Here you go: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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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

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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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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