Question 1131926
<pre>{{{drawing(500,500,-1.2,1.2,-1.2,1.2,

circle(0,0,1),
line(-cos(40*pi/180),sin(40*pi/180),cos(40*pi/180),sin(40*pi/180)),
line(0,0,0,sin(40*pi/180)), locate(0,0,O), locate(-.87,.7,A),
locate(.04,.4,c), locate(-.36,.4,r), locate(.35,.4,r),
line(-cos(40*pi/180),sin(40*pi/180),0,0),
line(0,0,cos(40*pi/180),sin(40*pi/180)),
locate(.03,.2,theta),locate(-.1,.2,theta),
locate(.2,.8,sqrt(r^2-c^2)),locate(-.5,.8,sqrt(r^2-c^2)),
locate(.8,.7,B),locate(0,.73,E), 
circle(cos(70*pi/180),sin(70*pi/180),.01),
circle(cos(70*pi/180),sin(70*pi/180),.02),
locate(.37,1,D)   )}}}

{{{matrix(1,7,

(matrix(4,1,area,of,segment,AEBDA)),
"",
""="",
"",
(matrix(4,1,area,of,sector,AOBDA)),
""-"",
(matrix(4,1,area,of,triangle,AOB)) )}}}

            {{{matrix(1,5,

""="",
"",
expr(1/2)r^2(2theta),
""-"",
expr(1/2)*(2*sqrt(r^2-c^2)*c)) )}}}

That's equivalent to what you have above since

{{{cos(theta)=c/r}}} and the inverse cosine is multiplied by {{{pi/"180°"}}} 
to change degrees to radians, for apparently degrees are assumed 
by the author of this problem.  If radians were assumed for the 
inverse cosine, we would not use the factor {{{pi/"180°"}}}.

Edwin</pre>