1. **State the problem:** Given a triangle with sides $(2x - 1)$ cm, $(x - 1)$ cm, and $(x + 1)$ cm, the largest angle is $120^\circ$. We need to find the value of $x$ and then the area of the triangle. 2. **Identify the largest side and angle:** The longest side is $(2x - 1)$ cm, opposite the largest angle $120^\circ$. 3. **Use the cosine rule for side $a = 2x - 1$ opposite angle $C =120^\circ$:** $$ a^2 = b^2 + c^2 - 2bc \cos C $$ where $b = x - 1$, $c = x + 1$. 4. Substitute values: $$ (2x - 1)^2 = (x - 1)^2 + (x + 1)^2 - 2(x - 1)(x + 1) \cos 120^\circ $$ 5. Compute cosines and squares: $$ \cos 120^\circ = -\frac{1}{2} $$ $$ (2x - 1)^2 = (x -1)^2 + (x +1)^2 - 2(x -1)(x +1)(-\frac{1}{2}) $$ 6. Expand squares: $$ (2x -1)^2 = (x^2 - 2x +1) + (x^2 + 2x +1) + (x -1)(x +1) $$ since $-2 \times -\frac{1}{2} = +1$. 7. Simplify the right side: $$ (2x -1)^2 = (x^2 - 2x +1) + (x^2 + 2x +1) + (x^2 -1) $$ $$ (2x -1)^2 = x^2 - 2x +1 + x^2 + 2x +1 + x^2 -1 $$ 8. Combine like terms: $$ (2x -1)^2 = 3x^2 +1 $$ 9. Expand left side: $$ (2x -1)^2 = 4x^2 -4x +1 $$ 10. Set equal and rearrange: $$ 4x^2 -4x +1 = 3x^2 + 1 $$ $$ 4x^2 -4x +1 - 3x^2 -1 = 0 $$ $$ x^2 -4x = 0 $$ 11. Factor: $$ x(x -4) = 0 $$ 12. Solutions: $$ x=0 \quad \text{or} \quad x=4 $$ Since sides must be positive lengths, discard $x=0$. **Answer for part (a):** $x=4$. 13. **Find the sides with $x=4$:** $$ 2x -1 = 2(4) -1 = 8 -1 = 7, \quad x -1 = 3, \quad x +1 = 5 $$ 14. **Find the area using formula for area with two sides and included angle:** $$ \text{Area} = \frac{1}{2}bc \sin C $$ Here, $b=3$, $c=5$, $C=120^\circ$. 15. Compute: $$ \sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} $$ 16. Calculate area: $$ \text{Area} = \frac{1}{2} \times 3 \times 5 \times \frac{\sqrt{3}}{2} = \frac{15 \sqrt{3}}{4} \approx 6.50 \text{ cm}^2 $$ **Answer for part (b):** Area = $\frac{15 \sqrt{3}}{4}$ cm$^2$ (approximately 6.50 cm$^2$).