What is the distance of segment BF Round to the nearest hundredth.

Answer:
Step-by-step explanation:
ΔBCE:
Use the Pythagorean theorem:
[tex]EB^2=EC^2+BC^2[/tex]
We have EC = 4cm and BC = 3cm. Substitute:
[tex]EB^2=4^2+3^2\\\\EB^2=16+9\\\\EB^2=25\to EB=\sqrt5\\\\EB=5\ cm[/tex]
BF is the diagonal of the square FEBA. The formula of a diagonal of a square with side a is:
[tex]d=a\sqrt2[/tex]
Therefore
[tex]BF=5\sqrt2\ cm[/tex]
Approximation:
[tex]BF\approx5(1.414)=7.07\ cm[/tex]