# 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 =  0.464

#### Solution:

$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}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 4 \pm \sqrt{ 12 } }{ 2 }=\dfrac{ 4 \pm 2 \sqrt{ 3 } }{ 2 } \ \\ x_{1,2}=2 \pm 1.7320508075689 \ \\ x_{1}=3.7320508075689 \ \\ x_{2}=0.26794919243112 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -3.7320508075689) (x -0.26794919243112)=0 \ \\ \ \\ 0

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