To determine the smallest diameter of the pipe that will fit the fiber optic line in the rectangular casing BCDE, we need to first understand the layout of the quadrilateral BCDE which is inscribed in a circle.
Since BE and DE both measure 3 cm each:
- Since BE is a vertical segment and DE is horizontal, we can assume that BCDE forms a right-angled triangle at points B and D.
- We can find the lengths of BD and CE, which are also sides of the rectangle.
By the properties of a right-angled triangle, let’s denote:
- BD = x (length of the segment)
- CE = x
In a rectangle, we can apply the Pythagorean theorem to find the diagonal, which can be derived from the segments. The diagonal AC can be expressed as:
\[ AC = \sqrt{(BE^2 + DE^2)} = \sqrt{(3^2 + 3^2)} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ cm} \]
The diameter of a circle that can fit a rectangle inscribed within would be equal to the diagonal length.
Thus, the smallest diameter of the pipe that will fit the fiber optic line is approximately \(4.24 , \text{cm}\).
Hence, rounding the answer to the nearest hundredth gives us:
4.24 cm (Choice: 4.24 cm).