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 de±ned 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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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.
 Fall '11
 Dr.Cheun
 Calculus, Angles, Pythagorean Theorem

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