document.write( "Question 1038957: A rectangular park measures 90m by 200m. The park wants to make several basketball courts in a rectangular area which is surrounded by a grassy area with uniform wiidth. If the basketball court is to take up one third of the total area, what musts its dimesions be? \n" ); document.write( "

Algebra.Com's Answer #653703 by Aldorozos(172) $\"\"$ $\"About$

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I solve this problem based on the assumption that we have a rectangle inside another rectangle.
\n" ); document.write( "The width of this rectangle should be 40 and the length should be 150. If we multiply 40 and 150 we get 6000 which is 1/3 of the area of the big rectangle which is 18000 = 90*200\r
\n" ); document.write( "\n" ); document.write( "Now we want to find out how we came out with a court that has a width of 40 and length of 150.
\n" ); document.write( "The with of this court should be 90-2x and its length should be 200-2x. x is the distance between the length of the smaller rectangle to the length of the large rectangle. This distance is the same as the distance between the widths.
\n" ); document.write( "(90-2x)(200-2x) = 6000 we simplify and get 18000 -400x -180x = 4x^2 = 6000
\n" ); document.write( "solving this quadratic equation we get x = 25
\n" ); document.write( "90-2x = 90-50 = 40 = width of the smaller rectangle
\n" ); document.write( "similarly 200-2x = 200-50 = 150 = length of the smaller rectangle = (basketball court)\r
\n" ); document.write( "\n" ); document.write( "The with of the smaller rectangle is (90 - 2x)and the length of the smaller rectangle is (200-2x)\r
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