Carcass wrote:
![Image](https://gmatclub.com/forum/download/file.php?id=43105)
Four semicircular arcs of length 2π are joined to make the figure above. What is the area of the enclosed region?
A. \(8\pi\)
B. \(8 + 8\pi\)
C. \(16 + 8\pi\)
D. \(16 + 16\pi\)
E. \(24\pi\)
If one SEMI-circle has a perimeter of 2π, then an ENTIRE circle must have a circumference of
4π ![Image](https://i.imgur.com/zop0ACg.png)
Circumference = (diameter)(π)
So, if the circumference of the circle is
4π, then
the diameter of each circle is 4 ![Image](https://i.imgur.com/Xoo88b1.png)
This means the square in the middle of the figure must have area 16.
![Image](https://i.imgur.com/4HQmEFa.png)
Now we need only find the area of 4 SEMIcircles.
If we combine 2 semicircles we get an entire circle.
So, if we combine 4 semicircles we get 2 entire circles.
Each circle has diameter 4, which means each circle has
RADIUS 2Area of a circle = π(radius²)
So, area of one circle = π(
2²) = 4π
So, the area of TWO circles = 8π
So, the TOTAL area = 16 + 8π
Answer: C
Cheers,
Brent