1. **Problem statement:** We need to show that the area $A$ of an equilateral triangle with side length $s$ is given by $$A = \frac{1}{4}s^2\sqrt{3}.$$\n\n2. **Recall the formula for the area of a triangle:** $$A = \frac{1}{2} \times \text{base} \times \text{height}.$$\n\n3. **Identify the base and height:** The base is $s$. The height $h$ is drawn from the opposite vertex perpendicular to the base, splitting the base into two equal segments each of length $\frac{s}{2}$.\n\n4. **Use the Pythagorean theorem to find $h$:** In the right triangle formed, the hypotenuse is $s$, one leg is $\frac{s}{2}$, and the other leg is $h$. So, $$h^2 + \left(\frac{s}{2}\right)^2 = s^2.$$\n\n5. **Solve for $h^2$: $$h^2 = s^2 - \left(\frac{s}{2}\right)^2 = s^2 - \frac{s^2}{4} = \frac{4s^2}{4} - \frac{s^2}{4} = \frac{3s^2}{4}.$$**\n\n6. **Take the square root to find $h$: $$h = \sqrt{\frac{3s^2}{4}} = \frac{s}{2} \sqrt{3}.$$**\n\n7. **Calculate the area using the base and height:** $$A = \frac{1}{2} \times s \times \frac{s}{2} \sqrt{3} = \frac{1}{2} \times \frac{s^2}{2} \sqrt{3} = \frac{s^2}{4} \sqrt{3}.$$\n\n8. **Final result:** $$\boxed{A = \frac{1}{4} s^2 \sqrt{3}}.$$\n\nThis completes the proof that the area of an equilateral triangle with side length $s$ is $\frac{1}{4} s^2 \sqrt{3}$.