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555-605 (Medium)|   Geometry|      
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Bunuel
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The figure above shows a circle inscribed in a square which is in turn 31 Oct 2018, 01:37
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77% (01:59) correct 23% (01:56) wrong based on 146 sessions

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The figure above shows a circle inscribed in a square which is in turn inscribed within a larger circle. What is the ratio of the area of the larger circle to that of the smaller circle?


A. √2

B. π/2

C. π^2/(4√2)

D. 2

E. π/√2


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D
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The figure above shows a circle inscribed in a square which is in turn 31 Oct 2018, 07:27
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Small circle radius = r,
Half of diagonal of square = sqrt(r^2 + r^2) = \(\sqrt{2}\)r = Radius of larger circle

Ratio = pi*\((sqrt(2)r)^2\)/pi*\(r^2\) = 2

Option D
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The figure above shows a circle inscribed in a square which is in turn 20 Nov 2018, 14:49
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nkin
Small circle radius = r,
Half of diagonal of square = sqrt(r^2 + r^2) = \(\sqrt{2}\)r = Radius of larger circle

Ratio = pi*\((sqrt(2)r)^2\)/pi*\(r^2\) = 2

Option D
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I'm a little lost. I thought the diagonal of a square was d=s√2 so wouldn't d=s√2/2 be the radius of the larger circle ? Where did you get sqrt(r^2 + r^2)
Please help me understand where I'm not seeing what's obvious
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The figure above shows a circle inscribed in a square which is in turn 14 Sep 2020, 07:07
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Bunuel

The figure above shows a circle inscribed in a square which is in turn inscribed within a larger circle. What is the ratio of the area of the larger circle to that of the smaller circle?


A. √2

B. π/2

C. π^2/(4√2)

D. 2

E. π/√2


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phd01.png
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Let the sides of the square be 2 , thus radius of the smaller circle is 1

So, Area of the smaller circle is \(π*1^2 = π\)

Diameter of the bigger circle is \(2\sqrt{2}\) , so radius of the bigger circle is \(\sqrt{2}\)

Hence, Area of the Larger circle is \(π*\sqrt{2}*\sqrt{2} = 2π\)

Quote:
What is the ratio of the area of the larger circle to that of the smaller circle?
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Thus, ratio of the area of the larger circle to that of the smaller circle is \(\frac{2π}{π} = 2\), Answer must be (D) 2
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The figure above shows a circle inscribed in a square which is in turn 14 Sep 2020, 09:36
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Let us assume that the small circle radius = r,
Diagonal of square = \(\sqrt{(r^2 + r^2)} = \sqrt{2}r\)

Half the diagonal is the radius of larger circle.

Ratio =\( \frac{pi*(\sqrt{2})^2*r^2}{pi*r^2} = 2\)

Option D
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The figure above shows a circle inscribed in a square which is in turn 14 Sep 2020, 16:14
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ratio = A_large / A_small = ?
A_large = pi*R^2
A_small = pi*r^2

pi cancels.
ratio = R^2 / r^2 = (R/r)^2

45-45-90 triangle with no values
R = rt(2)/2 -> allows square s=1 -> then r=1/2

R/r = [rt(2)/2] / [1/2] = rt(2)

(R/r)^2 = 2

Ans: D
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