Answer :
Consider the schematic diagram below,
Given that E is the mid point of BC,
[tex]BE=EC[/tex]The side BC measures 24 units, then it follows that,
[tex]\begin{gathered} BE=\frac{BC}{2}=\frac{24}{2}=12 \\ \Rightarrow BE=EC=12 \end{gathered}[/tex]It is known that the angle between two adjacent sides of a rectangle is 90 degrees. So the triangle BCD will be a right triangle.
In the triangle BCD,
[tex]\begin{gathered} \angle CBD=\tan ^{-1}(\frac{CD}{BC})^{} \\ \angle CBD=\tan ^{-1}(\frac{5}{24}) \\ \angle CBD\approx11.77^{\circ} \end{gathered}[/tex]Similarly, in the right triangle ABE,
[tex]\begin{gathered} \angle AEB=\tan ^{-1}(\frac{AB}{BE})^{} \\ \angle AEB=\tan ^{-1}(\frac{5}{12}) \\ \angle AEB\approx22.62^{\circ} \end{gathered}[/tex]Theorem: The sum of internal angles of a triangle is 180 degrees.
Applying this theorem in triangle BFE,
[tex]\begin{gathered} \angle BFE+\angle FBE+\angle FEB=180 \\ \angle BFE+11.77+22.62=180 \\ \angle BFE=180-22.62-11.77 \\ \angle BFE\approx145.61 \end{gathered}[/tex]Now, apply the sine rule in the triangle BFE,
[tex]\begin{gathered} \frac{BF}{\sin\angle BEF}=\frac{FE}{\sin\angle FBE}=\frac{EB}{\sin \angle BFE} \\ \frac{BF}{\sin 22.62}=\frac{FE}{\sin(11.77)}=\frac{12}{\sin(145.61)} \\ \frac{BF}{\sin22.62}=\frac{FE}{\sin(11.77)}=21.24 \end{gathered}[/tex]It follows that,
[tex]\begin{gathered} FE=21.24\times\sin (11.77) \\ FE=21.24\times0.204 \\ FE\approx4.33 \end{gathered}[/tex]Thus, the length of the side FE is 4.33 units, approximately.
