Sideways Pyramid

The figure above shows the sideways pyramid with apex at point A, and the submerged portion of the square base as rectangle BCDE. In addition we have point F and G indicating intersection of the water surface with the triangular faces of the pyramid. It is straightforward to calculate the height of the pyramid to be $7$ . The volume of our shape is the addition of two volumes anchored at the apex $A$ , namely,

$V = \frac{1}{3} ( 7 [BCDE] + y [CDGF] )$

BCDE is a rectangle with sides $12$ and $6$ , and CDGF is a trapezoid with bases $12$ and $6$ and height $x = \sqrt{ (\frac{11}{2})^2 - (\frac{(12 - 6)}{2})^2 } = \frac{1}{2} \sqrt{85}$ . Next, we need $y$ , and this is given by $y= 6 \sin \theta$ , where $\theta$ is the angle between the base of the pyramid and one of the trianglular faces, $\theta = \tan^{-1}(\frac{7}{6}) =\sin^{-1} ( \frac{7}{\sqrt{7^2+6^2} }) = \sin^{-1} (\dfrac{7}{\sqrt{85}})$ .

Putting it all together, we get,

$V = \frac{1}{3} ( 7 (6)(12) + 6 \frac{7}{\sqrt{85}} \cdot \frac{1}{2} (12 + 6) \cdot \frac{1}{2} \sqrt{85} )$

Simplifying, this becomes,

$V = \frac{1}{3} ( 504 + 189 ) = 231$

GeoGebra 3D calculator

$336-2\cdot42-21=231$