Find the area of an equilateral triangle with the given measurement of a 6-inch side in A = sq. in.

$${\mathtt{Area}} = {\frac{\left({\mathtt{base}}{\mathtt{\,\times\,}}{\mathtt{height}}\right)}{{\mathtt{2}}}}$$

$${\mathtt{base}} = {\mathtt{6}}$$

$${{\mathtt{base}}}^{{\mathtt{2}}} = {{\mathtt{height}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{base}}}{{\mathtt{2}}}}\right)}^{{\mathtt{2}}}$$

$${\mathtt{height}} = {\sqrt{{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{3}}}^{{\mathtt{2}}}}}$$

$${\mathtt{height}} = {\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}$$  

$${\mathtt{Area}} = {\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{6}}\right)}{{\mathtt{2}}}}$$

So the area is  $${\mathtt{9}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}$$  sq. in.