Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Aug 2710:00 AM EDT
-11:00 AM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0410:00 AM PDT
-11:00 AM PDT
What does test anxiety really look like on the GMAT? For many test takers, it does not look like a panic attack. Instead, it shows up as brain fog, rereading, negative comparisons, mental noise....
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 07
12:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT - Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
08 May 2008, 10:13
Kudos
Bookmarks
I ran across this question and think the answer provided is incorrect.
Sorry this image is so huge! (If anyone can help me shrink it, that'd be great.)
See the bottom of the page for the image, I'll post the question and info regarding the image as this image is not identical to the one in the quesiton.
**IMAGE NOTES**
The square is inside the circle. The corners of the square are points labeled A, B, C, and D on the circle. A=Top Left, B=Top Right, C = Bottom Right, D = Bottom Left. B & D are identified as right angles. A line connects AC. There is no center of the circle indicated. Here is the question:
**QUESTION**
If points A, B, C, and D are points on the circumference of the circle in the figure [below], what is the area of ABCD?
1) The radius of the circlie is \(\frac{\sqrt{2}}{2}\)
2) ABCD is a square.
**END QUESTION**
I was thinking the answer is A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient. (Note: while typing this, I realize that the answer is C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient, which was the answer given.)
Here is my reasoning:
When you know the radius of the circle, you also know the length of segment AC because AC is also the diameter of the circule. You're asked to find the area of the quadralateral ABCD. Using \(a^2 + b^2 = c^2\) doesn't quite get you all the way there. We don't know if it's a special traingle (i.e., \(1-1-sqrt{2}\)). If we did (i.e. a square), then we could figure the area of triangle and double it for the area of ABCD.
Double the radius, so \(2 * \frac{\sqrt{2}}{2}=\sqrt{2}\)[diameter of circle]. So if the hypotnuse of the triangle is \(\sqrt{2}\), then this is a special triangle, and know the base is 1, height = 1, so the area of the square is 1.
If ABCD is a non-square, we still only have the hypotnuse (sp?) of the triangle. We need 2 of 3 for the \(a^2 + b^2 = c^2\) to work. BOTH together are sufficient once we know that ABCD is a square and can use \(1-1-sqrt{2}\).
Even though I answered my own question while writing it, I hope this helps others.
Sorry this image is so huge! (If anyone can help me shrink it, that'd be great.)
See the bottom of the page for the image, I'll post the question and info regarding the image as this image is not identical to the one in the quesiton.
**IMAGE NOTES**
The square is inside the circle. The corners of the square are points labeled A, B, C, and D on the circle. A=Top Left, B=Top Right, C = Bottom Right, D = Bottom Left. B & D are identified as right angles. A line connects AC. There is no center of the circle indicated. Here is the question:
**QUESTION**
If points A, B, C, and D are points on the circumference of the circle in the figure [below], what is the area of ABCD?
1) The radius of the circlie is \(\frac{\sqrt{2}}{2}\)
2) ABCD is a square.
**END QUESTION**
I was thinking the answer is A) Statement (1) ALONE is sufficient, but statement (2) is not sufficient. (Note: while typing this, I realize that the answer is C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient, which was the answer given.)
Here is my reasoning:
When you know the radius of the circle, you also know the length of segment AC because AC is also the diameter of the circule. You're asked to find the area of the quadralateral ABCD. Using \(a^2 + b^2 = c^2\) doesn't quite get you all the way there. We don't know if it's a special traingle (i.e., \(1-1-sqrt{2}\)). If we did (i.e. a square), then we could figure the area of triangle and double it for the area of ABCD.
Double the radius, so \(2 * \frac{\sqrt{2}}{2}=\sqrt{2}\)[diameter of circle]. So if the hypotnuse of the triangle is \(\sqrt{2}\), then this is a special triangle, and know the base is 1, height = 1, so the area of the square is 1.
If ABCD is a non-square, we still only have the hypotnuse (sp?) of the triangle. We need 2 of 3 for the \(a^2 + b^2 = c^2\) to work. BOTH together are sufficient once we know that ABCD is a square and can use \(1-1-sqrt{2}\).
Even though I answered my own question while writing it, I hope this helps others.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
09 May 2008, 10:09
Kudos
Bookmarks
I think C is the right answer. In the question he said A,B,C and D are points on a circle. Consider two points A and B are the points of the diameter of the circle and B and C lie on the same side.I guess we cannot solve for area using statement 1
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
