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A circle is inscribed in a square, then a square is inscribed in this

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A circle is inscribed in a square, then a square is inscribed in this  [#permalink]

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New post 24 Mar 2019, 22:48
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A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?


(A) \(\frac{\pi}{16}\)

(B) \(\frac{\pi}{8}\)

(C) \(\frac{\pi}{316}\)

(D) \(\frac{\pi}{4}\)

(E) \(\frac{\pi}{2}\)

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Re: A circle is inscribed in a square, then a square is inscribed in this  [#permalink]

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New post 25 Mar 2019, 00:46
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Bunuel wrote:
A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?


(A) \(\frac{\pi}{16}\)

(B) \(\frac{\pi}{8}\)

(C) \(\frac{\pi}{316}\)

(D) \(\frac{\pi}{4}\)

(E) \(\frac{\pi}{2}\)


draw figure a per given description
let side of large square = 8
8 will be equal to diameter of large circle = radius = 4
and diameter of circle = diagonal of small square = 8 = s√2 ; s= 4√2
4√2 = diameter of small circle so ; radis = 2√2
area of small circle (2√2)^2 *pi = 8pi
and large square area = 8^2 = 64 so ratio
8pi/64 = pi/8 IMO B :cool:
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A circle is inscribed in a square, then a square is inscribed in this  [#permalink]

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New post 25 Mar 2019, 23:23
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let x be the side of the big square S1 => diagonal of S1 = diameter of inscribed circle C1 = \(\frac{x}{(\sqrt{2})}\)
side of inscribed square in CircleC1,S2 => diameter of Circle C1 =\(\frac{x}{(\sqrt{2})}\)
radius of CircleC2 inscribed in S2 ,C2 => Side of S2 =\(\frac{x}{2(\sqrt{2})}\)

Area of Small circle C2 = \(\pi(\frac{x}{2(\sqrt{2})})^2\)
Area of Square S1 = \(x^2\)

Ratio = Area of C2/ Area of S1
= \(\frac{\frac{\Pi x^2}{8}}{ x^2}\)
= \(\frac{\Pi}{8}\) =option B
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A circle is inscribed in a square, then a square is inscribed in this   [#permalink] 25 Mar 2019, 23:23
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