Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
AbimbolaOkeowo
 
Posts: 2
Joined: Wed Jun 11, 2008 4:07 am
 

In the figure, ABC is

by AbimbolaOkeowo Mon Jun 15, 2009 6:22 pm

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.
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: In the figure, ABC is

by RonPurewal Mon Jul 13, 2009 6:47 am

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!
agnitap
Students
 
Posts: 1
Joined: Sun Oct 04, 2009 8:08 am
 

Re: In the figure, ABC is

by agnitap Fri Feb 12, 2010 3:43 pm

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"
RonPurewal
Students
 
Posts: 19744
Joined: Tue Aug 14, 2007 8:23 am
 

Re: In the figure, ABC is

by RonPurewal Fri Feb 19, 2010 8:38 am

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.
chitrangada.maitra
Course Students
 
Posts: 75
Joined: Thu Feb 25, 2010 2:03 pm
 

Re: In the figure, ABC is

by chitrangada.maitra Thu Aug 19, 2010 10:19 pm

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,
mschwrtz
ManhattanGMAT Staff
 
Posts: 498
Joined: Tue Dec 14, 2004 1:03 pm
 

Re: In the figure, ABC is

by mschwrtz Fri Sep 03, 2010 2:56 am

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.
sureng.reddy
Course Students
 
Posts: 6
Joined: Mon Dec 07, 2009 6:47 pm
 

Re: In the figure, ABC is

by sureng.reddy Mon Apr 11, 2011 9:40 pm

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"
tim
Course Students
 
Posts: 5665
Joined: Tue Sep 11, 2007 9:08 am
Location: Southwest Airlines, seat 21C
 

Re: In the figure, ABC is

by tim Wed Apr 13, 2011 8:02 pm

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

Follow this link for some important tips to get the most out of your forum experience:
https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html