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Exercício de (Geometria Analítica, circunferência)
Exercício:
Considerando que o triângulo equilátero ABC está inscrito na circunferência de equação (x+3)² + (y-2)² = 27, então a medida do segjmento AB é:
a) 3.
b) 6
c) 9
d) 12
e) 15
Eu sei que o gabarito é letra C, mas não consigo chegar ao resultado. Queria saber como é a resolução, obrigada.
É dificil explicar porque é necessário desenhar a circunferencia e o triangulo inscrito, mas vou tentar.
Primeiro você determina o centro da circunferencia e o raio: C(-3 , 2) e r= 3raiz(3).
Como é um triangulo equilátero isncrito na circunferencia, o centro dela são os 4 pontos notáveis do triangulo, assim fica interessante a existencia do ponto chamado baricentro, pois é possilvel encontrar a ALTURA (h) do triangulo, que substituindo nesta equação: H = L . raiz(3)/2 , achamos o lado:
9.raiz(3)/2 = L . raiz(3)/2 —-> L = 9 . raiz(3) / raiz(3) —> L = 9
as retas r e s tem equaçoes (2k+3)x-3y-1=0 e y =3x/2+1/3,respectivamente.determine o numero real k para que r seja paralelo a s.
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