What is the area of an equilateral triangle whose side length is 4text{ cm}?

For an equilateral triangle of side $a$, the area is $\frac{\sqrt{3}}{4}a^2$. Substituting $a = 4$, we get $\frac{\sqrt{3}}{4}(4)^2 = 4\sqrt{3}\text{ cm}^2$. Alternatively, using Heron's formula with $s = 6$, Area = $\sqrt{6(6-4)(6-4)(6-4)} = \sqrt{6 \times 2 \times 2 \times 2} = \sqrt{48} = 4\sqrt{3}\text{ cm}^2$.