Question

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

Answer 1

BF ≈ 7.07 cm

Explanation:

ΔBCE:

Use the Pythagorean theorem:

`EB^2=EC^2+BC^2`

We have EC = 4cm and BC = 3cm. Substitute:

`EB^2=4^2+3^2\\\\EB^2=16+9\\\\EB^2=25\to EB=\sqrt5\\\\EB=5\ cm`

BF is the diagonal of the square FEBA. The formula of a diagonal of a square with side a is:

`d=a\sqrt2`

Therefore

`BF=5\sqrt2\ cm`

Approximation:

`BF\approx5(1.414)=7.07\ cm`