Find volume of parallelepiped where all faces are rhombuses with edge $a$ and angle $60^\circ$.

Question

Find volume of parallelepiped where all faces are rhombuses with edge $a$ and angle $60^\circ$.

I first noticed that $\dfrac{h}a=\sin60^\circ$, so I easly found that height of the rhombus is $h=\dfrac{a\sqrt3}2$. Then I easy calculated area of the base (and all edges): $B=\dfrac{a+a}2\cdot{h}=\dfrac{a^2\sqrt3}2$. To find volume, I need to find $H$ of whole prism. Then I wrote $\dfrac{H}h=\sin60^\circ$, so $H=\dfrac{3a}4$. From this, we can find that $V=B\cdot{H}=\dfrac{3\sqrt3a^3}8$, but in book it says that answer is $\dfrac{\sqrt2a^3}2$. Where did I made a mistake?

Answer 1

From here:

$$V = a^3 \sqrt{1 + 2 \cos^3 (\pi/3) - 3 \cos^2 (\pi/3)} = a^3 \sqrt{1 + (2/8) - (3/4)} = a^3/\sqrt{2}.$$