GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 May 2019, 12:08 ### 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
Your Progress

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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # In the circle above with center O, minor arc AC is equal in length to

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 55230
In the circle above with center O, minor arc AC is equal in length to  [#permalink]

### Show Tags

2
2 00:00

Difficulty:   45% (medium)

Question Stats: 68% (02:33) correct 32% (01:49) wrong based on 60 sessions

### HideShow timer Statistics In the circle above with center O, minor arc AC is equal in length to minor arc CF and line segment BD is parallel to line segment AF. If angle CAD = angle DAE = angle EAF, what is the measure of angle BDA?

A. 15
B. 30
C. 45
D. 60
E. 75

Attachment: Circle_O.png [ 13.63 KiB | Viewed 1767 times ]

_________________
Senior Manager  G
Joined: 24 Apr 2016
Posts: 331
Re: In the circle above with center O, minor arc AC is equal in length to  [#permalink]

### Show Tags

1
In this problem, we can use the property :

In a circle, if two minor arcs are equal in measure, then their corresponding chords are equal in measure

Now if we join C to E, we get the Right Angled Traingle ACF

Since the minor arcs AC and CF are equal in length, correspoding chords AC and CF are also equal.

Therefore Traingle ACF is an Isocelese Right Angled Triangle.

Angle ACF = 90, Angle CAF = Angle CFA = 45

angle CAD + angle DAE + angle EAF = Angle CAF = 45

As angle CAD = angle DAE = angle EAF, then angle CAD = angle DAE = angle EAF = 15

angle BDA = angle DAE + angle EAF = 30

Answer is B
Manager  S
Joined: 06 Jun 2013
Posts: 156
Location: India
Concentration: Finance, Economics
Schools: Tuck
GMAT 1: 640 Q49 V30 GPA: 3.6
WE: Engineering (Computer Software)
Re: In the circle above with center O, minor arc AC is equal in length to  [#permalink]

### Show Tags

the correct answer is B.

the angle subtended by arc AC and arc CF are equal, as a result of which angle subtended by both the acrs at the center is 90 (180 in total as AF is the diameter).

we know that the angle subtended at the circle is half the angle subtended at the centre.
also, it is given that angle CAD = angle DAE = angle EAF.

so angle subtended at the cirecumference of the circle by arc CF is 45.

since BD is parallel to AF, angle BDA is equal to angle DAF.

angle DAF = 30 (2/3rd of angle CAF )

hence angle BAD is 30. Re: In the circle above with center O, minor arc AC is equal in length to   [#permalink] 02 Jul 2017, 08:29
Display posts from previous: Sort by

# In the circle above with center O, minor arc AC is equal in length to

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics 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®.

#### MBA Resources  