1. **State the problem:** We are given a rectangle WXYZ with area 115.5 in$^2$. We need to find the perimeter of triangle $\triangle XYZ$.\n\n2. **Identify known information:** \n- Rectangle area formula: $\text{Area} = \text{length} \times \text{width}$.\n- The diagonals of a rectangle are equal, so $XZ = WY$.\n- Point P is the intersection of diagonals, and PQ is an altitude of triangle $\triangle XYZ$ with length 4 in.\n\n3. **Find the sides of the rectangle:**\nLet $XY = a$ and $YZ = b$. Then $a \times b = 115.5$.\n\n4. **Use the altitude PQ = 4 in:**\nSince P is the midpoint of diagonal XZ, and PQ is perpendicular to WZ, PQ is the altitude of $\triangle XYZ$ from vertex P to side YZ.\n\n5. **Use the Pythagorean theorem to find $XZ$:**\nSince $\triangle XYZ$ is right-angled at Y (rectangle corner), \n$$XZ = \sqrt{XY^2 + YZ^2} = \sqrt{a^2 + b^2}.$$\n\n6. **Express $a$ in terms of $b$:**\nFrom area, $a = \frac{115.5}{b}$.\n\n7. **Use altitude formula for right triangle:**\nAltitude from right angle vertex to hypotenuse is $PQ = \frac{ab}{XZ} = 4$.\nSubstitute $a = \frac{115.5}{b}$ and $XZ = \sqrt{a^2 + b^2}$:\n$$4 = \frac{\frac{115.5}{b} \times b}{\sqrt{\left(\frac{115.5}{b}\right)^2 + b^2}} = \frac{115.5}{\sqrt{\frac{115.5^2}{b^2} + b^2}}.$$\n\n8. **Simplify denominator:**\n$$\sqrt{\frac{115.5^2}{b^2} + b^2} = \sqrt{\frac{115.5^2 + b^4}{b^2}} = \frac{\sqrt{115.5^2 + b^4}}{b}.$$\n\n9. **Rewrite equation:**\n$$4 = \frac{115.5}{\frac{\sqrt{115.5^2 + b^4}}{b}} = \frac{115.5 b}{\sqrt{115.5^2 + b^4}}.$$\n\n10. **Square both sides:**\n$$16 = \frac{(115.5)^2 b^2}{115.5^2 + b^4}.$$\n\n11. **Cross multiply:**\n$$16(115.5^2 + b^4) = (115.5)^2 b^2.$$\n\n12. **Expand:**\n$$16 \times 115.5^2 + 16 b^4 = 115.5^2 b^2.$$\n\n13. **Rearrange:**\n$$16 b^4 - 115.5^2 b^2 + 16 \times 115.5^2 = 0.$$\n\n14. **Let $x = b^2$, then:**\n$$16 x^2 - 115.5^2 x + 16 \times 115.5^2 = 0.$$\n\n15. **Calculate constants:**\n$115.5^2 = 13340.25$, so equation is\n$$16 x^2 - 13340.25 x + 213444 = 0.$$\n\n16. **Use quadratic formula:**\n$$x = \frac{13340.25 \pm \sqrt{(13340.25)^2 - 4 \times 16 \times 213444}}{2 \times 16}.$$\n\n17. **Calculate discriminant:**\n$$D = 13340.25^2 - 4 \times 16 \times 213444 = 178000000 - 13676064 = 164323936.$$\n\n18. **Calculate roots:**\n$$x = \frac{13340.25 \pm 12822.5}{32}.$$\n\n19. **Two solutions:**\n- $x_1 = \frac{13340.25 + 12822.5}{32} = \frac{26162.75}{32} = 817.58$\n- $x_2 = \frac{13340.25 - 12822.5}{32} = \frac{517.75}{32} = 16.18$\n\n20. **Find $b$:**\n- $b = \sqrt{817.58} \approx 28.6$ (too large, check area)\n- $b = \sqrt{16.18} \approx 4.0$ (reasonable)\n\n21. **Find $a$:**\n$$a = \frac{115.5}{b} = \frac{115.5}{4.0} = 28.9.$$\n\n22. **Find $XZ$:**\n$$XZ = \sqrt{a^2 + b^2} = \sqrt{28.9^2 + 4.0^2} = \sqrt{835.2 + 16} = \sqrt{851.2} \approx 29.2.$$\n\n23. **Find perimeter of $\triangle XYZ$:**\n$$P = XY + YZ + XZ = 28.9 + 4.0 + 29.2 = 62.1 \text{ in.}$$\n\n**Final answer:** The perimeter of $\triangle XYZ$ is approximately $62.1$ inches.