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In the figure (attached), ABC is an equilateral triangle,

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In the figure (attached), ABC is an equilateral triangle,  [#permalink]

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New post 08 Aug 2009, 06:41
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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]

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New post 08 Aug 2009, 08:26
[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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New post 08 Aug 2009, 09:17
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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New post 08 Aug 2009, 09:30
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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New post 08 Aug 2009, 15:24
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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New post 08 Aug 2009, 19:07
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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New post 08 Aug 2009, 20:27
2
2
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.

\(DA = 4\)

\(\text{then } BD = 4 \times 2\)

\(\text {because } 30^\circ-60^\circ-90^\circ \triangle =\rightarrow 1-sqrt{3}-2 \text { sides}\)

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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New post Updated on: 09 Aug 2009, 00:40
nookway wrote:
T

\(\Rightarrow \angle BAC = \angle BDA = 60^\circ\)



I think it is
Rightarrow \angle BCA = \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, 20:50.
Last edited by crejoc on 09 Aug 2009, 00:40, edited 3 times in total.
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Re: equilateral triangle circumscribed circle  [#permalink]

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New post 08 Aug 2009, 22:23
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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New post 08 Aug 2009, 23:04
1
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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New post 09 Aug 2009, 00:39
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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Re: In the figure (attached), ABC is an equilateral triangle, &nbs [#permalink] 29 Dec 2017, 08:49
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