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The centres of the three circles form an equilateral triangle, so each of the sectors marked is $\frac{1}{6}$ of a circle. The shape below is a rectangle, so the sectors are $\frac 14$ of a circle. The curved portion of the perimeters is therefore the same as each of the circles: $2\pi \times 5 = 10\pi \text{cm}$.

The straight part is the same length as two radii, so is $10\text{cm}$ long.

The perimeter of the shaded shapes is therefore $10+10\pi\text{cm}$.

This problem is taken from the UKMT Mathematical Challenges.