Question 1043961
this is indeed a classic challenging problem the decimals make it lenghy and hairy ,cant be solved with out drawing,i dont know how to draw three tangent circles but i uploaded an image for you to refer to.


please refer to picture called "circles"

     *[illustration circles].




first lets find semi-perimeter S of the triangle
semi-perimeter S of the triangle ={{{(27.6+24.2+20.4)/(2)}}}
              ={{{36.1}}}



area of the triangle={{{sqrt(s(s-a)(s-b)(s-c))}}}
area of the triangle={{{sqrt(36.1(36.1-27.6)(36.1-24.2)(36.1-20.4))}}}
area of the triangle={{{239.4343}}} 

find angle A ,B and C

NOTE it is not right triangle,use cosine law

Cos A={{{(b^2+c^2-a^2)/(2bc)}}}                     
Cos A={{{((27.6)^2+(20.4)^2-(24.2)^2)/(2*27.6*20.4)}}} 
Cos A={{{(0.520069)}}} 
A={{{58.66}}}


Cos B={{{(a^2+c^2-b^2)/(2ac)}}}                     
Cos B={{{((27.6^2)+(24.2^2)-(20.4^2))/(2*27.6*24.2)}}}                     
Cos B={{{0.697}}}

B={{{45.81}}}

cos C={{{(180)-(58.66+45.81)}}}
cos C={{{75.53}}}


or use the {{{formula}}}




Cos C={{{(a^2+b^2-c^2)/(2ab)}}}
Cos C={{{((24.2^2)+(20.4^2)-(27.6^2))/(2*24.2*20.4)}}}                     
Cos C={{{0.2529}}}

C={{{75.34}}} very very minor difference


area sector a={{{(pi*r^2*angle A)/(360)}}}                     
area sector a={{{(pi*(11.9)^2*(58.66))/(360)}}}                     
area sector a={{{(72.45)}}}                     

area sector b={{{(pi*(15.7)^2*(45.8))/(360)}}}                     
area sector b={{{(98.467)}}}                     

area sector c={{{(pi*(8.5)^2*(75.34))/(360)}}}
area sector c={{{(47.477)}}}                     
                     

the area bounded by the three circles={{{((Area of the triangle) -(sum of   sector A B and C)))}}}

the area bounded by the three circles={{{(239.43) -(72.45+98.467+47.477)}}}
the area bounded by the three circles=~{{{(21cm^2)}}} is the answer

your feedback is welcome