$\begin{array}{l}
{h_1} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\
{h_2} = \frac{{{u^2}{{\sin }^2}\left( {90 – \theta } \right)}}{{2g}},R = \frac{{{u^2}\sin 2\theta }}{g}\\
Range\,R\,is\,same\,for\,angle\,\theta \,and\,\left( {{{90}^ \circ } – \theta } \right)\\
\therefore \,{h_1}{h_2} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \frac{{{u^2}{{\sin }^2}\left( {90 – \theta } \right)}}{{2g}}\\
 = \frac{{{u^4}\left( {{{\sin }^2}\theta } \right) \times {{\sin }^2}\left( {90 – \theta } \right)}}{{4{g^2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\sin \left( {90 – \theta } \right) = \theta \cos \theta } \right]
\end{array}$
$\begin{array}{l}
 = \frac{{{u^4}{{\left( {\sin \theta \cos \theta } \right)}^2} \times {{\cos }^2}\theta }}{{4{g^2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\sin 2\theta  = 2\sin \theta \cos \theta } \right]\\
 = \frac{{{u^4}{{\left( {\sin \theta \cos \theta } \right)}^2}}}{{4{g^2}}} = \frac{{{u^4}{{\left( {\sin 2\theta } \right)}^2}}}{{16g}}\\
 = \frac{{{{\left( {{u^2}\sin 2\theta } \right)}^2}}}{{16{g^2}}} = \frac{{{R^2}}}{{16}}\\
or,\,{R^2} = 16{h_1}{h_2}\,or\,R = 4\sqrt {{h_1}{h_2}} 
\end{array}$