Last visit was: 24 Apr 2026, 12:05 It is currently 24 Apr 2026, 12:05
Close
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

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.
Close
Request Expert Reply
Confirm Cancel
User avatar
rxs0005
Joined: 07 Jun 2004
Last visit: 21 Jun 2017
Posts: 436
Own Kudos:
3,310
 [10]
Given Kudos: 22
Location: PA
Posts: 436
Kudos: 3,310
 [10]
3
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
811,051
 [7]
Given Kudos: 105,873
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,051
 [7]
5
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
General Discussion
avatar
pbull78
Joined: 16 Dec 2011
Last visit: 13 Oct 2012
Posts: 28
Own Kudos:
Given Kudos: 12
GMAT Date: 04-23-2012
Posts: 28
Kudos: 25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Pretz
Joined: 30 Jul 2013
Last visit: 15 Sep 2019
Posts: 60
Own Kudos:
Given Kudos: 33
Concentration: Technology, General Management
GMAT Date: 07-03-2015
GPA: 3.8
WE:Information Technology (Computer Software)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 24 Apr 2026
Posts: 11,229
Own Kudos:
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,008
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Pretz
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?

ans C...
the square has one vertex on center of circle and other vertex on circle .. the square with max possible area will be the one with two opposite vertices on center and on circle..
so the radius of circle will become the diagonal of square..
that is diagonal=2=side*\(\sqrt{2}\).. or side =\(\sqrt{2}\)
area =\(\sqrt{2}\)*\(\sqrt{2}\)=2
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,442
Own Kudos:
79,404
 [2]
Given Kudos: 485
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,442
Kudos: 79,404
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Pretz
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?
Attachment:
Ques3.jpg
Ques3.jpg [ 6.46 KiB | Viewed 4576 times ]

Draw the circle with center O and radius 2. Now, O is one vertex of the square. How can you draw the square such that it has one vertex on the circle? Can you draw a vertex adjacent to O on the circle? Note that you cannot. In that case, the vertex opposite O will lie outside the circle. So the vertex opposite O should be on the circle and the vertices adjacent to O will lie in the circle. In this case, the radius of the circle will be the diagonal do the square.

Diagonal of square = 2
Side of square \(= 2/\sqrt{2} = \sqrt{2}\)
Area of square \(= \sqrt{2}^2 = 2\)
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
Given Kudos: 105,873
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,051
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Pretz
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4




Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?

Please search before posting and format properly (rules-for-posting-please-read-this-before-posting-133935.html#p1096628).
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,975
Own Kudos:
Posts: 38,975
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109818 posts
Tuck School Moderator
853 posts