Cube $ABCD-EFGH$. If $P$ midpoint of $HG$, $Q$ midpoint of $FG$, $R$ midpoint of $PQ$, $BS$ is projection $BR$ to $ABCD$, length of $BS$?

Question

The cube $ABCD-EFGH$ has side length $2$ unit. If $P$ is midpoint of $HG$ and $Q$ midpoint of $FG$, and $R$ is midpoint of $PQ$, and if $BS$ is the projection of $BR$ to $ABCD$, what is the length of $BS$?

---

Attempt:

By Pythagoras, we will find that $PR=QR=\sqrt{2}/2$. Then also $BQ = \sqrt{5}$. Then also $BR = \sqrt{5 + 2/4} = \sqrt{22}/2$. Since we know $SR=2$, then $BR = \sqrt{ (22/4) - 4 } = \sqrt{6/2} = \sqrt{6}/2$.

But another source writes the answer different, $\sqrt{10}/2$.

Answer 1

Let ZX be the projection of PQ to ABCD and SY$\perp$BC,

$$BS = \sqrt{BY^2+ SY^2} = \sqrt{\left(\frac32\right)^2+ \left(\frac12\right)^2}=\frac{\sqrt{10}}2$$