risolvi problema n.176
176a
sviluppo S = 8*(1,41+3,14/2) = 23,84 cm
area A = 8^2(1/2-3,14/16) = 64*(8-3,14)/16 = 19,44 cm^2
176b
sviluppo S' = 8(3,14+2) = 41,12 cm
area A' = 8^2/2 = 32 cm^2
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176)
1° figura:
perimetro $2p= l·\sqrt2+\dfrac{l·2π}{4} = 8×1,41+\dfrac{8×2×3,14}{4} =23,84~cm$;
area $A= \dfrac{l^2·π}{4}-\dfrac{l^2}{2} = \dfrac{8^2×3,14}{4}-\dfrac{8^2}{2} = 18,24~cm^2$.
2° figura:
perimetro $2p= \dfrac{\frac {l}{2}·2π}{4}×4+2·l = \dfrac{\frac {8}{2}×2×3,14}{4}×4+2×8 = 4×2×3,14+16 = 41,12~cm$;
area $A= \dfrac{l^2}{2} = \dfrac{8^2}{2} = 32~cm^2$.