|
Author |
Message |
|
TAGS:
|
|
|
Senior Manager
Joined: 11 Sep 2005
Posts: 331
Followers: 1
Kudos [?]:
3
[0], given: 0
|
In the figure, each side of square ABCD has length 1, the [#permalink]
 19 Oct 2007, 04:37
Question Stats:
0% (00:00) correct
100% (00:02) wrong based on 0 sessions
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
Attachments
set25q31.GIF [ 2.57 KiB | Viewed 1688 times ]
Last edited by Bunuel on 25 Jan 2012, 10:57, edited 2 times in total.
None of the answer choices shown is correct. See correct answer in my post below.
|
|
|
|
|
|
|
Senior Manager
Joined: 11 Sep 2005
Posts: 331
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Height of triangle BCE - 0.5
Base - 1
So area of BCE - 1/4
|
|
|
|
|
|
VP
Joined: 08 Jun 2005
Posts: 1172
Followers: 5
Kudos [?]:
78
[0], given: 0
|
If CE = 1 and BC = 1
so why is (1*1)/2 = 1/2 not the answer ?
this is way too easy
am I missing something ?
|
|
|
|
|
|
Intern
Joined: 02 Aug 2007
Posts: 38
Followers: 0
Kudos [?]:
2
[0], given: 0
|
singh_amit19 wrote: In the figure (attchd), 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
I got sqrt(2) / 4
I am wondering about ans. b and d in fact 2^-2 = 1/4 I think it should be sqrt(2) instead ... maybe I am wrong
Here is how I did it:
let's suppose O the intersection of ABCD diagonals. In tirangle BDE, the points A, C and O are alignes since C is the barycenter of the triangle (CE=CB=CD) and BED is isocel.
We can calculate the area of BCE by substracting the area of BOC from BOE.
area of BOE=BO*BE/2 = (1/sqrt(2)) * (1/sqrt(2) + 1) / 2 = (1 + sqrt(2))/4
area of BOC= 0.25 area of the square= 1/4
Thus area of BCE = (1 + sqrt(2))/4 - 1/4 = sqrt(2)/4
|
|
|
|
|
|
Senior Manager
Joined: 11 Sep 2005
Posts: 331
Followers: 1
Kudos [?]:
3
[0], given: 0
|
KillerSquirrel wrote: If CE = 1 and BC = 1 so why is (1*1)/2 = 1/2 not the answer ? this is way too easy am I missing something ? Area of a triangle = (1/2)XbaseXaltitude
So, how u getting (1*1)/2 = 1/2? R u assuming BCE a right angled triangle? if YES y n how???
|
|
|
|
|
|
Intern
Joined: 02 Aug 2007
Posts: 38
Followers: 0
Kudos [?]:
2
[0], given: 0
|
KillerSquirrel wrote: ronneyc wrote: KillerSquirrel wrote: singh_amit19 wrote: KillerSquirrel wrote: If CE = 1 and BC = 1 so why is (1*1)/2 = 1/2 not the answer ? this is way too easy am I missing something ? Area of a triangle = (1/2)XbaseXaltitude So, how u getting (1*1)/2 = 1/2? R u assuming BCE a right angled triangle? if YES y n how??? height = CE = 1 base = BC = 1 1*1*1/2 whats the OA ? I dont see why height = CE... What is the OA please ? the angle BCE should be 90° to use this formula which is not the case here
=> CE is not a height
|
|
|
|
|
|
Intern
Joined: 02 Aug 2007
Posts: 38
Followers: 0
Kudos [?]:
2
[0], given: 0
|
singh_amit19 wrote: OA is D....any takers????
if we calculate ans D which is (2^-2)/2 we get (1/4) * (1/2) = 1/8
Can you please provide the method ? I dont still dont see it ...
