1. **Stating the problem:** We are given a right prism with volume 480 cm³. The base ABCD is a quadrilateral where AD is parallel to BC. Given $\angle CDA = 90^\circ$, $BC = 4$ cm, and $AF = 8$ cm, we need to find the length of AD. 2. **Formula used:** The volume $V$ of a right prism is given by: $$V = \text{Area of base} \times \text{height}$$ 3. **Understanding the base:** Since $\angle CDA = 90^\circ$, triangle CDA is right-angled at D. 4. **Calculate the length of CD:** Since AD is parallel to BC and $BC = 4$ cm, and AF (height of prism) is 8 cm, the height of the prism is 8 cm. 5. **Calculate the area of the base ABCD:** The base is a trapezium with parallel sides AD and BC. Area of trapezium = $\frac{1}{2} (AD + BC) \times CD$ 6. **Find CD:** Since $\angle CDA = 90^\circ$, triangle CDA is right angled. Using Pythagoras theorem: $$CD = \sqrt{AD^2 + BC^2}$$ 7. **Express volume in terms of AD:** Volume = Area of base $\times$ height $$480 = \frac{1}{2} (AD + 4) \times CD \times 8$$ Simplify: $$480 = 4 (AD + 4) \times CD$$ 8. **Substitute $CD = \sqrt{AD^2 + 4^2} = \sqrt{AD^2 + 16}$:** $$480 = 4 (AD + 4) \times \sqrt{AD^2 + 16}$$ Divide both sides by 4: $$\cancel{4} \times (AD + 4) \times \sqrt{AD^2 + 16} = \frac{480}{\cancel{4}}$$ $$ (AD + 4) \times \sqrt{AD^2 + 16} = 120$$ 9. **Solve for AD:** Let $x = AD$, then: $$(x + 4) \sqrt{x^2 + 16} = 120$$ Square both sides: $$(x + 4)^2 (x^2 + 16) = 120^2 = 14400$$ Expand $(x + 4)^2 = x^2 + 8x + 16$: $$(x^2 + 8x + 16)(x^2 + 16) = 14400$$ Multiply out: $$x^4 + 8x^3 + 16x^2 + 16x^2 + 128x + 256 = 14400$$ Combine like terms: $$x^4 + 8x^3 + 32x^2 + 128x + 256 = 14400$$ Bring all terms to one side: $$x^4 + 8x^3 + 32x^2 + 128x + 256 - 14400 = 0$$ $$x^4 + 8x^3 + 32x^2 + 128x - 14144 = 0$$ 10. **Approximate solution:** By trial or numerical methods, approximate $x = AD \approx 8$ cm. **Final answer:** $$\boxed{AD \approx 8 \text{ cm}}$$