|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 25 Oct 2008
Posts: 619
Location: Kolkata,India
Followers: 6
Kudos [?]:
93
[0], given: 100
|
geometry triangles [#permalink]
06 Nov 2009, 05:58
Question Stats:
22% (02:25) correct
77% (00:40) wrong based on 0 sessions
Attachments

2.JPG [ 13.81 KiB | Viewed 1214 times ]
_________________
countdown-beginshas-ended-85483-40.html#p649902
|
|
|
|
|
|
|
Senior Manager
Joined: 31 Aug 2009
Posts: 426
Location: Sydney, Australia
Followers: 4
Kudos [?]:
76
[0], given: 20
|
Re: geometry triangles [#permalink]
06 Nov 2009, 06:27
1) Is insufficient - This length doesn't help us except to tell us that BCD is not a right isosceles triangle. 2) x = 60. This means that Angle BCD = 180-60 =120. The remaining angle CBD = 180-120-30 = 30. This tells us that the triangle is isosceles. BC = CD = 6.
ANS = B
|
|
|
|
|
|
Director
Joined: 25 Oct 2008
Posts: 619
Location: Kolkata,India
Followers: 6
Kudos [?]:
93
[0], given: 100
|
Re: geometry triangles [#permalink]
06 Nov 2009, 06:33
Nope..that's what I thought too!!please try again..
_________________
countdown-beginshas-ended-85483-40.html#p649902
|
|
|
|
|
|
Senior Manager
Joined: 18 Aug 2009
Posts: 340
Followers: 5
Kudos [?]:
120
[0], given: 13
|
Re: geometry triangles [#permalink]
06 Nov 2009, 06:56
According to me it should be D. Will explain in a while if this is the correct answer.
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[1] , given: 829
|
Re: geometry triangles [#permalink]
06 Nov 2009, 07:03
1
This post received KUDOS
tejal777 wrote: Nope..that's what I thought too!!please try again.. As it's DS question no need for actually finding the value of BC, rather than to determine that it's possible to determine it with either of statements: (1) BD=6\sqrt{3}, we know CD, BD and the angle between them the opposite side BC is fixed and has single value, meaning that you can not draw two or more triangles with given two sides and the angle between them. Sufficient. (2) x=60. Again we know x, hence we know all the angles in triangle BCD, plus we know one of the sides CD=6, again only one such triangle exists, hence BC value can be determined. Sufficient. Answer: D.
_________________
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
|
|
|
|
|
|
Director
Joined: 01 Apr 2008
Posts: 921
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 8
Kudos [?]:
124
[1] , given: 18
|
Re: geometry triangles [#permalink]
06 Nov 2009, 10:55
1
This post received KUDOS
Draw BE perpendicular to AD, let CE = y.
stmt1: sin30 = BE/BD, BD is known, so BE can be found as sin30 =1/2. cos30 = DE/BD, \sqrt{3}/2 = (6+y)/BD , so y can be found. Now we know BE and CE, apply pythagoras and find BC.
stmt2: sin30 = BE/BD, BD is known, so BE can be found as sin30 =1/2. Now apply sin60 = BE/BC, so BC can be found.
D
|
|
|
|
|
|
Senior Manager
Joined: 31 Aug 2009
Posts: 426
Location: Sydney, Australia
Followers: 4
Kudos [?]:
76
[0], given: 20
|
Re: geometry triangles [#permalink]
06 Nov 2009, 15:06
Kudos to both of you. Had not thought about that.
So just for my knowledge, if we know 2 sides of a triangle, and the angle in between we can safely determine that the information is sufficient?
When answering I did think about what you guys said, but thought that we a) couldn't assume that if we have 2 fixed sides and an angle we can derive the 3rd (althgouh I know we don't need to actaully derive it) and b) that trigonometry wasn't really required knowledge for the GMAT?
