document.write( "Question 1122020: Pease help me solve this equation:
\n" ); document.write( "Find the dimensions of a rectangle whose area is 247 cm2 and whose perimeter is 64 cm. (Enter your answers as a comma-separated list.) \n" ); document.write( "

Algebra.Com's Answer #738052 by ikleyn(52787) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Let me show you a quick, elegant and unexpected method of solving such problems.\r
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document.write( "You are given that the perimeter of the rectangle is 64 cm;  hence, the sum of the length and the width is one half of that:\r\n" );
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document.write( "x + y = 32,  where x is the length, and y is the width.\r\n" );
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document.write( "Then the average of the length and the width is one half of 32, i.e. 16.\r\n" );
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document.write( "It is clear that the values of x and y are remoted at the same value/distance \"u\" from 16, so we can write\r\n" );
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document.write( "x = 16 + u,\r\n" );
document.write( "y = 16 - u.\r\n" );
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document.write( "Then the area is  xy = (16+u)*(16-u) = , and it is equal to 247, according to the condition.\r\n" );
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document.write( "Hence,   = 247,  which gives   = 256 - 247 = 9,  and then  u =  = 3.\r\n" );
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document.write( "Thus the length is  x = 16 + u = 16 + 3 = 19,\r\n" );
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document.write( "and  the width  is  x = 16 - u = 16 - 3 = 13.\r\n" );
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document.write( "Answer.  The dimensions of the rectangle are  13 cm  and  19 cm.\r\n" );
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\n" ); document.write( "\n" ); document.write( "See the lesson\r
\n" ); document.write( "\n" ); document.write( " - Three methods to find the dimensions of a rectangle when its perimeter and the area are given\r
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