![We\ are\ given\ that,\\Sides\ of\ the\ formed\ square=90\ ft\\As\ we\ know\ that,\\s^2+s^2=d^2[Pythagoras\ Property]\\2s^2=d^2\\Hence,\\\sqrt{2s^2} = \sqrt{d^2} \\\sqrt{2}s=d\\Hence,\\As\ the\ distance\ between\ Base\ 1\ and\ 3\ is\ the\ diagonal.\\Hence,\\The\ distance\ between\ Base\ 1\ and\ 3=\sqrt{2}*Side\\ The\ distance\ between\ Base\ 1\ and\ 3=\sqrt{2}*90\\Hence,\\](/tpl/images/0997/5218/6d585.png)
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