It is currently 26 Sep 2017, 07:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# GMAT CLUB TEST m18 - question 15

Author Message
Intern
Joined: 18 Dec 2009
Posts: 37

Kudos [?]: 8 [0], given: 7

Location: California
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

Can someone explain how angle EAD = angle BDA as mentioned in the solution?

Kudos [?]: 8 [0], given: 7

Manager
Joined: 16 Apr 2010
Posts: 214

Kudos [?]: 138 [0], given: 12

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

Kudos [?]: 138 [0], given: 12

Manager
Joined: 22 Jun 2010
Posts: 56

Kudos [?]: 73 [0], given: 10

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!

Kudos [?]: 73 [0], given: 10

Intern
Joined: 18 Dec 2009
Posts: 37

Kudos [?]: 8 [1], given: 7

Location: California
Re: GMAT CLUB TEST m18 - question 15 [#permalink]

### Show Tags

23 Jul 2010, 11:23
1
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]

Last edited by WithHopesToWin on 23 Jul 2010, 11:25, edited 1 time in total.

Kudos [?]: 8 [1], given: 7

Intern
Joined: 18 Dec 2009
Posts: 37

Kudos [?]: 8 [0], given: 7

Location: California
Re: GMAT CLUB TEST m18 - question 15 [#permalink]

### Show Tags

23 Jul 2010, 11:24
btw, thanks Jakolik!!

Kudos [?]: 8 [0], given: 7

Manager
Joined: 16 Apr 2010
Posts: 214

Kudos [?]: 138 [0], given: 12

Re: GMAT CLUB TEST m18 - question 15 [#permalink]

### Show Tags

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

You're welcome

Kudos [?]: 138 [0], given: 12

Re: GMAT CLUB TEST m18 - question 15   [#permalink] 23 Jul 2010, 22:15
Display posts from previous: Sort by

# GMAT CLUB TEST m18 - question 15

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.