Question

a company is replacing cables with fiber optic lines in rectangular casing bcde. if de = 3 cm and be = 3.5 cm, what is the smallest diameter of pipe that will fit the fiber optic line? round your answer to the nearest hundredth.

Explanation:

Step1: Recall the Pythagorean theorem in a rectangle - inscribed in a circle.

In rectangle BCDE inscribed in a circle, the diameter of the circle is equal to the length of the rectangle's diagonal. For right - triangle BDE, by the Pythagorean theorem \(d^{2}=DE^{2}+BE^{2}\), where \(d\) is the diameter of the circle.

Step2: Substitute the given values.

Given \(DE = 3\mathrm{cm}\) and \(BE=3.5\mathrm{cm}\), we have \(d^{2}=3^{2}+3.5^{2}\). First, calculate \(3^{2}=9\) and \(3.5^{2}=12.25\). Then \(d^{2}=9 + 12.25=21.25\).

Step3: Find the value of \(d\).

Take the square root of both sides: \(d=\sqrt{21.25}\approx4.61\mathrm{cm}\).

Answer:

\(4.61\mathrm{cm}\)