Question 763263

you need to use the area of "Sector of the circle" formula
 
{{{A=(n/360)r^2*pi}}} where n is the number of degrees in the central angle of the sector 

your shaded area is equal to area of the triangle minus area of three sectors of the circle

area of the triangle:

sides length: {{{a=6}}},{{{b=6}}},{{{c=6}}}

find height {{{h}}} which divides triangle into two right angle triangles whose sides are: hypotenuse {{{6}}}, one leg is {{{a/2=3}}} and other leg is  {{{h}}}

so,  {{{h^2=6^2-3^2}}}

{{{h^2=36-9}}}

{{{h^2=27}}}

{{{h=sqrt(27)}}}

{{{h=5.196}}}

now we can find the area of triangle {{{A[t]}}}:

{{{A[t]=(1/2)a*h}}}

{{{A[t]=(1/2)6*5.196}}}

{{{A[t]=3*5.196}}}

{{{A[t]=15.588}}}


now find area of circle sector:

{{{A[cs]=(n/360)r^2*pi}}}........{{{n=60}}}, {{{r=3}}}

{{{A[cs]=(60/360)3^2*3.14}}}

{{{A[cs]=(1/6)9*3.14}}}

{{{A[cs]=28.26/6}}}

{{{A[cs]=4.71}}}

you have three of these sectors, so total area is {{{3A[cs]=3*4.71}}}

{{{3A[cs]=14.13}}}

now we can find the area of the shaded portion: it is equal to difference between the area of the triangle and three sectors


{{{A[shaded_portion]=A[t]-3A[cs]=15.588-14.13}}}

{{{A[shaded_portion]=A[t]-3A[cs]=1.458}}}