III gimnazjum

Matematyka wokół nas 3

Liczba przekątnych w n-kącie wynosi: 2n(n−3)​ Skoro każdy n-kąt ma n boków, to musi być zatem spełnione równanie: 2n(n−3)​=n                n&gt;0 2n(n−3)​=n                n&gt;0 n(n−3)=2n n2−3n−2n=0 n2−5n=0 n(n−5)=0 n=5 <section class="answer"><h4 class="answer-heading">Odpowiedź:</h4> Szukanym wielokątem jest pięciokąt.</section>

Rozwiązanie 1

Liczba przekątnych w n-kącie wynosi: 2n(n−3)​ Szukamy takiego n, dla którego spełnione jest równanie

Liczba przekątnych w n-kącie wynosi:

Liczba przekątnych w n-kącie wynosi: 2n(n−3)​ Szukamy takiego n, dla ktoˊrego spełnione jest roˊwnanie: 2n(n−3)​=20      n&gt;0

a) pięciokąt (n−2)⋅180=540 n−2=180540​ n−2=3 n=5

Miara kąta wewnętrznego ośmiokąta foremnego wynosi: 180∘−(360∘:8)=180∘−45∘=135o

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Promienˊ okręgu opisanego na troˊjkącie jest roˊwny 32​ wysokosˊci. Mamy zatem: r=32​h=32​⋅a23<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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c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
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H400000v40H845.2724
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Promienˊ okręgu wpisanego w troˊjkąt roˊwnoboczny jest roˊwny 31​ wysokosˊci troˊjkąta. Mamy zatem: r=31​h=31​⋅a23<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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x - długosˊcˊ ramienia 1,2⋅x - długosˊcˊ dłuz˙szej podstawy

3x - długosˊcˊ kroˊtszego boku deltoidu 7x - długosˊcˊ dłuz˙szego boku deltoidu

x - jeden bok roˊwnoległoboku y - drugi bok roˊwnoległoboku

x - długosˊcˊ dłuz˙szej przekątnej 14−x - długosˊcˊ kroˊtszej przekątnej

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