A chord with a length of #5 # runs from #pi/12 # to #pi/8 # radians on a circle. What is the area of the circle?

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Answer 1 · 2018-05-18T15:18:11

#color(blue)(pi((5sin((23pi)/48))/(sin(pi/24)))^2#

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From diagram:

#theta=pi/8-(pi)/12=pi/24#

We need to find radius #bbr#:

We have an isosceles triangle with angle at the apex of #pi/24#

Two remaining angles are:

#(pi-(pi/24))/2=(23pi)/48#

Using sine rule:

#sin(pi/24)/5=sin((23pi)/48)/r#

#r=(5sin((23pi)/48))/(sin(pi/24))#

Area of circle is:

#pi((5sin((23pi)/48))/(sin(pi/24)))^2~~4590.212958#