MHB Find the diameter of one circle

[anemone]

Five identical semicircles are arranged as shown. Find the diameter of one circle.
[TIKZ]
\draw (0,0) -- (16.5, 0);
\begin{scope}
\clip (0,0) rectangle (4.5,4.5);
\draw (2.25,0) circle(2.25);
\draw (0,0) -- (4.5,0);
\end{scope}
\begin{scope}
\clip (6,0) rectangle (10.5,4.5);
\draw (8.25,0) circle(2.25);
\draw (6,0) -- (10.5,0);
\end{scope}
\begin{scope}
\clip (12,0) rectangle (16.5,4.5);
\draw (14.25,0) circle(2.25);
\draw (12,0) -- (16.5,0);
\end{scope}
\begin{scope}
\clip (2.75,0) rectangle (7.25,-4.5);
\draw (5,0) circle(2.25);
\draw (2.75,0) -- (7.25,0);
\end{scope}
\begin{scope}
\clip (9.25,0) rectangle (13.75,-4.5);
\draw (11.5,0) circle(2.25);
\draw (9.25,0) -- (13.75,0);
\end{scope}
\draw [<->] (4.5, 0.5) -- (6, 0.5);
\draw [<->] (10.5, 0.5) -- (12, 0.5);
\draw [<->] (0, -0.5) -- (2.75, -0.5);
\draw [<->] (7.25, -0.5) -- (9.25, -0.5);
\draw [<->] (13.75, -0.5) -- (16.5, -0.5);
\coordinate[label=left:12] (A) at (5.5,0.8);
\coordinate[label=left:12] (B) at (11.5,0.8);
\coordinate[label=left:22] (C) at (1.6,-0.8);
\coordinate[label=left:16] (D) at (8.6,-0.8);
\coordinate[label=left:22] (D) at (15.6,-0.8);
[/TIKZ]

As this is a Singapore primary math problem, it is understandable that one can solve it without the use of algebra method (form an equation and solve the equation is what I mean by algebra method). I enjoyed this problem quite a bit, therefore I wanted to post it here to let others to try to solve it without the use of algebra method...:)

 

[castor28]

I would say that the sums of the gaps are 24 (above) and 60 (below). As there is one more semicircle above, its diameter is equal to the difference 36.

 

[anemone]

Thanks castor28 for your reply!

But, I still count that as an algebra method, hehehe...I will let others have a chance to take a stab at it before I post the so called without-algebra solution. Please stay tuned! :)

 

[anemone]

Hi castor28!

I don't know where my head was when I made the previous reply (Tmi) , I'm so sorry!(Sadface) Your answer is spot on!

Here is a diagram to illustrate a slightly different approach than castors28's method:

[TIKZ]
\draw (0,0) -- (16.5, 0);
\begin{scope}
\clip (0,0) rectangle (4.5,4.5);
\draw (2.25,0) circle(2.25);
\draw (0,0) -- (4.5,0);
\end{scope}
\begin{scope}
\clip (6,0) rectangle (10.5,4.5);
\draw (8.25,0) circle(2.25);
\draw (6,0) -- (10.5,0);
\end{scope}
\begin{scope}
\clip (12,0) rectangle (16.5,4.5);
\draw (14.25,0) circle(2.25);
\draw (12,0) -- (16.5,0);
\end{scope}
\begin{scope}
\clip (0,0) rectangle (4.5,-4.5);
\draw (2.25,0) circle(2.25);
\draw (0,0) -- (4.5,0);
\end{scope}
\begin{scope}
\clip (12,0) rectangle (16.5,-4.5);
\draw (14.25,0) circle(2.25);
\draw (12,0) -- (16.5,0);
\end{scope}
\draw [<->] (4.5, 0.5) -- (6, 0.5);
\draw [<->] (10.5, 0.5) -- (12, 0.5);
\draw [<->] (4.5, -0.5) -- (6, -0.5);
\draw [<->] (7.25, -0.5) -- (9.25, -0.5);
\draw [<->] (10.5, -0.5) -- (12, -0.5);
\draw [<->] (6, -0.5) -- (7.25, -0.5);
\draw [<->] (9.25, -0.5) -- (10.5, -0.5);
\coordinate[label=left:12] (A) at (5.5,0.8);
\coordinate[label=left:12] (B) at (11.5,0.8);
\coordinate[label=left:12] (C) at (5.5,-0.8);
\coordinate[label=left:16] (D) at (8.6,-0.8);
\coordinate[label=left:12] (D) at (11.5,-0.8);
\coordinate[label=left:10] (E) at (6.9,-0.8);
\coordinate[label=left:10] (F) at (10.1,-0.8);
[/TIKZ]

$\therefore \text{diameter}=10+16+10=36$