document.write( "Question 120198: I don't understand this word problem. The problem says \"The perimeter of a rectangle is 60 inches., and its area is 200 inches^2. Find the dimensions of the rectangle.\" Could you please help me? \n" ); document.write( "

Algebra.Com's Answer #88077 by checkley71(8403) $\"\"$ $\"About$

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PERIMETER=2L+2W
\n" ); document.write( "AREA=L*W
\n" ); document.write( "60=2L+2W
\n" ); document.write( "200=L*W OR L=200/W NOW SUBSTITUTE (200/W) FOR L IN THE FIRST EQUATION & SOLVE FOR W
\n" ); document.write( "60=2(200/W)+2W
\n" ); document.write( "60=400/W+2W ADD THE 2 TERMS ON THE LEFT SIDE OF THE EQUATION BY FINDING A COMMON DENOMINATOR (W) THUS:
\n" ); document.write( "60=(400+2W*W)/W
\n" ); document.write( "60=(400+2W^2)/W NOW CROSS MULTIPLY
\n" ); document.write( "400+2W^2=60W
\n" ); document.write( "2W^2-60W+400=0
\n" ); document.write( "2(W^2-30W+200)=0
\n" ); document.write( "2(W-20)(W-10)=0
\n" ); document.write( "W-20=0
\n" ); document.write( "W=20 ANSWER FOR THE WIDTH & 10 FOR THE LENGTH.
\n" ); document.write( "OR W-10=0
\n" ); document.write( "W=10 FOR THE WIDTH & 20 FOR THE LENGTH.
\n" ); document.write( "PROOF
\n" ); document.write( "60=2*10+2*20
\n" ); document.write( "60=20+40
\n" ); document.write( "60=60
\n" ); document.write( "&
\n" ); document.write( "200=10*20
\n" ); document.write( "200=200
\n" ); document.write( " \n" ); document.write( "

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