A Cube labeled $ABCD$(base)-$EFGH$(top) has edge $8$cm. It has a pyramid with base $ABCD$ inside it, with the top-point labeled as $P$. It is also known that the distance from $Q$ (the midpoint of $FG$) to the plane $P-CD$ is $4$cm. What is the height of the pyramid?
When I draw the cube and a pyramid inside, I see that the distance from $Q$ to the center of the top square is $4$cm, thus the distance from $Q$ to any plane of the pyramid will be longer than $4$cm.
But the answer key says $3$cm.
Let's look at the plane $BCFG$. And, also, assume that the pyramid is regular (square symmetric.)
I am going to use the OP's notations below.
Centered at $Q$, consider a circle of radius $4$ cm. Determine the two lines (going through $B$ and $C$, respectively) that are tangent to this circle. These lines correspond to the planes $DCP$ and $ABP$. The common point of these lines corresponds to the top of the pyramid. The pyramid can be seen as $\triangle BCP$ in the figure above.
The altitude of the pyramid is, indeed, $3$ cm.