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GMAT CLUB TEST m18 - question 15

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Intern
Joined: 18 Dec 2009
Posts: 37
Location: California
Schools: Kellogg (2Y - Waitlisted), Anderson(ADMITTED),UCI(ADMITTED with $$), Booth R2 ADMITTED) GMAT CLUB TEST m18 - question 15 [#permalink] Show Tags 22 Jul 2010, 16:14 In rectangle ABCD ( BC >AB] ), E is the point of intersection of diagonals. If angle Angle ABD is twice the angle EAD , what is the value of angle CED ? * 30 degrees * 45 degrees * 60 degrees * 90 degrees * 120 degrees angle EAD = angle BDA . angle ABD + angle BDA + 90 = 2*angle BDA + angle BDA + 90 = 180. From this equation, angle BDA = 30 . angle BDC = 90 - angle BDA = 60 . Thus, triangle CED is equilateral and angle CED = 60 The correct answer is C. Can someone explain how angle EAD = angle BDA as mentioned in the solution? Manager Joined: 16 Apr 2010 Posts: 221 Re: GMAT CLUB TEST m18 - question 15 [#permalink] Show Tags 23 Jul 2010, 01:09 Hi, Diagonal of a rectangle are equal and bisect each other. Thus EA=ED and angle EAD = angle EDA = angle BDA regards, Jack Manager Joined: 22 Jun 2010 Posts: 57 Re: GMAT CLUB TEST m18 - question 15 [#permalink] Show Tags 23 Jul 2010, 04:53 can someone post a graph maybe explaining the relations? The forumulas have me totally confused, as I keep mixing up the angels and lose track of where which angle was in what relation... Thanks! Intern Joined: 18 Dec 2009 Posts: 37 Location: California Schools: Kellogg (2Y - Waitlisted), Anderson(ADMITTED),UCI(ADMITTED with$$$), Booth R2 ADMITTED) Re: GMAT CLUB TEST m18 - question 15 [#permalink] Show Tags 23 Jul 2010, 11:23 1 This post received KUDOS Hi, Consider the rectangle ABCD (BC>AD) in the doc attached. Given facts: angle ABD = 2 * angle EAD Since diagonals of a rectangle are equal and bisect each other, we have EA=ED. Implying angle EAD = EDA. Lets consider angle EAD = angle EDA = x. We already know all angles in a rectangle are 90. Therefore angle BAD = 90 Considering the triangle BAD, sum of angles in a triangle is 180. So, angle BAD + angle ABD + angle BDA = 180 90 + 2x + x = 180 => 3x = 90 => X = 30 Now, angle EDC = 90-30 = 60 Consider triangle ACD, again angle CAD + angle ADC + angle DCA = 180 30 + 90 +angle DCA = 180 = > angle DCA =60 Knowing 2 angles of triangle EDC to be 60, makes it a equilateral triangle, hence angle CED =60. Hope its not confusing! Attachments gmatclub.docx [11.4 KiB] Downloaded 203 times  To download please login or register as a user Last edited by WithHopesToWin on 23 Jul 2010, 11:25, edited 1 time in total. Intern Joined: 18 Dec 2009 Posts: 37 Location: California Schools: Kellogg (2Y - Waitlisted), Anderson(ADMITTED),UCI(ADMITTED with$), Booth R2 ADMITTED)
Re: GMAT CLUB TEST m18 - question 15 [#permalink]

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23 Jul 2010, 11:24
btw, thanks Jakolik!!
Manager
Joined: 16 Apr 2010
Posts: 221
Re: GMAT CLUB TEST m18 - question 15 [#permalink]

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23 Jul 2010, 22:15
WithHopesToWin wrote:
btw, thanks Jakolik!!

You're welcome
Re: GMAT CLUB TEST m18 - question 15   [#permalink] 23 Jul 2010, 22:15
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GMAT CLUB TEST m18 - question 15

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