CAT 2019 Shift 2 Paper

From the diagram, it is obvious that $\mathit{A}\mathit{B}$ is the height of the equilateral triangle and is also the slant height of the pyramid.Therefore, $\mathit{A}\mathit{B}=\frac{\sqrt{3}}{2}\times$ side $=\frac{\sqrt{3}}{2}\times 20=10\sqrt{3}$
And $\mathit{A}\mathit{O}=\frac{1}{2}\times$ side $=\frac{1}{2}\times 20=10$
Applying Pythagoras theorem in triangle AOB
$\mathit{O}{\mathit{B}}^2=\mathit{A}{\mathit{B}}^2-\mathit{O}{\mathit{A}}^2$
$={\left(10\sqrt{3}\right)}^2-{10}^2$
$=200$
Hence, the height of the pyramid (OB) $=10\sqrt{2}$