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Re: DS: Triangle [#permalink]
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12 Jul 2012, 06:20
mymba99 wrote: GMAT TIGER wrote: seems you missed the posts above yours. Read them again as D is incorrect.. buffdaddy wrote: The answer is infact D, here is how
1) this is sufficient (as agreed by all)
2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13  x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient I disagree. The answer must be D. S2: Sqrt [S(Sa)(Sb)(Sc)] =30 Where S=(a+b+c)/3. @GMATTIGER, so what you are saying is that you can draw other triangle with two sides 5, and 13, with area = 30. And you are saying you can draw other triangle with the third side greater than 12 but can keep the area and other two sides constant.I think it is not possible. Can you draw a triangle with sides 5, 13, 1million, with area 30? AND at the same time can you draw other triangle with sides 5,13, 20 with area 30? Answer should be D.Please help me if I missed anything. A is Correct . About 2) the other side could be 12 or SQRT (244) these are the roots of this equation driven from Sqrt [S(Sa)(Sb)(Sc)] =30 that gives us : suppose that X= 3rd side then you have X^4388*x^2+35136=0 solved to X=12 or X= SQRT (244) could be 15.6204993518133
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Re: What is the length of segment BC? (1) Angle ABC is 90 [#permalink]
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12 Jul 2012, 06:58
+1 for A. The second statement gives two answers 12 and 15.62 as outlined above.



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Re: What is the length of segment BC? [#permalink]
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13 Jul 2012, 00:35
Guys I understand that solving this complex equation proves we can't get a get a single value...but what's wrong with the figure described by @buffdaddy...drawing a perpendicular and getting a x, then y and then third side...i actually followed the same logic... i found lot of people are telling its wrong but plz pin point what's wrong with that approach... Also i tried to calculate the area with 15.63 is around 27...somebody plz recalculate ... one more thing even if we r not using A, option B ans can not contradict with A (plz correct me if wrong)
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Re: DS: Triangle [#permalink]
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13 Jul 2012, 00:45
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buffdaddy wrote: The answer is infact D, here is how
2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13  x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient this gives us z=12 (X=1.92, Y= 11.07 , H=4.61) the problem is with the diagram , if you consider that h could be outside of the triangle then you may have this: H= 4.61 , X= 1.92 and Y= X+13 ( not 13X )! then Y= 14.92 which gives us Z= SQRT (244) or Z= 15.620 ...)
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Re: What is the length of segment BC? [#permalink]
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18 Jul 2012, 04:11
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Re: What is the length of segment BC? [#permalink]
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03 Sep 2012, 18:43
Hey Bunuel,
God I was torn on this question. I took 5 as the base and found the height to be 12. Wasn't sure if we could conclude since, two sides were 5,13 and a height of 12, that we can conclude 5,12,13. I guess the better question is under what condition can we conclude a height to be the side of 90* triangle?
Can , I say this with confidence?
1) If one of the heights of a triangle is equal to one of its sides, the triangle must be a 90* right triangle. 2) If a triangle is a right 90* triangle, one its heights is equal to one of the triangles sides.
3) If three sides of a triangle satisfy, \(a^2+b^2=c^2\) it must be a right 90* triangle. 4) If a triangle is a right 90* triangle, it must satisfy \(a^2+b^2=c^2\)



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Re: DS: Triangle [#permalink]
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12 Oct 2013, 20:08
hoping_for_stern wrote: Contradicting statement 2 : We can keep the fact in mind that triangles between two parallel lines have same area.(draw a line ll to AC / BC's value will vary and still satisfy the area being constant at 30 ,as base and height in this case are constant (AC) and distance between ll lines remain the same , which in this case would be the height of the triangle)
Heron's formula makes one jump to a quick conclusion about statement 2 being sufficient but an important point is overlooked by the fact using it to find the value of BC would give some polynomial ( x3 or x4 or whatever ....) and that will lead to multiple values of x (at least 2 in this case ). Hence insufficient.
Statement 1 is sufficient Hence A Answer Should Be "D" since area is equals= 0.5*A*B*SinQ; so from here we can get the included angle between the given two sides and then use Cosine law to get the third side easily..... +1 KUDDUS if I'm Right



