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Unformatted text preview: a 2 2 bc = 2 bc b 2 c 2 + a 2 2 bc = a 2 ( b c ) 2 2 bc = ( a ( b c ))( a + ( b c )) 2 bc = ( a b + c )( a + b c ) 2 bc . Thus, substituting these expressions into equation (2.32), we have K 2 = 1 4 b 2 c 2 ( a + b + c )( b + c a ) 2 bc ( a b + c )( a + b c ) 2 bc = a + b + c 2 b + c a 2 a b + c 2 a + b c 2 , and since we dened s = 1 2 ( a + b + c ), we see that K 2 = s ( s a )( s b )( s c ) , so upon taking square roots we get K = r s ( s a )( s b )( s c ) . QED...
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 Fall '11
 Dr.Cheun
 Calculus, Angles, Pythagorean Theorem

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