Using pythagorean theorem to find AC,
![\begin{gathered} (AB)^2=(AC)^2+(BC)^2 \\ \text{Where AB=5 and BC=4} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8b54k8fd4d01edms0m8ovnfk0gvrjobxkj.png)
![\begin{gathered} 5^2=(AC)^2+4^2 \\ 25=(AC)^2+16 \\ (AC)^2=25-16 \\ (AC)^2=9 \\ (AC)=\sqrt[]{9} \\ (AC)=3\text{ units} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kqh2xjticvdm219kfoyomcjc49k0iq1oph.png)
The right-angled triangles ACB and EDB,
Using the scale ratio of the sides,
![(AC)/(DE)=(CB)/(DB)=(AB)/(EB)](https://img.qammunity.org/2023/formulas/mathematics/college/dfv4vjbsjvow3w8g14rydv8accvbnmuc2v.png)
![\begin{gathered} \text{Where AC=3 and DE=1} \\ AB=5\text{ and EB=unknown} \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/f5lq7g7s3pr4q87r7nrkxn7uzrwwibe6oh.png)
Substituting the variables,
![\begin{gathered} (3)/(1)=(5)/(EB) \\ \text{Crossmultiply} \\ 3* EB=5*1 \\ EB=(5)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9er3440yo0e7m17dk6mhykshkohfejott5.png)
![\begin{gathered} Where\text{ CB=4 and DB= unknown} \\ (3)/(1)=(4)/(DB) \\ DB=(4)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gcq9qci5splh4yigbj60s4prmp4nfi7jii.png)
Since the ratio of the sides is AC:DE is 3:1,
Hence, the reasonable length of DE is 1 and correct option is B.