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In the coordinate plane, rectangle WXYZ has vertices at (–2

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In the coordinate plane, rectangle WXYZ has vertices at (–2 [#permalink] New post 24 May 2017, 01:47
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In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, y), (4, y), and (4, –1). If the area of WXYZ is 18, what is the length of its diagonal?

A) \(3 \sqrt{2}\)

B) \(3 \sqrt{3}\)

C) \(3 \sqrt{5}\)

D) \(3 \sqrt{6}\)

E) \(3 \sqrt{7}\)
[Reveal] Spoiler: OA

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Re: In the coordinate plane, rectangle WXYZ has vertices at (–2 [#permalink] New post 10 Oct 2017, 09:15
Since we have two points as (-1, -2) and (-1, 4) who are in line, we easily compute the length of one of the sides of the rectangle as 4-(-2) = 6.

Then, given the area, we can compute the other side of the rectangle as 18/6 = 3.

Know, the diagonal of the rectangle is the hypotenuse of the right triangle of legs 3 and 6, thus it can be computed as 6^2+3^9 = 45 = 3sqrt(5).

Answer C
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Re: In the coordinate plane, rectangle WXYZ has vertices at (–2 [#permalink] New post 02 Nov 2018, 12:42
Answer should be root(40) i.e. 2 * root(10)
Re: In the coordinate plane, rectangle WXYZ has vertices at (–2   [#permalink] 02 Nov 2018, 12:42
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In the coordinate plane, rectangle WXYZ has vertices at (–2

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