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# Figure ABCD (below) is a square with sides of length x. Arcs AB, AD

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Figure ABCD (below) is a square with sides of length x. Arcs AB, AD  [#permalink]

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Updated on: 06 Feb 2019, 04:49
3
10
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Difficulty:

85% (hard)

Question Stats:

51% (02:32) correct 49% (02:47) wrong based on 88 sessions

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Figure ABCD (below) is a square with sides of length x. Arcs AB, AD, BC, and DC are all semicircles. What is the area of the red region, in terms of x?

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semicircle.JPG [ 28.39 KiB | Viewed 15502 times ]

Originally posted by PathFinder007 on 02 Aug 2014, 08:30.
Last edited by Bunuel on 06 Feb 2019, 04:49, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Figure ABCD (below) is a square with sides of length x. Arcs AB, AD  [#permalink]

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02 Aug 2014, 15:39
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PathFinder007 wrote:
Figure ABCD (below) is a square with sides of length x. Arcs AB, AD, BC, and DC are all semicircles. What is the area of the red region, in terms of x?

Not as hard as it seems. Consider the image below:

If we subtract the area of two blue semicircles, so the area of the whole circle, from the area of the square we get the combined area of two white regions, so the area of the four white regions we have in the original image, would be twice of that.

Now, since the length of the side of the square is x, then so is the diameter of the circle, which makes the radius equal to x/2 and the area of the circle (two blue semicircles) equal to $$\pi{r^2}=\frac{\pi{x^2}}{4}$$. The area of two white regions is therefore $$x^2 - \frac{\pi{x^2}}{4}$$ and the area of four white regions is $$2x^2 - \frac{\pi{x^2}}{2}$$.

The area of the red region equals to the area of the square minus the area of four white regions: $$x^2 - (2x^2 - \frac{\pi{x^2}}{2})= \frac{\pi{x^2}}{2} - x^2$$.

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Re: Figure ABCD (below) is a square with sides of length x. Arcs AB, AD  [#permalink]

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06 Feb 2019, 04:50
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Re: Figure ABCD (below) is a square with sides of length x. Arcs AB, AD   [#permalink] 06 Feb 2019, 04:50
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