The side length of the square, a=4.
An equilateral triangle has equal sides. Since the base of the triangle is one side of the square, we get
[tex]AB=BC=AC=4[/tex]
Since BQ⊥AC, BP is perpendicular to AC, and hence, BP is the altitude the equilateral triangle ABC.
The altitude BP of the equilateral triangle ABC is,
[tex]\begin{gathered} BP=\frac{\sqrt[]{3}}{2}a \\ BP=\frac{\sqrt[]{3}}{2}\times4 \\ BP=3.464 \\ \end{gathered}[/tex]
Also, since BQ⊥ED, BQ⊥AC and P is midpoint of AC, Q is midpoint of ED.
[tex]\begin{gathered} EQ=\frac{1}{2}ED=\frac{1}{2}a \\ EQ=\frac{1}{2}\times4 \\ EQ=2\text{ } \end{gathered}[/tex]
[tex]\begin{gathered} BQ=BP+PQ \\ BQ=3.464+4 \\ BQ=7.464 \end{gathered}[/tex]
Now using Pythagoras theorem in triangle BQE,
[tex]\begin{gathered} BE=\sqrt[]{BQ^2+EQ^2} \\ BE=\sqrt[]{(7.464)^2+2^2} \\ BE=7.7 \end{gathered}[/tex]
Therefore, BE=7.7.
