Outside point

The square ABCD and the point E lying outside the given square are given. What is the area of the square when the distance | AE | = 2, | DE | = 5 a | BE | = 4?

Correct result:

S =  9.1973

Solution:

$$\begin{gathered} AE=2 \\ DE=5 \\ BE=4 \\ a,x,y>0 \\ 1. \\ A{E}^2=y^2+x^2 \\ B{E}^2=(a-y{)}^2+x^2 \\ D{E}^2=y^2+(x+a{)}^2 \\ 2. \\ A{E}^2=y^2+x^2 \\ B{E}^2=(a+y{)}^2+x^2 \\ D{E}^2=y^2+(x+a{)}^2 \\ 3. \\ A{E}^2=y^2+x^2 \\ B{E}^2=(a+y{)}^2+x^2 \\ D{E}^2=y^2+(x-a{)}^2 \\ 2: \\ a=3.03271\doteq 3.0327 \\ x=1.94589\doteq 1.9459 \\ y=\sqrt{A{E}^2-x^2}=\sqrt{2^2-1.9459^2}\doteq 0.4621 \\ S=a^2=3.0327^2\doteq 9.1973=9.1973 \\ \text{ Verifying Solution: } \\ {l}_1=\sqrt{y^2+x^2}=\sqrt{0.4621^2+1.9459^2}=2 \\ {l}_2=\sqrt{(a+y{)}^2+x^2}=\sqrt{(3.0327+0.4621{)}^2+1.9459^2}\doteq 4 \\ {l}_3=\sqrt{y^2+(x+a{)}^2}=\sqrt{0.4621^2+(1.9459+3.0327{)}^2}\doteq 5 \end{gathered}$$

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