We have to find the length of the circumference and the perimeter of the rectangle.
The circumference is defined by
[tex]C=\pi d[/tex]According to the graph, the diameter is 12 units long.
[tex]C=3.14\cdot12=37.68[/tex]Hence, there are needed 37.68 units to put a fence on the circumference.
On the other hand, we have to find the width and length of the rectangle with the distance formula.
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]For the base, we use the points (-2, -10) and (10, -1)
[tex]\begin{gathered} d=\sqrt[]{(10-(-2))^2+(-1-(-10))^2} \\ d=\sqrt[]{(10+2)^2+(-1+10)^2} \\ d=\sqrt[]{12^2+9^2}=\sqrt[]{144+81}=\sqrt[]{225} \\ d=15 \end{gathered}[/tex]The base of the rectangle is 15 units long.
For the height, we use the points (-2, -10) and (-5, -6).
[tex]\begin{gathered} d=\sqrt[]{(-5-(-2))^2+(-6-(-10))^2} \\ d=\sqrt[]{(-5+2)^2+(-6+10)^2} \\ d=\sqrt[]{(-3)^2+4^2}=\sqrt[]{9+16}=\sqrt[]{25} \\ d=5 \end{gathered}[/tex]The height of the rectangle is 5 units.
Then, we find the perimeter.
[tex]P=2(w+l)=2(15+5)=2(20)=40[/tex]The perimeter of the rectangle is 40 units.
Hence, there would be needed more fencing for the rectangle than the circle.