0,08(3)=1/12
0,0(6)=1/15
M
D.................C
A......M'........C'........B
CC'/12=BC/15⇒CC'=12K;BC=15K
AC=BD=40
BC'=√(15K)²-12K²=9K⇒BC²=BC'×AB⇒AB=225K²/9K=25K
CD=C'D'=25K-2×9K=7K
T.Pitagora ΔACB⇒AC=40=√AB²-BC²=20 K⇒K=AC/20=2
a)AB=25×2=50
CD=7×2=14
CC'=12×2=24 h
b)Aria=(AB+CD)×CC'/2
c)ΔMM'C≈ΔMM'B