|
|
|
|
|
|
Director
Joined: 25 Oct 2008
Posts: 619
Location: Kolkata,India
Followers: 6
Kudos [?]:
93
[0], given: 100
|
Re: geometry triangles [#permalink]
06 Nov 2009, 17:44
Bunuel's way is absolutly correct!! And no trig is not reqd:)
_________________
countdown-beginshas-ended-85483-40.html#p649902
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11628
Followers: 1802
Kudos [?]:
9618
[2] , given: 829
|
Re: geometry triangles [#permalink]
06 Nov 2009, 20:26
2
This post received KUDOS
yangsta8 wrote: So just for my knowledge, if we know 2 sides of a triangle, and the angle in between we can safely determine that the information is sufficient? This is a very good question. Well, I think everybody agrees that knowing such tips is very important for GMAT. Especially in DS as it helps to avoid time wasting by not calculating an exact numerical values. When can we say that information given is sufficient to calculate some unknown value in triangle? Think it's the same as determining congruency. If we are given some data and we can conclude that ONLY one triangle with given measurements exists, it should mean also that with given data we can calculate anything regarding this triangle. Determining congruency: 1. SAS (Side-Angle-Side): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent. 2. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. 3. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. So, knowing SAS or ASA is sufficient to determine unknown angles or sides.NOTE IMPORTANT EXCEPTION:
The SSA condition (Side-Side-Angle) which specifies two sides and a non-included angle (also known as ASS, or Angle-Side-Side) does not always prove congruence, even when the equal angles are opposite equal sides. Specifically, SSA does not prove congruence when the angle is acute and the opposite side is shorter than the known adjacent side but longer than the sine of the angle times the adjacent side. This is the ambiguous case. In all other cases with corresponding equalities, SSA proves congruence. The SSA condition proves congruence if the angle is obtuse or right. In the case of the right angle (also known as the HL (Hypotenuse-Leg) condition or the RHS (Right-angle-Hypotenuse-Side) condition), we can calculate the third side and fall back on SSS. To establish congruence, it is also necessary to check that the equal angles are opposite equal sides. So, knowing two sides and non-included angle is NOT sufficient to calculate unknown side and angles. Angle-Angle-Angle
AAA (Angle-Angle-Angle) says nothing about the size of the two triangles and hence proves only similarity and not congruence. So, knowing three angles is NOT sufficient to determine lengths of the sides.In our original question we had had SAS situation with (1), and ASA situation in (2) so each alone was indeed sufficient to calculate any other unknown value in this triangle.
_________________
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
|
|
|
|
|
|
Director
Joined: 01 Apr 2008
Posts: 921
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 8
Kudos [?]:
124
[0], given: 18
|
Re: geometry triangles [#permalink]
06 Nov 2009, 20:47
Yup. Bunuel is bang on...And yes, trig knowledge is NOT required for GMAT, but if you have a basic understanding, it helps.
|
|
|
|
|
|
VP
Joined: 05 Mar 2008
Posts: 1489
Followers: 10
Kudos [?]:
164
[0], given: 31
|
Re: geometry triangles [#permalink]
15 Nov 2009, 19:44
Economist wrote: Yup. Bunuel is bang on...And yes, trig knowledge is NOT required for GMAT, but if you have a basic understanding, it helps. wow..very good tip..never knew that... would a question like this actually appear on an exam? if so, I got some studying to do
|
|
|
|
|
|
Senior Manager
Status: Do and Die!!
Joined: 15 Sep 2010
Posts: 337
Followers: 1
Kudos [?]:
27
[1] , given: 192
|
1
This post received KUDOS
If CD = 6, what is the length of BC? (1) 6 \sqrt{3}(2) x = 60 how 1 is enough ?
Attachments

g2.JPG [ 4.99 KiB | Viewed 678 times ]
_________________
I'm the Dumbest of All !!
|
|
|
|
|
|
Manager
Joined: 02 Sep 2010
Posts: 50
WE 1: Business Development Manger
WE 2: Assistant Manager-Carbon Trading
WE 3: Manager-Carbon Trading
Followers: 2
Kudos [?]:
3
[0], given: 16
|
Re: geometry triangles [#permalink]
18 Nov 2010, 09:43
From statement 1 we can know that the triangle BDC is 30-60-90 degree because cd=6 and bd =6root3 so for a 30-60-90 triangle x-xroot3-2x=6-6root3-12 so the length of BC =12 From Statement 2 I dont have asolution yet
|
|
|
|
|
|
Verbal GMAT Forum Moderator
Joined: 31 Jan 2010
Posts: 500
WE 1: 4 years Tech
Followers: 7
Kudos [?]:
67
[0], given: 149
|
Re: geometry triangles [#permalink]
20 Nov 2010, 06:21
tejal777 wrote: :cry: Using Premise 1) when BD is given,BC is given ,use cosine formula BC ^ 2 = BD ^ 2 + + CB ^ 2 - 2 CB.BC Cos 30 2) When x = 60, BCD=120 , Triangle is isoceles , CB=6 therefore D, either premise satisfies
_________________
My Post Invites Discussions not answers Try to give back something to the Forum.I want your explanations, right now ! Please let me know your opinion about the Chandigarh Gmat Centrehttp://gmatclub.com/forum/gmat-experience-at-chandigarh-india-centre-111830.html
|
|
|
|
|
|
|
Re: geometry triangles
[#permalink]
20 Nov 2010, 06:21
|
|
|
|
|
|
|
|
|
|
|