document.write( "Question 55097: Can someone help me on this problem. I have gotten two different results and not sure which was is right.\r
\n" ); document.write( "\n" ); document.write( " The length of a rectangle is 2 cm more than twice its width. f the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
\n" ); document.write( " Thanks,
\n" ); document.write( " Sher \n" ); document.write( "
Algebra.Com's Answer #37335 by anjulasahay(30)\"\" \"About 
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Can someone help me on this problem. I have gotten two different results and not sure which was is right.\r
\n" ); document.write( "\n" ); document.write( " The length of a rectangle is 2 cm more than twice its width. f the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle\r
\n" ); document.write( "\n" ); document.write( "ans:
\n" ); document.write( " let length of the rectangle = l
\n" ); document.write( " width of the rectangle = w\r
\n" ); document.write( "\n" ); document.write( "hence according to the question ,
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\n" ); document.write( " l = 2+2*w ----I
\n" ); document.write( "and perimeter of rectangle = 2(l+w)
\n" ); document.write( " = 2((2+2w)+w) (putting l from I)
\n" ); document.write( " =2(2+3w)=52
\n" ); document.write( " or 2+3w = 21
\n" ); document.write( " or 3w=19
\n" ); document.write( " or w = 19/3
\n" ); document.write( " hence l = 2+2*19/3 = 44/3
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