document.write( "Question 1054756: The perimeter of a rectangle is 27 cm,and its area is 35 square cm.Find the length and then width of the rectangle.
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Algebra.Com's Answer #669958 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The perimeter of a rectangle is 27 cm,and its area is 35 square cm.Find the length and then width of the rectangle. \n" ); document.write( "----------- \n" ); document.write( "P = 2(L + W) \n" ); document.write( "L*W = 35 \n" ); document.write( "--------------------------- \n" ); document.write( "L = (P/2)-W \n" ); document.write( "-------- \n" ); document.write( "Substitute for \"L\" and solve for \"W\": \n" ); document.write( "[(P/2)-W]W = 35 \n" ); document.write( "----------------------------- \n" ); document.write( "W^2 - (27/2)W +35 = 0 \n" ); document.write( "---------------------------- \n" ); document.write( "W = [(27/2) +- sqrt[13.5^2 - 4*35]]/(2 \n" ); document.write( "---------------------------------------- \n" ); document.write( "W = [13.5 +- 6.5] \n" ); document.write( "Width = 7 cm \n" ); document.write( "--- \n" ); document.write( "Since L*W = 35, Length = 5 cm \n" ); document.write( "-------------------------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "-------------- \n" ); document.write( " \n" ); document.write( " |