Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 09 Feb 2013
Posts: 121

In the figure above, showing circle with center O and points [#permalink]
Show Tags
28 Feb 2013, 05:23
12
This post was BOOKMARKED
Question Stats:
63% (03:06) correct
37% (01:45) wrong based on 293 sessions
HideShow timer Statistics
Attachment:
1.jpg [ 6.23 KiB  Viewed 6146 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Kudos will encourage many others, like me. Good Questions also deserve few KUDOS.
Last edited by Bunuel on 28 Feb 2013, 06:57, edited 1 time in total.
Edited the question.



Math Expert
Joined: 02 Sep 2009
Posts: 39662

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
28 Feb 2013, 07:07
3
This post received KUDOS
Expert's post
13
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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\). Answer: A. For more check here: mathtriangles87197.html
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 15 May 2013
Posts: 7

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
11 Jun 2013, 03:54
3
This post received KUDOS
1
This post was BOOKMARKED
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) Answer is  A



Verbal Forum Moderator
Joined: 16 Jun 2012
Posts: 1127
Location: United States

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
12 Jun 2013, 00:56
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
GPA: 3.7
WE: Analyst (Consulting)

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
12 Jun 2013, 01:42
Thank you for another great geometry question, emmak. Now let's solve it properly Attachment:
1.jpg [ 6.23 KiB  Viewed 4953 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: 15956

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
08 Oct 2014, 03:06
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Intern
Joined: 15 Jul 2012
Posts: 37

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
12 Oct 2014, 12:18
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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\). Answer: A. For more check here: mathtriangles87197.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: 39662

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
12 Oct 2014, 12:30
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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\). Answer: A. For more check here: mathtriangles87197.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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15956

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
26 Oct 2015, 04:07
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15956

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
01 Feb 2017, 04:15
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
01 Feb 2017, 05:25
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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 AOC is the diameter so the angle subtended in semi circle ABC will be 90 degrees. So triangle ABC is a right triangle at B. If \(AB = 12\) and \(BC = 4*\sqrt{3}\), this is a ratio of \(\sqrt{3}:1\) which means that the hypotenuse will be 2 on the ratio scale. So \(AC = 2*4*\sqrt{3} = 8\sqrt{3}\) Area of triangle ABC = (1/2) * AB * BC = (1/2) * Altitude from B to AC * AC Altitude from B to AC \(= 12 * (4\sqrt{3}) / (8\sqrt{3}) = 6\) Area of triangle AOB = (1/2) * AO * Altitude from B to AC = \((1/2) * (8\sqrt{3})/2 * 6 = 12\sqrt{3}\) Answer (A)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Director
Joined: 07 Dec 2014
Posts: 711

Re: In the figure above, showing circle with center O and points [#permalink]
Show Tags
01 Feb 2017, 11:18
emmak wrote: Attachment: The attachment 1.jpg is no longer available 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
Attachments
triangle.png [ 36.06 KiB  Viewed 818 times ]



Intern
Joined: 13 Jan 2015
Posts: 16

In the figure above, showing circle with center O and points [#permalink]
Show Tags
12 Feb 2017, 23:04
I only calculated the area of triangle ABC which came out to be 24sqrt3 . Now triangle AOB will have an area smaller than triangle ABC .Went through the options and 12sqrt3 is the only value that is less than 24sqrt3 . Took less than 30 secs.




In the figure above, showing circle with center O and points
[#permalink]
12 Feb 2017, 23:04








Similar topics 
Author 
Replies 
Last post 
Similar Topics:




The figure above shows three circles, all centered on point O. What is

PareshGmat 
3 
26 Oct 2016, 10:57 

1


The figure above shows three circles, all centered on point O. What i

Bunuel 
6 
08 Jan 2015, 09:04 

6


In the figure above, point O is the center of the circle

alex1233 
7 
02 Jun 2016, 09:48 

23


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

emmak 
19 
31 Jul 2016, 00:24 

182


In the figure above, point O is the center of the circle and

Bunuel 
43 
25 May 2017, 11:35 



