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# M24 8

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Manager
Joined: 13 May 2010
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M24 8 [#permalink]

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21 Mar 2012, 08:17
In quadrilateral $$ABCD$$ , $$AB = CD$$ and $$BC = AD$$ . If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$ ?

(C) 2008 GMAT Club - m24#8

* 30 degrees
* 50 degrees
* 70 degrees
* 100 degrees
* 120 degrees

Because $$AB = CD$$ and $$BC = AD$$ , $$ABCD$$ is a parallelogram. $$\angle ADC = \angle BDA + \angle BDC = \angle CBD + \angle ABD = \angle CBD + (180 - \angle BAD - \angle BDA) = 30 + (180 - 80 - 30) = 100$$

Can we conclusively say that a quadrilateral whose opposite sides are equal will always be a parallelogram? Don't we need to know if the opposite sides are parallel as well to determine if it is a parallelogram? I can't think of any other scenario than parallelogram but want to confirm.
Manager
Joined: 10 Jan 2011
Posts: 244
Location: India
GMAT Date: 07-16-2012
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WE: Consulting (Consulting)
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Kudos [?]: 41 [0], given: 25

Re: M24 8 [#permalink]

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27 Mar 2012, 22:08
teal wrote:
In quadrilateral $$ABCD$$ , $$AB = CD$$ and $$BC = AD$$ . If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$ ?

(C) 2008 GMAT Club - m24#8

* 30 degrees
* 50 degrees
* 70 degrees
* 100 degrees
* 120 degrees

Because $$AB = CD$$ and $$BC = AD$$ , $$ABCD$$ is a parallelogram. $$\angle ADC = \angle BDA + \angle BDC = \angle CBD + \angle ABD = \angle CBD + (180 - \angle BAD - \angle BDA) = 30 + (180 - 80 - 30) = 100$$

Can we conclusively say that a quadrilateral whose opposite sides are equal will always be a parallelogram? Don't we need to know if the opposite sides are parallel as well to determine if it is a parallelogram? I can't think of any other scenario than parallelogram but want to confirm.

IMO, if opposite sides of a quadrilateral are equal we can safely assume that it is a parallelogram.
_________________

-------Analyze why option A in SC wrong-------

Re: M24 8   [#permalink] 27 Mar 2012, 22:08
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# M24 8

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