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cristhian



Registrado: 27 Mar 2011
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MensajePublicado: Mie Mar 30, 2011 7:11 am    Título del mensaje: poligono equiangulo Responder citando

holaaa..... alguien si me ayuda a resolver este problema....porfaa[/img]


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Si alguien me ayuda con este problema.... porfa
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cv



Registrado: 25 Sep 2010
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MensajePublicado: Jue Mar 31, 2011 5:47 pm    Título del mensaje: Responder citando

En realidad no se si es una demostracion pero tal vez te ayude:

tg(tita/2) = (l/2)/a = l/(2*a)

donde l= longitud de un lado del poligono y a=apotema

cot(tita/2) = 2*a/l

por otro lado el semi-perimetro es

P = n*l/2, donde n=numero de lados.

La cte. a la que se hace referencia = P*cot(tita/2) = n*l/2 * 2*a/l = n*a

asi a1+a2+a3 +....+an = n*a (numero de lados por su apotema)

Saludos.
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judy9146



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MensajePublicado: Jue Nov 01, 2012 9:18 am    Título del mensaje: jordan 11 concords Responder citando

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hayes



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MensajePublicado: Vie Sep 27, 2013 9:03 am    Título del mensaje: Coach Outlet Online Responder citando

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