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In the figure above, ABCD is a quadrilateral in which sides AB and CD

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In the figure above, ABCD is a quadrilateral in which sides AB and CD  [#permalink]

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New post 24 Nov 2018, 09:15
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In the figure above, ABCD is a quadrilateral in which sides AB and CD are parallel. What is the area of the quadrilateral?
(1) The shortest distance between sides AB and CD is 4 units.
(2) The length of side AD is √17 units.

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Re: In the figure above, ABCD is a quadrilateral in which sides AB and CD  [#permalink]

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New post 24 Nov 2018, 09:18
A statement 1 alone is sufficient, by pythagoras theorem, base can be determined, using whoch area (base *height) can be found
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Re: In the figure above, ABCD is a quadrilateral in which sides AB and CD  [#permalink]

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New post 24 Nov 2018, 10:04
1
Let us drop perpendiculars from B onto DC meeting DC at E and from D onto AB meeting AB at F.

(1) Perpendicular length = 4, so BE = 4, EC = 3 by Pythagoras theorem, so DE = 8 - 3 = 5. So, BF = 5. AF = 6 - 5 = 1, so AD = sqrt(1^2 + 4^2) = sqrt(17), area can be found by calculating areas of triangles ADF, BCE and rectangle DEBF. Hence Sufficient.

(2) Given AD = sqrt(17), Assume DE = BF = x, so EC = 8 - x and AF = 6 - x. Equating the perpendicular lengths for both triangles we get:

17 - (6-x)^2 = 25 - (8-x)^2
From this we can find the value of x and hence can calculate the areas of the two triangles mentioned above and the rectangle to get the area of the quadrilateral. Hence Sufficient!

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Re: In the figure above, ABCD is a quadrilateral in which sides AB and CD  [#permalink]

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New post 24 Nov 2018, 10:23
Option E.
We have to find area by dropping perpendicular. But we wont know the length of side AB between perpendicular and AD and similarly length of side CD between perpendicular and BC.

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Re: In the figure above, ABCD is a quadrilateral in which sides AB and CD &nbs [#permalink] 24 Nov 2018, 10:23
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In the figure above, ABCD is a quadrilateral in which sides AB and CD

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