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What is the greatest possible area of a square that is

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rxs0005
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What is the greatest possible area of a square that is [#permalink] New post   Updated on: 15 Jul 2014, 14:27
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What is the greatest possible area of a square that is completely contained within a circle with radius 2, with one vertex at the center of the circle and one other vertex on the circle?

A. \(\frac{\sqrt{2}}{2}\)

B. \(\sqrt{2}\)

C. 2

D. \(2\sqrt{2}\)

E. 4

can some one draw this visually

thanks
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C

Originally posted by rxs0005 on 14 Feb 2012, 06:41.
Last edited by Bunuel on 15 Jul 2014, 14:27, edited 3 times in total.
Edited the question and added the answer choices.
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Re: What is the greatest possible area of a square that is [#permalink] New post   14 Feb 2012, 07:21
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rxs0005 wrote:
What is the greatest possible area of a square that is completely contained within a circle with radius 2, with one vertex at the center of the circle and one other vertex on the circle?

A. root(2) / 2
B. root(2)
C. 2
D. 2root(2)
E. 4

can some one draw this visually

thanks


Here you go:
Attachment:
Circle.PNG
Circle.PNG [ 15.64 KiB | Viewed 4996 times ]
As you can see the radius of the circle is the diagonal of the square. The area of a square: \(\frac{diagonal^2}{2}=\frac{2^2}{2}=2\).

Answer: C.

Notice that if another vertex on the circle is not the endpoint of the diagonal but is the endpoint of another side then the circle won't be completely contained within a circle.
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pbull78
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Re: What is the greatest possible area of a square that is [#permalink] New post   23 Feb 2012, 06:42
thanks bunuel for nice visual explanation,

we can solve it by writting
Diagonal of square = root2side
side = diagonal/ rt2
area = side sqr = square of 2/rt2
that equals to 2
hence answer is C
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What is the greatest possible area of a square that is completely cont [#permalink] New post   01 Mar 2015, 21:01
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?
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What is the greatest possible area of a square that is completely cont [#permalink] New post   01 Mar 2015, 23:02
Expert Reply
Pretz wrote:
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?


ans C...
the square has one vertex on center of circle and other vertex on circle .. the square with max possible area will be the one with two opposite vertices on center and on circle..
so the radius of circle will become the diagonal of square..
that is diagonal=2=side*\(\sqrt{2}\).. or side =\(\sqrt{2}\)
area =\(\sqrt{2}\)*\(\sqrt{2}\)=2
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Re: What is the greatest possible area of a square that is completely cont [#permalink] New post   02 Mar 2015, 01:47
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Pretz wrote:
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?

Attachment:
Ques3.jpg
Ques3.jpg [ 6.46 KiB | Viewed 4003 times ]


Draw the circle with center O and radius 2. Now, O is one vertex of the square. How can you draw the square such that it has one vertex on the circle? Can you draw a vertex adjacent to O on the circle? Note that you cannot. In that case, the vertex opposite O will lie outside the circle. So the vertex opposite O should be on the circle and the vertices adjacent to O will lie in the circle. In this case, the radius of the circle will be the diagonal do the square.

Diagonal of square = 2
Side of square \(= 2/\sqrt{2} = \sqrt{2}\)
Area of square \(= \sqrt{2}^2 = 2\)
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Re: What is the greatest possible area of a square that is [#permalink] New post   02 Mar 2015, 02:47
Expert Reply
Pretz wrote:
What is the greatest possible area of a square that is
completely contained within a circle with radius 2, with one
vertex at the center of the circle and one other vertex on the
circle?
(A)\sqrt{2}/2
(B) \sqrt{2}
(C) 2
(D) 2\sqrt{2}
(E) 4



Looks like the OA is incorrect. I am unable to justify the answer. Could someone please help?


Please search before posting and format properly (rules-for-posting-please-read-this-before-posting-133935.html#p1096628).
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Re: What is the greatest possible area of a square that is [#permalink] New post   05 Jun 2023, 03:50
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Re: What is the greatest possible area of a square that is [#permalink]
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