Perimeter of Shape Fomed By Four Circles Equally Spaced About an Inner Circle
The triangle formed by the points ADB is equilateral, as are the triangles BEC, DCF and ADG. The point ABCD forms a square, so the angles DBE, ECF, FDG and GAD are all equal to
\[360-90-2 \times 60=150\]
degrees.The perimeter of the shape is then
\[4 \times \frac{150}{360} \times 2 \pi \times 6= 20 \pi\]
.