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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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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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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