A circle has a chord that goes from #pi/3 # to #pi/8 # radians on the circle. If the area of the circle is #25 pi #, what is the length of the chord?

Question

1297 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-22T18:51:15

#5 sqrt 2 sqrt {1 - cos frac{5 pi}{24}}#

Put #x# and #y# axes adequately so that

#x^2 + y^2 = R^2#

Area = #pi R^2 = 25 pi Rightarrow R = 5#

The chord is #AB#, such that

#A = 5 (cos frac{pi}{8}, sin frac{pi}{8})#

#B = 5 (cos frac{pi}{3}, sin frac{pi}{3})#

#|AB|^2 = (x_A - x_B)^2 + (y_A - y_B)^2#

# = 25 (cos a - cos b)^2 + 25 (sin a - sin b)^2#

# = 25 (cos^2 a + cos^2 b - 2 cos a cos b + sin^2 a + sin^2 b - 2 sin a sin b)#

# = 25 [1 + 1 - 2 cos(a - b) ]#

# = 50 [1 - cos(frac{pi}{3} - frac{pi}{8}) ]#

#|AB| = 5 sqrt 2 sqrt {1 - cos (pi (8 - 3)/24)}#