In the figure above, showing circle with center O and points : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 04:18

### 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

# In the figure above, showing circle with center O and points

Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Feb 2013
Posts: 120
Followers: 1

Kudos [?]: 852 [0], given: 17

In the figure above, showing circle with center O and points [#permalink]

### Show Tags

28 Feb 2013, 04:23
8
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

63% (03:10) correct 37% (01:34) wrong based on 207 sessions

### HideShow timer Statistics

Attachment:

1.jpg [ 6.23 KiB | Viewed 5089 times ]
In the figure above, showing circle with center O and points A, B, and C on the circle, given that length (AB)= 12 and length (BC)=$$4\sqrt{3}$$, what is the area of triangle AOB?

A. $$12\sqrt{3}$$

B. $$24\sqrt{3}$$

C. 48

D. $$48\sqrt{3}$$

E. 72
[Reveal] Spoiler: OA

_________________

Kudos will encourage many others, like me.
Good Questions also deserve few KUDOS.

Last edited by Bunuel on 28 Feb 2013, 05:57, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93506 [3] , given: 10568

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

28 Feb 2013, 06:07
3
KUDOS
Expert's post
10
This post was
BOOKMARKED

In the figure above, showing circle with center O and points A, B, and C on the circle, given that length (AB)= 12 and length (BC)=$$4\sqrt{3}$$, what is the area of triangle AOB?

A. $$12\sqrt{3}$$

B. $$24\sqrt{3}$$

C. 48

D. $$48\sqrt{3}$$

E. 72

Important property:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B. Thus the area of ABC is $$\frac{1}{2}*AB*BC=\frac{1}{2}*12*4\sqrt{3}=24\sqrt{3}$$.

Now, notice that since OA=OC=radius, then BO turns out to be the median of triangle ABC.

Another important property:
Each median divides the triangle into two smaller triangles which have the same area.

So, (area of AOB) = (area of COB) = 1/2 (area of ABC) = $$\frac{1}{2}*24\sqrt{3}=12\sqrt{3}$$.

For more check here: math-triangles-87197.html
_________________
Intern
Joined: 15 May 2013
Posts: 7
Followers: 0

Kudos [?]: 17 [3] , given: 7

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

11 Jun 2013, 02:54
3
KUDOS
As AC (diameter of the cirlcle) is one side of the triangle, ABC is right angle triangle and B is 90 degrees.

Area of ABS is 1/2 * 12*4*sqrt(3) = 24 Sqrt(3)
The are of AOB must be less than ABC.
in the given options only option A is less than 24 Sqrt(3)
Verbal Forum Moderator
Joined: 15 Jun 2012
Posts: 1153
Location: United States
Followers: 260

Kudos [?]: 2872 [0], given: 123

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

11 Jun 2013, 23:56
emmak wrote:
Attachment:
1.jpg
In the figure above, showing circle with center O and points A, B, and C on the circle, given that length (AB)= 12 and length (BC)=$$4\sqrt{3}$$, what is the area of triangle AOB?

A. $$12\sqrt{3}$$

B. $$24\sqrt{3}$$

C. 48

D. $$48\sqrt{3}$$

E. 72

Area of ABC = 1/2*AB*BC = 1/2*12*4\sqrt{3} = 24\sqrt{3}
Area of AOB = 1/2 Area of ABC = 12\sqrt{3}

Hence, A is correct.
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

Chris Bangle - Former BMW Chief of Design.

Intern
Status: Currently Preparing the GMAT
Joined: 15 Feb 2013
Posts: 31
Location: United States
GMAT 1: 550 Q47 V23
GPA: 3.7
WE: Analyst (Consulting)
Followers: 1

Kudos [?]: 16 [0], given: 11

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

12 Jun 2013, 00:42
Thank you for another great geometry question, emmak. Now let's solve it properly

Attachment:

1.jpg [ 6.23 KiB | Viewed 4054 times ]

The first thing you'll notice in the figure above is that the hypothenuse (the biggest side in a triangle - in this case AC) is the diameter of the circle. Which automatically means that the ABC triangle is a right triangle. Therefore, the Pythagorean theorem become applicable.

We know the lengths of AB and BC, we can therefore deduce the length of AC by using the Pythagorean theorem.

