Answer: 3.66 cm
Step-by-step explanation: Given a rectangular casing BCDE with segment DE = 3 cm and segment BE = 3.5 cm.
The area A of a rectangle is length multiply by width.
Where length L = 3.5 cm and
width W = 3 cm
Area A = 3.5 × 3 = 10.5 cm^2
The pipe that will fit the fiber optic line is in cylindrical shape. Where area of a cylinder = πr^2.
But area A = 10.5. Substitute the values for the area of the cylinder
10.5 = πr^2
10.5 = 3.143 × r^2
Make r^2 the subject of formula
r^2 = 10.5/3.143
r = sqrt ( 3.34225 )
r = 1.828
Diameter = 2 × radius
Diameter = 2 × 1.829
Diameter = 3.656 cm
Therefore, the smallest diameter of pipe that will fit the fiber optic line is 3.66 cm approximately.
