1. **State the problem:** We need to find the area of a triangle with sides 12 mm, 8.4 mm, and 19 mm. 2. **Formula used:** Since we know all three sides, we use Heron's formula for the area of a triangle: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $a$, $b$, and $c$ are the side lengths, and $s$ is the semi-perimeter: $$s = \frac{a+b+c}{2}$$ 3. **Calculate the semi-perimeter:** $$s = \frac{12 + 8.4 + 19}{2} = \frac{39.4}{2} = 19.7$$ 4. **Calculate the area:** $$A = \sqrt{19.7(19.7 - 12)(19.7 - 8.4)(19.7 - 19)}$$ $$= \sqrt{19.7 \times 7.7 \times 11.3 \times 0.7}$$ 5. **Simplify inside the square root:** $$19.7 \times 7.7 = 151.69$$ $$11.3 \times 0.7 = 7.91$$ $$151.69 \times 7.91 = 1199.06$$ 6. **Find the square root:** $$A = \sqrt{1199.06} \approx 34.6$$ 7. **Final answer:** The area of the triangle is approximately **34.6 mm²** to 1 decimal place.