A chord with a length of #5 # runs from #pi/8 # to #pi/3 # radians on a circle. What is the area of the circle?

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Answer 1 · 2018-01-07T02:23:23

Area of circle #color(blue)(A = 190.0334#

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Length of chord #AB = 5#

#AM = (AB)/2 = 5/2#

Angle subtended by the chord 5 at the center = #/_(AOM) = theta#

Given #theta = pi/3 - pi/8 = (5pi)/24#

#/_(AOM) = theta / 2 = (5pi)/48#

In right angle triangle AOM,

#OA = r = (AM) / sin ((theta)/2)#

#r = (5/2) / sin ((5pi) / 48) = 5 / (2 * sin ((5pi)/48)) = 7.7775#

Area of the circle #A_c = pi r^2 = pi * (7.7775)^2#

Area of circle #color(blue)(A = 190.0334#