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

 It is currently 15 Dec 2019, 02:48

### 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

# Arcs DE, EF, FD are centered at C, B and A, in equilateral

Author Message
TAGS:

### Hide Tags

Intern
Joined: 18 Feb 2011
Posts: 33
GPA: 3.91
Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

10 Feb 2012, 07:09
1
13
00:00

Difficulty:

55% (hard)

Question Stats:

71% (02:46) correct 29% (02:52) wrong based on 215 sessions

### HideShow timer Statistics

Arcs DE, EF, FD are centered at C, B and A, in equilateral triangle ABC, as shown above. If the area of the triangle is 6, what is the area of the shaded region formed by the intersection of the arcs and the triangle?

A) 6 - $$\frac{\sqrt{3}}{3}$$$$\pi$$
B) 6 - $$\frac{\sqrt{3}}{2}$$$$\pi$$
C) $$\frac{\pi}{\sqrt{3}}$$
D) 6 - $$\sqrt{3}\pi$$
E) $$\frac{\pi}{6}$$
Math Expert
Joined: 02 Sep 2009
Posts: 59726

### Show Tags

10 Feb 2012, 07:35
1
1
nafishasan60 wrote:

Arcs DE, EF, FD are centered at C, B and A, in equilateral triangle ABC, as shown above. If the area of the triangle is 6, what is the area of the shaded region formed by the intersection of the arcs and the triangle?

A) 6 - $$\frac{\sqrt{3}}{3}$$$$\pi$$
B) 6 - $$\frac{\sqrt{3}}{2}$$$$\pi$$
C) $$\frac{\pi}{\sqrt{3}}$$
D) 6 - $$\sqrt{3}\pi$$
E) $$\frac{\pi}{6}$$

The area of an equilateral triangle is $$area=a^2*\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side. Thus $$a^2*\frac{\sqrt{3}}{4}=6$$ --> $$a^2=8*\sqrt{3}$$ --> radius of the circles (which make arcs) is a/2, so $$r^2=\frac{a^2}{4}$$. Now, as the triangle is equilateral each arc is 60 degrees, so the area of white region (non-shaded) is $$\frac{60*3}{360}*\pi{r^2}=\frac{1}{2}*\pi{\frac{8*\sqrt{3}}{4}=\pi{\sqrt{3}}$$ (three 60 degrees arcs).

The area of the shaded region is $$6-\pi{\sqrt{3}}$$.

_________________
Manager
Joined: 19 Apr 2011
Posts: 162
Schools: Booth,NUS,St.Gallon
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

10 Feb 2012, 08:00
option D 6 - \sqrt{3}\pi
_________________
+1 if you like my explanation .Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 59726
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

06 Mar 2014, 02:34
Bumping for review and further discussion.

_________________
Manager
Joined: 06 Mar 2014
Posts: 82
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

27 Jul 2015, 08:50
Bunuel:

Can you please tell me how did u get the radius as a/2.
Is there a formula for calculating radius inside a equilateral triangle or u did just by some calculation??
Can u also post some links having same concept tested.

Thanks
Math Expert
Joined: 02 Aug 2009
Posts: 8324
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

27 Jul 2015, 09:00
Shree9975 wrote:
Bunuel:

Can you please tell me how did u get the radius as a/2.
Is there a formula for calculating radius inside a equilateral triangle or u did just by some calculation??
Can u also post some links having same concept tested.

Thanks

since all three arcs of circle touch each other externally, all three have same radius as its an equilateral triangle..
if each side is a, the radius of each circle is a/2
_________________
CEO
Joined: 20 Mar 2014
Posts: 2560
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

27 Jul 2015, 09:07
Shree9975 wrote:
Bunuel:

Can you please tell me how did u get the radius as a/2.
Is there a formula for calculating radius inside a equilateral triangle or u did just by some calculation??
Can u also post some links having same concept tested.

Thanks

Look at the attached picture. There is no "formula" as such. It is calculated by looking at the figure as shown below.

From the figure, if 'a' is the side of the equilateral triangle and 'r' is the radius of the 2 arcs,

r+r = a ---> r =a/2 (this is true as all the 3 sectors created by the 3 arcs are equal).

Hope this helps.
Attachments

2015-07-27_12-01-00.jpg [ 11.07 KiB | Viewed 2881 times ]

Non-Human User
Joined: 09 Sep 2013
Posts: 13737
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral  [#permalink]

### Show Tags

16 Jan 2019, 10:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Arcs DE, EF, FD are centered at C, B and A, in equilateral   [#permalink] 16 Jan 2019, 10:02
Display posts from previous: Sort by