document.write( "Question 427041: How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54. \n" ); document.write( "
Algebra.Com's Answer #297000 by Alan3354(69443) You can put this solution on YOUR website! How does a 4th grader find the dimensions of a rectangle when the perimeter is 58 and the area is 54. \n" ); document.write( "--------------- \n" ); document.write( "Perimeter = 2L + 2W = 58 --> L + W = 29 \n" ); document.write( "Area = L*W = 54 \n" ); document.write( "------------------ \n" ); document.write( "L = 54/W \n" ); document.write( "(54/W) + W = 29 \n" ); document.write( "54 + W^2 = 29W \n" ); document.write( " \n" ); document.write( "(W - 2)*(W - 27) = 0 \n" ); document.write( "W = 2 \n" ); document.write( "L = 27 \n" ); document.write( "----------- \n" ); document.write( "I'm not sure that'll work in the 4th grade, it's been awhile since I was there. \n" ); document.write( "---------------------- \n" ); document.write( "Try this: \n" ); document.write( "Factor 54 and look for a pair that adds to 29 \n" ); document.write( "54 = \n" ); document.write( "1 x 54 \n" ); document.write( "2 x 27 \n" ); document.write( "3 x 18 \n" ); document.write( "6 x 9\r \n" ); document.write( "\n" ); document.write( "Only 2 by 27 fits. \n" ); document.write( "If the solution is not integers, that won't work. \n" ); document.write( " |