FIGURE 5.1 below shows a shaped lamina with a right-angled triangular hole - NSC Civil Technology Civil Services - Question 5 - 2016 - Paper 1

The area of part 2 can be calculated as:

$ext{Area} = ext{Width} imes ext{Height} = 60 ext{ mm} imes 60 ext{ mm} = 3600 ext{ mm}^2$

For part 3, the area can be computed using:

ext{Area} = rac{1}{2} imes ext{Base} imes ext{Height} = rac{1}{2} imes 15 ext{ mm} imes 30 ext{ mm} = 225 ext{ mm}^2

The total area of the lamina is the sum of all parts:

= 2700 ext{ mm}^2 + 3600 ext{ mm}^2 - 450 ext{ mm}^2 = 5850 ext{ mm}^2$$

The centroid of part 3 can be found using the formula:

ext{Centroid} = rac{ ext{Base}}{3} = rac{15 ext{ mm}}{3} = 5 ext{ mm}

So, the centroid from A–A is 55 mm.

For part 1, the centroid is located at:

ext{Centroid} = rac{ ext{Height}}{2} = rac{30 ext{ mm}}{2} = 15 ext{ mm}

Thus, from B–B, it is at 55 mm.

For part 3's centroid from B–B, we calculate:

$ext{Position from B} = 30 ext{ mm} + 5 ext{ mm} = 35 ext{ mm}$

The centroid of part 2 from B–B can be found as:

ext{Centroid} = 30 ext{ mm} + rac{60 ext{ mm}}{2} = 30 ext{ mm} + 30 ext{ mm} = 60 ext{ mm}