E, F, G, and H are the vertices of a polygon. Is polygon EFGH a square?
(1) EFGH is a parallelogram.
(2) The diagonals of EFGH are perpendicular bisectors of one another.
Square?
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 131
- Joined: Mon May 05, 2008 10:46 am
- Thanked: 1 times
-
- Senior | Next Rank: 100 Posts
- Posts: 38
- Joined: Thu Jun 19, 2008 4:54 am
- Location: Pune, India
I think the asnwer can be E..
Statement 1 is not sufficient, since a rectangle can also be called as a parallelogram.
Statement 2 alone is also not sufficient, since the same conditions can be said for a rectangle.
Both together do not give any additional information..hence E
Statement 1 is not sufficient, since a rectangle can also be called as a parallelogram.
Statement 2 alone is also not sufficient, since the same conditions can be said for a rectangle.
Both together do not give any additional information..hence E
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
A rectangle doesn't have perpendicular bisectors (unless it's a square, of course).microke wrote:I think the asnwer can be E..
Statement 1 is not sufficient, since a rectangle can also be called as a parallelogram.
Statement 2 alone is also not sufficient, since the same conditions can be said for a rectangle.
Both together do not give any additional information..hence E
However, both statements could be describing a rhombus, so the answer is in fact (e).
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course