As shown in the figure ,the length of the chord AB is 5, which runs from #pi/4# to #pi/2#. This would lie in the 1st Quadrant. Now divide the circumference of the circle in eight equal parts , from 0 to#pi/4#, #pi/4# to #pi/2#, #pi/2# to #(3pi)/4#, #(3pi)/4# to #pi#, #pi# to #(5pi)/4#, #(5pi)/4# to #(3pi)/2#, #(3pi)/2# to #(7pi)/4# and #(7pi)/4# to #2pi#. The chord lengths joining these points on the circumference would all be equal to 5. If all these points are joined the resultant figure would be a regular octagon with side 5, as shown in the figure along side.
Now, let the radius of the circle be 'r'. In triangle OAB, angle O is #360/8# or #45^o#. The base angles A and B would thus be each #67 1/2 # degrees.
Draw perpendicular OD from O to side AB. This would bisect AB because OAB is an isosceles triangle. This means AD=BD= 2.5. Since ODB is a rt triangle #r cos67 1/2 = 2.5# . This would give #r= 2.5 sec 67 1/2#
= 6.53
Area of circle would be #pi r^2= 3.14 (6.53)^2#= 134.066
