$A^2+B^2 +AB=36. B^2+C^2+BC=49. C^2+A^2+AC=64.$ Find $(A+B+C)^2$ 
$$A^2+B^2 +AB=36.\\ B^2+C^2+BC=49.\\ C^2+A^2+AC=64.$$ Find $(A+B+C)^2$.

I have tried it by using geometry I.e constructing a triangle and marking a point inside it which is making 120 ° and then using cosine rule 
But have difficulty in solving further
Please use geometry 

Here is a purely algebraic way. Subtract the equations in pairs to get$$(C-B)(A+B+C)=28,$$$$(C-A)(A+B+C)=13,$$$$(A-B)(A+B+C)=15.$$Let us write $x=(A+B+C)^2$ and $y=BC+CA+AB$. Then squaring and adding the above three equations yields$$x(x-3y)=589.$$Summing the original equations gives$$2x-3y=149.$$Eliminating $3y$ between these equations results in the quadratic equation$$x^2-149x+589=0.$$Solving this, we get$$x=\tfrac12(149\pm63\surd5),$$where the larger root corresponds to the case when $A$, $B$, and $C$ are all positive.
