A chord with a length of #5 # runs from #pi/12 # to #pi/2 # radians on a circle. What is the area of the circle?

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Answer 1 · 2016-10-16T22:15:41

#A ~~ 52.98#

If one draws two radii from the center to each end of the chord, one has a triangle.

Compute the angle formed by the two radii:

#theta = pi/2 - pi/12 = (6pi)/12 - pi/12 = (5pi)/12#

We can use the Law of Cosines to find the length of #r^2#:

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

where #c = 5, a = b = r# and #theta = (5pi)/12#

#5^2 = r^2 + r^2 - 2(r)(r)cos((5pi)/12)#

#r^2 = 5^2/(2 - 2cos((5pi)/12)#

The area of the circle is the above multiplied by #pi#:

#A = (5^2pi)/(2 - 2cos((5pi)/12)#

#A ~~ 52.98#