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What is b in the circle shown above? [#permalink]
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Updated on: 27 Feb 2017, 22:50
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What is b in the circle shown above? (1) bc = 5a (2) The length of PQ is \(\sqrt{84}\) Attachment:
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Originally posted by godot53 on 27 Feb 2017, 22:12.
Last edited by Bunuel on 27 Feb 2017, 22:50, edited 1 time in total.
Renamed the topic and edited the question.



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Re: What is b in the circle shown above? [#permalink]
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27 Feb 2017, 22:50



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Re: What is b in the circle shown above? [#permalink]
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From the given data, 1) b is a common side to both the right angled triangles 2) Both the triangles have one common angle (90 degrees)
So, c/a = b/b => c=a
And by using Pythagoras theorem, 84= (b*b) + (5 * 5)
Hence Statement 2 alone is sufficient.
Kudos +1 guys!



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Re: What is b in the circle shown above? [#permalink]
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12 Mar 2017, 09:49
hey, I am confused as to c=a in this case. could you explain this?



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12 Mar 2017, 20:46
Confused as well. Detailed explanation will be helpful.



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12 Mar 2017, 21:36
vmelgargalan wrote: hey, I am confused as to c=a in this case. could you explain this? The STE 1 can be made from the figure using the theory of similar triangles. using Pythagorean theory we can make an equation with only one variable. B is sufficient
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What is b in the circle shown above? [#permalink]
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14 Mar 2017, 05:55
JanarthCj wrote: From the given data, 1) b is a common side to both the right angled triangles 2) Both the triangles have one common angle (90 degrees)
So, c/a = b/b => c=a
And by using Pythagoras theorem, 84= (b*b) + (5 * 5)
Hence Statement 2 alone is sufficient.
Kudos +1 guys! per your logic, each triangle inscribed in a circle is a isosceles right triangle. I think you are confusing it with another property or similar triangles i.e ratio of area = square of the ratio of sides



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Re: What is b in the circle shown above? [#permalink]
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15 Mar 2017, 05:15
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from the similar triangle we can say that b/5 = a/c or bc =5a and c^2 = b^2 + 5^2
St 1: bc=5a. Mentioned above. INSUFFICIENT St 2: a = √84 therefore bc = 5√84 and c^2 = b^2 +25. two variable, two equation. SUFFICIENT
Option B



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Re: What is b in the circle shown above? [#permalink]
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12 Jul 2017, 14:59
JanarthCj wrote: From the given data, 1) b is a common side to both the right angled triangles 2) Both the triangles have one common angle (90 degrees)
So, c/a = b/b => c=a
And by using Pythagoras theorem, 84= (b*b) + (5 * 5)
Hence Statement 2 alone is sufficient.
Kudos +1 guys! Could you please explain how do you know that the triangles are similar? Based on the given information, don't you need to confirm that an additional pair of sides is similar, or that an additional pair of angles is equal? Thank you!



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What is b in the circle shown above? [#permalink]
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21 Jul 2017, 12:13
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godot53 wrote: What is b in the circle shown above? (1) bc = 5a (2) The length of PQ is \(\sqrt{84}\) Attachment: The attachment Capture.PNG is no longer available From the information and attached figure we can derive: angle PQR = angle QOR = 90 degrees (1) angle PRQ = angle QRO = x degrees (2) From 1 and 2 and AAA similarity ===> angle QPR = angle QQR = y degrees Therefore the triangles PQR and QOR are similar PQ/QR = QO/OR ==> a/c = b/5 ==> bc =5a.........(3) Statement 1 ==> Same as can be derived from given information and insufficient since various combinations exist Statement 1 ==> From (3), a = √84 and b^2 + 5^2 = C^2 (since QOR is a right angle triangle) we can get the value of b
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Re: What is b in the circle shown above? [#permalink]
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21 Jul 2017, 20:29
Supermaverick wrote: godot53 wrote: What is b in the circle shown above? (1) bc = 5a (2) The length of PQ is \(\sqrt{84}\) From the information and attached figure we can derive: angle PQR = angle QOR = 90 degrees (1) angle PRQ = angle QRO = x degrees (2) From 1 and 2 and AAA similarity ===> angle QPR = angle QQR = y degrees Therefore the triangles PQR and QOR are similar PQ/QR = QO/OR ==> a/c = b/5 ==> bc =5a.........(3) Statement 1 ==> Same as can be derived from given information and insufficient since various combinations exist Statement 1 ==> From (3), a = √84 and b^2 + 5^2 = C^2 (since QOR is a right angle triangle) we can get the value of b Please explain how do we find the value of 'b'



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Re: What is b in the circle shown above? [#permalink]
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06 Sep 2017, 09:39
with a value as rt(84), we can find the value of diameter in terms of x+5 for triangle PQR and using the A value we can find length of x in terms of b and so on create a equation and by solving it we will get b x as 7 and b as rt(35).



