A chord with a length of #15 # runs from #pi/3 # to #pi/2 # radians on a circle. What is the area of the circle?

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Answer 1 · 2017-07-07T14:44:07

The area of the circle is #=2638u^2#

The angle subtended at the center of the circle is

#theta=pi/2-pi/3=1/6pi#

The length of the chord is

#l=2*r*sin(theta/2)#

So,

#r=l/(2sin(theta/2))#

#=15/(2sin(1/12pi))#

#=29#

The area of the circle is

#A=pi*r^2=pi*29^2=2638u^2#