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

Question

Untitled.png 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} Screen Shot 2013-09-29 at 5.38.24 PM.png 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.

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=21219 
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} .

Answer: C.

Hope it's clear.

Marcab · Answer

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

aakrity · Answer

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.

LakerFan24 · Answer

is there any way of solving this by plugging in #s (ex: for the radius?)

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

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

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