Question 104247
For both figures, you have to find out the relationship between its perimeter (the length of wire) and the area. 
For a square of side L. The perimeter is 4L. The area is L*L. 
{{{P=4L}}}
{{{L=P/4}}}
{{{A=L*L}}}
{{{A=P^2/16}}}
For a circle of radius, R, the perimeter is 2*pi*R and the area is pi*R*R.
{{{P=2(pi)R}}}
{{{R=P/(2(pi))}}}
{{{A=(pi)R^2}}}
{{{A=(pi)(P/(2(pi))^2)}}}
{{{A=P^2/(4(pi))}}}
The perimeter of the square is the length of wire used in the square. 
{{{P[1]}}}
The perimeter of the circle is the length of wire used in the circle. 
{{{P[2]}}}
The areas of the square and circle equal each other. 
1.{{{P[1]^2/16=P[2]^2/(4(pi))}}}
and the length of the two cut wires equals 360 inches.
2.{{{P[1]+P[2]=360}}}
From equation 1,
{{{P[1]^2/16=P[2]^2/(4(pi))}}}
{{{P[1]^2=16*P[2]^2/(4(pi))}}}
{{{P[1]=2*P[2]/sqrt((pi))}}}
Substitute into equation 2,
{{{P[1]+P[2]=360}}}
{{{2*P[2]/sqrt((pi))+P[2]=360}}}
{{{P[2](1+2/sqrt((pi)))=360}}}
{{{P[2]=360/(1+2/sqrt((pi)))}}}
{{{P[2]}}}=169.1 inches
From equation 2, 
{{{P[1]+P[2]=360}}}
{{{P[1]=360-P[2]}}}
{{{P[1]}}}=190.9 inches
Check your answer
The area of the square is 
{{{A=P^2/16}}}
{{{A=2278}}}
The area of the square is 
{{{A=P^2/(4(pi))}}}
{{{A=P^2/(4(pi))}}}
{{{A=2276}}}
The answers are off a little from roundoff error but good enough. Good answer.