= Solution
At height $z$, the horizontal <triangle> has half-width $L\sqrt{1-z/(2H)}$ along $x$ and depth $D(1-z/H)$ along $y$. These shrink with $z$, so its <orthogonal projection> onto the $(x,y)$ <plane> is the base <triangle> with vertices $(-L,0)$, $(L,0)$ and $(0,D)$:
$$
0\leq y\leq D,\qquad |x|\leq L(1-y/D).
$$
The <orthogonal projection> onto the $(y,z)$ <plane> is the <triangle> with vertices $(0,0)$, $(D,0)$ and $(0,H)$:
$$
0\leq z\leq H,\qquad0\leq y\leq D(1-z/H).
$$
The <orthogonal projection> onto the $(x,z)$ <plane> is bounded by $z=0$, $z=H$ and the two curves
$$
\boxed{x=\pm L\sqrt{1-z/(2H)},\qquad0\leq z\leq H.}
$$
Equivalently, these curves lie on the <parabola> $z=2H(1-x^2/L^2)$, truncated at height $H$. In particular the top width is $\sqrt2L$, so this projection has a horizontal upper edge from $-L/\sqrt2$ to $L/\sqrt2$; it is not triangular. The top horizontal cross-section has zero depth, consistent with this edge.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1-q12-projections.png]
{title=The three coordinate projections of the triangular-section solid in Question 12}
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