document.write( "Question 976442: A Rectangular lot measures (5w+32)cm by (5w-32)cm is added to both dimensions how is area affected : EXPLAIN
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Algebra.Com's Answer #598026 by Boreal(15235)\"\" \"About 
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The dimensions are (5w+32)cm by (5w-32)cm . I don't know what is added.\r
\n" ); document.write( "\n" ); document.write( "The area as it stands is 25w^2-1024. If w=10, the area is 2500-1024=1476, an 82X18 rectangle\r
\n" ); document.write( "\n" ); document.write( "If you add 10 cm, the lot is (5w+42)(5w-22)=25w^2+100w-964. If w=10, area is 3500-964=2536, a 92X28 rectangle. It is quadratic, however, and not simply linear, like the perimeter.\r
\n" ); document.write( "\n" ); document.write( "Areas increase as the square, but it is more complicated in a rectangle because of internal products. Double the side length of a square and you increase the area 4 times. Triple it and you increase the area 9 times. With a rectangle, it is not that simple, but the This is basically how you do these problems. Again, without knowing what is added, I can't tell.\r
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\n" ); document.write( "\n" ); document.write( "Example:
\n" ); document.write( "2*5 lot =10 square units.
\n" ); document.write( "add x to each.
\n" ); document.write( "(2+x)(5+x)=10 + 7x +x^2\r
\n" ); document.write( "\n" ); document.write( "if x =1, the lot's new area is 3*6=18
\n" ); document.write( "x^2+7x+10=18.
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