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M22-25

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Bunuel
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M22-25 [#permalink] New post   16 Sep 2014, 01:17
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If a circle is inscribed in a square, what is the ratio of the diagonal of the square to the radius of the circle?

A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{2}}\)
C. \(\sqrt{2}\)
D. 2
E. \(2\sqrt{2}\)
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E
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Bunuel
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Re M22-25 [#permalink] New post   16 Sep 2014, 01:17
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Official Solution:

If a circle is inscribed in a square, what is the ratio of the diagonal of the square to the radius of the circle?

A. \(\frac{1}{2}\)
B. \(\frac{1}{\sqrt{2}}\)
C. \(\sqrt{2}\)
D. 2
E. \(2\sqrt{2}\)

If \(x\) is the side of the square, then the diagonal of the square is \(\sqrt{2}x\) and the radius of the circle is \(\frac{x}{2}\). The required ratio \(= \frac{\sqrt{2}x}{\frac{x}{2}} = 2\sqrt{2}\).

Answer: E
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Re: M22-25 [#permalink] New post   17 Mar 2017, 12:00
I didn't get it :(
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adkikani
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Re: M22-25 [#permalink] New post   18 Dec 2017, 10:08
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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Re: M22-25 [#permalink] New post   18 Dec 2017, 10:15
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adkikani wrote:
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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Re: M22-25 [#permalink] New post   26 Jan 2018, 08:40
+1 for option E
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Re: M22-25 [#permalink]
26 Jan 2018, 08:40
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