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In the circle above, chord AC = 26. What is the area of the circle? (

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In the circle above, chord AC = 26. What is the area of the circle? ( [#permalink]

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In the circle above, chord AC = 26. What is the area of the circle?

(1) Angle ABC = 120°
(2) AB = BC = 15

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Re: In the circle above, chord AC = 26. What is the area of the circle? ( [#permalink]

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New post 13 May 2018, 08:44
Bunuel

Please help in solving this question.

gmatbusters wrote:
In the circle above, chord AC = 26. What is the area of the circle?

(1) Angle ABC = 120°
(2) AB = BC = 15

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Re: In the circle above, chord AC = 26. What is the area of the circle? ( [#permalink]

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Solution:
St1) As angle ABC = 120, it is clear AC is not the diameter.
Moreover the triangle/Circle cannot be drawn uniquely by the given information.
NOT SUFFICIENT.

St2) Using statement 2 and the question stem, we get all the sides of the triangle ABC. So the triangle can be drawn uniquely.
Now since the three vertices of the triangle are now fixed, only a unique circle can be drawn.
So the area can be determined.
hence SUFFICIENT.

Answer B.
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Re: In the circle above, chord AC = 26. What is the area of the circle? ( [#permalink]

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New post 13 May 2018, 09:38
Statement 1 alone is sufficient as the central reflex angle would be 240 and the central angle will be 120. Drop a perpendicular from centre of the circle to the chord and then you have a 30-60-90 triangle where you can find all the sides and thus the radius and hence the area.

Statement 2 is sufficient as the triangle will be an isosceles triangle and for given dimensions, there will be a unique circumcircle that can be drawn.

Each alone is Sufficient to answer. Hence C


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Re: In the circle above, chord AC = 26. What is the area of the circle? (   [#permalink] 13 May 2018, 09:38
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