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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% (01:45) wrong based on 287 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?

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

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 = \((1/2) * (8\sqrt{3})/2 * 6 = 12\sqrt{3}\)

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.
_________________

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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 4848 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) = \(\frac{1}{2}*24\sqrt{3}=12\sqrt{3}\).

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

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

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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?

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

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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.

gmatclubot

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

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