A chord with a length of #18 # runs from #pi/4 # to #pi/2 # radians on a circle. What is the area of the circle?

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Answer 1 · 2017-05-07T03:27:00

#"Area" = 162pi(2+sqrt2)#

The chord and radii drawn to each end of the chord form an isosceles triangle. The angle, #theta#, between the two radii is:

#theta = pi/2-pi/4#

#theta = pi/4#

We can use the Law of Cosines:

#c^2 = a^2 + b^2 - 2(a)(b)cos(theta)#

to find the value of #r^2#

Let #c = 18#, #a = r#, and #b = r#

#18^2 = r^2 + r^2 - 2(r)(r)cos(pi/4)#

#18^2 = 2r^2 - 2r^2cos(pi/4)#

#r^2 = 18^2/(2-2cos(pi/4)#

#r^2 = 18^2/(2-sqrt2)#

#r^2 = 18^2(2+sqrt2)/2#

#r^2 = 162(2+sqrt2)#

To find the area of a circle, multiply by #pi#:

#"Area" = 162pi(2+sqrt2)#