document.write( "Question 674411: draw and label a rectangle that has a perimeter of 60 cm and an area of 200 square centimeters \n" ); document.write( "

Algebra.Com's Answer #419227 by Edwin McCravy(20055) $\"\"$ $\"About$

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draw and label a rectangle that has a perimeter of 60 cm and an area of 200 square centimeters
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document.write( "       Perimeter = 2×Length + 2×Width\r\n" );
document.write( "            Area = Length×Width\r\n" );
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document.write( "              60 = 2L + 2W\r\n" );
document.write( "             200 = LW\r\n" );
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document.write( "Splve the first equation for W and substitute in the second equation:\r\n" );
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document.write( "              60 = 2L + 2W\r\n" );
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document.write( "Divide through by 2\r\n" );
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document.write( "              30 =  L +  W\r\n" );
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document.write( "          30 - L = w \r\n" );
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document.write( "Substitute (30 - L) for W in\r\n" );
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document.write( "             200 = LW\r\n" );
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document.write( "             200 = L(30 - L)\r\n" );
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document.write( "             200 = 30L - L²\r\n" );
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document.write( "Get 0 on the right:\r\n" );
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document.write( "  L² - 30L + 200 = 0\r\n" );
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document.write( "Factor the left side:\r\n" );
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document.write( "(L - 20)(L - 10) = 0\r\n" );
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document.write( "Use the zero-factor property:\r\n" );
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document.write( "  L - 20 = 0;   L - 10 = 0\r\n" );
document.write( "       L = 20;       L = 10\r\n" );
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document.write( "Subdtitute in\r\n" );
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document.write( "  30 - L = w    30 - L = w    \r\n" );
document.write( " 30 - 20 = W   30 - 10 = W \r\n" );
document.write( "      10 = W        20 = W\r\n" );
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document.write( "We usually think of the length as the longer side, so\r\n" );
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document.write( "the length is 20 and the width is 10\r\n" );
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document.write( "Width = 10 centimeters\r\n" );
document.write( "                  Length = 20 centimeters\r\n" );
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document.write( "Edwin

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