A chord with a length of #4 # runs from #pi/8 # to #pi/6 # radians on a circle. What is the area of the circle?

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Answer 1 · 2016-12-21T18:13:26

#Area = 935pi#

The chord and two radii, each drawn from the center to its respective end of the chord form an isosceles triangle.

The angle, #theta# between the two radii is:

#theta = pi/6 - pi/8#

#theta = pi/24#

let side #c = 4#
Let sides #a = b = r#

Use the Law of Cosines

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

#4^2 = r^2 + r^2 - 2(r)(r)cos(theta)#

#4^2 = r^2(1 + 1 - 2cos(theta))#

#r^2 = 4^2/(2 - 2cos(pi/24)#

#r^2 ~~ 935 #

The area of a circle is:

#Area = pir^2#

#Area = 935pi#