The picture of an irregular figure with straight side AB is shown below - NSC Technical Mathematics - Question 11 - 2024 - Paper 2

To find the width of each part when AB is divided into five equal parts, we simply divide the total length of AB by 5:

$\text{Width of each part} = \frac{AB}{5} = \frac{15}{5} = 3\text{ cm}$

Therefore, each part measures 3 cm.

To determine the area of the irregular figure using the mid-ordinate rule, we take the mid-ordinates of the given ordinates.

Given ordinates: 2,96 cm, 6,99 cm, 3,98 cm, 2 cm, 6,05 cm, 3,99 cm.

Calculating the mid-ordinates:

• Mid-ordinate 1 = 2,96
• Mid-ordinate 2 = 6,99
• Mid-ordinate 3 = 3,98
• Mid-ordinate 4 = 2,00
• Mid-ordinate 5 = 6,05
• Mid-ordinate 6 = 3,99

Total sum of these mid-ordinates:

$\text{Total} = 2,96 + 6,99 + 3,98 + 2 + 6,05 + 3,99 = 25,97$

Now calculating the area:

$\text{Area} = \text{Width} \times \text{Total Sum of Mid-ordinates} = 3 \times 25,97 = 77,91\text{ cm}^2$

Thus, the area of the irregular figure is approximately 77,91 cm².

The radius of the cone can be found from the diameter of the cylindrical part. Given diameter = 3,67 metres, the radius is:

$r = \frac{\text{diameter}}{2} = \frac{3,67}{2} = 1,835\text{ m}$

The total height of the container is 9,56 meters, which includes the heights of both cones. Each cone has a height of 1,5 meters, making the combined height of the cones:

$\text{Height of both cones} = 2 \times 1,5 = 3\text{ m}$

Thus, the height of the cylindrical section is:

$\text{Height of cylinder} = 9,56 - 3 = 6,56\text{ m}$

The volume of the container consists of the volumes of the cylinder and the two cones. The volume of the cylinder is:

$V_{cylinder} = \pi r^2 h = \pi (1,835)^2 (6,56) \approx 76,61\text{ m}^3$

The volume of one cone is:

$V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1,835)^2 (1,5) \approx 5,66\text{ m}^3$

Thus, the total volume of the container is:

$V_{total} = V_{cylinder} + 2 \times V_{cone} \approx 76,61 + 11,32 \approx 87,93\text{ m}^3$

To determine if 100 m² is sufficient, we need to calculate the total surface area of the container. The total surface area consists of the surface area of the cylindrical part and the two cones:

$\text{Surface Area} = 2\pi rh + 2\pi r^2 = 2\pi (1,835)(6,56) + 2\pi (1,835)^2 \approx 102,96\text{ m}^2$

Since 102,96 m² is greater than 100 m², we conclude that 100 m² of material is not sufficient to manufacture the container.