Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: Inscribed Triangle [#permalink]
03 Mar 2012, 13:31
5
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
Attachment:
Inscribed Triangles.gif [ 2.27 KiB | Viewed 6831 times ]
In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?
You should know the following properties to solve this question: • All inscribed angles that subtend the same arc are equal. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle. Hence, all inscribed angles that subtend the same arc are equal. • 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 hypotenuse, then that triangle is a right triangle. • In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). For more check Circles Triangles and chapters of Math Book: math-circles-87957.html and math-triangles-87197.html
So, from above we'll have that as DAB=90 degrees then DB must be a diameter of the circle. Next, as angles ACB and ADB subtend the same arc AB then they must be equal and since ACB=60 (remeber ACB is an equilateral triangle) then ADB=60 too. Thus DAB is 30-60-90 triangle and its sides are in ratio \(1 : \sqrt{3}: 2\).
(1) DA = 4 --> the side opposite 30 degrees is 4, then hypotenuse DB=diameter=4*2=8 --> radius=4 --> \(area=\pi{r^2}=16\pi\). Sufficient.
(2) Angle ABD = 30 degrees --> we knew this from the stem, so nothing new. Not sufficient.
Re: Geometry: Circle in [#permalink]
03 Apr 2012, 15:46
Bunuel wrote:
Merging similar topics. Please ask if anything remains unclear.
Thanks for merging! I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?
Re: Geometry: Circle in [#permalink]
04 Apr 2012, 00:15
1
This post received KUDOS
Expert's post
KGG88 wrote:
Bunuel wrote:
Merging similar topics. Please ask if anything remains unclear.
Thanks for merging! I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?
All inscribed angles that subtend the same arc are equal. Since angles ACB and ADB subtend the same arc AB then they must be equal. Next, there is no central angle shown on the diagram, I just mentioned Central Angle Theorem to explain why is above property true: angles ACB and ADB have the same central angle AOB (O is the center of the circle) and since they both equal to half of it then they must be equal.
Re: Why is this angle 60degrees? [#permalink]
03 Apr 2013, 12:43
So I chose both statements together are sufficient but here is the explanation provided: Since both angles ACB and ADB are opposite side AB, and angle ACB is part of an equilateral triangle, we can say both ACB and ADB are 60 degrees.
Why does that hold true? I get that : 1. DAB is 90 2. CAB = ACB = BAC = 60
and based on that all i can say is ABD < 60 so if we arent told ABD's value, cant ABD = 70 and ADB = 20?
In any case, if we can deduce ABD from just statement 2, then the rest is easy. if ABD = 30 degrees (as told by statement 1), then DB = diameter. we can figure out the length of DB using the 30-60-90 relationship and solve for area.
Re: Why is this angle 60degrees? [#permalink]
03 Apr 2013, 12:59
2
This post received KUDOS
wown wrote:
So I chose both statements together are sufficient but here is the explanation provided: Since both angles ACB and ADB are opposite side AB, and angle ACB is part of an equilateral triangle, we can say both ACB and ADB are 60 degrees.
Why does that hold true? I get that : 1. DAB is 90 2. CAB = ACB = BAC = 60
and based on that all i can say is ABD < 60 so if we arent told ABD's value, cant ABD = 70 and ADB = 20?
In any case, if we can deduce ABD from just statement 2, then the rest is easy. if ABD = 30 degrees (as told by statement 1), then DB = diameter. we can figure out the length of DB using the 30-60-90 relationship and solve for area.
thanks.
There is a thing you are not considering: in any right triangle inscribed in a circle, the hypotenuse coincides with a diameter of the circle. So we know that DB is the diameter, that the bisectors of the ACB triangle intersect at the center (ancient math theory if I remember...). ABC=60° is divided in two 30°angles, DAB=90° and so ADB=60° And now as you said "we can figure out the length of DB using the 30-60-90 relationship and solve for area." _________________
It is beyond a doubt that all our knowledge that begins with experience.
Re: Why is this angle 60degrees? [#permalink]
03 Apr 2013, 13:28
1
This post received KUDOS
Wown,
This is very good trap C DS questions and if running out of time on exam you want to guess anything but C.
That said, the problem is using the relationship between an inscribed angle and central angle. central angle is the angle formed by an arc at the center and inscribed angle is the angle formed by the same arc at the circle.
central angle = 2 * inscribed angle - Here <ACB and <ADB are inscribed angles formed by arc AB.
From the stem, it is given that <ACB = 60 so the central angle for arch AB is 120. Make O as the center of the circle, so <AOB = 120
Now <ADB is the inscribed angle for <AOB so it will be 60 degrees. Triangle ADB is a 30-60-90 type right angle triangle and stmt 2 gives DA = 8 using which you can calculate the diameter of the circle as 16 and circumference as 16pi. Hence stmt B is sufficient.
Thanks for sharing this problem...
//Kudos please, if the above explanation is good. _________________
Re: Why is this angle 60degrees? [#permalink]
03 Apr 2013, 13:43
nt2010 wrote:
Wown,
This is very good trap C DS questions and if running out of time on exam you want to guess anything but C.
That said, the problem is using the relationship between an inscribed angle and central angle. central angle is the angle formed by an arc at the center and inscribed angle is the angle formed by the same arc at the circle.
central angle = 2 * inscribed angle - Here <ACB and <ADB are inscribed angles formed by arc AB.
From the stem, it is given that <ACB = 60 so the central angle for arch AB is 120. Make O as the center of the circle, so <AOB = 120
Now <ADB is the inscribed angle for <AOB so it will be 60 degrees. Triangle ADB is a 30-60-90 type right angle triangle and stmt 2 gives DA = 8 using which you can calculate the diameter of the circle as 16 and circumference as 16pi. Hence stmt B is sufficient.
Thanks for sharing this problem...
//Kudos please, if the above explanation is good.
Also did not consider this central angle-inscribed angle relationship... very interesting
Re: Why is this angle 60degrees? [#permalink]
03 Apr 2013, 20:40
Expert's post
wown wrote:
I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?
We know that angle subtended by the same arc, on the circumference are always equal.
From F.S 1, the data given is of no value. It is so because Angle ACB = Angle ADB = 60 degrees. Thus, angle ABD has to be (90-60) = 30 degrees.Insufficient.
From F.S 2, we know the length of AD. Thus, in the case of a 60-30-90 triangle, we can easily calculate the hypotenuse = the diameter =BD and hence the area.Sufficient.
Re: In the figure, ABC is an equilateral triangle, and DAB is [#permalink]
11 Oct 2015, 08:22
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. _________________
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...