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Re: DS: Triangle [#permalink]
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13 Oct 2013, 02:40
himang wrote: hoping_for_stern wrote: Contradicting statement 2 : We can keep the fact in mind that triangles between two parallel lines have same area.(draw a line ll to AC / BC's value will vary and still satisfy the area being constant at 30 ,as base and height in this case are constant (AC) and distance between ll lines remain the same , which in this case would be the height of the triangle)
Heron's formula makes one jump to a quick conclusion about statement 2 being sufficient but an important point is overlooked by the fact using it to find the value of BC would give some polynomial ( x3 or x4 or whatever ....) and that will lead to multiple values of x (at least 2 in this case ). Hence insufficient.
Statement 1 is sufficient Hence A Answer Should Be "D" since area is equals= 0.5*A*B*SinQ; so from here we can get the included angle between the given two sides and then use Cosine law to get the third side easily..... +1 KUDDUS if I'm Right That's not correct. The answer to this question is A, not D. The formula you posted is correct: \(Area=\frac{1}{2}BA*CA*sinA\) > \(sinA=\frac{12}{13}\), but from this we cannot calculate angle \(A\), as \(sin A=Sin (180A)\). Meaning that this won't give us ONLY one value for angle A, thus we won't have ONLY one value for CB > A can be acute angle, making B right angle AND making CB equal to 12 OR A can be obtuse angle, making B acute angle AND making CB greater then 12. Check here: whatisthelengthofsegmentbc6363920.html#p1105437Hope it helps.
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Re: What is the length of segment BC? [#permalink]
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19 Oct 2013, 04:12
Bunuel wrote: Responding to a pm. What is the length of segment BC? (1) Angle ABC is 90 degrees. \(BC=\sqrt{13^25^2}\). Sufficient. (2) The area of the triangle is 30. Consider the diagram below: Attachment: ABC.png Notice that segment \(AB_2\) is the mirror reflection of segment \(AB_1\) around the vertical line passing through point A. Now, if the height of triangles \(ACB_1\) and \(ACB_2\) is \(\frac{60}{13}\), then the area of both triangles is \(area=\frac{1}{2}*base*height=\frac{1}{2}*13*\frac{60}{13}=30\). So, as you can see we can have different lengths of segment CB (\(CB_1\) and \(CB_2\)). Not sufficient. Answer: A. Hope it's clear. Oh, now i know why it is not D. Thank you Math God



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Re: What is the length of segment BC? [#permalink]
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06 Dec 2013, 08:51
While I don't have a specific example, I can't help but think I have solved similar problems (referring specifically to in the information provided in (2) with the information. I see why the height could be different  we cannot say for certain what the height is because we are given no defined angle measurements, correct?



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Re: What is the length of segment BC? [#permalink]
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27 Feb 2014, 00:21
use hero's formula to get and equation of the degree of x^4 which has 4 roots. So ans must be A



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Re: What is the length of segment BC? [#permalink]
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18 May 2014, 09:04
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(1) Angle ABC is 90 degrees. Since ABC is right angled at B, we can easily find the length of BC using Pythagoras Theorem; SUFFICIENT.
(2) The area of the triangle is 30. Although Heron's formula is out of GMAT scope, but still let me clarify it here that why statement 2 is NOT sufficient.
Let us assume that BC = x Then s = (5 + 13 + x)/2 = (18 + x)/2 = 9 + (x/2) Now area of triangle, A = √[s(s  a)(s  b)(s  c)] or A² = [s(s  a)(s  b)(s  c)] (30)² = [9 + (x/2)] * [9 + (x/2)  5] * [9 + (x/2)  13] * [9 + (x/2)  x] 900 = [9 + (x/2)] * [4 + (x/2)] * [4 + (x/2)] * [9  (x/2)] 900 = [81  (x/2)²][(x/2)²  16] Let (x/2)² = y 900 = [81  y][y  16] 900 = 81y  y² + 16y  1296 y²  97y + 2196 = 0 y²  36y  61y + 2196 = 0 y(y  36)  61(y  36) = 0 (y  61)(y  36) = 0 y = 36, 61 (x/2)² = 36, (x/2)² = 61, which clearly implies that we are getting 2 values of x, which means 2 values of BC. So, statement 2 is NOT sufficient.
The correct answer is A.
I hope that helps.



