A piece of wood which was 50 cm long, was in the shape of a uniform cylinder.
A prism is cut out from the wood such that its cross-section is a minor segment of a circle, with centre $O$, radius $r$ cm and $\angle AOB = \frac{7}{9}\pi$ radians.

1. Convert the given area from square metres to square centimetres to match the units of $r$.

$\text{Area} = 3 \times 10^{-3} \text{ m}^2 = 3 \times 10^{-3} \times 100^2 \text{ cm}^2 = 30 \text{ cm}^2$

2. Set up the equation for the area of the minor segment.

$\text{Area of minor segment} = \frac{1}{2}r^2(\theta - \sin\theta)$
$\frac{1}{2}r^2\left(\frac{7}{9}\pi - \sin\left(\frac{7}{9}\pi\right)\right) = 30$

3. Solve for $r$.

$r^2 = \frac{60}{\frac{7}{9}\pi - \sin\left(\frac{7}{9}\pi\right)}$
$r^2 = \frac{60}{2.44346 - 0.642787}$
$r^2 = \frac{60}{1.80067} \approx 33.3208$
$r = \sqrt{33.3208} \approx 5.7724$
$r = 5.772 \text{ (correct to 3 decimal places)}$

1. Identify the six edges of the prism and their respective lengths.

The 6 edges consist of:

• 2 straight edges along the length of the prism ($50$ cm each)
• 2 curved edges (minor arcs $AB$ on both cross-sections)
• 2 straight edges (chords $AB$ on both cross-sections)

2. Calculate the length of the minor arc $AB$.

Length of one arc $= r\theta = 5.772 \times \frac{7}{9}\pi \approx 14.103$ cm

3. Calculate the length of the chord $AB$.

Using the cosine rule in $\triangle AOB$:
$AB^2 = r^2 + r^2 - 2r^2 \cos(\frac{7}{9}\pi)$
$AB = \sqrt{5.772^2 + 5.772^2 - 2(5.772)^2 \cos(\frac{7}{9}\pi)} \approx 10.848$ cm
(Alternatively, $AB = 2r \sin(\frac{7\pi}{18}) \approx 10.848$ cm)

4. Calculate the total length of the edges and conclude.

Total length $= 2(50) + 2(14.103) + 2(10.848) = 100 + 28.206 + 21.696 = 149.902$ cm
$149.902 \text{ cm} = 1.499 \text{ m}$
Since $1.499 \text{ m} < 2 \text{ m}$, there is sufficient ribbon.