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

 It is currently 16 Oct 2019, 17:26

### 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

# O is the center of the semicircle. If angle BCO = 30 and BC

Author Message
TAGS:

### Hide Tags

SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1749
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

### Show Tags

26 Mar 2015, 19:56
2
9
00:00

Difficulty:

25% (medium)

Question Stats:

79% (02:59) correct 21% (02:54) wrong based on 156 sessions

### HideShow timer Statistics

Attachment:

triacirc_q1.png [ 14.51 KiB | Viewed 6119 times ]

O is the center of the semicircle. If angle BCO = 30 and BC $$= 6\sqrt{3}$$, what is the area of triangle CBO?

A. $$4\sqrt{3}$$
B.$$6\sqrt{3}$$
C. $$9\sqrt{3}$$
D. $$12\sqrt{3}$$
E. $$24\sqrt{3}$$

_________________
Kindly press "+1 Kudos" to appreciate
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

### Show Tags

26 Mar 2015, 21:32
3
PareshGmat wrote:
Attachment:
triacirc_q1.png

O is the center of the semicircle. If angle BCO = 30 and BC $$= 6\sqrt{3}$$, what is the area of triangle CBO?

A. $$4\sqrt{3}$$
B.$$6\sqrt{3}$$
C. $$9\sqrt{3}$$
D. $$12\sqrt{3}$$
E. $$24\sqrt{3}$$

Since BCO = 30 degrees and angle ABC = 90 degrees (angle in a semi circle), angle CAB = 60 degrees and we are dealing with 30-60-90 triangle (ABC). So sides are in the ratio $$1:\sqrt{3}:2$$.
Side BC which is opposite 60 degrees in $$6\sqrt{3}$$ so AB = 6 and AC = 12. Hence radius of the circle OC (= OA = OB) is 6.
Since OA = OB, angle OAB = angle ABO = 60 degrees. Triangle ABO is equilateral with area $$= (\sqrt{3}/4) * 6^2 = 9\sqrt{3}$$

Area of triangle ABC $$= (1/2)*AB*BC = (1/2)*6*6\sqrt{3} = 18\sqrt{3}$$

Area of triangle BCO = Area of ABC - Area of ABO $$= 18\sqrt{3} - 9\sqrt{3} = 9\sqrt{3}$$

Another Method:

Note that the two triangles which form the triangle ABC will have equal area because their altitude will be the same (from A to BC) and theirs bases have the same length (OB - OC). So half the area of ABC will be the required area.
_________________
Karishma
Veritas Prep GMAT Instructor

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15263
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

### Show Tags

26 Mar 2015, 23:45
Hi PareshGmat,

The "key" to this Multi-Shape Geometry question is the rule that when a triangle has it's vertices on the circumference of the triangle AND one of its sides is the diameter of the circle, then that triangle is a RIGHT TRIANGLE. From there, you can use a variety of right triangle rules (depending on whatever additional information you're given) to answer whatever question is asked.

From the additional information, you know that we're dealing with a 30/60/90 triangle and you know the length of one of the sides. You can now figure out the lengths of the other sides and the area of the triangle.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Director
Joined: 23 Jan 2013
Posts: 525
Schools: Cambridge'16
Re: O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

### Show Tags

31 Dec 2016, 01:24
can find directly area of CBO

attitude^2=6^2-(3*sqrt3)^2=9, so attitude=3

A=(6*sqrt3*3)/2=9*sqrt3

C
Non-Human User
Joined: 09 Sep 2013
Posts: 13210
Re: O is the center of the semicircle. If angle BCO = 30 and BC  [#permalink]

### Show Tags

30 Oct 2018, 19:29
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: O is the center of the semicircle. If angle BCO = 30 and BC   [#permalink] 30 Oct 2018, 19:29
Display posts from previous: Sort by