In the figure, polygons $A$, $E$, and $F$ are isosceles right triangles; $B$, $C$, and $D$ are squares with sides of length $1$; and $G$ is an equilateral triangle. The figure can be folded along its edges to form a polyhedron having the polygons as faces. The volume of this polyhedron is
$\textbf{(A)}\ 1/2\qquad\textbf{(B)}\ 2/3\qquad\textbf{(C)}\ 3/4\qquad\textbf{(D)}\ 5/6\qquad\textbf{(E)}\ 4/3$
Since there are three squares, the figure may appear to be based upon a unit cube.
Since there is an equilateral triangle of side length $\sqrt{2}$, this triangle has the length of a facial (non-spatial) diagonal of a unit cube.
Since there are three isosceles right triangles that are half the area of a face, they might represent faces of the cube that have been cut in half along the diagonal.
Putting these three clues together, we can mentally fold the solid back together to form a unit cube that has had one vertex removed.  Along with this vertex, the triangular half of each of the three adjoining faces has been removed.  When the unit cube is sliced like this, it will cut three of the faces into isosceles triangles like $A$, $E$, and $F$.  Additionally, it will create an equilateral triangle that has as its sides three facial diagonals, just like $G$.
So the volume is a unit cube with a tetrahedron removed.  The tetrahedron has three right triangular faces like $A$, $E$, and $F$, and one equilateral triangular face like $G$.  The volume of this removed tetrahedron can be calculated as follows:
Place one of the three right triangles on the ground as the base.  It has an area of $\frac{1}{2}$.  The height is the height of one of the other two right triangles - the equilateral triangle is slanted and is not a height.  Thus, the volume of the tetrahedron is $\frac{1}{3}\cdot \frac{1}{2}\cdot 1 = \frac{1}{6}$, and when this volume is removed, the remaining volume is $\frac{5}{6}$, which is option $\boxed{D}$.