Pole trapezu o podstawach 7cm i 11cm wynosi 432 cm kwadratowe. Jaka jest wysokosc tego trapezu . str.58 zad 5
P=1/2*h*8*(a+b)
432=1/2*h*(7+11)
432=1/2*h*18
432=9*h
h=432:9
h=48
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P=1/2*h*8*(a+b)
432=1/2*h*(7+11)
432=1/2*h*18
432=9*h
h=432:9
h=48