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

 It is currently 23 May 2019, 12:39

### 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

# What is the area of the portion of the circle above created by arc ACB

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55266
What is the area of the portion of the circle above created by arc ACB  [#permalink]

### Show Tags

12 Sep 2018, 01:03
00:00

Difficulty:

45% (medium)

Question Stats:

80% (01:43) correct 20% (01:39) wrong based on 17 sessions

### HideShow timer Statistics

What is the area of the portion of the circle above created by arc ACB and cord AB if the radius of the circle is 6 and the length of arc ACB is 2π?

A. $$6\pi$$

B. $$6\pi - 6$$

C. $$6\pi - 9$$

D. $$6\pi - \frac{25}{2}$$

E. $$6\pi - 9\sqrt{3}$$

Attachment:

image035.jpg [ 1.85 KiB | Viewed 357 times ]

_________________
VP
Status: Learning stage
Joined: 01 Oct 2017
Posts: 1008
WE: Supply Chain Management (Energy and Utilities)
What is the area of the portion of the circle above created by arc ACB  [#permalink]

### Show Tags

12 Sep 2018, 01:26
Bunuel wrote:

What is the area of the portion of the circle above created by arc ACB and cord AB if the radius of the circle is 6 and the length of arc ACB is 2π?

A. $$6\pi$$

B. $$6\pi - 6$$

C. $$6\pi - 9$$

D. $$6\pi - \frac{25}{2}$$

E. $$6\pi - 9\sqrt{3}$$
Attachment:
image035.jpg

Let the centre of circle is at O. Join OA and OB.
Arc ACB=$$2\pi*6(\frac{AngleAOB}{2*\pi})$$
Or, $$2*\pi=6*AngleAOB$$ Or, Angle AOB=60 degree

Now, OAB is an equilateral triangle. OA=OB=AB=6

Required area= Are aof sector AOBCA- Area of triangle AOB
=$$\pi*6^2(60/360)-\sqrt{3}/4 * 6^2$$
=$$6\pi - 9\sqrt{3}$$
_________________
Regards,

PKN

Rise above the storm, you will find the sunshine
What is the area of the portion of the circle above created by arc ACB   [#permalink] 12 Sep 2018, 01:26
Display posts from previous: Sort by