1. **Problem statement:** Show that the area $A$ of an equilateral triangle with side length $x$ is given by $$A = \frac{\sqrt{3}}{4} x^2.$$\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 $x$. To find the height $h$, drop a perpendicular from the top vertex to the base, splitting the equilateral triangle into two right triangles each with base $\frac{x}{2}$ and hypotenuse $x$.\n\n4. **Use the Pythagorean theorem to find $h$:**\n$$h = \sqrt{x^2 - \left(\frac{x}{2}\right)^2} = \sqrt{x^2 - \frac{x^2}{4}} = \sqrt{\frac{3x^2}{4}} = \frac{\sqrt{3}}{2} x.$$\n\n5. **Calculate the area:**\n$$A = \frac{1}{2} \times x \times h = \frac{1}{2} \times x \times \frac{\sqrt{3}}{2} x = \frac{\sqrt{3}}{4} x^2.$$\n\n6. **Conclusion:** We have shown that the area of an equilateral triangle with side length $x$ is $$A = \frac{\sqrt{3}}{4} x^2,$$ as required.