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

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

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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

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Last edited by Bunuel on 28 Feb 2013, 06:57, edited 1 time in total.
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Re: In the figure above, showing circle with center O and points [#permalink]

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New post 28 Feb 2013, 07:07
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Image
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
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Re: In the figure above, showing circle with center O and points [#permalink]

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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

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

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

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

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New post 12 Jun 2013, 01:42
Thank you for another great geometry question, emmak. Now let's solve it properly :)

Attachment:
1.jpg
1.jpg [ 6.23 KiB | Viewed 5569 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

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

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New post 12 Oct 2014, 12:18
Bunuel wrote:
Image
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?

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

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New post 12 Oct 2014, 12:30
saggii27 wrote:
Bunuel wrote:
Image
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.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | 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?
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Re: In the figure above, showing circle with center O and points [#permalink]

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New post 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
triangle.png [ 36.06 KiB | Viewed 1435 times ]

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

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New post 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. :-D

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