1. **Problem Statement:** Determine the centre of gravity (centroid) of the composite shape from point P. 2. **Approach:** We find the centroid by dividing the shape into simpler parts, calculating each part's area and centroid coordinates, then using the formula for composite centroids: $$\bar{x} = \frac{\sum A_i x_i}{\sum A_i}, \quad \bar{y} = \frac{\sum A_i y_i}{\sum A_i}$$ where $A_i$ is the area and $(x_i, y_i)$ is the centroid of part $i$. 3. **Identify parts and dimensions:** - Part 1: Left trapezoid, base1 = 1.0 m, base2 = 2.0 m (assumed from shape), height = 3.0 m - Part 2: Rectangle, width = 2.0 m, height = 3.0 m - Part 3: Right trapezoid, base1 = 2.0 m, base2 = 1.0 m, height = 3.0 m - Part 4: Semicircle, radius = 0.5 m - Part 5: Right triangle, base = 2.0 m, height = 2.0 m 4. **Calculate areas:** - $A_1 = \frac{(1.0 + 2.0)}{2} \times 3.0 = 4.5$ m$^2$ - $A_2 = 2.0 \times 3.0 = 6.0$ m$^2$ - $A_3 = \frac{(2.0 + 1.0)}{2} \times 3.0 = 4.5$ m$^2$ - $A_4 = \frac{1}{2} \pi (0.5)^2 = 0.3927$ m$^2$ - $A_5 = \frac{1}{2} \times 2.0 \times 2.0 = 2.0$ m$^2$ 5. **Locate centroids of each part relative to P:** - Part 1 (trapezoid): centroid $x_1 = \frac{1.0 + 2.0 + (2.0 - 1.0)}{3} = 1.5$ m (approx), $y_1 = \frac{3.0}{2} = 1.5$ m - Part 2 (rectangle): centroid $x_2 = 1.0 + \frac{2.0}{2} = 2.0$ m, $y_2 = 3.0 + \frac{3.0}{2} = 4.5$ m - Part 3 (trapezoid): centroid $x_3 = 3.0 + \frac{2.0 + 1.0 + (1.0 - 2.0)}{3} = 5.5$ m (approx), $y_3 = \frac{3.0}{2} = 1.5$ m - Part 4 (semicircle): centroid $x_4 = 1.0 + 1.0 + 0.5 = 2.5$ m, $y_4 = 6.0 + \frac{4 \times 0.5}{3 \pi} = 6.212$ m - Part 5 (triangle): centroid $x_5 = 4.0 + \frac{2.0}{3} = 4.667$ m, $y_5 = 6.0 + \frac{2.0}{3} = 6.667$ m 6. **Calculate total area:** $$A = 4.5 + 6.0 + 4.5 + 0.3927 + 2.0 = 17.3927$$ 7. **Calculate composite centroid coordinates:** $$\bar{x} = \frac{4.5 \times 1.5 + 6.0 \times 2.0 + 4.5 \times 5.5 + 0.3927 \times 2.5 + 2.0 \times 4.667}{17.3927} = \frac{6.75 + 12 + 24.75 + 0.9817 + 9.334}{17.3927} = \frac{53.815}{17.3927} = 3.09 \text{ m}$$ $$\bar{y} = \frac{4.5 \times 1.5 + 6.0 \times 4.5 + 4.5 \times 1.5 + 0.3927 \times 6.212 + 2.0 \times 6.667}{17.3927} = \frac{6.75 + 27 + 6.75 + 2.438 + 13.334}{17.3927} = \frac{56.272}{17.3927} = 3.23 \text{ m}$$ **Final answer:** $$\boxed{\bar{x} = 3.09 \text{ m}, \quad \bar{y} = 3.23 \text{ m}}$$ This is the centre of gravity of the composite shape from point P.