Z szesciokata foremnego o boku 4pierwiasteka z 3.Wycieto kolo wpisane w ten szesciokat.jakie pole ma pozostala czesc szesciokata?
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Dane :
a= 4√3
P₁ = 3a²√3 /2
P₁ = 3 *(4√3)²/2 = 3*36/2 = 39
promień koło wpisanego w sześciokąt :
r = h
h = a√3/2
h = 4√3*√3 /2 = 6
r = 6
P = πr²
P = π * 6² = 36π
P₁ - P = 39 – 36 = 3