Questions: Plumbing Find the offset distance d of the length of pipe shown in the diagram below. The total length of the pipe is 62 in.

Solution Steps

The horizontal length is the sum of $20\frac{3}{4}$ inches and $31\frac{1}{2}$ inches. Converting to improper fractions: $20\frac{3}{4}=\frac{83}{4}$ and $31\frac{1}{2}=\frac{63}{2}=\frac{126}{4}$. Adding these gives $\frac{83}{4}+\frac{126}{4} = \frac{209}{4}=52\frac{1}{4}$ inches.

The total length of the pipe is 62 inches, and the horizontal length is $52\frac{1}{4}$ inches. Thus, the bent section has a length of $62 - 52\frac{1}{4} = 62-\frac{209}{4} = \frac{248-209}{4}=\frac{39}{4} = 9\frac{3}{4}$ inches.

The bent section of the pipe, the offset distance $d$, and the 9-inch horizontal section shown in the diagram form a right triangle. The bent section is the hypotenuse. Therefore, $d^2 + 9^2 = (9\frac{3}{4})^2$. So, $d^2 + 81 = (\frac{39}{4})^2 = \frac{1521}{16}$. Thus, $d^2 = \frac{1521}{16}-81 = \frac{1521}{16} - \frac{1296}{16} = \frac{225}{16}$. Taking the square root of both sides: $d=\sqrt{\frac{225}{16}} = \frac{\sqrt{225}}{\sqrt{16}} = \frac{15}{4}=3\frac{3}{4}$.

Final Answer: