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Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?

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Giant Holly Leaf

Find the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr).

Holly

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

James Page of Hethersett High School, Norfolk solved the Holly problem, well done James.



Length $AB = 2 + 2\sqrt2$ cm and length $BC = 4 + 2\sqrt 2$ cm.

To find this answer I first took a diagram of the shape and put in the angles. I took 45 from 180 degrees to find the interior angle so I could find the inner curve. I then found that this angle is used 8 times in the diagram. Each circle has a radius of 1 cm and a circumference of $2\pi$ cm. To find the length of the edges of these pieces of the holly I did this:

$\begin{eqnarray} \\ 180 - 45 &=& 135 \\ 135/360 &=& 0.375 \\ (135/360) \times 8 \times (\pi \times 2) &=& 6\pi {\rm cm} \end{eqnarray}$

Then there are the two middle ones of which I had to find the circumference and divide by 2 then multiply by 2 for the two of them, giving a total $2\pi$ cm. I then added the two answers to get the perimeter of the holly leaf which is $8\pi = 25.13$ cm (to 2 decimal places).

If you wished to do a 16 spike holly of the same sort the ends would be the same and it would have an extra 3 circles on each side making the perimeter an extra $6\pi$ cm making a total of $14\pi = 43.98$ cm (to 2 decimal places).

The rectangle $ABCD$ has area

$$(2 + 2\sqrt 2)(4 + 2\sqrt 2) = (16 + 12 \sqrt 2).$$

From this we have to subtract the areas of the four triangles at the corners, a total area of 4 sq. cm., and also the areas of 8 sectors of circles with angles of 135 degrees and the areas of the two semicircles, in all a total area of $4\pi$ sq. cm. The area of the 10 spike holly leaf is

$$16 + 12\sqrt 2 - 4 - 4\pi = 12 + 12\sqrt 2 - 4\pi = 16.40 {\rm cm}^2$$