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

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28 Feb 2013, 05:23

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37% (01:41) wrong based on 186 sessions

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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?

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

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28 Feb 2013, 07:07

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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

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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\).

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

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11 Jun 2013, 03:54

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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

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

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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."

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

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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 3363 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.

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 :

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

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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).

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

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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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\).

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

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12 Oct 2014, 12:30

Expert's post

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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\).

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. _________________

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