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:

$$\begin{gathered} a^2=1^2+x^2 \\ {a}^2=(1-x{)}^2+(1-x{)}^2 \\ 1+x^2=(1-x{)}^2+(1-x{)}^2 \\ {1}^2+x^2=(1-x{)}^2+(1-x{)}^2 \\ -x^2+4x-1=0 \\ {x}^2-4x+1=0 \\ a=1;b=-4;c=1 \\ D=b^2-4ac=4^2-4\cdot 1\cdot 1=12 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{4\pm \sqrt{12}}{2}}={\displaystyle \frac{4\pm 2\sqrt{3}}{2}} \\ {x}_{1,2}=2\pm 1.73205080757 \\ {x}_1=3.73205080757 \\ {x}_2=0.267949192431 \\ \text{ Factored form of the equation: } \\ (x-3.73205080757)(x-0.267949192431)=0 \\ 0<x<1 \\ x=x_2=0.2679\doteq 0.2679 \\ a=\sqrt{1+x^2}=\sqrt{1+0.2679^2}\doteq 1.0353 \\ S={\displaystyle \frac{\sqrt{3}}{4}}\cdot a^2={\displaystyle \frac{\sqrt{3}}{4}}\cdot 1.0353^2=0.4641 \end{gathered}$$

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