SOLUTION: a circle is divided into two parts by a chord, 3cm away from the center, find the area of the smaller part if the circle has a total area of 201

OA = OB = r = 7.99876785

cos(∠AOC) =  =  = 

∠AOC = 67.97211683° 

∠AOB = ∠AOC + ∠BOC = 2(67.97211683°) = 135.9442337°


Area of Sector AXBO       ∠AOB      
--------------------  =  ------
Area of whole circle      360°

Area of Sector AXBO       135.9442337°      
--------------------  =  -------------
      201                     360°  

Area of Sector AXBO   =   

Area of Sector AXBO   =   

Area of Sector AXBO   =   75.90219713

Area of circle Segment AXBC = Area of Sector AXBO - Area of ΔABO

Now we must find the area of ΔABO

It is twice the area of ΔACO

We have its height OC = 3

We need its base AC

tan(∠AOC) = 

OC·tan(∠AOC) = AC

3·2.471623109 = AC

7.414869326 = AC

Area of ΔACO =  =  = 11.12230399

Area of ΔABO = 2·Area of ΔACO = 2(11.12230399) = 22.24460798 

Area of circle Segment AXBC = Area of Sector AXBO - Area of ΔABO

Area of circle Segment AXBC = 75.90219713 - 22.24460798 = 53.65758918

Answer: 53.65758918 round off as your teacher instructed. The numbers
        here are as far as calculator gives them.  

-----------------------------------------------------

There is a formula that gives the area of a sector of a circle 
directly from the central angle, which would have been easier, but
I think your teacher expected you to do it the above longer way.
But, anyway, that formula is: 

Area of sector =  = the central angle = ∠AOB = 135.9442337° 

Area of sector =  = 53.65758915

Notice that the very last digit differs from the very last digit when
calculated above.  But you can expect that because a tiny bit of error
is made whenever more calculations are made, and there were more 
calculations made in the above than when substituting directly into the
formula.  But they are the same when rounded off, even to 7 decimal places.

Edwin