A company makes biodegradable paper cups in the shape of a right circular cone - Leaving Cert Mathematics - Question 7 - 2020

The curved surface area (CSA) of a cone is calculated using the formula:

$CSA = \pi r l$

Substituting the values: $CSA = \pi \times 3.3 \times 9$

Calculating this gives: $CSA \approx 93.31 \text{ cm}^2$ (correct to 2 decimal places).

To find the angle θ, we first calculate the circumference of the cup using:

$\text{Circumference} = 2\pi r = 2 \times \pi \times 3.3$

The arc length of the sector is given for the slant angle, and we calculate:

$\theta = \frac{\text{arc length}}{\text{circumference}} \times 360$

Substituting: $\theta = \frac{2 \times 3.3 \times 90}{2\pi \cdot 3.3} \times 360 = 132^{\circ}$.

To find the volume of water up to the line F, we need to calculate the volume of the smaller cone formed by the top portion:

The height of this smaller cone is: $7.37 = 8.37 - 1$ $v = \frac{1}{3}\pi r^2 h$ $v = \frac{1}{3}\pi (2.905)^2(7.37) \approx 65.16 \text{ cm}^3$ (correct to 1 decimal place).

The volume of water flowing in is: $\text{Volume} = \pi r^2 h = \pi(0.8)^2(2.5)$

Thus the volume filled in one second is: $v = 5.0265 \text{ cm}^3$.

The total time to fill to line F is: $time = \frac{65.16}{5.0265} \approx 13 ext{ seconds}.$

To limit the capacity to 60 cm³, set up the volume equation:

$v = \frac{1}{3}\pi r^2 h\text{ where } h = 8.37 - x$

Setting the volume to target: $60 = \frac{1}{3}\pi (3.3)^2(8.37 - x)$

Solving for x, we find: $x \approx 1.2 ext{ cm (correct to 1 decimal place)}.$