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In the figure (attached), ABC is an equilateral triangle, [#permalink]
 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
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Re: equilateral triangle circumscribed circle [#permalink]
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]
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]
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]
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]
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]
08 Aug 2009, 21:27
The answer is A. Given: \triangle ABC is Equilateral \Rightarrow \angle ABC = \angle BCA = \angle CAB = 60^\circAlso given \triangle BAD is a right triangle \Rightarrow \angle BAD = 90^\circNow if you notice Arc AB is intercepted by inscribed angles \angle BACand \angle BDA\Rightarrow \angle BCA = \angle BDA = 60^\circFor \triangle BAD we know \angle BAD = 90^\circ\angle BDA = 60^\circ\Rightarrow \angle ABD = 30^\circAsking \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^\circ-60^\circ-90^\circ \triangle =\rightarrow 1-sqrt{3}-2 \text { sides}Sufficient Statement 2 \angle ABD = 30^\circThis 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]
08 Aug 2009, 21:50
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..
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]
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]
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]
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: equilateral triangle circumscribed circle
[#permalink]
09 Aug 2009, 01:39
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