Using pythagorean theorem to find AC,
[tex]\begin{gathered} (AB)^2=(AC)^2+(BC)^2 \\ \text{Where AB=5 and BC=4} \end{gathered}[/tex][tex]\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}[/tex]
The right-angled triangles ACB and EDB,
Using the scale ratio of the sides,
[tex]\frac{AC}{DE}=\frac{CB}{DB}=\frac{AB}{EB}[/tex]
[tex]\begin{gathered} \text{Where AC=3 and DE=1} \\ AB=5\text{ and EB=unknown} \\ \end{gathered}[/tex]
Substituting the variables,
[tex]\begin{gathered} \frac{3}{1}=\frac{5}{EB} \\ \text{Crossmultiply} \\ 3\times EB=5\times1 \\ EB=\frac{5}{3} \end{gathered}[/tex][tex]\begin{gathered} Where\text{ CB=4 and DB= unknown} \\ \frac{3}{1}=\frac{4}{DB} \\ DB=\frac{4}{3} \end{gathered}[/tex]
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.
