5.1 Calculate the area of part 1 - NSC Civil Technology Construction - Question 5 - 2016 - Paper 1

Part 2 is a rectangle as well, so we use the same area formula: $\text{Area}_2 = 60 \text{ mm} \times 60 \text{ mm} = 3600 \text{ mm}^2$

Part 3 is a smaller rectangle, and its area can be calculated as: $\text{Area}_3 = 15 \text{ mm} \times 30 \text{ mm} = 450 \text{ mm}^2$

To find the total area of the lamina, we can subtract the area of part 3 from the sum of areas of part 1 and part 2: $\text{Total Area} = \text{Area}_1 + \text{Area}_2 - \text{Area}_3$ $\text{Total Area} = 2700 \text{ mm}^2 + 3600 \text{ mm}^2 - 450 \text{ mm}^2 = 5850 \text{ mm}^2$

The centroid of part 3 can be determined based on its dimensions. Its distance from A–A is calculated as: $\text{Centroid} = \frac{15 \text{ mm}}{2} = 7.5 \text{ mm}$ Since it is 30 mm from A to the bottom, the centroid distance from A would be: $\text{Distance}_{A-A} = 30 - 7.5 = 22.5 \text{ mm}$

For part 1, the centroid can be found: $\text{Centroid}_1 = \frac{30 \text{ mm}}{2} = 15 \text{ mm}$ Given that part 1 is 30 mm wide, the centroid will thus be: $\text{Distance}_{B-B} = 30 - 15 = 15 \text{ mm}$

For part 3, we consider its width: $\text{Centroid}_{3,B-B} = \frac{30 \text{ mm}}{2} = 15 \text{ mm}$ Therefore, the distance from B would be: $\text{Distance}_{B-B} = 15 + 15 = 30 \text{ mm}$

Using the same method, for part 2: $\text{Centroid}_{2} = \frac{60 \text{ mm}}{2} = 30 \text{ mm}$ So the distance from B will be: $\text{Distance}_{B-B} = 60 - 30 = 30 \text{ mm}$