In the figure (attached), ABC is an equilateral triangle, : DS Archive
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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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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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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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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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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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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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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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08 Aug 2009, 20:27
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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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08 Aug 2009, 20:50
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..

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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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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08 Aug 2009, 23:04
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, 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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23 Aug 2015, 13:37
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Re: In the figure (attached), ABC is an equilateral triangle, [#permalink]

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14 Oct 2016, 00:41
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Re: In the figure (attached), ABC is an equilateral triangle,   [#permalink] 14 Oct 2016, 00:41
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# In the figure (attached), ABC is an equilateral triangle,

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