document.write( "Question 404844: The area of a rectangle is 36 square inches. The perimeter of the rectangle is 36 inches. Find the dimensions of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #286131 by Tatiana_Stebko(1539) $\"\"$ $\"About$

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The area of a rectangle is $\"S=L%2AW\"$ , when L is the length, W is the width, so $\"L%2AW=36\"$
\n" ); document.write( "The perimeter af a rectangle is $\"P=2%2A%28L%2BW%29\"$ , so $\"2%2A%28L%2BW%29=36\"$ , $\"L%2BW=18\"$
\n" ); document.write( "So, the system of equations
\n" ); document.write( " $\"L%2AW=36\"$
\n" ); document.write( " $\"L%2BW=18\"$ From the second equation $\"W=18-L\"$ , put it in the first equation
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\n" ); document.write( " $\"L%2A%2818-L%29=36\"$
\n" ); document.write( " $\"18L-L%5E2=36\"$
\n" ); document.write( " $\"L%5E2-18L%2B36=0\"$
\n" ); document.write( " $\"L+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$
\n" ); document.write( " $\"L+=+%28-%28-18%29+%2B-+sqrt%28+%28-18%29%5E2-4%2A1%2A36+%29%29%2F%282%2A1%29+\"$
\n" ); document.write( " $\"L+=+%2818+%2B-+sqrt%28+180%29%29%2F2+\"$
\n" ); document.write( " $\"L1+=9+%2B15sqrt%28+0.2%29=15.7\"$ or $\"L2=9-15sqrt%280.2%29=2.3\"$
\n" ); document.write( " $\"W1=18-%289%2B15sqrt%280.2%29%29=9-15sqrt%280.2%29=2.3\"$ or $\"W2=18-%289-15sqrt%280.2%29%29=9%2B15sqrt%280.2%29=15.7\"$
\n" ); document.write( "Answer 15.7inches and 2.3inches
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\n" ); document.write( "Are you sure that numbers in your problem is correct?The area of a rectangle is 36 square inches. The perimeter of the rectangle is 36 inches. \n" ); document.write( "

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