Question 1104854
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<pre>
{{{S[circle]}}} = {{{pi*r^2}}} = {{{25*pi^2}}}  ====>


{{{r^2}}} = {{{25*pi}}}    ====>    r = {{{sqrt(25*pi)}}} = {{{5*sqrt(pi)}}}    ====>  D= diameter = 2r = {{{10*sqrt(pi)}}}.
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