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.
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2211:00 AM EDT
-11:59 PM EDT
Take 30% off all study plans with code PICNIC - Expires on Monday, May 25th
May 2512:01 AM PDT
-11:59 PM PDT
On the 25th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PST- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
27 Nov 2007, 14:42
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 1
sessions
History
Date
Time
Result
Not Attempted Yet
In the figure, each side of square ABCD has length 1, the length of line
Segment CE is 1, and the length of line segment BE is equal to the length
Of line segment DE. What is the area of the triangular region BCE?
a. 1/3
b. (2^-2 )/4
c. 1/2
d. (2^-2)/2
e. 3/4
See attached file for figure
Segment CE is 1, and the length of line segment BE is equal to the length
Of line segment DE. What is the area of the triangular region BCE?
a. 1/3
b. (2^-2 )/4
c. 1/2
d. (2^-2)/2
e. 3/4
See attached file for figure
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 Problem Solving (PS) 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!
27 Nov 2007, 15:26
Kudos
Bookmarks
area(BCE)=area(BGE)-area(BGC)
area(BCE)=1/2*1/√2*(1+h)-1/2*1/√2*h=1/2*1/√2=2^(-3/2)
but 2^(-3/2) is not among possible solutions. What is wrong?
area(BCE)=1/2*1/√2*(1+h)-1/2*1/√2*h=1/2*1/√2=2^(-3/2)
but 2^(-3/2) is not among possible solutions. What is wrong?
Attachments
GC56301.gif [ 11.57 KiB | Viewed 3059 times ]
27 Nov 2007, 15:37
Kudos
Bookmarks
walker
Your figure is right. I was confused with the figure. The answer is D.
You are on right track. BG=GC = root(2)/2 (half of diagonal of square)
Once we know BG and GC,rest is as you said.
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 Problem Solving (PS) 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!
