Question 1006244
<pre>
{{{drawing(400,250,-10,22,-2,18,

locate(-.8,2.5,4),green(locate(3.5,3,8)),
locate(0,5.5,A),locate(-.2,0,B),locate(6.7,0,C),
line(-12,0,24,0),locate(6.7,9.2,O),
line(-6,0,-6,4), line(-6,4,0,4), line(0,0,0,4), line(0,0,sqrt(48),0),
green(line(sqrt(48),0,0,4),line(0,4,sqrt(48),8),line(sqrt(48),8,sqrt(48),0)),
circle(sqrt(48),8,8) )}}}

sin(&#8736;ACB) = 4/8 = 1/2

therefore &#8736;ACB = 30°

Therefore since &#8736;BCO = 90°, &#8736;ACO = 60°

Therefore &#916;OAC is an equilateral triangle because the two radii, 
OA and OC are equal, making it isosceles, and it has one 60° angle.  

So all the sides of &#916;OAC are equal, so OC=8.

So the radius of the wheel is 8, its diameter is 16 and its
area is

<font face="symbol">p</font>r² = <font face="symbol">p</font>(8)² = 64<font face="symbol">p</font>.

Edwin</pre>