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Re: In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, [#permalink]
Bunuel wrote:
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) \(3 \sqrt{7}\)



Given, Area of the rectangle=Length*Breadth=18 sq. unit
or,{4-(-2)}*Breadth=18 (Since the the ordinate 'y' remains constant and position of abscissa varies from -2 to 4)
or, Breadth=18/6=3 unit
Now. length of diagonal=\(\sqrt(Length^2\)+\(breadth^2\))=\(\sqrt(6^2\)+\(3^2\))=\(\sqrt(36\)+\(9\))=\(\sqrt(45)\)=3\(\sqrt5\)

So, the correct answer choice is (C)
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Re: In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, [#permalink]
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Re: In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, [#permalink]
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