A chord of length $6$ units divides a circle into two distinct areas. If the circle has a radius of 6 units, what is the area of the larger region, in square units? Express your answer in simplest radical form in terms of $\pi$.
Nope, still not correct Ooops.... back to the drawing board......
Draw a line from the center to each of the endpoints of the chord
this forms a equilateral triangle the central angle is 60 degrees (pi/3 radians)
the area of this sector is 60/360 of the TOTAL area of the circle
pi r^2 x 60/360 = 6 pi
Now subtract the area of the equilateral triangle to get the area enclosed by the chord
area of equalateral triangle = r^2 (sqrt3) / 4 = 9 sqrt3
6pi - 9 sqrt3 = area of chord
Total area of cicle - area enclosed by chord = larger area
36 pi - ( 6pi - 9 sqrt3)
30 pi + 9 sqrt3 units2
(there has to be a better way! )
The smaller area is the difference between the area of a sector of 1/6 of the circle less the area of a equilateral triangle with a side of 6.....this is given by
[(1/6) pi * 6^2 - (1/2)6^2 * √3/2 ] (1)
So....the larger area =
Area of circle - (1)
So
pi*6^2 - [(1/6) pi * 6^2 - (1/2)6^2 * √3/2 ] =
(5/6) pi * 6^2 + 9√3 =
[30pi + 9√3] units^2
I felt like , if anyone knew one, YOUR answer would 'learn it to me' Haha....