1. **State the problem:** Find the area of triangle $\triangle ABC$ where sides $AB = BC = 40$ m and base $AC = 50$ m. 2. **Formula used:** The area of a triangle can be found using the formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ We need to find the height from vertex $B$ perpendicular to base $AC$. 3. **Important rule:** Since $AB = BC$, $\triangle ABC$ is isosceles, so the height from $B$ bisects $AC$. 4. **Calculate half of base $AC$:** $$\frac{50}{2} = 25$$ 5. **Use Pythagoras theorem to find height $h$:** $$h = \sqrt{AB^2 - \left(\frac{AC}{2}\right)^2} = \sqrt{40^2 - 25^2} = \sqrt{1600 - 625} = \sqrt{975}$$ 6. **Simplify $\sqrt{975}$:** $$\sqrt{975} = \sqrt{25 \times 39} = 5\sqrt{39}$$ 7. **Calculate area:** $$\text{Area} = \frac{1}{2} \times 50 \times 5\sqrt{39} = 25 \times 5\sqrt{39} = 125\sqrt{39}$$ 8. **Approximate numerical value:** $$\sqrt{39} \approx 6.245$$ $$\text{Area} \approx 125 \times 6.245 = 780.625$$ **Final answer:** $$\boxed{125\sqrt{39} \text{ m}^2 \approx 780.63 \text{ m}^2}$$