thx
|
|
|
|
|
|
Director
Status: GMAT Learner
Joined: 14 Jul 2010
Posts: 672
Followers: 21
Kudos [?]:
108
[0], given: 31
|
Re: In the figure (attchd), each side of square ABCD has length [#permalink]
25 Jan 2012, 09:15
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[2] , given: 829
|
Re: In the figure (attchd), each side of square ABCD has length [#permalink]
25 Jan 2012, 10:53
2
This post received
KUDOS
Baten80 wrote: I still in trouble with this problem. My document shows OA is B. Please help. Responding to a pm. Attachment: 
Square ABCD.JPG [ 12.11 KiB | Viewed 1242 times ]
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?This proble can be solved in many ways. One of the approaches: Note that as BE=DE the triangles BCE and CDE are congruent (all three sides are equal) --> \angle{BCE}=\angle{DCE}. So if we continue the line segment CE it will meet with the crosspoint of the diagonals, let's call it point O. Also note that EO will be perpendicular to BD, as diagonals in square make a right angle. The area BCE=BOE-BOC. Area of BOC is one fourth of the square's = \frac{1}{4}. Area BOE, \frac{1}{2}BO*EO. BO=\frac{\sqrt{2}}{2}, half of the diagonal of a square. EO=CE+CO=1+\frac{\sqrt{2}}{2}=\frac{2+\sqrt{2}}{2}, CO is also half of the diagonal of a square. So AreaBOE=\frac{1}{2}BO*EO=\frac{1}{2}*\frac{\sqrt{2}}{2}*\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{4}. Area BCE=BOE-BOC=\frac{\sqrt{2}+1}{4}-\frac{1}{4}=\frac{\sqrt{2}}{4}None of the answer choices shown is correct.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Manager
Joined: 12 Nov 2011
Posts: 145
Followers: 0
Kudos [?]:
1
[0], given: 24
|
Re: In the figure, each side of square ABCD has length 1, the [#permalink]
25 Jan 2012, 11:03
very hard one for me
thanx for explanation
|
|
|
|
|
|
Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 190
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)
Followers: 3
Kudos [?]:
14
[0], given: 1
|
Re: In the figure, each side of square ABCD has length 1, the [#permalink]
26 Jan 2012, 06:34
Bunnel thanks for the explanation +1
|
|
|
|
|
|
Intern
Joined: 17 Oct 2011
Posts: 29
Location: Taiwan, Province of China
GMAT 1: 590 Q39 V34
GMAT 2: 680 Q47 V35
Followers: 0
Kudos [?]:
9
[0], given: 10
|
Re: In the figure, each side of square ABCD has length 1, the [#permalink]
08 Feb 2012, 02:18
The answer choices given are wrong.
The correct ACs are:
A. 1/3
B. (sq.root 2)/4
C. 1/2
D. (sq.root 2)/2
E. 3/4
OA: B
|
|
|
|
|
|
Manager
Joined: 10 Jan 2010
Posts: 194
Location: Germany
Concentration: Strategy, General Management
GMAT 1: Q V
GPA: 3
WE: Consulting (Telecommunications)
Followers: 2
Kudos [?]:
15
[0], given: 7
|
Re: In the figure, each side of square ABCD has length 1, the [#permalink]
08 Feb 2012, 06:07
Could not solve it. But thanks for the explanation. One level to hard for me.
|
|
|
|
|
|
Senior Manager
Joined: 19 Apr 2011
Posts: 288
Schools: Booth,NUS,St.Gallon
Followers: 3
Kudos [?]:
27
[0], given: 47
|
Re: In the figure, each side of square ABCD has length 1, the [#permalink]
10 Feb 2012, 10:53
the figure shown above is a square pyramid .Only then BE and DE will be the same .Now we are asked to find the are of one of the slant faces . Area of a pyramid =1/2 * perimeter of the base * slant height =1/2 *(4)*1 We need the area of one of the sides only .Hence Area = 1/2. Option B is the answer .Sorry if i had brought in some new formulas and concepts .Hope you are aware .
_________________
+1 if you like my explanation .Thanks
|
|
|
|
|
|
|
Re: In the figure, each side of square ABCD has length 1, the
[#permalink]
10 Feb 2012, 10:53
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
Figure ABCD is a rectangle with sides of length x
|
dynocomet |
3 |
15 Jul 2007, 19:45 |
|
|
|
 |
In the figure, each side of square ABCD has length 1, the
|
gregspirited |
2 |
27 Nov 2007, 14:42 |
|
|
|
|
each side of square ABCD has length 1, the length of line
|
Jcpenny |
1 |
19 Nov 2008, 21:13 |
|
|
1
|
|
each side of square ABCD has length 1, the length of line
|
haichao |
3 |
20 Nov 2008, 09:12 |
|
|
2
|
|
In the figure, each side of square ABCD has length 1, the
|
vaivish1723 |
5 |
25 Jan 2010, 10:05 |
|
|
|
|
|
|
close
