Grazie
area A = (25+15)*15*4/3 / 2 = 40*10 = 400 cm^2
Trapezio isoscele.
altezza $DH= \frac{4}{3}CD = \frac{4}{3}×15 = 20~cm$;
area $A= \frac{(AB+CD)×DH}{2} = \frac{(25+15)×20}{2} = \frac{40×20}{2}= 400~cm^2$.
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