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

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2 Answers
Oct 30, 2016

#0.5036* r^2#

Explanation:

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#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#

Oct 30, 2016

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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#