1. **State the problem:** We have a circular storm drain pipe partially filled with water. The chord length across the water surface is 48 inches, and the water depth (distance from the chord to the bottom of the pipe) is 4.25 inches. We need to find the diameter of the pipe. 2. **Identify the knowns and unknowns:** - Chord length $c = 48$ inches - Depth of water $d = 4.25$ inches - Radius of the pipe $r$ (unknown) - Diameter $D = 2r$ (what we want to find) 3. **Use the formula relating chord length, radius, and sagitta (depth):** The sagitta (depth) $d$ of a chord in a circle of radius $r$ and chord length $c$ is given by: $$ d = r - \sqrt{r^2 - \left(\frac{c}{2}\right)^2} $$ 4. **Rearrange the formula to solve for $r$: ** $$ d = r - \sqrt{r^2 - \left(\frac{c}{2}\right)^2} \implies \sqrt{r^2 - \left(\frac{c}{2}\right)^2} = r - d$$ Square both sides: $$ \left(\sqrt{r^2 - \left(\frac{c}{2}\right)^2}\right)^2 = (r - d)^2 $$ $$ r^2 - \left(\frac{c}{2}\right)^2 = r^2 - 2rd + d^2 $$ 5. **Simplify and solve for $r$: ** $$ r^2 - \frac{c^2}{4} = r^2 - 2rd + d^2 $$ Cancel $r^2$ on both sides: $$ - \frac{c^2}{4} = - 2rd + d^2 $$ Multiply both sides by $-1$: $$ \frac{c^2}{4} = 2rd - d^2 $$ Add $d^2$ to both sides: $$ 2rd = \frac{c^2}{4} + d^2 $$ Divide both sides by $2d$: $$ r = \frac{\frac{c^2}{4} + d^2}{2d} $$ 6. **Plug in the values $c=48$, $d=4.25$: ** $$ r = \frac{\frac{48^2}{4} + 4.25^2}{2 \times 4.25} = \frac{\frac{2304}{4} + 18.0625}{8.5} = \frac{576 + 18.0625}{8.5} = \frac{594.0625}{8.5} $$ 7. **Calculate $r$: ** $$ r \approx 69.89 \text{ inches} $$ 8. **Find the diameter $D = 2r$: ** $$ D = 2 \times 69.89 = 139.78 \text{ inches} $$ **Final answer:** The diameter of the storm drain pipe is approximately **139.78 inches**.