gmatt1476 wrote:
Patricia purchased x meters of fencing. She originally intended to use all of the fencing to enclose a square region, but later decided to use all of the fencing to enclose a rectangular region with length y meters greater than its width. In square meters, what is the positive difference between the area of the square region and the area of the rectangular region?
(1) xy = 256
(2) y = 4
DS37571.01
1. Let the side of the square be \(b\). Now, as the fencing was bought to enclose the square hence we can safely conclude that \(4b = x\) i.e. perimeter of the square is equal to \(x\)
2. Let the width of the rectangle be \(a\) and length be \(y + a\). Now, as the fencing was bought to enclose the rectangle hence we can safely conclude that \(4a + 2y = x\) i.e. perimeter of the rectangle is equal to \(x\)
3. The difference between the areas:
\(b^2 - a(y+a) = b^2 - a^2 - ay\)
4. Replacing \(b\) with \(\frac{x}{4}\) and \(a\) with \(\frac{x-2y}{4}\)
\((\frac{x}{4})^2 - (\frac{x-2y}{4})^2 - (\frac{x-2y}{4})*y\)
\(\frac{x^2}{16} - (\frac{x^2 - 4xy + 4y^2}{16}) - (\frac{xy - 2y^2}{4})\)
Equalizing the denominators to work out the fractions:-\(\frac{x^2}{16} - (\frac{x^2 - 4xy + 4y^2}{16}) - (\frac{4xy - 8y^2}{16})\)
\(\frac{x^2 - x^2 - 4y^2 + 4xy - 4xy + 8y^2}{16}\)
Cancelling out terms:-\(\frac{x^2 - x^2 + 4xy - 4xy + 8y^2 - 4y^2}{16}\) \(=\) \(\frac{4y^2}{16}\)
Multiplying the numerator and denominator with 4:-\(\frac{y^2}{4}\)
We need the value of \(y\) and option B provides it. Ans.
B _________________
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