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Re: The figure above shows a circle inscribed in a square which [#permalink]
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The figure above shows a circle inscribed in a square which [#permalink]
Let the diameter of the larger circle be \(x\).

Its radius will be \(\frac{x}{2}\)

Area will be \( \pi (\frac{x}{2})^2 = \frac{\pi}{4} \times x^2\)

Now the diameter of the larger circle is the diagonal of the inscribed square.

Diagonal of the inscribed square = \(x\)

The side of the inscribed square = \(\frac{x}{\sqrt{2}}\)

Now the side of the inscribed square is the diameter of the smaller circle

Diameter of the smaller circle = \(\frac{x}{\sqrt{2}}\)

Radius of the smaller circle = \(\frac{x}{2\sqrt{2}}\)

Area of the smaller circle = \(\pi \times (\frac{x}{2\sqrt{2}})^2 = \frac{\pi}{8} \times x^2\)

The ratio of the Area of the Larger Circle to the Area of the Smaller Circle = \(\frac{\frac{\pi}{4} \times x^2}{\frac{\pi}{8} \times x^2} = 2\)

Answer is Choice D
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The figure above shows a circle inscribed in a square which [#permalink]
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