10.101. Find the area of a regular triangle (Fig. 10.97) inscribed in a square with side $a$, given that one of the vertices of the triangle coincides with a vertex of the square.
Solution.

$CD=a; BE=FD=x; AF=AE=a-x; CE=EF=CF=y. \mathrm{Pe}-$ solving the system $\left\{\begin{array}{l}y^{2}=a^{2}+x^{2}, \\ y^{2}=2(a-x)^{2} ;\end{array} a^{2}+x^{2}=2(a-x)^{2}\right.$. Solving this equation, we get $x=a(2-\sqrt{3})$. Let's find $y^{2}=a^{2}+a^{2}(2-\sqrt{3})^{2}=a^{2}(8-4 \sqrt{3})$. The area of the desired equilateral triangle is found using the formula $S=\frac{y^{2} \sqrt{3}}{4}=\frac{4(2-\sqrt{3}) \sqrt{3} a^{2}}{4}=a^{2}(2 \sqrt{3}-3)$.

Answer: $\quad a^{2}(2 \sqrt{3}-3)$