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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.

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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
Area of $T_{1}$ = $\frac{(\prod *4)-(2* \prod+4)}{4}$
Difference in area between $T_{1}$ and $(M_{1}+M_{2})$ = $\frac{\prod - 2}{2}$-$\frac{(\prod *4)-(2* \prod+4)}{4}$ = $0$

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It's about Venn diagrams.

Denote by $T$ the area of a $T_i\,$, by $S$ the area of a scraping tool, by $L=M_1+M_2$ the area of a lens, and by $K$ the area of the surrounding circle. Then $$4T+4S+4L=K$$ and $$S+2L={K\over4}\ .$$ From these two equations one at once gets $T=L$.

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