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In the figure shown, the circle has a diameter of 10, which

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In the figure shown, the circle has a diameter of 10, which [#permalink]

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New post Updated on: 12 Jun 2013, 02:32
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In the figure shown, the circle has a diameter of 10, which is also the side length of the implied square. If an outline is formed by half the circle, one quarter of the square, and a straight line connecting the perpendicular bisectors of the circle, what is the perimeter of this shape?

A. \(5+5(\pi + \sqrt{2})\)
B. \(10 + 10\pi +5 \sqrt{2}\)
C. \(10 + 5(\pi + \sqrt{2})\)
D. \(15 +\pi \sqrt{2}\)
E. \(10\pi\sqrt{3}\)

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Originally posted by fozzzy on 12 Jun 2013, 02:24.
Last edited by Bunuel on 12 Jun 2013, 02:32, edited 1 time in total.
Edited the stem and the answer choices.
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Re: In the figure shown, the circle has a diameter of 10, which [#permalink]

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New post 12 Jun 2013, 02:37
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In the figure shown, the circle has a diameter of 10, which is also the side length of the implied square. If an outline is formed by half the circle, one quarter of the square, and a straight line connecting the perpendicular bisectors of the circle, what is the perimeter of this shape?

A. \(5+5(\pi + \sqrt{2})\)
B. \(10 + 10\pi +5 \sqrt{2}\)
C. \(10 + 5(\pi + \sqrt{2})\)
D. \(15 +\pi \sqrt{2}\)
E. \(10\pi\sqrt{3}\)

Diameter = 10 --> radius =5.

The perimeter of half the circle = \(\frac{2\pi{r}}{2}=5\pi\);
The perimeter of half the square = \(r+r=10\);
The length of the straight line = \(\sqrt{5^2+5^2}=5\sqrt{2}\).

Total = \(5\pi+10+5\sqrt{2}\).

Answer: C.
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Re: In the figure shown, the circle has a diameter of 10, which [#permalink]

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New post 12 Jun 2013, 02:39
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Given \(radius = 5\), the circumference is \(2*5*\pi\), so half circumference is \(5\pi\).

Then we have two sides of a 5-side square, \(5+5=10\)

Then we have an hypostnus of a right triangle with two 5 sides, \(\sqrt{5^2+5^2}=2\sqrt{5}\)

Tot sum=\(5\pi+10+5\sqrt{2}\)
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Re: In the figure shown, the circle has a diameter of 10, which [#permalink]

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Re: In the figure shown, the circle has a diameter of 10, which   [#permalink] 17 Apr 2017, 17:58
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