4. 216 Each of the squares has a side-length of $18 \mathrm{~cm}$. Therefore each of the triangles has two sides of length $18 \mathrm{~cm}$. Hence the triangles are isosceles. Let the angle contained by the two $18 \mathrm{~cm}$ sides of these triangle be $x^{\circ}$. The interior angles of a square and a regular hexagon are $90^{\circ}$ and $120^{\circ}$ respectively. By considering angles at a point we have $x+90+90+120=360$. Therefore $x=60$ and the triangles are equilateral. All twelve outer edges of the figure are $18 \mathrm{~cm}$ in length. Therefore $P=12 \times 18=216$.
4. The figure shows a shape consisting of a regular hexagon of side $18 \mathrm{~cm}$, six triangles and six squares. The outer perimeter of the shape is $P \mathrm{~cm}$. What is the value of $P$ ?