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Arc ABC is a semicircle. The perimeter of region ABC is x.

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Joined: 08 Oct 2011
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Arc ABC is a semicircle. The perimeter of region ABC is x.  [#permalink]

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Updated on: 29 Sep 2013, 11:28
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Arc ABC is a semicircle. The perimeter of region ABC is x. What is the area of region ABC in terms of x?

A. $$\frac{\pi{x^2}}{8(2+\pi)^2}$$

B. $$\frac{x^2}{64}$$

C. $$\frac{\pi{x^2}}{2(2+\pi)^2}$$

D. $$\frac{\pi{x^2}}{(2+\pi)^2}$$

E. $$\frac{\pi{x^2}}{(4+\pi)^2}$$

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Screen Shot 2013-09-29 at 5.38.24 PM.png [ 23.76 KiB | Viewed 6198 times ]

My approach to the question:

If ABC is a semicircle with circumference x, then (2πr)/2=x
πr=x, r=x/π

Area of the circle = 1/2 π r^2 = 1/2 π (x/π) ^ 2 = x^2/2π

Unfortunately, I did not see any such answer option hence randomly picked C. Where did I go wrong in the calculation? Expert opinion appreciated.

Originally posted by aakrity on 29 Sep 2013, 05:22.
Last edited by Bunuel on 29 Sep 2013, 11:28, edited 1 time in total.
Edited the question.
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Re: Arc ABC is a semicircle  [#permalink]

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29 Sep 2013, 06:04
aakrity wrote:
Arc ABC is a semicircle. The perimeter of region ABC is x. What is the area of region ABC in terms of x?

U didn't take the diameter of the semicircle.
On taking the diameter of the semi-circle, the radius in terms of x would come out as r=x/(2+pi).
Rest is simple.
Area=pi*r^2/2.
Hence +1C
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Re: Arc ABC is a semicircle  [#permalink]

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29 Sep 2013, 06:21
Marcab wrote:
aakrity wrote:
Arc ABC is a semicircle. The perimeter of region ABC is x. What is the area of region ABC in terms of x?

U didn't take the diameter of the semicircle.
On taking the diameter of the semi-circle, the radius in terms of x would come out as r=x/(2+pi).
Rest is simple.
Area=pi*r^2/2.
Hence +1C

Thanks for your response but this is where the Kaplan explanation got me. Isn't the circumference / perimeter of the circle 2πr only? Hence, πr for a semi circle.
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Re: Arc ABC is a semicircle. The perimeter of region ABC is x.  [#permalink]

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29 Sep 2013, 11:27
2

Arc ABC is a semicircle. The perimeter of region ABC is x. What is the area of region ABC in terms of x?

A. $$\frac{\pi{x^2}}{8(2+\pi)^2}$$

B. $$\frac{x^2}{64}$$

C. $$\frac{\pi{x^2}}{2(2+\pi)^2}$$

D. $$\frac{\pi{x^2}}{(2+\pi)^2}$$

E. $$\frac{\pi{x^2}}{(4+\pi)^2}$$

The perimeter of region ABC is half of the circumference (arc ABC) + the diameter of the circle (AC) = $$\frac{2\pi{r}}{2}+2r=x$$ --> $$r=\frac{x}{\pi+2}$$.

The area of the semicircle = $$\frac{\pi{r^2}}{2}=\frac{\pi{x^2}}{2(\pi+2)^2}$$.

Hope it's clear.
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Re: Arc ABC is a semicircle. The perimeter of region ABC is x.  [#permalink]

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27 Feb 2017, 18:06
is there any way of solving this by plugging in #s (ex: for the radius?)
Re: Arc ABC is a semicircle. The perimeter of region ABC is x. &nbs [#permalink] 27 Feb 2017, 18:06
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