1. **State the problem:** We need to find the area of a triangle with sides 13, 40, and 49. 2. **Formula used:** We use Heron's formula to find the area of a triangle when all three sides are known. The formula is: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ 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:** $$s = \frac{13 + 40 + 49}{2} = \frac{102}{2} = 51$$ 4. **Calculate the area using Heron's formula:** $$A = \sqrt{51(51-13)(51-40)(51-49)} = \sqrt{51 \times 38 \times 11 \times 2}$$ 5. **Simplify inside the square root:** $$51 \times 38 = 1938$$ $$11 \times 2 = 22$$ So, $$A = \sqrt{1938 \times 22} = \sqrt{42636}$$ 6. **Find the square root:** $$\sqrt{42636} = 206.5$$ (approximately) 7. **Final answer:** The area of the triangle is approximately **206.5** square units.