document.write( "Question 290178: A rectangle has a perimeter of 50 cm. Is it possible to have an area of 136 cm^2? If so, what are the dimensions? \n" ); document.write( "
Algebra.Com's Answer #209995 by stanbon(75887)\"\" \"About 
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A rectangle has a perimeter of 50 cm.
\n" ); document.write( "Is it possible to have an area of 136 cm^2? If so, what are the dimensions?
\n" ); document.write( "-------------------------------------------
\n" ); document.write( "Equations:
\n" ); document.write( "2(L+W) = 50
\n" ); document.write( "LW = 136
\n" ); document.write( "-------------------
\n" ); document.write( "Modify:
\n" ); document.write( "L = 25-W
\n" ); document.write( "---
\n" ); document.write( "Substitute:
\n" ); document.write( "(25-W)W = 136
\n" ); document.write( "25W - W^2 = 136
\n" ); document.write( "W^2 - 25W +136 = 0
\n" ); document.write( "---
\n" ); document.write( "Check the discrimant to see if there is a solution:
\n" ); document.write( "(25)^2 -4*1*136 is greater than zero so there is a solution.
\n" ); document.write( "---
\n" ); document.write( "What are the dimensions?
\n" ); document.write( "w = [25 +- sqrt(25^2-4*136)]/2
\n" ); document.write( "W = [25 +- sqrt(81)]/2
\n" ); document.write( "W = [25 +- 9]/2
\n" ); document.write( "w = (34/2) or W = 16/2
\n" ); document.write( "W = 17 or W = 8
\n" ); document.write( "---
\n" ); document.write( "If W = 17, L = 8
\n" ); document.write( "If W = 8, L = 17
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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