Question #09700

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Answer 1 · 2016-10-30T16:23:33

Please see the explanation.

Please notice that, if you know the area and the value of s, then you have a quadratic with r as the variable:

#pir^2 + pisr - A = 0#

Divide everything by #pi#

#r^2 + sr - A/pi = 0#

#r = {-s +-sqrt(s^2 + 4A/pi)}/2#

Negative numbers do not make sense for a radius, therefore, you should change the #+-# to only #+#:

#r = {-s +sqrt(s^2 + 4A/pi)}/2#