1. **State the problem:** We have a triangle ABC with sides AB = 4.6 cm, BC = 8.3 cm, and an acute angle ABC. The area of the triangle is 12 cm². We need to find the perimeter of triangle ABC. 2. **Recall the formula for the area of a triangle:** $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are two sides enclosing angle $C$. 3. **Identify the sides and angle:** Here, angle $ABC$ is between sides $AB$ and $BC$, so: - $a = AB = 4.6$ cm - $b = BC = 8.3$ cm - $C = \angle ABC$ 4. **Use the area formula to find $\sin(\angle ABC)$:** $$12 = \frac{1}{2} \times 4.6 \times 8.3 \times \sin(\angle ABC)$$ $$12 = 19.09 \times \sin(\angle ABC)$$ $$\sin(\angle ABC) = \frac{12}{19.09} \approx 0.6285$$ 5. **Find $\cos(\angle ABC)$ using Pythagorean identity:** $$\cos(\angle ABC) = \sqrt{1 - \sin^2(\angle ABC)} = \sqrt{1 - 0.6285^2} = \sqrt{1 - 0.395} = \sqrt{0.605} \approx 0.778$$ 6. **Use the cosine rule to find side AC:** $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)$$ $$AC^2 = 4.6^2 + 8.3^2 - 2 \times 4.6 \times 8.3 \times 0.778$$ $$AC^2 = 21.16 + 68.89 - 59.38 = 30.67$$ $$AC = \sqrt{30.67} \approx 5.54 \text{ cm}$$ 7. **Calculate the perimeter:** $$\text{Perimeter} = AB + BC + AC = 4.6 + 8.3 + 5.54 = 18.44 \text{ cm}$$ 8. **Round to 3 significant figures:** $$\boxed{18.4 \text{ cm}}$$