document.write( "Question 1181295: The length of the rectangle is one more than the width. If the dimensions are both decreased by 2 units, the area of the new rectangle is 30 sq. units less than the area of the original rectangle. Find the area of the original rectangle. \n" ); document.write( "

Algebra.Com's Answer #811181 by ikleyn(52909) $\"\"$ $\"About$

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\n" ); document.write( "The length of the rectangle is one more than the width.
\n" ); document.write( "If the dimensions are both decreased by 2 units, the area of the new rectangle
\n" ); document.write( "is 30 sq. units less than the area of the original rectangle.
\n" ); document.write( "Find the area of the original rectangle.
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document.write( "Let w be the width of the rectangle; then its length is (w+1) unit, according to the condition.\r\n" );
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document.write( "The hypothetical rectangle has the width of (w-2) units and the length of ((w+1)-2) = (w-1) units;\r\n" );
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document.write( "so its area is this product  (w-2)*(w-1).\r\n" );
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document.write( "And now, we have THIS equation for the areas\r\n" );
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document.write( "    (w-2)*(w-1) = w*(w+1) - 30.\r\n" );
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document.write( "Simplify; then solve for w\r\n" );
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document.write( "    w^2 - 2w - w + 2 = w^2 + w - 30\r\n" );
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document.write( "            -3w + 2  = w - 30\r\n" );
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document.write( "             2 + 30  = w + 3w\r\n" );
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document.write( "              32     = 4w\r\n" );
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document.write( "               w     = 32/4 = 8\r\n" );
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document.write( "Thus we found that the dimensions of the original rectangle are  w= 8 (the width)  and  8+1 = 9 (the length).\r\n" );
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document.write( "Hence, the area of the original triangle is  8*9 = 72 square units.\r\n" );
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document.write( "    To CHECK : the dimensions of the smaller rectangle are 6 and 7 units;  its area is 6*7 = 42 square units;\r\n" );
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document.write( "    it is 30 units less than 72 square units, the area of the original rectangle.   ! Checked !\r\n" );
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