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A square with edge length 2 cm has semicircles drawn on each side.
Find the total area of the shaded region.

Here is an image of the diagram shown :

enter image description here

Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)

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5 Answers 5

The square has area $4$.

The top and bottom white region together have area $4$ minus $2$ times the area of the semi circle, that is, $4-2\times \pi r^2/2=4-\pi$ (since $r=1$).

enter image description here

So, dividing by $2$, the top blue region has area $2-\pi/2$.

If you remove this area from the area of the top semi circle you obtain the area of the two top red parts.

enter image description here

Double it to obtain the red area.

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One sees four semi-circles. Each delineates a semi-disk. Summing the areas of these four semi-disks, one counts twice each purple region and once each white region.

Hence the purple area plus the area of the square with side $2$ is twice the area of a disk with diameter $2$, that is, $\color{purple}{\mathbf{purple\ area}}+2^2=2\cdot\frac14\pi\cdot2^2$.

Finally, $\color{purple}{\mathbf{purple\ area}}=2\pi-4$.

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You lost me at "Hence the purple area...". Could you explain further please? I'm confused. –  mowwwalker Jan 29 '13 at 3:35
    
(2 x Purple) + White = Purple + Full square. –  Did Jan 29 '13 at 7:51
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There are $8$ circular segments, each of area

$$\frac{\pi}{2} r^2 (\theta - \sin{\theta}) $$

where $r$ is the radius, and $\theta$ is the angle subtended at the center by the segment. Here, from the geometry, $r$ is clearly $1 \, \mathrm{cm}$ and $\theta = \frac{\pi}{2}$, each segment spanning half a semicircle.

Therefore the shaded area is

$$8 \frac{\pi}{2} (1 \, \mathrm{cm}^2) \left ( \frac{\pi}{2} - 1 \right ) = (2 \pi - 4) \, \mathrm{cm}^2$$

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A layman's (mine) approach to solving it:

There are 4 half-circles contained in the square. The 4 half-circles have the same area as two whole circles with the same radius. The radius (half the length of the diameter, which is the length of a side of the cube) is 1 and the area of a circle is $\pi r^2$, so the area of the four half-circles is $2 ( \pi * (1)^2)$, which is $2\pi$. The square can only hold an area of $4$ and all its space is used, so the excess area of the circles' area must be contained in overlap. The the area not able to be contained be the square without overlap is given by $2\pi - 4$.

This, of course, only works out nicely because ALL the area inside the square is being used. Were part of the square unused, you would have to add the area of the unused portion into the calculation of the overlap.

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I did not know I was a layman. This is good news. Or not. Oh, whatever. –  Did Jan 29 '13 at 8:59
    
@Did, Haha, sorry if I offended you. I know that some of the mathematicians on here are really experienced, so I thought the explanations might go over the OP's head as they often do mine. –  mowwwalker Jan 29 '13 at 19:38
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Move one half of a leaf to the circle:

$\hspace{3cm}$enter image description here

The area of the quarter circle is $\frac\pi4$; the area of the triangle is $\frac12$. Thus, the area of the half-leaf (in red) is $\frac\pi4-\frac12$. There are two per leaf, therefore the area of four leaves is $$ 8\left(\frac\pi4-\frac12\right)=2\pi-4 $$

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