question 10 (worth 1 point)
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.

Question

Accepted Answer

# Explanation:
## Step1: Apply Pythagorean theorem
In a circle, if $BCDE$ is a cyclic quadrilateral and $\angle BCD = 90^{\circ}$ (angle inscribed in a semi - circle), then $BD$ is the diameter of the circle. Let $BD$ be the diameter. In right - triangle $BDE$, by the Pythagorean theorem $BD^{2}=DE^{2}+BE^{2}$.
## Step2: Substitute the given values
Given $DE = 3\mathrm{cm}$ and $BE=3.5\mathrm{cm}$. Then $BD^{2}=3^{2}+3.5^{2}$.
$$
\begin{align*}
BD^{2}&=9 + 12.25\
BD^{2}&=21.25\
BD&=\sqrt{21.25}\
BD&=\sqrt{\frac{2125}{100}}=\sqrt{\frac{85}{4}}=\frac{\sqrt{85}}{2}\approx4.61\mathrm{cm}
\end{align*}
$$

# Answer:
$4.61\mathrm{cm}$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply Pythagorean theorem

Step2: Substitute the given values

Answer: