MHB How Can You Calculate the Area of Shaded Regions in Complex Geometric Figures?

[anemone]

Hello all!

It is so embarrassing to ask because I would think there is a trick to solve this problem without going through the trigonometric formulas like sine rule for example (because this is a primary math problem) but for some reason, I can't see through it...if you can solve it without using any of the trigonometric formulas, can you please enlighten me? Many thanks!

The figure is made up of a circle, identical semicircles and a square of side 12 cm. What is the area of the regions with all those number labels?
[TIKZ]
\draw (0,-3) rectangle (6,3);
\begin{scope}
\draw (3,0) circle(3);
\end{scope}
\begin{scope}
\draw (0,-3) arc (-90:90:3cm);
\end{scope}
\begin{scope}
\draw (6,-3) arc (-90:-270:3cm);
\end{scope}
\begin{scope}
\draw (0,3) arc (-180:0:3cm);
\end{scope}
\begin{scope}
\draw (6,-3) arc (0:180:3cm);
\end{scope}
\coordinate[label=above: 1] (1) at (0.15,-1.8);
\coordinate[label=above: 1] (1) at (1.5,2.6);
\coordinate[label=above: 1] (1) at (4.5,2.6);
\coordinate[label=above: 1] (1) at (5.8,1.3);
\coordinate[label=above: 1] (1) at (5.8,-1.7);
\coordinate[label=above: 2] (2) at (3,2);
\coordinate[label=above: 2] (2) at (5,-0.3);
\coordinate[label=above: 3] (3) at (4.2,0.8);
\coordinate[label=above: 4] (4) at (5.5,2.3);
[/TIKZ]

 

[Opalg]

Use the notation $w = $ area of region 1, $x = $ area of region 2, $y = $ area of region 3, $z =$ are of region 4.

We want four equations for these four unknowns, and it's easy to get three equations:

Area of the whole square gives $8w + 4x + 4y + 4z = 144$, so $2w+x+y+z = 36$;
Area of the whole circle gives $4x+4y = 36\pi$, so $x+y=9\pi$;
Area of one of the semicircles gives $2w+x+2y+2z = 18\pi$.

But we need a fourth equation, and that is less obvious. I think the best route is to calculate the area between the whole circle and one of the semicircles. The triangle whose vertices are the centre of the semicircle, the centre of the whole circle, and one of their points of intersection, is equilateral with side $6$ (the red triangle in the diagram below). So its angles are $60^\circ$. The sector of the semicircle between the blue and red radii is therefore one-third of a whole circle and so has area $12\pi$. The triangle with vertices at the centre of the semicircle and the two points of intersection with the whole circle (with two blue sides and one red side in the diagram) has area $9\sqrt3$. Subtracting that from the area of the sector, and multiplying by $2$, you see that the area between the semicircle and the whole circle is $24\pi - 18\sqrt3$. That gives you the fourth equation, $x+2y = 24\pi - 18\sqrt3$.

Now all you have to do is to solve the four equations. I got the answers as
$w = 36 - 9\sqrt3 - 6\pi$,
$x = 18\sqrt3 - 6\pi$,
$y = 15\pi - 19\sqrt3$,
$z = 3\pi - 18(2-\sqrt3)$.
But I haven't checked those calculations so I may well have made errors.

[TIKZ]\draw (0,-3) rectangle (6,3);
\begin{scope}
\draw (3,0) circle(3);
\end{scope}
\begin{scope}
\draw (0,-3) arc (-90:90:3cm);
\end{scope}
\begin{scope}
\draw (6,-3) arc (-90:-270:3cm);
\end{scope}
\begin{scope}
\draw (0,3) arc (-180:0:3cm);
\end{scope}
\begin{scope}
\draw (6,-3) arc (0:180:3cm);
\end{scope}
\draw[red] (0,0) -- (1.5,2.6) -- (3,0) -- cycle ;
\draw[blue] (1.5,2.6) -- (1.5,-2.6) -- (0,0) ;
\coordinate[label=above: 1] (1) at (0.15,-1.8);
\coordinate[label=above: 1] (1) at (1.5,2.6);
\coordinate[label=above: 1] (1) at (4.5,2.6);
\coordinate[label=above: 1] (1) at (5.8,1.3);
\coordinate[label=above: 1] (1) at (5.8,-1.7);
\coordinate[label=above: 2] (2) at (3,2);
\coordinate[label=above: 2] (2) at (5,-0.3);
\coordinate[label=above: 3] (3) at (4.2,0.8);
\coordinate[label=above: 4] (4) at (5.5,2.3);[/TIKZ]

 

[anemone]

Awesome, Opalg! (Cool) Thanks for the insightful solution, and your answers for $w,\,x$ and $z$ are correct, but I believe you made a typo in the value for $y$, the $-19\sqrt{3}$ should be $-18\sqrt{3}$.