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In the figure above, AB || CD and AD || BC. If BC is 4 units
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24 Mar 2020, 00:03
24 Mar 2020, 00:03
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In the figure above, AB || CD and AD || BC. If BC is 4 units long and CD is 2 units long, what is the area of quadrilateral ABCD?
(A) 4
(B) 4\(\sqrt{2}\)
(C) 6
(D) 8
(E) 6\(\sqrt{2}\)
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Re: In the figure above, AB || CD and AD || BC. If BC is 4 units
[#permalink]
12 Feb 2021, 06:32
12 Feb 2021, 06:32
2
The figure above is a Parallelogram, where, AB || CD and AD || BC. Also, BC = AD = 4 and AB = CD = 2.
Area of a Parallelogram = Base x Height
Drop a perpendicular from point C to line segment AD (let us name it is as point E)
Now, triangle CED is an Isosceles Right Triangle [45-45-90 triangle]
Ratio of sides in 45-45-90 triangle = \(x : x : \sqrt{2}x\)
opposite to 45 = \(x\)
opposite to 90 = \(\sqrt{2}x\) = CD = 2
So, \(x\) = \(\sqrt{2}\)
i.e. Height = \(\sqrt{2}\)
Therefore, Area of parallelogram = 4 x \(\sqrt{2}\) = \(4\sqrt{2}\)
Hence, option B
Area of a Parallelogram = Base x Height
Drop a perpendicular from point C to line segment AD (let us name it is as point E)
Now, triangle CED is an Isosceles Right Triangle [45-45-90 triangle]
Ratio of sides in 45-45-90 triangle = \(x : x : \sqrt{2}x\)
opposite to 45 = \(x\)
opposite to 90 = \(\sqrt{2}x\) = CD = 2
So, \(x\) = \(\sqrt{2}\)
i.e. Height = \(\sqrt{2}\)
Therefore, Area of parallelogram = 4 x \(\sqrt{2}\) = \(4\sqrt{2}\)
Hence, option B
gmatclubot
Re: In the figure above, AB || CD and AD || BC. If BC is 4 units [#permalink]
12 Feb 2021, 06:32
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