boxed{ (96 \sqrt{3} + 72) {cm}^{2} }[/tex]
