Question 1201843
<pre>
In the figure below, ADC is a chord of a circle centre O and passing
through the points A, B and C. BD is a perpendicular bisector 
of the chord AC.  AD=8 cm and BD=2 cm. Calculate the area of the 
minor segment ABCD.

I'll just do the first one for you.

Here are the steps to find the area of a segment of a circle.

1. Identify the radius of the circle and label it 'r'.
2. Identify the central angle AOC made by the arc of the segment and label it 
{{{theta}}}.
3. Find the area of triangle AOC using the formula {{{A=expr(1/2)base*height}}} or {{{A=expr(1/2)r^2*sin(theta)}}}.
4. Find the area of the sector OABC using the formula
{{{expr(theta^""/360^o)*pi*r^2}}}, if θ is in degrees (or)
{{{expr(1/2)r^2*theta}}}, if θ is in radians.
5. Subtract the area of the triangle OAC from the area of the sector OABC to
find the area of the segment ABCD.

So we need to find radius r and angle &theta;.

{{{drawing(400,420,-10,10,-4,17, locate(-8.6,.5,A),locate(8.2,.5,C),
locate(-.2,13.6,theta),
locate(-.2,15.8,O),locate(-.2,-2.1,B), locate(.15,.83,D), 

locate(-4,.83,matrix(1,2,8,cm)),

locate(-5.1,7.5,r),
arc(0,15,6,-6,242,298),
arc(0,15,34,-34,242,298), locate(.2,-.6,matrix(1,2,2,cm)), 

line(-8,0,8,0), line(-8,0,0,15), line(8,0,0,15), line(0,0,0,-2) )}}}

Draw in OD (in green).  Since OA, OB, OC are all radii, with length r,
and since BD=2 cm, OD = OB-BD = r-2 

{{{drawing(400,420,-10,10,-4,17, locate(-8.6,.5,A),locate(8.2,.5,C),
green(line(0,0,0,15),locate(.2,7.2,r-2),locate(-.8,13.3,theta/2)),
locate(-.2,15.8,O),locate(-.2,-2.1,B), locate(.15,.83,D), 

locate(-4,.83,matrix(1,2,8,cm)),

locate(-5.1,7.5,r),

arc(0,15,34,-34,242,298), locate(.2,-.6,matrix(1,2,2,cm)), 

line(-8,0,8,0), line(-8,0,0,15), line(8,0,0,15), line(0,0,0,-2) )}}}

To find radius r, we use the Pythagorean theorem on right triangle OAD:

{{{matrix(7,3,

r^2,""="",8^2+(r-2)^2,
r^2,""="",64+r^2-4r+4,
0, ""="",64-4r+4,
0, ""="",68-4r,
4r,""="",68,
r,""="",68/4,
r,""="",17)}}}

To find &theta;, we use:

{{{sin(theta/2)=opposite/hypotenuse=AD/OA=8/17}}}
{{{theta/2=28.07248694^o=matrix(1,2,0.4899573263,radians)}}}
{{{theta=56.14497388^o=matrix(1,2,0.9799146526,radians)}}}

Now we go back to the given figure:

{{{drawing(400,420,-10,10,-4,17, locate(-8.6,.5,A),locate(8.2,.5,C),
locate(-.2,13.6,theta),
locate(-.2,15.8,O),locate(-.2,-2.1,B), locate(.15,.83,D), 

locate(-4,.83,matrix(1,2,8,cm)),

locate(-5.1,7.5,r),
arc(0,15,6,-6,242,298),
arc(0,15,34,-34,242,298), locate(.2,-.6,matrix(1,2,2,cm)), 

line(-8,0,8,0), line(-8,0,0,15), line(8,0,0,15), line(0,0,0,-2) )}}}

For step 3, we find the area of triangle OAC either by

{{{A=expr(1/2)base*height}}}
{{{A=expr(1/2)AC*OD=expr(1/2)(16)(15)=matrix(1,2,120,cm^2)}}}

or {{{A=expr(1/2)r^2*sin(theta)=expr(1/2)(17^2)sin(56.14497388^o)=matrix(1,2,120,cm^2)}}}.

For step 4, we find the area of the entire sector ABCO either by 

{{{expr(theta^""/360^o)*pi*r^2}}}, if θ is in degrees

{{{expr(theta^""/360^o)*pi*r^2}}}^""/360^o)*pi*r^2}}}{{{""=""}}}{{{expr(56.14497388^o/360^o)*3.141582654*17^2}}}{{{""=""}}}{{{matrix(1,2,141.5972166,cm^2)}}}
(or)
{{{expr(1/2)r^2*theta}}}, if θ is in radians.

{{{expr(1/2)r^2*theta}}}{{{""=""}}}{{{expr(1/2)17^2*0.9799146526}}}{{{""=""}}}{{{matrix(1,2,141.5976673,cm^2)}}}

Notice there's a slight difference between those two values.

Finally we do step 6 and subtract the area of the triangle from the area of the
sector

{{{141.5972166-120=matrix(1,2,21.5972166,cm^2)}}}

(or)

{{{141.5976673-120=matrix(1,2,21.5976673,cm^2)}}}

Edwin</pre>