Question

Six circular biscuits of diameter 10 cm are arranged on a circular plate as shown below. What is the circumference of the plate in centimetres?

Hint:

When dealing with problems involving circles packed together, connecting the centers of the circles is often the key. This usually reveals simple geometric shapes like equilateral triangles or regular hexagons, which simplifies the problem significantly.

Answer 1

Correct Answer: 92.9

Step 1: Recall the formula for a ring of touching circles.
For \(n\) equal circles of radius \(r\), arranged in a ring so that each touches its two neighbours, the distance from the ring's centre to any circle's centre is \(R' = \dfrac{r}{\sin(\pi/n)}\).

Step 2: Plug in the numbers.
Here \(r = 5\) cm and \(n = 6\), and \(\sin(\pi/6) = \sin(30^{\circ}) = 0.5\).
\[ R' = \frac{5}{0.5} = 10 \text{ cm} \]

Step 3: Add the biscuit radius to get the plate's radius.
\[ R_{\text{plate}} = R' + r = 10 + 5 = 15 \text{ cm} \]

Step 4: Final Answer.
\[ C = 2\pi(15) = 30\pi \approx \boxed{94.25 \text{ cm}} \]

Answer 2

The same distance from the plate's centre to a biscuit's centre can also be pinned down using plain coordinates, without appealing to any hexagon or general ring formula.

Let the plate's centre be the origin \(O\), and let \(R'\) be the (unknown) distance from \(O\) to each biscuit's centre.
Place one biscuit's centre at \(M_1 = (R', 0)\), and its neighbour, sitting \(60^{\circ}\) around the circle, at \(M_2 = (R'\cos 60^{\circ}, R'\sin 60^{\circ})\).
The distance between these two centres, using the standard distance formula, is \[ M_1M_2 = \sqrt{(R' - R'\cos 60^{\circ})^2 + (R'\sin 60^{\circ})^2} \]
Expanding, and using \(\cos 60^{\circ} = 0.5\) and \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\): \[ M_1M_2 = \sqrt{(R' - 0.5R')^2 + \left(\frac{\sqrt{3}}{2}R'\right)^2} = \sqrt{(0.5R')^2 + \left(\frac{\sqrt{3}}{2}R'\right)^2} = \sqrt{0.25R'^2 + 0.75R'^2} = \sqrt{R'^2} = R' \]
So the distance between two neighbouring biscuit centres works out to simply \(R'\). Since the biscuits touch, this distance must equal twice the biscuit radius, \(2r = 10\) cm. \[ R' = 10 \text{ cm} \]
Adding the biscuit's own radius gives the plate's radius, and from there the circumference. \[ R_{\text{plate}} = 10 + 5 = 15 \text{ cm}, \qquad C = 2\pi(15) = 30\pi \approx 94.25 \text{ cm} \]

So the correct answer is about 94.25 cm.