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In the figure shown, a square is inscribed in a circle. Within the squ

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Bunuel
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In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   20 Nov 2020, 01:15
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In the figure shown, a square is inscribed in a circle. Within the square are four circles, each with two points of tangency on the square and with two points of tangency with two other circles. The diameter of each smaller circle is 2x. What is the circumference of the outer circle in terms of x?


A. \(\pi x \sqrt{2}\)

B. \(2\pi x \sqrt{2}\)

C. \(3\pi x \sqrt{2}\)

D. \(4\pi x \sqrt{2}\)

E. \(5\pi x \sqrt{2}\)


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Re: In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   20 Nov 2020, 01:37
Explanation:

Side of Square = 2x + 2x = 4x
Diagonal of square = Side*√2 = 4x*√2

The radius of the Larger circle = Side of Square/2 = 2x*√2

Circumference of Larger circle = 2πr = 2π*2x*√2 = 4πx√2

IMO-D
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Re: In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   20 Nov 2020, 02:58
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IMO: D
Summation of small dia of two circles = Side of Square = 2x + 2x = 4x
Dia of Big circle (D) = diagonal of square = Square root ((4x)^2 + (4x)^2) = 4X√2
Therefore, Circumference of big circle = Pi D = Pi (4X√2)
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Re: In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   20 Nov 2020, 05:01
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Diameter of the small circle is \(2x\), therefore the side of the square is equal to the addition of the diameters of two circle which is 2x+2x=4x.

Diagonal of the square is equal to the diameter of the outer circle.
Therefore, the diameter of the square is equal to \(side\)*\(\sqrt{2}\) = \(4x\)*\(\sqrt{2}\).

Therefore, the circumference of the outer circle =\( 2*Pi*r= 2*Pi*\)\(\frac{4x*\sqrt{2}}{2}\).

Therefore, the OA is D.
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Re: In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   21 Nov 2020, 00:23
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ANSWER IMO D
Side of Square = 2x + 2x = 4x
Diagonal of square = Side*√2 = 4x*√2 = diameter of a larger circle

Circumference of Larger circle = 2π*2x*√2 = 4πx√2



Bunuel wrote:

In the figure shown, a square is inscribed in a circle. Within the square are four circles, each with two points of tangency on the square and with two points of tangency with two other circles. The diameter of each smaller circle is 2x. What is the circumference of the outer circle in terms of x?


A. \(\pi x \sqrt{2}\)

B. \(2\pi x \sqrt{2}\)

C. \(3\pi x \sqrt{2}\)

D. \(4\pi x \sqrt{2}\)

E. \(5\pi x \sqrt{2}\)


Project PS Butler


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In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   21 Nov 2020, 06:05
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Hi,

OA is D.

Since the diameter of 1 small circle is 2x, therefore the side of the square is equal to 2x+2x=4x.

Diagonal of the square is equal to the diameter of the outer circle.

Diameter of the circle is equal to Side*√2 = 4x*√2.

And finally to find the Circumference we just need to use the Diameter of the outer circle

Circumference = 2πr
In this case= 2*π*2x*√2
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Re: In the figure shown, a square is inscribed in a circle. Within the squ [#permalink] New post   28 Nov 2020, 10:28
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Bunuel wrote:

In the figure shown, a square is inscribed in a circle. Within the square are four circles, each with two points of tangency on the square and with two points of tangency with two other circles. The diameter of each smaller circle is 2x. What is the circumference of the outer circle in terms of x?


A. \(\pi x \sqrt{2}\)

B. \(2\pi x \sqrt{2}\)

C. \(3\pi x \sqrt{2}\)

D. \(4\pi x \sqrt{2}\)

E. \(5\pi x \sqrt{2}\)



Solution:

Since the diameter of each smaller circle is 2x and the side length of the square is twice that, a side of the square is 4x. Therefore, the diagonal of the square is 4x√2. Since the diagonal of the square is also the diameter of the outer circle, the circumference of the outer circle is 4x√2 * π = 4πx√2.

Answer: D
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Re: In the figure shown, a square is inscribed in a circle. Within the squ [#permalink]
28 Nov 2020, 10:28
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