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 Post subject: In the figure, ABC is
 Post Posted: Mon Jun 15, 2009 6:22 pm 
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Posts: 2
I am curious to learn how adding the line segment OA helped us. Was it not already evident that both ABD and ACD were inscribed?

Just want to make sure I am not missing a key point.

Thanks!

In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees



ANSWER:In order to find the area of the circle, we need to know its radius or a related quantity, such as the diameter or the circumference. Let us first see how much we can conclude based on the initial givens before we consider the statements.

We are told that DAB is a right triangle. A right triangle inscribed in a circle will have a diameter of the circle as its hypotenuse. Therefore DB is a diameter. Denoting the center of the circle with O, and adding the auxiliary line segment (radius) OA, we get the following picture:



We now see that both angle ADB and angle ACB are inscribed angles in the circle, and that they both span the same arc, ACB. Therefore, they must be equal in measure (and each equal to half the corresponding central angle, AOB. But we know how big angle ACB is—it’s 60 degrees, because ABC is an equilateral triangle. Thus angle ADB is 60 degrees as well, and angle ABD must be 30 degrees. This is because the three internal angles of right triangle DAB add up to 180 degrees. At this point, we can recognize triangle DAB as a 30-60-90 triangle.

(1) SUFFICIENT: knowing that the short side of the 30-60-90 triangle is equal to 4, we can solve for the hypotenuse DB, which is 8. We know that DB is the diameter of the circle, so we can calculate the desired area.

(2) INSUFFICIENT: this statement tells us nothing about the radius, diameter or circumference of the circle. In fact, in light of the advance work we did, we can see that it tells us nothing that we do not already know.

The correct answer is A.


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 Post subject: Re: In the figure, ABC is
 Post Posted: Mon Jul 13, 2009 6:47 am 
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ManhattanGMAT Staff


Posts: 6765
hi -

could you do us a favor, please, and post an image (screen shot) along with the question?
this thread is viewed by other posters who won't know what is happening unless the problem is accompanied by a diagram.

if you don't know how to post image files, you can upload them to image hosting websites (such as postimage.org), and then post a direct link to the hosting site here.
directions:
(1) go to postimage.org
(2) click "browse"
(3) find the file on your computer and double click it
(4) click "upload"
(5) when the uploading is complete, copy the "direct link" (which should be displayed on the page, underneath some other codes) and paste it into this thread.

thanks!


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 Post subject: Re: In the figure, ABC is
 Post Posted: Fri Feb 12, 2010 3:43 pm 
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Students


Posts: 1
http://www.postimage.org/image.php?v=aVRSRli

Here is the image for this question.

This is a good tricky question. I got this on my manhattan practice test. The whole answer relies on the fact that 2 angles in 2 separate triangles are equal. I dont know about this rule, can you please explain. Below is the snapshot of the official answer given in the manhattan exam key.

"We are told that DAB is a right triangle. A right triangle inscribed in a circle will have a diameter of the circle as its hypotenuse. Therefore DB is a diameter. Denoting the center of the circle with O, and adding the auxiliary line segment (radius) OA, we get the following picture.
We now see that both angle ADB and angle ACB are inscribed angles in the circle, and that they both span the same arc, ACB. Therefore, they must be equal in measure"


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 Post subject: Re: In the figure, ABC is
 Post Posted: Fri Feb 19, 2010 8:38 am 
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ManhattanGMAT Staff


Posts: 6765
AbimbolaOkeowo wrote:
I am curious to learn how adding the line segment OA helped us. Was it not already evident that both ABD and ACD were inscribed?


if you already know that two inscribed angles are automatically equal if they happen to cut off the same arc, then, no, you don't need segment OA.

on the other hand, some students haven't memorized quite so many rules -- given an inscribed angle, they may only know how to relate it to a central angle. if that's the case, then these students are going to need the central angle in order to make the required connections.


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 Post subject: Re: In the figure, ABC is
 Post Posted: Thu Aug 19, 2010 10:19 pm 
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Course Students


Posts: 75
Can we say that diameter DB bisects angle ABC because triangle ABC is equilateral?

If yes, do we still need to introduce the central angle?

If no, why not?

Thanks,


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 Post subject: Re: In the figure, ABC is
 Post Posted: Fri Sep 03, 2010 2:56 am 
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ManhattanGMAT Staff


Posts: 506
Can we say that diameter DB bisects angle ABC because triangle ABC is equilateral?

I'm not sure that I follow this question. Do you mean to ask whether we can say this without considering S2? Then the answer is no. Though trinagle ABD must be a right triangle if BD is a diameter, angle ABD need not be 30 degrees. In fact, 0<ABD in degrees<90.


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 Post subject: Re: In the figure, ABC is
 Post Posted: Mon Apr 11, 2011 9:40 pm 
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Course Students


Posts: 6
I have same doubt as agnitap. Reposting the same question. Can some one pls. clarify how the 2 angles in 2 separate triangles are equal?



http://www.postimage.org/image.php?v=aVRSRli

Here is the image for this question.

This is a good tricky question. I got this on my manhattan practice test. The whole answer relies on the fact that 2 angles in 2 separate triangles are equal. I dont know about this rule, can you please explain. Below is the snapshot of the official answer given in the manhattan exam key.

"We are told that DAB is a right triangle. A right triangle inscribed in a circle will have a diameter of the circle as its hypotenuse. Therefore DB is a diameter. Denoting the center of the circle with O, and adding the auxiliary line segment (radius) OA, we get the following picture.
We now see that both angle ADB and angle ACB are inscribed angles in the circle, and that they both span the same arc, ACB. Therefore, they must be equal in measure"


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 Post subject: Re: In the figure, ABC is
 Post Posted: Wed Apr 13, 2011 8:02 pm 
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ManhattanGMAT Staff


Posts: 1779
Location: Southwest Airlines, seat 21C
welcome to something i like to call "The Rules". :) this is how it is; you need to memorize this Rule. both the angles in question are called inscribed angles; the Rule is that the measure of the inscribed angle is half the arc the angle cuts off. in this case, both the angles cut off the same arc, so they both have measure equal to half the arc. this means they are the same size..

_________________
Tim Sanders
Manhattan GMAT Instructor


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