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Topic: Area problems
Replies: 1   Last Post: Jul 7, 2008 10:37 AM

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Alexander Bogomolny

Posts: 406
Registered: 12/4/04
Re: Area problems
Posted: Jul 7, 2008 10:37 AM
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First, the function is nit "sen" but "sin".

Second, sin(45) = sqrt(2)/2 which is only approximately equals 0.707106781.

So, your formula gives:

A = 1/2 (4)(5)(sen 45) = 10 sqrt(2) / 2 = 5 sqrt(2),

as is declared in the book.

A. Bogomolny

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