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
George C.
Dec 16, 2016

#2sqrt(3)-2#

Explanation:

The area of a circle is given by the formula:

#"area" = pi r^2#

where #r# is the radius.

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 #2sqrt(2)#.

Bisecting the chord with a line joining the origin to the midpoint at #pi/6# we obtain two right angled triangles, each with hypotenuse of length #2sqrt(2)# and acute angle #pi/12#.

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 #2sqrt(3)-2#

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]}

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