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Bunuel niks18

Quote:

If \(x\) is the side of the square, then the diagonal of the square is \(\sqrt{2}x\)
This is simple application of Pythagoras theorem


Quote:
and the radius of the circle is \(\frac{x}{2}\). The required ratio \(= \frac{\sqrt{2}x}{\frac{x}{2}} = 2\sqrt{2}\).
Is this from the property that side of square = diameter for circle inscribed in triangle?
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Bunuel niks18

Quote:

If \(x\) is the side of the square, then the diagonal of the square is \(\sqrt{2}x\)
This is simple application of Pythagoras theorem


Quote:
and the radius of the circle is \(\frac{x}{2}\). The required ratio \(= \frac{\sqrt{2}x}{\frac{x}{2}} = 2\sqrt{2}\).
Is this from the property that side of square = diameter for circle inscribed in triangle?

The diagonal of the square is: \(diagonal^2=side^2+side^2=x^2 + x^2\) --> \(diagonal=\sqrt{2}x\)

When a circle is inscribed in a square, the side of a square = the diameter of the circle, so the radius of the circle is \(\frac{side}{2}\).
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+1 for option E
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