A circle has a chord that goes from #( 5 pi)/6 # to #(5 pi) / 4 # radians on the circle. If the area of the circle is #39 pi #, what is the length of the chord?

1 Answer
sankarankalyanam
Oct 19, 2017

Chord length #= **39/2 or 19.5**#

Explanation:

Angle sub tended at the center #theta# by the chord is
# (5pi)/4 - (5pi) /6 = (15pi - 10pi) / 6 = (10pi)/6 = (5pi)/3#
#theta / 2 = (5pi)/6#

Circumference #= 2pir = 39pi#
#r = (39pi) / (2pi) = 39/2#

If chord length is c,
#sin (theta/2) = (c/2)/r #

#c = 2r sin (theta/2) = 2* (39/2) * sin ((5pi)/6)#
#color(red)(sin ((5pi)/6) = sin (pi - ((5pi)/6) = sin (pi/6)))#

#c = 39 *sin(pi/6) = 39 * (1/2) = 39/2 = 19.5#

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