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 $\pi$.

Guest Nov 7, 2019

#1**0 **

Nope, still not correct Ooops.... back to the drawing board......

ElectricPavlov Nov 7, 2019

edited by
ElectricPavlov
Nov 7, 2019

edited by ElectricPavlov Nov 7, 2019

edited by ElectricPavlov Nov 7, 2019

edited by ElectricPavlov Nov 7, 2019

edited by ElectricPavlov Nov 7, 2019

#2**0 **

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 units^{2}

^{(there has to be a better way! )}

ElectricPavlov Nov 7, 2019

#3**+1 **

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

CPhill Nov 7, 2019

#6**0 **

I felt like , if anyone knew one, YOUR answer would 'learn it to me' Haha....

ElectricPavlov
Nov 7, 2019