Oblicz pole ośmiokąta foremnego wpisanego w okrąg o promieniu 10cm.
Prosze o pomoc :)
r=a√(2+√2)/2
10=a√(2+√2)/2/²
a²(1+√2/2)=100
a²=100:[(1+√2/2)]
a²=[100(1-√2/2)/(1-½)
a²=200-100√2
pole=2a²(1+√2)=2(200-100√2)(1+√2)=
400+400√2-200√2-400=
200√2cm²
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r=a√(2+√2)/2
10=a√(2+√2)/2/²
a²(1+√2/2)=100
a²=100:[(1+√2/2)]
a²=[100(1-√2/2)/(1-½)
a²=200-100√2
pole=2a²(1+√2)=2(200-100√2)(1+√2)=
400+400√2-200√2-400=
200√2cm²