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In the figure, O is the center of the circle of radius 3, an

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smartyman
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In the figure, O is the center of the circle of radius 3, an [#permalink] New post   Updated on: 24 Jun 2014, 07:58
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In the figure, O is the center of the circle of radius 3, and ABCD is a square. If angle AOP = x and the side BC of the square is tangent to the circle, then what is the area of the square ABCD ?

(1) PC = 3
(2) x = 60
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D

Originally posted by smartyman on 24 Jun 2014, 07:53.
Last edited by Bunuel on 24 Jun 2014, 07:58, edited 1 time in total.
Edited the question.
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Re: In the figure, O is the center of the circle of radius 3, an [#permalink] New post   24 Jun 2014, 08:05
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In the figure, O is the center of the circle of radius 3, and ABCD is a square. If angle AOP = x and the side BC of the square is tangent to the circle, then what is the area of the square ABCD ?

You can solve this question without actually computing anything there.

(1) PC = 3 --> the length of OC, which is the hypotenuse in triangle BCO, equals to OP + PC = radius + PC = 3 + 3 = 6. Thus we know the lengths of the two sides (OB and OC) in that triangles which enables us to find the third one, BC. Since BC is the side of the square then its area is BC^2. Sufficient.

(2) x = 60. Triangle BOC is 30-60-90 right triangle. We know one of the sides there (BO), hence we can find BC. Since BC is the side of the square then its area is BC^2. Sufficient.

Answer: D.
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Re: In the figure, O is the center of the circle of radius 3, an [#permalink] New post   24 Jun 2014, 08:09
smartyman wrote:
Attachment:
2014-06-24_1857.png
In the figure, O is the center of the circle of radius 3, and ABCD is a square. If angle AOP = x and the side BC of the square is tangent to the circle, then what is the area of the square ABCD ?

(1) PC = 3
(2) x = 60



pc=3 then by pythagores theorem you can calculate the one side of square and there by area.
x=60 by tan formula= tan(angle)= p/b you can calculate one side of square and there by area.
so both independently give you answer.
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Re: In the figure, O is the center of the circle of radius 3, an [#permalink] New post   10 Jul 2014, 04:39
Could you explain more about finding the length of BC using tan(60)?
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Re: In the figure, O is the center of the circle of radius 3, an [#permalink] New post   10 Jul 2014, 12:49
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evdo wrote:
Could you explain more about finding the length of BC using tan(60)?


You don't need trigonometry for the GMAT. Every GMAT geometry question can be solved without it. Check my post above.
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Re: In the figure, O is the center of the circle of radius 3, an [#permalink] New post   20 Oct 2016, 22:53
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evdo wrote:
Could you explain more about finding the length of BC using tan(60)?


When one angle of a right triangle is 60 degrees, we know that the third angle will be 30 degrees.
In a 30-60-90 triangle, the ratio of the sides is always \(1:\sqrt{3}:2\) (the side opposite 30, opposite 60 and opposite 90 respectively).
So if you know any one side, you know the other two too.

In triangle BOC, BO = 3 (side opposite 30 degree angle).
So BC (side opposite 60 degree angle) is \(3*\sqrt{3}\).

Area of square = BC^2 = 27

Tan 60 does the same thing. Tan 60 is \(\sqrt{3}\) so it gives that BC = \(3*\sqrt{3}\)
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Re: In the figure, O is the center of the circle of radius 3, an [#permalink]
20 Oct 2016, 22:53
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