Question 1190163
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A square of area 15 can't be inscribed in a circle of diameter 6; so I will assume that the area of 15 is of a rectangle -- as suggested in the first phrase of your post -- instead of the area of a square.<br>
Let x be one dimension of the rectangle
Then 15/x is the other dimension<br>
The diameter of the circle is the diagonal of the rectangle:<br>
{{{sqrt(x^2+(15/x)^2)=6}}}
{{{x^2+225/x^2=36}}}
{{{x^2-36+225/x^2=0}}}
{{{x^4-36x^2+225=0}}}<br>
The equation does not factor over the integers, so use a numerical method or a graphing calculator to find the possible values of x.  I leave that to you.<br>
Note the equation is an even function.  It has four zeros -- two positive values and the opposites of those two.<br>
Obviously the negative solutions make no sense in the problem.<br>
The two positive solutions are the two dimensions of the rectangle.<br>
Use those dimensions to find the perimeter.<br>