Question

Given the radius of the circle inscribing the hexagon is `r` express the shaded area in terms of `r`?

Answer 1

`0.5036* r^2`

Explanation:

`OA=OC=OD=OB=OE=OF=r`
`angleOCD=360^@/6=60^@ => Delta OCD` is equilateral
Area `DeltaOCD=sqrt3/4*r^2`

The 3 green areas in sector `OCD` are the same.

One green area `A_G=pir^2*60/360-sqrt3/4*r^2`
`=> A_G=(2pi-3sqrt3)/12*r^2`

The black area `A_B` in sector `OCD = pir^2*60/360- 3A_G`
`=> A_B=(pir^2)/6-3((2pi-3sqrt3)/12)*r^2`
`=((2pi-6pi+9sqrt3)/12)*r^2`
`=((9sqrt3-4pi)/12)*r^2`

Now, let the shaded area in your diagram be `A_S`
`=> A_S=2*A_B`
`=> A_S=2*((9sqrt3-4pi)/12)*r^2`

`=> A_S=((9sqrt3-4pi)/6)*r^2=0.5036*r^2`

Answer 2

Given the radius of the circle inscribing the hexagon is r

The area of the circle `Delta_c=pir^2`

The area of the each of 6 equilateral triangles having length of each side r is `Delta_t=sqrt3/4r^2`

Area of each of the segment of circle marked X
`=Delta_x=1/6Delta_c-Delta_t`

From the figure it is obvious that the area of Yellow shaded region in the given figure is

`Delta_y=2Delta_t-4xxDelta_x`

`=>Delta_y=2Delta_t-4xx(1/6Delta_c-Delta_t)`

`=>Delta_y=6Delta_t-2/3Delta_c`

`=>Delta_y=6xxsqrt3/4r^2-2/3xxpir^2`

`=>Delta_y=(3sqrt3)/2r^2-2/3xxpir^2`

`=>Delta_y=((9sqrt3-4pi))/6r^2`