pomocy!!!!!!pilne sylwiaaa: dlugosc krawedzi graniastoslupa trojkatnego sa kolejnymi wyrazzami ciagu arytmetycznego o roznicy 2.najdluzsza krawedz bedaca wysokoscia graniastoslupa jest rowna 12.oblicz objetosc graniastoslupa i pole jego powierzchni

Eta:

Janek 191: x, x + 2 , x + 4, x + 6 zatem x + 6 = 12 x = 6 x + 2 = 8 x + 4 = 10 Trójkąt jest prostokątny, bo 6² + 8² = 10² Objętość V = Pp *h = 0,5 *6*8* 12 = 24*12 = 288 j³ −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− Pole Pc = 2 Pp + Pb = 2 *24 + ( 6 + 8 + 10)*12 = 48 + 24*12 = 48+ 288 = 336 j² −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− −−−−−−−−−−−−−−−−−−−−−−−

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