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# M22-25

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:17
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Difficulty:

35% (medium)

Question Stats:

59% (00:43) correct 41% (00:35) wrong based on 128 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:17
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 $$= \frac{\sqrt{2}x}{\frac{x}{2}} = 2\sqrt{2}$$.

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17 Mar 2017, 12:00
I didn't get it
Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 1196
Location: India
WE: Engineering (Other)

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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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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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18 Dec 2017, 10:15
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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26 Jan 2018, 08:40
+1 for option E
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Re: M22-25 &nbs [#permalink] 26 Jan 2018, 08:40
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# M22-25

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