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Rectangle ABCD has a perimeter of 26. The half circle with diameter AD

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Rectangle ABCD has a perimeter of 26. The half circle with diameter AD [#permalink]

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New post 12 May 2016, 03:42
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Rectangle ABCD has a perimeter of 26. The half circle with diameter AD has an area of 8π. What is the perimeter of the part of the figure that is not shaded?

A. 26 + 4π
B. 18 + 8π
C. 18 + 4π
D. 14 + 4π
E. 14 + 2π

[Reveal] Spoiler:
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2016-05-12_1536.png
2016-05-12_1536.png [ 1.99 KiB | Viewed 2396 times ]
[Reveal] Spoiler: OA

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Re: Rectangle ABCD has a perimeter of 26. The half circle with diameter AD [#permalink]

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New post 12 May 2016, 05:59
(pi * r^2)/2 = 8pi
r^2 = 16
r = 4 --> Diameter = Length of the rectangle = 8

16 + 2(Width) = 26
Width = 5

Perimeter of the part of the figure thats not shaded = 8 + 10 + (2 * pi * 4)/2 = 18 + 4pi

Answer: C
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Re: Rectangle ABCD has a perimeter of 26. The half circle with diameter AD [#permalink]

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New post 25 May 2017, 18:43
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Re: Rectangle ABCD has a perimeter of 26. The half circle with diameter AD [#permalink]

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New post 25 May 2017, 20:56
Area of half circle = πr^2/2, where r is the radius.
equating, 8π = πr^2/2, we get r^2 = 16 or r = 4.

Thus the side of the rectangle (having half circle) is 4*2 = 8. Let the other side be x.
Then perimeter = 2(8+x) = 26, this gives x=5

Perimeter of part that is not shaded = AB + BC + CD + semi circular arc

Now, AB = CD = 5, BC = 8.
Semi circular arc = 2πr/2 = πr = π*4 = 4π

Hence required perimeter = 5+5+8+4π = 18+4π

Thus answer is C
Re: Rectangle ABCD has a perimeter of 26. The half circle with diameter AD   [#permalink] 25 May 2017, 20:56
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Rectangle ABCD has a perimeter of 26. The half circle with diameter AD

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