Last visit was: 23 Apr 2026, 17:34 It is currently 23 Apr 2026, 17:34
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
emmak
Joined: 09 Feb 2013
Last visit: 08 Jul 2014
Posts: 103
Own Kudos:
4,856
 [36]
Given Kudos: 17
Posts: 103
Kudos: 4,856
 [36]
4
Kudos
Add Kudos
32
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,870
 [31]
10
Kudos
Add Kudos
21
Bookmarks
Bookmark this Post
avatar
vigrah
Joined: 15 May 2013
Last visit: 25 Sep 2022
Posts: 5
Own Kudos:
98
 [5]
Given Kudos: 7
Posts: 5
Kudos: 98
 [5]
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
pqhai
User avatar
Retired Moderator
Joined: 16 Jun 2012
Last visit: 26 Nov 2015
Posts: 864
Own Kudos:
Given Kudos: 123
Location: United States
Posts: 864
Kudos: 8,939
Kudos
Add Kudos
Bookmarks
Bookmark this Post
emmak
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.
User avatar
Virgilius
Joined: 15 Feb 2013
Last visit: 25 May 2016
Posts: 25
Own Kudos:
Given Kudos: 11
Status:Currently Preparing the GMAT
Location: United States
GMAT 1: 550 Q47 V23
GPA: 3.7
WE:Analyst (Consulting)
GMAT 1: 550 Q47 V23
Posts: 25
Kudos: 61
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thank you for another great geometry question, emmak. Now let's solve it properly :)

Attachment:
1.jpg
1.jpg [ 6.23 KiB | Viewed 15519 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 :-D
User avatar
saggii27
Joined: 15 Jul 2012
Last visit: 31 Mar 2016
Posts: 23
Own Kudos:
Given Kudos: 245
Posts: 23
Kudos: 62
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

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: 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?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,870
 [1]
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,870
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
saggii27
Bunuel

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: 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.
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,396
 [3]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
 [3]
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
emmak
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)
User avatar
gracie
Joined: 07 Dec 2014
Last visit: 11 Oct 2020
Posts: 1,028
Own Kudos:
Given Kudos: 27
Posts: 1,028
Kudos: 2,022
Kudos
Add Kudos
Bookmarks
Bookmark this Post
emmak
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
triangle.png [ 36.06 KiB | Viewed 11026 times ]

User avatar
ram186
Joined: 13 Jan 2015
Last visit: 08 Dec 2021
Posts: 10
Own Kudos:
Given Kudos: 187
Posts: 10
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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. :-D
avatar
vicky2019
Joined: 22 Sep 2018
Last visit: 18 Jan 2022
Posts: 22
Own Kudos:
Given Kudos: 65
Posts: 22
Kudos: 32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
emmak
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


AB=12 and BC=4\sqrt{3}

AC is the diameter of the circle and also the side of triangle ABC. So the angle at B is 90.

AC^2=AB^2 + BC^2
from this we get AC=8\sqrt{3}

Area of ABC= 1/2 * AB * BC

= 24\sqrt{3}

now

BC=OC=OB=4\sqrt{3}

Area of triangle OBC is \sqrt{3}/4 * (4\sqrt{3})^2..i.e., 12\sqrt{3}

Area of AOB = Area of ABC- Area of OBC

= 24\sqrt{3} - 12\sqrt{3}

= 12\sqrt{3}

IMO: option A
avatar
pragya007
Joined: 21 May 2020
Last visit: 23 Oct 2020
Posts: 10
Own Kudos:
Given Kudos: 224
Posts: 10
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Question is good but tge option make it 600 difficulty one. Because for the whole triangle ABC is are is 24√3 so any how area of AOB will be less than 24√3 and only one smaller than that option A

Posted from my mobile device
User avatar
Kritisood
Joined: 21 Feb 2017
Last visit: 19 Jul 2023
Posts: 488
Own Kudos:
Given Kudos: 1,090
Location: India
GMAT 1: 700 Q47 V39
Products:
GMAT 1: 700 Q47 V39
Posts: 488
Kudos: 1,315
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

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: https://gmatclub.com/forum/math-triangles-87197.html

could you please elaborate on how BO is the median?
"Now, notice that since OA=OC=radius, then BO turns out to be the median of triangle ABC"
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,870
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kritisood
Bunuel

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: https://gmatclub.com/forum/math-triangles-87197.html

could you please elaborate on how BO is the median?
"Now, notice that since OA=OC=radius, then BO turns out to be the median of triangle ABC"

The median of a triangle is a line from a vertex to the midpoint of the opposite side. O is the midpoint of AC, so BO is the median.
avatar
Pranjal3107
avatar
Current Student
Joined: 13 Apr 2020
Last visit: 06 Jul 2022
Posts: 133
Own Kudos:
Given Kudos: 1,709
Location: India
GMAT 1: 710 Q45 V41
GPA: 3
GMAT 1: 710 Q45 V41
Posts: 133
Kudos: 18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shouldn't A0=B0=OC. In that case since the length of AC=8 root 3 then A0=OC=4 root 3 and also B0=4 root 3 ?
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,964
Own Kudos:
Posts: 38,964
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109785 posts
Tuck School Moderator
853 posts