Question

using bcde. if de = 3 cm and be = 3 cm, what is the smallest diameter of pipe that will fit the fiber optic line? round your answer to the nearest hundredth.
3.54 cm
3.91 cm
4.24 cm

Explanation:

Step1: Apply Pythagorean theorem

In right - triangle $\triangle DEB$, since $\angle D EB = 90^{\circ}$ (angle inscribed in a semi - circle), and $DE = 3$ cm, $BE = 3$ cm. According to the Pythagorean theorem $DB^{2}=DE^{2}+BE^{2}$.

Step2: Calculate the length of $DB$

Substitute $DE = 3$ and $BE = 3$ into the formula: $DB=\sqrt{3^{2}+3^{2}}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\approx4.24$ cm. The diameter of the circle (which represents the smallest pipe that can fit the fiber - optic line) is the length of $DB$.

Answer:

C. $4.24$ cm