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H=2a
Pc= Pp+ Pb
V= ⅓*Pp*H
V= ⅓*a²*2a= 2a³/3
5⅓ = 2a³/3
16/3 = 2a³/3
16 = 2a³
a³ = 8
a = 2
Pp = 2 * 2 = 4
hb- wys. ściany bocznej
Pb = 4 * ½ * a * hb = 2a * hb
H = 2a = 4
½a = 1
4² +1² = (hb)²
hb = √17
Pb = 2 * 2 * √17 = 4√17
Pc = 4 + 4√17
Odp. Pole powierzchni całkowitej tego ostrosłupa wynosi 4+4√17