11.1 The diagram below shows rectangular wall art with a partially shaded irregular shape - NSC Technical Mathematics - Question 11 - 2022 - Paper 2

If the horizontal side is increased by 80%, the new length will be:

$L = 21 \text{ m} \times 1.8 = 37.8 \text{ m}$

The number of equal parts remains the same (7 parts), so each segment will now have a width of:

$a' = \frac{37.8 \text{ m}}{7} = 5.4 \text{ m}$

Using the same heights:

The area can be calculated similarly as before:

$A' = \frac{5.4}{2} (0.9 + 2 \times 1.1 + 2 \times 2.8 + 2 \times 3.2 + 2 \times 2.2 + 2 \times 1.8 + 2 \times 1.1 + 1.1)$

The final area calculated is: $A' = 74.79 \text{ m}^2$

To calculate the exterior surface area of the half-cylindrical container, use the formula:

$TSA = 2\pi r^2 + 2\pi rh$

For the half-cylinder, we can use:

• Radius $r = 50 \text{ cm}$
• Height $h = 70 \text{ cm}$

Calculating:

$TSA = 2\pi(50)^2 + 2\pi(50)(70)$

Evaluating: $= 18,849.56 ext{ cm}^2$

To calculate the volume of the cone-shaped tank, we use:

$V_{cone} = \frac{1}{3}\pi r^2 h$

Where:

• Radius $r = 200 \text{ cm}$
• Height $h = 150 \text{ cm}$

Calculating: $V_{cone} = \frac{1}{3} \pi (200)^2(150) = 12,566,370.61 \text{ cm}^3$

Next, we calculate the volume of the half-cylindrical tank:

$V_{cylinder} = \pi r^2 h$

For the half-cylinder:

• Radius $r = 50 \text{ cm}$
• Length $h = 70 \text{ cm}$

Calculating: $V_{cylinder} = \pi(50)^2(70) = 1,837,078.55 \text{ cm}^3$

Finally, to determine how many times the cylinder can be filled:

$ext{Number of fillings} = \frac{12,566,370.61}{1,837,078.55} \approx 6.84$

Rounding down, the cone can fill the cylinder 6 times.