III gimnazjum

Matematyka z plusem 3. Zbiór zadań 2001

Odległość między środkami okręgów  ∣S1​S2​∣ spełnia nierówność: |R-r|&lt; `|S_1S_2| to okręgi przecinają się. Odległość ∣S1​S2​∣ wynosi ∣S1​S2​∣=(3−−3)2+(2−−1)2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.88em' viewBox='0 0 400000 1944' preserveAspectRatio='xMinYMin slice'><path d='M983 90
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M1001 80h400000v40h-400000z'/></svg>​=36+9<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​=45<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​=35<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​≈6,7   |R-r|=|6| a |R+r|=9 więc okręgi przecinają się, czyli odpowiedź D<section class="answer"><h4 class="answer-heading">Odpowiedź:</h4> D</section>

Rozwiązanie 1

Promień okręgu wpisanego dany jest wzorem: r2​

Kąt wewnętrzny dziesięciokąta foremnego: 360o:10=36o

Przekątna kwadratu to średnica okręgu opisanego na kwadracie czyli: d=a2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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