Four circles & a square in a circle 
Radius of the big triangle is $2$. ABCD is a square. What is the difference between $T_{1}$ and $(M_{1}+M_{2})$.
I have solved it already though I don't know if my answer is right or wrong.  My solution is very long.  Is there any short/quick/easy method to solve this problem?  Please check it out.

Area of ALS=$\frac{\prod}{8}$
area of $X_{1}$ = (area of ALD-area of ALS)
          = $\frac{1}{2} - \frac{\prod}{8}$
Area of $M_{1}$ =(area of ALD-(area of $X_{1}$+area of $X_{2}$))
          = $\frac{1}{2} - \frac{4- \prod}{4}$
Area of $(M_{1}+M_{2})$ = $\frac{\prod - 2}{2}$
Area of small triangles=A+C+D+B
                       =$2\prod+4$
Area of $T_{1}$ = $\frac{(\prod *4)-(2* \prod+4)}{4}$
Difference in area between $T_{1}$ and $(M_{1}+M_{2})$ = $\frac{\prod - 2}{2}$-$\frac{(\prod *4)-(2* \prod+4)}{4}$ = $0$                                              
