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In the figure (attached), ABC is an equilateral triangle, [#permalink]
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08 Aug 2009, 07:41
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In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle? (1) DA = 4 (2) Angle ABD = 30 degrees OA later after discussion == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 09:26
[quote="crejoc"]In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?
(1) DA = 4 (2) Angle ABD = 30 degrees
FROM STEM : DB = DIAMETER
from 1
da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff
from 2 we dont have any side length...insuff
A



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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 10:17
yezz wrote: crejoc wrote: In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?
(1) DA = 4 (2) Angle ABD = 30 degrees
FROM STEM : DB = DIAMETER
from 1
da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff
from 2 we dont have any side length...insuff
A I could not get where the stem says DB is the diameter?? For me it is C. We need to find the length of a side of equilateral triangle in order to find the radius and hence the area of the circle. Combining 1 and 2 we can get the length of AB.



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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 10:30
Economist wrote: yezz wrote: crejoc wrote: In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?
(1) DA = 4 (2) Angle ABD = 30 degrees
FROM STEM : DB = DIAMETER
from 1
da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff
from 2 we dont have any side length...insuff
A I could not get where the stem says DB is the diameter?? For me it is C. We need to find the length of a side of equilateral triangle in order to find the radius and hence the area of the circle. Combining 1 and 2 we can get the length of AB. try to draw any right angle triangle with 3 verticies on the circle and the right angle is one of these verticies without the hyp being the diameter!



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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 16:24
yezz wrote: db bisects cba,we can know angles of the right triangle
How can we Know the angles of the right triangle, can you explain that..



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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 20:07
yezz wrote: da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff
A The OA is A , you are right, but how we can know the angles of the right triangle by knowing that db bisects cba.?? explain please..



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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 21:27
The answer is A. Given: \(\triangle ABC\) is Equilateral \(\Rightarrow \angle ABC = \angle BCA = \angle CAB = 60^\circ\) Also given \(\triangle BAD\) is a right triangle \(\Rightarrow \angle BAD = 90^\circ\) Now if you notice Arc AB is intercepted by inscribed angles \(\angle BAC\)and \(\angle BDA\) \(\Rightarrow \angle BCA = \angle BDA = 60^\circ\) For \(\triangle BAD\) we know \(\angle BAD = 90^\circ\) \(\angle BDA = 60^\circ\) \(\Rightarrow \angle ABD = 30^\circ\) Asking \(\text{Area of circle} = \pi r^2\) \(\Rightarrow r=? \text{ or } 2r=d=BD=?\) Lets look at Statement 1. \(DA = 4\) \(\text{then } BD = 4 \times 2\) \(\text {because } 30^\circ60^\circ90^\circ \triangle =\rightarrow 1sqrt{3}2 \text { sides}\) Sufficient Statement 2 \(\angle ABD = 30^\circ\) This does not tell us anything about the length of any sides. Hence insufficient. The answer is A



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Re: equilateral triangle circumscribed circle [#permalink]
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Updated on: 09 Aug 2009, 01:40
nookway wrote: T
\(\Rightarrow \angle BAC = \angle BDA = 60^\circ\)
I think it is Rightarrow \angle B CA = \angle BDA = 60^\circ I too reasoned the problem this way, but i cant understand "yezz's" explanation of yezz wrote: db bisects cba,we can know angles of the right triangle.
I thought if there is some we can deduce the angles other than quoted by nookway, it would be also helpful..
Originally posted by crejoc on 08 Aug 2009, 21:50.
Last edited by crejoc on 09 Aug 2009, 01:40, edited 3 times in total.



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Re: equilateral triangle circumscribed circle [#permalink]
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08 Aug 2009, 23:23
crejoc wrote: I think it is Rightarrow \angle BCA = \angle BDA = 60^\circ
Thanks for pointing out the typo.



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Re: equilateral triangle circumscribed circle [#permalink]
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09 Aug 2009, 00:04
crejoc wrote: yezz wrote: db bisects cba,we can know angles of the right triangle
How can we Know the angles of the right triangle, can you explain that.. any equilateral triangle drawn inside a circle (60,60,60 angles) and if you draw the diameter it will bisect one side and one angle.ie: the triangle is exactly in the middle of the circle and the diameter cut it into halves ( similar triangles). if we draw the 3 perpendicular bisectors of the triangle's 3 angles they will intersect at the center of the circle.



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Re: equilateral triangle circumscribed circle [#permalink]
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09 Aug 2009, 01:39
yezz wrote: crejoc wrote: yezz wrote: db bisects cba,we can know angles of the right triangle
How can we Know the angles of the right triangle, can you explain that.. any equilateral triangle drawn inside a circle (60,60,60 angles) and if you draw the diameter it will bisect one side and one angle.ie: the triangle is exactly in the middle of the circle and the diameter cut it into halves ( similar triangles). if we draw the 3 perpendicular bisectors of the triangle's 3 angles they will intersect at the center of the circle. Got it, that was really nice .. thanks for letting know it..



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Re: In the figure (attached), ABC is an equilateral triangle, [#permalink]
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29 Dec 2017, 09:49
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Re: In the figure (attached), ABC is an equilateral triangle,
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