A vertical triangular plane area, submerged in water, with one side in the free surface, vertex downward and latitude ‘h’ was the pressure centre below the free surface by
Centre of pressure
\(\begin{array}{l} {{\rm{h}}^{\rm{*}}} = {\rm{\bar X}} + \frac{{{{\rm{I}}_{\rm{G}}}}}{{{\rm{A\bar X}}}} = \frac{{\rm{h}}}{3} + \frac{{\frac{{{\rm{b}}{{\rm{h}}^3}}}{{36}}}}{{\frac{{{\rm{bh}}}}{2}.\frac{{\rm{h}}}{3}}}\\ = \frac{{\rm{h}}}{3} + \frac{{\rm{h}}}{6} = \frac{{\left( {2 + 1} \right){\rm{h}}}}{6} = \frac{{\rm{h}}}{2} \end{array}\)
Important point:
Geometry 
Centre of pressure 
\(\frac{2h}{3}\) 

\(\frac{h}{2}\) 

\(\frac{{3h}}{4}\) 

\(\frac{{5h}}{8}\) 

\(\frac{{3\pi D}}{{32}}\) 

\(\frac{{3\pi D}}{{32}}\) 

\(\frac{{h\left( {a + 3b} \right)}}{{2\left( {a + 2b} \right)}}\) 