A circle has a chord that goes from #pi/12 # to #pi/4 # radians on the circle. If the area of the circle is #8 pi #, what is the length of the chord?
1 Answer
Explanation:
The area of a circle is given by the formula:
#"area" = pi r^2#
where
So in our example:
#8pi = pi r^2#
Hence:
#r = sqrt(8) = 2sqrt(2)#
So the chord is the base of an isoceles triangle with the other two sides of length
Bisecting the chord with a line joining the origin to the midpoint at
Hence the length of half of the chord is the length of the smaller leg of the right angled triangle:
#2sqrt(2)*sin(pi/12) = 2sqrt(2)*1/4(sqrt(6)-sqrt(2)) = sqrt(3)-1#
So the length of the chord is
graph{(x^2+y^2-8)(y-x)(y - (1/4(sqrt(6)-sqrt(2)))x)(y-1/(sqrt(3))x)((y-2)+tan(pi/3)(x-2)) = 0 [-1.055, 3.945, -0.26, 2.24]}