Pressure on isosceles triangular plate

The plate is vertical and its axis of symmetry (the altitude) is horizontal, at 6 m depth. So the 4 m base is vertical, and the centroid lies on the axis at depth 6 m.

 free surface ==========================
                 : 6 m
        |\       :
   4 m  |  \     v
  base  |----G-----> axis (horizontal)
 (vert.)|  /   6 m altitude
        |/

Given: base $b = 4$ m, altitude $h = 6$ m, water.

Area and centroid

$$A = \frac{1}{2}\times 4 \times 6 = 12\ \text{m}^2, \qquad \bar h = 6\ \text{m}$$

The centroid lies on the axis, at $h/3 = 2$ m from the base (4 m from the apex).

Total pressure

$$F = \rho g A\bar h = 1000 \times 9.81 \times 12 \times 6 = 706\,320\ \text{N} = 706.32\ \text{kN}$$

Second moment about the horizontal centroidal axis (the axis of symmetry)

The triangle is two right triangles, each of base 6 m (along the axis) and height 2 m, on either side of the axis:

$$I_G = 2 \times \frac{6 \times 2^3}{12} = \frac{h b^3}{48} = \frac{6 \times 4^3}{48} = 8\ \text{m}^4$$

Centre of pressure (depth)

$$\begin{aligned} h^* &= \bar h + \frac{I_G}{A\bar h} = 6 + \frac{8}{12 \times 6} \\ &= 6 + 0.111 = 6.111\ \text{m} \end{aligned}$$

Horizontally, the plate is symmetric about the axis (product of inertia about the centroidal axes is zero), so the centre of pressure lies at the same horizontal position as the centroid: 2 m from the base.

Answer: $F = 706.32$ kN, acting at 6.111 m depth (0.111 m below the axis), 2 m from the vertical base (4 m from the apex).