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Re: What is the length of segment BC? [#permalink]
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16 Jun 2014, 10:11
GMAT TIGER wrote: Attachment: Triangle.jpg What is the length of segment BC? (1) Angle ABC is 90 degrees. (2) The area of the triangle is 30. Here 1st is self sufficient, using Pythagoras theorem. 2nd is also self sufficient. We can take base as 13 m and area is given as 30. that way we will get the height suppose x which is distance from B to AC. let AD be the altitude. Now in triangle ABD applying Pythagoras theorem, we will get measure of AD as we have already calculated the height. knowing AD we will get CD. Knowing CD, we will apply Pythagoras theorem in triangle BDC and will get the measure of side BC.



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Re: What is the length of segment BC? [#permalink]
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26 May 2015, 10:26
We dont know whether it is a right angle triangle Hence stmt 1 ab is 9 not suff as we dont know the length of the other side Stmt 2 also insuff similarly Combining we get The triangle can be 4 4 9 or 9 9 4 but 4 4 9 is not possible since the 3rd side of the triangle cannot be greater than the sum of the other two sides
Hence C



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Re: What is the length of segment BC? [#permalink]
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27 May 2015, 21:07
Hi kasturi72, In Fact 1, we're told that angle ABC is a RIGHT ANGLE. With the two sides that we're given in the prompt (5 and 13), we CAN determine the length of the missing side (it's 12), so THAT information is SUFFICIENT to answer the question. GMAT assassins aren't born, they're made, Rich
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Re: What is the length of segment BC? [#permalink]
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28 May 2015, 01:02
The first is suff as we can apply Pythagoras theorem The second is insuff as applying Heron's Formula makes a quadratic eqn and gives two values of x..
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Re: What is the length of segment BC? [#permalink]
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02 Dec 2015, 06:52
GMAT TIGER wrote: Attachment: Triangle.jpg What is the length of segment BC? (1) Angle ABC is 90 degrees. (2) The area of the triangle is 30. (1) straight sufficient, it gives us a right trianlge and therefir BC=12 (2) Theory: Height must be perpendicular to the base of a triangle > in a right trianlge that could be just to legs or an altitude to hypotenuse BUT here we are not told that it's a right triangle, so don't assume that BC is perpendicular to AB, bcs we are not given such information. So you can not find BC, as it's not a height of this triangle. Answer A
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Re: What is the length of segment BC? [#permalink]
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21 Mar 2016, 23:32
just one thing that confuses me : so for A) we are assuming ABC is right angled at B making AC hypo  and then we get the triplet  5,12,13 SO there is no way that ABC can be right angled at A making BC as hypo and then length of BC changes .
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Re: What is the length of segment BC? [#permalink]
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29 Aug 2016, 12:31
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GMAT TIGER wrote: Attachment: Triangle.jpg What is the length of segment BC? (1) Angle ABC is 90 degrees. (2) The area of the triangle is 30. This is such a shrewd trap type. I hope to God I don't fall for it in the exam. A alone is sufficient but when you come down to B you tend to use the information that you got from A. Also I was thinking how smart I am to already know the pythagorean triplet and Bam!
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Re: What is the length of segment BC? [#permalink]
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29 Oct 2017, 06:56
Bunuel wrote: Responding to a pm. What is the length of segment BC? (1) Angle ABC is 90 degrees. \(BC=\sqrt{13^25^2}\). Sufficient. (2) The area of the triangle is 30. Consider the diagram below: Attachment: ABC.png Notice that segment \(AB_2\) is the mirror reflection of segment \(AB_1\) around the vertical line passing through point A. Now, if the height of triangles \(ACB_1\) and \(ACB_2\) is \(\frac{60}{13}\), then the area of both triangles is \(area=\frac{1}{2}*base*height=\frac{1}{2}*13*\frac{60}{13}=30\). So, as you can see we can have different lengths of segment CB (\(CB_1\) and \(CB_2\)). Not sufficient. Answer: A. Hope it's clear. Hi Bunuel. I think second statement is also sufficient. Both triangles are based on same base. so area of both triangles should be same. There will be change in sin angle only. But sin(x) =sin(180x) Sent from my ONE A2003 using GMAT Club Forum mobile app




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