Na jutro! k.king199: Dlugość krawedzi prostopadloscianu wychodzacych z tego samego wierzchołka sa kolejnymi wielokrotnościami liczby 3. Pole powierzchni calkowitej tego prostopadlościanu wyosi 468cm². Wyznacz: a) objętść tego prostopadloscianu b) dlugosc jego przekątnej.

dero2005: a = 3n b = 3(n+1) = 3n + 3 c = 3(n+2) = 3n + 6 P_c = 2(a*b) + 2(a*c) + 2(b*c) = 468 P_c = 2[(3n(3n+3)] + 2[3n(3n+6)] + 2[(3n+3)(3n+6)] = 468 −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− po policzeniu −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− n = 2 a = 6 b = 9 c = 12 V = a*b*c = 648 cm³ d = √a² + b² = 3√13 D = √d² + c² = 3√29