Hello N,
Let's develop some general formulas which we can use to answer the questions. First, let's find the hydrostatic force on a vertically submerged rectangle, whose width is $w$ and height is $h$. If we orient a vertical $x$-axis along one side of the rectangle, with the origin at the surface and the positive direction down, we may compute the force along one horizontal elemental rectangle having area $A(x)$ making up the entire rectangle as:
$$dF=\rho xA(x)$$
The area of this elemental rectangle is its width at $x$ times its height $dx$, hence:
$$dF=\rho xw(x)\,dx$$
where:
$$w(x)=w$$ (the width is constant for a rectangle)
And so we have:
$$dF=\rho w x\,dx$$
Now, if the top edge of the rectangle is at $x=x_1$ and the bottom edge is at $x=x_2$, then by summing all of the elements of the force, we find:
$$F=\rho w\int_{x_1}^{x_2}x\,dx$$
Applying the FTOC, we obtain:
$$F=\frac{\rho w}{2}\left[x^2 \right]_{x_1}^{x_2}=\frac{\rho w}{2}\left(x_2^2-x_1^2 \right)$$
Since the height $h$ of the rectangle is $$h=x_2-x_1$$, we may write:
(1) $$F=\frac{\rho wh}{2}\left(2x_1+h \right)$$
Next, let's consider the hydrostatic force on a right triangle whose vertical leg lies along the $x$-axis. Let the horizontal leg, the base of the triangle, be of length $b$, and the vertical leg, the altitude of the triangle be of length $h$.
We may begin as before, computing the elemental force:
$$dF=\rho xw(x)\,dx$$
This time, the width of the triangle varies with the depth, and does so linearly. At a depth of $x=x_1$, we have $w(x)=0$ and at a depth of $x=x_1+h$, we have $w(x)=b$. Hence:
$$w(x)=\frac{b}{h}\left(x-x_1 \right)$$
And this allows us to write:
$$dF=\frac{\rho b}{h} x\left(x-x_1 \right)\,dx$$
$$dF=\frac{\rho b}{h}\left(x^2-x_1x \right)\,dx$$
Summing all of these elements of force, we find:
$$F=\frac{\rho b}{h}\int_{x_1}^{x_1+h} x^2-x_1x\,dx$$
Applying the FTOC, we find:
$$F=\frac{\rho b}{6h}\left[x^2\left(2x-3x_1 \right) \right]_{x_1}^{x_1+h}=\frac{\rho b}{6h}\left(\left(x_1+h \right)^2\left(2\left(x_1+h \right)-3x_1 \right)-\left(x_1 \right)^2\left(2\left(x_1 \right)-3x_1 \right) \right)=$$
$$\frac{\rho b}{6h}\left(\left(x_1+h \right)^2\left(2h-x_1 \right)+x_1^3 \right)=\frac{\rho bh}{6}\left(3x_1+2h \right)$$
And so we have for the right triangle:
(2) $$F=\frac{\rho bh}{6}\left(3x_1+2h \right)$$
Finally, let's consider a rectangle of width $w$ and height $h$ submerged at an angle such that the rectangle is an inclined plane. We may orient an $x$-axis that runs down the surface of the plane, parallel to its height, and whose origin is at the upper edge. Let $d(x)$ be the depth of the fluid, where $d(0)=d_1$ is the depth of the upper edge and $d\left(h \right)=d_2$ is the depth of the lower edge. We may then state:
$$d(x)=\frac{d_2-d_1}{h}x+d_1$$
We may begin as before, computing the elemental force:
$$dF=\rho w \left(\frac{d_2-d_1}{h}x+d_1 \right)\,dx$$
Summing the elements, we obtain:
$$F=\rho w\int_0^{h} \frac{d_2-d_1}{h}x+d_1\,dx$$
Application of the FTOC gives us:
$$F=\rho w\left[\frac{d_2-d_1}{2h}x^2+d_1x \right]_0^{h}=\rho w\left(\frac{d_2-d_1}{2h}h^2+d_1h \right)=\rho w\left(\frac{d_2h+d_1h}{2} \right)=\frac{\rho wh\left(d_1+d_2 \right)}{2}$$
If $\ell$ is the horizontal distance through which the plane occupies, then by Pythagoras, we may state:
$$h=\sqrt{\ell^2+\left(d_1-d_1 \right)^2}$$
And so we may state:
(3) $$F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}$$
