An equilateral

An equilateral triangle is inscribed in a square of side 1 unit long so that it has one common vertex with the square. What is the area of the inscribed triangle?

Correct result:

S =  0.4641

Solution:

$$\begin{gathered} a^2=1^2+x^2 \\ {a}^2=(1-x{)}^2+(1-x{)}^2 \\ 1+x^2=(1-x{)}^2+(1-x{)}^2 \\ {1}^2+x^2=(1-x{)}^2+(1-x{)}^2 \\ {1}^2+0.267949192431^2=(1-0.267949192431{)}^2+(1-0.267949192431{)}^2 \\ 0<x<1 \\ x=x_2=0.2679\doteq 0.2679 \\ a=\sqrt{1+x^2}=\sqrt{1+0.2679^2}\doteq 1.0353 \\ S={\displaystyle \frac{\sqrt{3}}{4}}\cdot a^2={\displaystyle \frac{\sqrt{3}}{4}}\cdot 1.0353^2=0.4641 \end{gathered}$$

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