document.write( "Question 949093: the length of a rectangle is twice its width. if the perimeter of the rectangle is 64 ft, find its area \n" ); document.write( "

Algebra.Com's Answer #579366 by macston(5194) $\"\"$ $\"About$

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W=width; L=length=2W; P=perimeter=2(L+W)=64 ft; Area =L*W
\n" ); document.write( "P=2(L+W)
\n" ); document.write( "64 ft=2(L+W) Divide each side by 2
\n" ); document.write( "32 ft=L+W Substitute for L
\n" ); document.write( "32 ft=2W+W
\n" ); document.write( "32 ft=3W Divide each side by 3
\n" ); document.write( "(32/3)ft=W
\n" ); document.write( "Area =L*W Substitute for L
\n" ); document.write( "Area=2W*W
\n" ); document.write( " $\"Area=2W%5E2\"$
\n" ); document.write( " $\"Area=2%2832ft%2F3%29%5E2\"$ = $\"2%281024%2F9%29sq+ft\"$ =227.6 sq ft
\n" ); document.write( "ANSWER Area of the rectangle is 227.6 sq ft.
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