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If rectangle \(ABCD\) is not a square, what is its perimeter??
(1) The longer side of the rectangle is 2 meters shorter than its diagonal
(2) The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\)
(1) The longer side of the rectangle is 2 meters shorter than its diagonal
(2) The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\)
16 Sep 2014, 00:59
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Bookmarks
Official Solution:
If rectangle \(ABCD\) is not a square, what is the perimeter of rectangle \(ABCD\)?
(1) The longer side of the rectangle is 2 meters shorter than its diagonal.
Given: \(a = d - 2\), where \(a\) is the length of the longer side and \(d\) is the length of the diagonal. Not sufficient on its own to get the perimeter.
(2) The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\).
Given: \(\frac{b}{d} = \frac{1}{3}\), so \(d=3b\), where \(b\) is the length of the shorter side. Not sufficient on its own to get the perimeter.
(1)+(2) Combining both statements, we get \(a = d - 2\) and \(d = 3b\), which leads to \(a = 3b - 2\). Using the Pythagorean theorem, we have \(a^2 + b^2 = d^2\), which gives \(a^2 - 32a - 32 = 0\). From the quadratic equation, there will be two solutions for \(a\): one positive and another negative. Given the context, the negative solution is invalid. Using the valid positive value for \(a\), we can determine the value of \(b\) and, subsequently, the perimeter. Sufficient.
Answer: C
If rectangle \(ABCD\) is not a square, what is the perimeter of rectangle \(ABCD\)?
(1) The longer side of the rectangle is 2 meters shorter than its diagonal.
Given: \(a = d - 2\), where \(a\) is the length of the longer side and \(d\) is the length of the diagonal. Not sufficient on its own to get the perimeter.
(2) The ratio of the shorter side of the rectangle to its diagonal is \(\frac{1}{3}\).
Given: \(\frac{b}{d} = \frac{1}{3}\), so \(d=3b\), where \(b\) is the length of the shorter side. Not sufficient on its own to get the perimeter.
(1)+(2) Combining both statements, we get \(a = d - 2\) and \(d = 3b\), which leads to \(a = 3b - 2\). Using the Pythagorean theorem, we have \(a^2 + b^2 = d^2\), which gives \(a^2 - 32a - 32 = 0\). From the quadratic equation, there will be two solutions for \(a\): one positive and another negative. Given the context, the negative solution is invalid. Using the valid positive value for \(a\), we can determine the value of \(b\) and, subsequently, the perimeter. Sufficient.
Answer: C
