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: Understanding the Concept: 
The problem asks for the circumference of a large circular plate that perfectly encloses six smaller, identical circular biscuits. To find the circumference of the plate, we first need to determine its radius. 
Step 2: Key Formula or Approach: 


The radius of each biscuit (\(r\)) is half of its diameter. 
The centers of the six identical, touching circles form a regular hexagon. 
The radius of the large plate (\(R\)) is the distance from the center of the arrangement to the outer edge of any biscuit. 
The circumference of the plate is given by the formula \(C = 2 \pi R\). 
Step 3: Detailed Explanation: 


Calculate the biscuit radius: The diameter of a biscuit is 10 cm. \[ r_{\text{biscuit}} = \frac{\text{Diameter}}{2} = \frac{10 \text{ cm}}{2} = 5 \text{ cm} \] 
Determine the geometry of the arrangement: The six biscuits are arranged symmetrically around a central point. The centers of these six biscuits form the vertices of a regular hexagon. The distance from the central point of the plate to the center of any biscuit is equal to the side length of this hexagon. Since the biscuits are touching, the side length of the hexagon is the sum of the radii of two adjacent biscuits, which is equal to the diameter of one biscuit. \[ \text{Distance from plate center to biscuit center} = \text{Diameter of biscuit} = 10 \text{ cm} \] 
Calculate the radius of the large plate: The radius of the large plate (\(R\)) is the distance from its center to the outermost edge of one of the biscuits. This is the sum of the distance from the plate's center to a biscuit's center, and the radius of that biscuit. \[ R_{\text{plate}} = (\text{Distance from center to biscuit center}) + r_{\text{biscuit}} \] \[ R_{\text{plate}} = 10 \text{ cm} + 5 \text{ cm} = 15 \text{ cm} \] 
Calculate the circumference of the plate: Now, use the formula for the circumference. \[ C = 2 \pi R_{\text{plate}} = 2 \pi (15) = 30\pi \text{ cm} \] Using the approximation \(\pi \approx 3.14159\): \[ C \approx 30 \times 3.14159 = 94.2477 \text{ cm} \] 
Step 4: Final Answer: 
The calculated circumference is approximately 94.25 cm, which falls within the accepted answer range of 92.9 to 94.6 cm. 
 

Answer 2

There's a quicker route to the plate's radius using the shape formed by the biscuits' centres, without first working out the whole hexagon's side-length rule.

Each biscuit has diameter 10 cm, so radius \(r = 5\) cm.
Take the plate's centre \(O\) and the centres of any two neighbouring biscuits, \(M_1\) and \(M_2\). Since the six biscuits are arranged symmetrically all the way around \(O\), the angle \(M_1 O M_2\) is \(\frac{360^{\circ}}{6} = 60^{\circ}\).
Because \(O\) is equally far from every biscuit's centre, \(OM_1 = OM_2\), so triangle \(OM_1M_2\) is isosceles. But an isosceles triangle with its apex angle equal to \(60^{\circ}\) must have its two remaining base angles add up to \(120^{\circ}\) and be equal to each other, giving \(60^{\circ}\) each, so all three angles are \(60^{\circ}\). That makes triangle \(OM_1M_2\) equilateral, so all three of its sides are equal. \[ OM_1 = OM_2 = M_1M_2 \]
Since the two neighbouring biscuits touch each other, the distance between their centres equals twice the biscuit radius. \[ M_1M_2 = 2r = 2(5) = 10 \text{ cm} \]
By the equilateral triangle property just shown, this is also the distance from the plate's centre to any biscuit's centre. \[ OM_1 = 10 \text{ cm} \]
The plate's radius reaches from \(O\) all the way to the outer edge of a biscuit, so it equals this distance plus the biscuit's own radius. \[ R_{\text{plate}} = 10 + 5 = 15 \text{ cm} \]
The plate's circumference follows directly. \[ C = 2\pi R_{\text{plate}} = 2\pi(15) = 30\pi \approx 94.25 \text{ cm} \]

So the correct answer is about 94.25 cm.