<s>[INST] Height of a cube edge from the floor 
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?

I'm sorry for my bad English, but I try to illustrate it as follows.

The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?
 [/INST] Let $p = \angle ACA'$, $q = \angle ECA$ (in the original figure), where $A,C,A'$ are the same as in Math Lover's answer.
Then $\tan p = \frac{1}{\sqrt 7}, p = \tan^{-1} \frac{1}{\sqrt 7}$ and $\tan q = \frac{2}{2 \sqrt 2} = \frac{1}{\sqrt 2}, q = \tan^{-1} \frac{1}{\sqrt2}$. Now using the arctangent addition formula:
$$p + q = \tan^{-1} \frac{1/\sqrt7 + 1/\sqrt2}{1 - (1/\sqrt7)(1/\sqrt2)} = \tan^{-1} \frac{\sqrt{2} + \sqrt{7}}{\sqrt{14}-1} = \tan^{-1} \frac{\sqrt{2\cdot2\cdot7}+\sqrt{2\cdot7\cdot7}+\sqrt{2}+\sqrt{7}}{13} = \frac{1}{13} \left(8 \sqrt2 + 3 \sqrt7 \right).$$
So $\sin(p+q) = EN / EC \Rightarrow EN = 2\sqrt{3}a \sin(p+q)$. Now since $\sin(\tan^{-1} x) = \frac{x}{\sqrt{x^2+1}}$ (draw a triangle), then we have the monstrous task of simplifying this expression:
$$\sin(p+q) = \frac{(1/13) \left(8 \sqrt2 + 3 \sqrt7 \right)}{\sqrt{(1/169)(191+48\sqrt{14})+1}}$$
and after incorporating the $1$, denesting the square root, and dividing, I can confirm that this eventually gives $EN = a(1 + \sqrt{7/2})$.
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