Question 1189420
<pre>

{{{drawing(400,400,-22,22,-22,22,
locate(-20,.7,A),locate(19.2,.7,B),locate(-3.9,20.4,D),
locate(-10.9,0,O),locate(-16,-11,C),circle(0,0,.4),locate(0,-.5,P),
locate(-7.2,9.5,20),locate(-12.9,-4.2,12),locate(-15,1.7,8),
green(line(0,0,-169/11,-11.17849174)), locate(-5.75,1.7,11),locate(-8.5,-5.5,19),
circle(0,0,19), line(-19,0,19,0), line(-41/11,18.63081957,-169/11,-11.17849174)
)}}}

We will calculate the area of sector APC, then calculate the area of ΔCOP 
and subtract it from the sector and that will leave the area of OCA.  
Go here:

http://www.algebra.com/tutors/students/your-answer.mpl?question=1189421

where I have already calculated ∠APC = 0.6290064738 radians.  The formula 
for a sector is {{{A}}}{{{""=""}}}{{{expr(1/2)*r^2*theta}}} where &theta; is in radians.  

So the area of the sector APC is

{{{A}}}{{{""=""}}}{{{expr(1/2)*19^2*0.6290064738}}}{{{""=""}}}{{{113.5356685}}}

Now we'll find the area of ΔCOP by Heron's formula:

{{{Area}}}{{{""=""}}}{{{sqrt(s(s-a)(s-b)(s-c))}}}}

where s represents the semi-perimeter.

The perimeter = CO+OP+CP = 12+11+19 = 42

The semi-perimeter is half of that or s = 21.

{{{Area}}}{{{""=""}}}{{{sqrt(21(21-12)(21-11)(21-19))}}}} 

{{{Area}}}{{{""=""}}}{{{sqrt(21(9)(10)(2))}}}}

{{{Area}}}{{{""=""}}}{{{sqrt(3780)}}}} 

{{{Area}}}{{{""=""}}}{{{61.4817046}}}}

Subtracting the area of ΔCOP from the area of the
sector APC, we get

{{{113.5356685}}}{{{""-""}}}{{{61.4817046}}}}{{{""=""}}}{{{52.0539639}}} cm<sup>2</sup>.

 Edwin</pre>