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Re: [obm-l] P.A.
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] P.A.
- From: "Andre Araujo" <alsdearaujo@xxxxxxxxx>
- Date: Thu, 6 Sep 2007 08:58:30 -0300
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Bruna,
seja a PA (a, a*sqrt(2), a^2), onde sqrt( ) representa a raiz quadrada. Assim,
a + a^2 = 2*a*sqrt(2) => a^2 -[2*sqrt(2) - 1]*a = 0 => a*{a - [2*sqrt(2) - 1]} = 0 , como a é diferente de zero (medida do lado do quadrado), então:
a = 2*sqrt(2) - 1.
André Araújo.
Em 06/09/07, Bruna Carvalho <bruna.carvalho.pink@xxxxxxxxx
> escreveu:Os números que exprimem o lado, a diagonal e a área
de um quadrado estão em PA, nessa ordem, então, qual é o
perímetro do quadrado?
--
Bjos,
Bruna