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The perimeter of the large triangle is $24 \; \text{cm}$, so it has side of length $8 \; \text{cm}$, so each small triangle has side of length $1 \; \text{cm}$. So each small triangle has perimeter $3 \; \text{cm}$
\begin{eqnarray}\text{Total length of black lines} &=& \text{Total perimeter of shaded triangles} \\ &=& \text{Number of shaded triangles} \times 3\; \text{cm} \\ &=& 27 \times 3\; \text{cm}\\ &=& 81\; \text{cm} \end{eqnarray}
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
Can you predict, without drawing, what the perimeter of the next shape in this pattern will be if we continue drawing them in the same way?