III gimnazjum

Matematyka wokół nas 3

2Πr=9cm 6r=9cm r=23​cm P=Πr2 P=3⋅(23​)2=3⋅49​=427​=6,75cm2 Odp. Pole tego koła wynosi 6,75cm2 .

Rozwiązanie 1

Pk​=52Π=25Πcm2

Πr2=36Π r2=36

2Πr=4a a=21​Πr

r=0,1m=10cm Ob=2⋅10Π≈62,8cm

Ob=2⋅37,5Π≈235,5cm

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Zauważamy, że duży kwadrat można podzielić na pięć mniejszych. Zamalowany na zielono kwadratu to 1 z 5 kwadratów, stąd jego pole stanowi 1/5 pola dużego kwadratu.

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