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In the figure above, square ABCD has an area of 25.

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by Gmat_mission » Mon Jun 10, 2019 5:16 am

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In the figure above, square ABCD has an area of 25. What is the area of the circle with center O?

\(A. \dfrac{5\sqrt{2}\pi}{2}\)
\(B. \dfrac{25\pi}{4}\)
\(C. \dfrac{25\pi}{2}\)
\(D. 5\sqrt{2}\pi\)
\(E. 25\pi\)

[spoiler]OA=C[/spoiler]

Source: Princeton Review
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by Ian Stewart » Mon Jun 10, 2019 6:21 am
The diameter of the circle is the diagonal of the square, and the diagonal of the square is the hypotenuse of a 45-45-90 triangle with legs of length 5. Since sides in a 45-45-90 triangle are in a 1 to 1 to √2 ratio, the diameter of the circle is 5√2, so the radius is (5√2)/2, and the area is π r^2 = π (5√2/2)^2 = π (25)(2)/4 = 25π/2.
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by Scott@TargetTestPrep » Tue Jun 11, 2019 7:15 pm
Gmat_mission wrote:Image
In the figure above, square ABCD has an area of 25. What is the area of the circle with center O?

\(A. \dfrac{5\sqrt{2}\pi}{2}\)
\(B. \dfrac{25\pi}{4}\)
\(C. \dfrac{25\pi}{2}\)
\(D. 5\sqrt{2}\pi\)
\(E. 25\pi\)

[spoiler]OA=C[/spoiler]

Source: Princeton Review
Since the area of the square is 25, the side of the square is 5, and thus the diagonal of the square = diameter of the circle = 5√2.

Thus, the radius is (5√2)/2, and the area of the circle is [(5√2)/2]^2 x π = 50π/4 = 25π/2.

Answer: C

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