AB=x
CD=1/5*x
—————————
areaABCD=1/2*(x+1/5*x)*9=54
1/2*6/5x*9=54
x=540/54————->x=10cm=AB
CD=1/5*10=2cm
somma basi b+B = 2A/h = 108/9 = 12 cm = b+5b
12 = 6b
b = 12/6 = 2,0 cm
B = 2*5 = 10,0 cm
Somma delle basi $AB~+DC= \frac{2A}{CH} = \frac{2~×54}{9} = 12~cm$ (formula inversa dell'area);
base maggiore $AB= \frac{12}{1+5}~×5 = \frac{12}{6}~×5 =10~cm$;
base minore $DC= \frac{1}{5}~ ×10 = 2~cm$.