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

Step 1: Recall the standard result.
For 6 equal circles of radius \( r \) packed in a ring, each touching its two neighbours and all touching one enclosing circle from the inside, the enclosing circle's radius is always \( 3r \).

Step 2: Apply it directly.
\[ r = 5 \text{ cm} \implies R = 3r = 15 \text{ cm} \]

Step 3: Compute the circumference.
\[ C = 2\pi R = 2\pi(15) = 30\pi \]
\[ \boxed{30\pi \approx 94.25 \text{ cm}} \]

Answer 2

A third way is to place the six biscuit centres directly on a coordinate grid and measure the chord between two neighbours.

Put the plate's centre at the origin, with the biscuit centres sitting at equal angles of \( 60^{\circ} \) around a circle of unknown radius \( R_c \). The straight-line distance (chord) between two adjacent centres, separated by an angle \( \theta = 60^{\circ} \), is given by the chord-length formula

\[ \text{chord} = 2R_c\sin\left(\frac{\theta}{2}\right) = 2R_c\sin(30^{\circ}) = 2R_c\left(\frac{1}{2}\right) = R_c \]

Since adjacent biscuits touch, this chord equals twice the biscuit radius, \( 2r = 10 \) cm, so \( R_c = 10 \) cm.

The plate's own radius is this value plus one biscuit radius: \( R = 10 + 5 = 15 \) cm, giving circumference \( C = 2\pi(15) = 30\pi \approx 94.25 \) cm.

So the correct answer is 94.25 cm (approximately).