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: Use the formula for the area of a cyclic quadrilateral (Brahmagupta's formula is not needed here as we can use the fact that \( \angle BCD = 90^{\circ} \) and all vertices lie on a circle. The diameter \( BD \) can be found using the Pythagorean theorem in the right - angled triangle \( BCD \) (but wait, we can also use the formula for the area of a triangle and then relate it to the circum - diameter. Another approach: Since \( BCDE \) is cyclic, and for a cyclic quadrilateral \( BCDE \), we can use the formula \( BD=\sqrt{DE^{2}+BE^{2}}\) (by the Pythagorean theorem in the right - angled triangle \( BDE \) if we consider the fact that the angle subtended by a diameter is a right - angle. Wait, no, more accurately, for a cyclic quadrilateral \( BCDE \) with \( \angle BCD = 90^{\circ} \), the hypotenuse \( BD \) is the diameter of the circum - circle.

We know that in a right - angled triangle (if we consider the property that an angle inscribed in a semi - circle is a right - angle. So, if \( \angle BCD = 90^{\circ} \), then \( BD \) is the diameter of the circle.
Using the Pythagorean theorem \( BD=\sqrt{DE^{2}+BE^{2}}\) (because in a cyclic quadrilateral \( BCDE \), if we assume \( \angle BED=\angle BCD = 90^{\circ} \) (angles in the same segment subtended by the diameter \( BD \)).
Given \( DE = 3\mathrm{cm}\) and \( BE=3.5\mathrm{cm}\).

$$BD=\sqrt{3^{2}+3.5^{2}}$$

Step2: Calculate the value inside the square root

$$3^{2}+3.5^{2}=9 + 12.25=21.25$$

$$BD=\sqrt{21.25}\approx4.61\mathrm{cm}$$

Answer:

\(4.61\mathrm{cm}\)