document.write( "Question 73092: The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #52353 by rmromero(383) $\"\"$ $\"About$

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document.write( "The length of a rectangle is 2 cm more than twice its width.  
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document.write( "If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.\r
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document.write( "What is asked in the problem?
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document.write( " find the dimensions of the rectangle.\r
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document.write( "Given:
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document.write( " The length of a rectangle is 2cm more than twice its width
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document.write( " The perimeter of the rectangle is 52cm\r
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document.write( "Representation:
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document.write( "  Let x = the width of the rectangle
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document.write( "     2x+2 = the length of the rectangle\r
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document.write( "Equation
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document.write( "          2l + 2w = P
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document.write( "     2(2x+2) + 2x = 52        Multiply (Distributive Property)
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document.write( "      4x + 4 + 2x = 52        Combine like terms
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document.write( "           6x + 4 = 52
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document.write( "       6x + 4 - 4 = 52 - 4    subtract 4 both sides
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document.write( "               6x = 48        divide both sides by 6
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document.write( "                x = 8 cm -----> Width\r
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document.write( "     length = 2x + 2 
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document.write( "            = 2(8) + 2  
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document.write( "            = 16 + 2
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document.write( "            = 18 cm    ----- length\r
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document.write( "Checking\r
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document.write( "         2L + 2W = P
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document.write( "    2(18) + 2(8) = 52
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document.write( "         36 + 16 = 52
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document.write( "              52 = 52\r
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