A circle is inscribed in equilateral triangle ABC such that : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 22 Jan 2017, 23:32

### 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

# A circle is inscribed in equilateral triangle ABC such that

Author Message
Manager
Joined: 01 Jun 2006
Posts: 77
Followers: 1

Kudos [?]: 1 [0], given: 0

A circle is inscribed in equilateral triangle ABC such that [#permalink]

### Show Tags

01 Oct 2006, 10:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1. A circle is inscribed in equilateral triangle ABC such that point D lies on the circle and on line segment AC, point E lies on the circle and on line segment AB and point F lies on the cirlce and on line segment BC. If line segment AB = 6, what is the area of the figure created by line segments AD, AE and the minor arc DE?

a) 3*underroot(3) - 9/4pi
b) 3*underroot(3) - pi
c) 6*underroot(3) - pi
d) 9*underroot(3) - 3pi
e)cannot be determined from the info given

Gmat Gurus help needed....

usman
Manager
Joined: 01 Oct 2006
Posts: 242
Followers: 1

Kudos [?]: 11 [0], given: 0

### Show Tags

01 Oct 2006, 17:30
Is the answer (b) -> 3*underroot(3) - pi
Manager
Joined: 28 Aug 2006
Posts: 244
Location: Albuquerque, NM
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

02 Oct 2006, 20:56
Draw perpendicular bisectors from each vertex to the other side of triangle

For each of these bisectors lenght is sqrt(6^2-3^2) = 3sqrt(3)

All three perpendicualr bisectors meet at the center of circle and divide themselves in ratio of 2:1

Thus the radius of circle is sqrt(3)

Area of Eq. triangle is {sqrt(3)/4}* 6^2 = 9sqrt(3) and area of circle is 3*pi

Now Triangle -circle area is 9sqrt(3) - 3*pi

to find area of what is asked divide this by 3 since by symmetry of the problems we have 3 such equal areas

Answewr is 3*sqrt(3) -pi
Manager
Joined: 01 Jun 2006
Posts: 77
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

03 Oct 2006, 01:42
Great explanation jainan but can you explain this point that you made....
"All three perpendicualr bisectors meet at the center of circle and divide themselves in ratio of 2:1"

Can you please explain this point....
Thanks for much.....
Manager
Joined: 28 Aug 2006
Posts: 244
Location: Albuquerque, NM
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

03 Oct 2006, 04:28
Since the triangle formed by the radius of the circle (perpendicular),

half the base of original equilateral triangle (base) and distance from one of the vertices to the center of circle (hypotenuse) form a 30-60-90 triangle and we also know that sum of base and hypotenuse of that triangle by symmetry is 3sqrt(3), thus you can find hypotenuse = 2sqrt(3) and perpendicualr (radius) = sqrt(3).

If you draw the figure accurately, you can see the symmetry
Manager
Joined: 01 Jun 2006
Posts: 77
Followers: 1

Kudos [?]: 1 [0], given: 0

### Show Tags

08 Oct 2006, 11:01
thanks jainan....much appreciated
Senior Manager
Joined: 28 Aug 2006
Posts: 306
Followers: 13

Kudos [?]: 150 [0], given: 0

### Show Tags

08 Oct 2006, 11:19
Simple way is

1/3( area of triangle- area of circle)

1/3( 9sqrt(3)- 3 pi)

3sqrt(3)-pi
_________________
08 Oct 2006, 11:19
Display posts from previous: Sort by