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Re: geometry triangles [#permalink]
06 Nov 2009, 05:27
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.
Re: geometry triangles [#permalink]
06 Nov 2009, 06:03
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tejal777 wrote:
Nope..that's what I thought too!!please try again..
If CD = 6, what is the length of BC?
As it's a DS question no need to actually find the value of BC, rather than to determine that it's possible to find 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 cannot 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 the length of BC can be determined. Sufficient.
Re: geometry triangles [#permalink]
06 Nov 2009, 09:55
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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.
Re: geometry triangles [#permalink]
06 Nov 2009, 14:06
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?
Re: geometry triangles [#permalink]
06 Nov 2009, 19:26
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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. _________________
Re: geometry triangles [#permalink]
18 Nov 2010, 08:43
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
Re: geometry triangles [#permalink]
20 Nov 2010, 05:21
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 _________________
Re: If CD = 6, what is the length of BC? [#permalink]
21 Jun 2013, 10:23
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tejal777 wrote:
Attachment:
The attachment g2.JPG is no longer available
If CD = 6, what is the length of BC?
(1) \(BD=6\sqrt{3}\) (2) x = 60
One way to solve this would be to draw a perpendicular line from B to A to make a 90 deg triangle and then solve.
Statement 1: \(BD=6\sqrt{3}\) - Sufficient
For Triangle BAD we know <C = 30, <A = 90, so <B = 60. We now have a 30:60:90 triangle with sides in the ratio x:x\(\sqrt{3}\):2x. Knowing BD allows us to calculate the value of x = 3\(\sqrt{3}\). So AD = 9 and hence AC = 3 and BA = 3\(\sqrt{3}\).
Now for triangle BAC we know the 2 sides we can calculate BC the 3rd side. Sufficient.
Statement 2: x = 60 - Sufficient If x = 60 then <C = 120 and <B = 30. We now have an isosceles triangle with two equal sides making BC and CD equal (corresponding angles being equal) so BC = 6. Sufficient.
Attachments
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_________________
___________________________________________ Consider +1 Kudos if my post helped
Re: If CD = 6, what is the length of BC? [#permalink]
10 Dec 2013, 12:46
If CD = 6, what is the length of BC?
(1) BD=6\sqrt{3}
We know the length of BD and CD. We also know the angle that lies between them. BC is fixed. We don't have to find the solution, just confirm that there is only one possible solution. The only way that BC's angle or length could change is if line DC were extended to the left but as the diagram states, that isn't possible because we are given it's length. Sufficient.
(2) x = 60
If x = 60 then the interior angle c = 120. Seeing as angle d = 30, angle b also = 60. The shape and side lengths of this triangle are fixed into position and cannot be changed. Given side length CD = 6 then there is only one possible answer for BC. Sufficient.
If CD = 6, what is the length of BC? [#permalink]
03 Mar 2015, 07:17
(1) we know that CD = 6, and angle CDB = 30 we can draw a perpendicular line, and get 2 right triangles (30-60-90), since the new angle CFD must be 90 degrees, we can conclude that angle C must be 60 degrees. Thus, we have CD = 6, CF = 3, and FD = 3 sqrt 3. Hm, this is interesting, 3 sqrt 3 is half of BD. That means that CF is the median of BD. Knowing FD & BF, we can calculate for BC. but that is not needed. Statement 1 Sufficient
(2) x = 60, thus we can conclude that we have an isosceles triangle, and BC = CD.
Re: If CD = 6, what is the length of BC? [#permalink]
03 Mar 2015, 08:27
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Hi bunnel pls correct me if my approach is wrong. Here angle D is 30
In statement 1 BD 6root3
Draw a perpendicular from Cto BD at O then angle at O 90 angle D 30 angle C 60. If CD =6 then side opposite to D 30 degree is half then CO is 3 then OD is 3root3 then OB= 3root3. Trianlge OCD are congruent OCB then BC =CD. Value of BC=6
gmatclubot
Re: If CD = 6, what is the length of BC?
[#permalink]
03 Mar 2015, 08:27
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