A pole stands vertically on the centre of a square when  $\text{'}\alpha \text{'}$ is the elevation of the Sun, its shadow just reaches the sides of the square and is at a distance  $\text{'}x\text{'}$ and  $\text{'}y\text{'}$ from the ends of that side.  The height of the pole is

Let  $\text{'}O\text{'}$ be the centre of the square  $OP$

The pole  $OQ$ is shadow and  $OP=h,CQ=x,QB=y$

Let  $OR\perp BC$

$OR=\frac{x+y}{2},BR=\frac{x+y}{2}$

$RQ=BR-BQ$

$=\frac{x+y}{2}-y$

$RQ=\frac{x-y}{2}$

In  $\Delta OPQ$

$\tan \alpha =\frac{OP}{OQ}$

$OQ=h\cot \alpha$

From  $\Delta OQR$

$O{Q}^2=O{R}^2+R{Q}^2$

$h^2{\cot }^2\alpha ={\left(\frac{x+y}{2}\right)}^2+{\left(\frac{x-y}{2}\right)}^2$

$h^2{\cot }^2\alpha =\frac{2(x^2+y^2)}{4}$

$h^2=\frac{x^2+y^2}{2{\cot }^2\alpha }$

$h^2=\frac{x^2+y^2}{2}{\tan }^2\alpha$

$h=\sqrt{\frac{x^2+y^2}{2}}\tan \alpha$