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What is b in the circle shown above? [#permalink]
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10 Nov 2017, 09:19
I got it wrong but I'll provide my approach after thinking about it, to see if I can help.
Problem: DS to find b The information provided by the problem is: 1. Two triangles with a common side and adjacent angles of 90º angle (lets name it L). 2. Triangle QRL measures: 5bc, angles Q,R,90º. Looks like 306090 but there is no certainty, still pitagoras can be applied. 3. One big triangle PQR that is isoceles Right triangle (because internal angles theory, the diameter shows that the angle is 90º). And is inside a circle so pitagoras is possible. Mental note: Get a side then can apply "306090, x x√3, 2x" or Pitagoras.
Statements: (1) bc=5a Can't apply pitagoras or "30,60,90" Not Sufficient > BCE
2) PQ = √84 One side! So can apply pitagoras and "306090" Can find C! with C then can find b with pitagoras. Sufficient > B



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Re: What is b in the circle shown above? [#permalink]
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16 Nov 2017, 22:16
ByjusGMATapp wrote: from the similar triangle we can say that b/5 = a/c or bc =5a and c^2 = b^2 + 5^2
St 1: bc=5a. Mentioned above. INSUFFICIENT St 2: a = √84 therefore bc = 5√84 and c^2 = b^2 +25. two variable, two equation. SUFFICIENT
Option B Seems like you made your mind bc = 5√84 this is not right You are taking bc=5a from the 1st statement and assuming that statement 2 alone is enough. Can somebody please provide a clear answer, why B and not C?



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What is b in the circle shown above? [#permalink]
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16 Nov 2017, 22:19
PaterD wrote: I got it wrong but I'll provide my approach after thinking about it, to see if I can help.
Problem: DS to find b The information provided by the problem is: 1. Two triangles with a common side and adjacent angles of 90º angle (lets name it L). 2. Triangle QRL measures: 5bc, angles Q,R,90º. Looks like 306090 but there is no certainty, still pitagoras can be applied. 3. One big triangle PQR that is isoceles Right triangle (because internal angles theory, the diameter shows that the angle is 90º). And is inside a circle so pitagoras is possible. Mental note: Get a side then can apply "306090, x x√3, 2x" or Pitagoras.
Statements: (1) bc=5a Can't apply pitagoras or "30,60,90" Not Sufficient > BCE
2) PQ = √84 One side! So can apply pitagoras and "306090" Can find C! with C then can find b with pitagoras. Sufficient > B Your entire solution is based that it's 306090, triangle, which we don't know



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Re: What is b in the circle shown above? [#permalink]
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17 Nov 2017, 06:50
cbh wrote: PaterD wrote: I got it wrong but I'll provide my approach after thinking about it, to see if I can help.
Problem: DS to find b The information provided by the problem is: 1. Two triangles with a common side and adjacent angles of 90º angle (lets name it L). 2. Triangle QRL measures: 5bc, angles Q,R,90º. Looks like 306090 but there is no certainty, still pitagoras can be applied. 3. One big triangle PQR that is isoceles Right triangle (because internal angles theory, the diameter shows that the angle is 90º). And is inside a circle so pitagoras is possible. Mental note: Get a side then can apply "306090, x x√3, 2x" or Pitagoras.
Statements: (1) bc=5a Can't apply pitagoras or "30,60,90" Not Sufficient > BCE
2) PQ = √84 One side! So can apply pitagoras and "306090" Can find C! with C then can find b with pitagoras. Sufficient > B Your entire solution is based that it's 306090, triangle, which we don't know I thought that kind o trick was more known or used... Is a shortcut that can be used on right triangle that has the following angle measures has: Angle measures: 90 60  30 Side measures: 2x  x√3  x Is similar for equilateral: Angle measures: 90 45  45 Side measures: x√2  x  x Still the more possible approaches the better, this one is a hard one



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What is b in the circle shown above? [#permalink]
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17 Nov 2017, 07:08
PaterD wrote: I thought that kind o trick was more known or used... Is a shortcut that can be used on right triangle that has the following angle measures has: Angle measures: 90 60  30 Side measures: 2x  x√3  x
Is similar for equilateral: Angle measures: 90 45  45 Side measures: x√2  x  x
Still the more possible approaches the better, this one is a hard one I meant you can't know if it's 90 60 30 or 90 45 45 or any other angle combination https://en.wikipedia.org/wiki/Thales%27 ... heorem.gifYou just know that this is right triangle. That's it Actually there are 3 right triangles, which are similar to each other, so basically that kind of question could be solved through similarity but I don't see how that can be solved via option B only Bunuel wrote: can you please elaborate it? I'm confused a bit, much appreciated!



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Re: What is b in the circle shown above? [#permalink]
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17 Nov 2017, 09:23
cbh wrote: PaterD wrote: I thought that kind o trick was more known or used... Is a shortcut that can be used on right triangle that has the following angle measures has: Angle measures: 90 60  30 Side measures: 2x  x√3  x
Is similar for equilateral: Angle measures: 90 45  45 Side measures: x√2  x  x
Still the more possible approaches the better, this one is a hard one I meant you can't know if it's 90 60 30 or 90 45 45 or any other angle combination https://en.wikipedia.org/wiki/Thales%27 ... heorem.gifYou just know that this is right triangle. That's it Actually there are 3 right triangles, which are similar to each other, so basically that kind of question could be solved through similarity but I don't see how that can be solved via option B only Bunuel wrote: can you please elaborate it? I'm confused a bit, much appreciated! Dude you are right.... I assumed the angle measures I'll call some cavalry to solve this as I can't see a proper way to crack this nut VeritasPrepKarishma could you please help us with this?




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