1. **State the problem:** We have a right-angled triangle with the vertical side length $x - 2$ cm and the horizontal side length $x + 4$ cm. 2. **Known information:** The area of the triangle is given as 27.5 cm². 3. **Formula for area of a triangle:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ For this triangle: $$27.5 = \frac{1}{2} \times (x - 2) \times (x + 4)$$ 4. **Solve for $x$:** Multiply both sides by 2: $$55 = (x - 2)(x + 4)$$ Expand the right-hand side: $$55 = x^2 + 4x - 2x - 8$$ Simplify: $$55 = x^2 + 2x - 8$$ Add $8$ to both sides: $$55 + 8 = x^2 + 2x$$ $$63 = x^2 + 2x$$ Rearrange to standard quadratic form: $$x^2 + 2x - 63 = 0$$ 5. **Factor the quadratic equation:** We look for two numbers that multiply to $-63$ and add to $2$. Those numbers are $9$ and $-7$: $$(x + 9)(x - 7) = 0$$ 6. **Find the roots:** $$x + 9 = 0 \implies x = -9$$ $$x - 7 = 0 \implies x = 7$$ Since side lengths must be positive and $x-2$ must be positive: - If $x = -9$, then $x-2 = -11$, which is negative (not valid). - If $x = 7$, then $x-2 = 5$ cm and $x+4 = 11$ cm (both positive). 7. **Determine the shortest side:** Between the two sides $5$ cm and $11$ cm, the shortest side is $$\boxed{5\text{ cm}}$$ Therefore, the length of the shortest side of the triangle is 5 cm.