First of all : $$AC^2 = AB^2 + BC^2$$

By plugging in the values of AB and BC, we get :

$$AC^2 = 12^2 + (4\sqrt{3})^2$$
$$AC^2 = 144 + 16*3 = 144 + 48 = 192$$
$$AC = \sqrt{192}$$

By breaking down 192 into prime factors we get : $$192 = 2^6 * 3$$

So $$AC = 8\sqrt{3}$$

Since O is the center of the circle, then $$AO = OC = OB = \frac{AC}{2} = 4\sqrt{3}$$

Notice that the triangle AOB is an isoceles triangle (since AO = OB) and the formula for calculating the area of a triangle is : $$Area = \frac{base * height}{2}$$

The base here is AB. We need to find the height. Since AOB is an isoceles triangle, then the height will cut the base in half. So, noting H as the height of AOB, we can apply the Pythagorean theorem to deduce the value of the height :

$$H^2 + \frac{AB^2}{4} = OB^2$$

$$H = \sqrt{OB^2 - \frac{AB^2}{4}}$$

By plugging in the values of AB and OB we get :

$$H = \sqrt{16*3 - 6^2} = \sqrt{48 - 36} = \sqrt{12} = 2*\sqrt{3}$$

We can finally apply the formula to calculate the area of AOB :

$$Area of AOB = \frac{H*AB}{2} = 2*12*\sqrt{3}/2$$ = $$12*\sqrt{3}$$

Which corresponds to answer choice A.

Hope that helped
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13509
Followers: 577

Kudos [?]: 163 [0], given: 0

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

08 Oct 2014, 02:06
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.
_________________
Intern
Joined: 15 Jul 2012
Posts: 38
Followers: 0

Kudos [?]: 4 [0], given: 245

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

12 Oct 2014, 11:18
Bunuel wrote:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that length (AB)= 12 and length (BC)=$$4\sqrt{3}$$, what is the area of triangle AOB?

A. $$12\sqrt{3}$$

B. $$24\sqrt{3}$$

C. 48

D. $$48\sqrt{3}$$

E. 72

Important property:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B. Thus the area of ABC is $$\frac{1}{2}*AB*BC=\frac{1}{2}*12*4\sqrt{3}=24\sqrt{3}$$.

Now, notice that since OA=OC=radius, then BO turns out to be the median of triangle ABC.

Another important property:
Each median divides the triangle into two smaller triangles which have the same area.

So, (area of AOB) = (area of COB) = 1/2 (area of ABC) = $$\frac{1}{2}*24\sqrt{3}=12\sqrt{3}$$.

For more check here: math-triangles-87197.html

Hey Bunuel,
Each median divides the triangle into two smaller triangles which have the same area

is it that according to this property as mentioned above by you mean that the two smaller triangles are similiar?
Math Expert
Joined: 02 Sep 2009
Posts: 36609
Followers: 7099

Kudos [?]: 93506 [0], given: 10568

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

12 Oct 2014, 11:30
saggii27 wrote:
Bunuel wrote:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that length (AB)= 12 and length (BC)=$$4\sqrt{3}$$, what is the area of triangle AOB?

A. $$12\sqrt{3}$$

B. $$24\sqrt{3}$$

C. 48

D. $$48\sqrt{3}$$

E. 72

Important property:
A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

According to the above, we have that ABC is a right triangle, with a right angle at B. Thus the area of ABC is $$\frac{1}{2}*AB*BC=\frac{1}{2}*12*4\sqrt{3}=24\sqrt{3}$$.

Now, notice that since OA=OC=radius, then BO turns out to be the median of triangle ABC.

Another important property:
Each median divides the triangle into two smaller triangles which have the same area.

So, (area of AOB) = (area of COB) = 1/2 (area of ABC) = $$\frac{1}{2}*24\sqrt{3}=12\sqrt{3}$$.

For more check here: math-triangles-87197.html

Hey Bunuel,
Each median divides the triangle into two smaller triangles which have the same area

is it that according to this property as mentioned above by you mean that the two smaller triangles are similiar?

Where did I say that the two smaller triangles are similar? They are not.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13509
Followers: 577

Kudos [?]: 163 [0], given: 0

Re: In the figure above, showing circle with center O and points [#permalink]

### Show Tags

26 Oct 2015, 03:07
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: In the figure above, showing circle with center O and points   [#permalink] 26 Oct 2015, 03:07
Similar topics Replies Last post
Similar
Topics:
The figure above shows three circles, all centered on point O. What is 3 30 Dec 2014, 19:59
1 The figure above shows three circles, all centered on point O. What i 6 30 Dec 2014, 06:50
6 In the figure above, point O is the center of the circle 7 05 Mar 2013, 10:38
21 In the figure above, showing circle with center O and points 19 25 Feb 2013, 01:21
151 In the figure above, point O is the center of the circle and 39 16 Jan 2011, 15:56
Display posts from previous: Sort by