1. **State the problem:** Find the centroid $(\bar{x}, \bar{y})$ of the composite area consisting of a right triangle, a rectangle, and a semicircle as described. 2. **Identify individual areas and centroids:** - Triangle (A1): base = 4 m, height = 3 m - Rectangle (A2): width = 4 m, height = 6 m - Semicircle (A3): radius = 2 m 3. **Calculate areas:** $$A_1 = \frac{1}{2} \times 4 \times 3 = 6 \text{ m}^2$$ $$A_2 = 4 \times 6 = 24 \text{ m}^2$$ $$A_3 = \frac{1}{2} \pi \times 2^2 = 2\pi \approx 6.283 \text{ m}^2$$ 4. **Find centroids of each shape relative to origin O:** - Triangle centroid $g_1$: located at $\left(\frac{4}{3}, \frac{3}{3}\right) = (1.333, 1)$ m - Rectangle centroid $g_2$: center at $\left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3)$ m - Semicircle centroid $g_3$: for a semicircle on the x-axis, centroid is at $\left(4 + 0, 6 + \frac{4r}{3\pi}\right)$ where $r=2$ Calculate $y$ coordinate: $$6 + \frac{4 \times 2}{3\pi} = 6 + \frac{8}{3\pi} \approx 6 + 0.849 = 6.849 \text{ m}$$ So $g_3 = (4, 6.849)$ m 5. **Calculate composite centroid coordinates:** $$\bar{x} = \frac{A_1 x_1 + A_2 x_2 + A_3 x_3}{A_1 + A_2 + A_3} = \frac{6 \times 1.333 + 24 \times 2 + 6.283 \times 4}{6 + 24 + 6.283}$$ Calculate numerator: $$6 \times 1.333 = 8$$ $$24 \times 2 = 48$$ $$6.283 \times 4 = 25.132$$ Sum numerator: $$8 + 48 + 25.132 = 81.132$$ Sum denominator: $$6 + 24 + 6.283 = 36.283$$ Therefore: $$\bar{x} = \frac{81.132}{36.283} \approx 2.237 \text{ m}$$ $$\bar{y} = \frac{A_1 y_1 + A_2 y_2 + A_3 y_3}{A_1 + A_2 + A_3} = \frac{6 \times 1 + 24 \times 3 + 6.283 \times 6.849}{36.283}$$ Calculate numerator: $$6 \times 1 = 6$$ $$24 \times 3 = 72$$ $$6.283 \times 6.849 \approx 43.028$$ Sum numerator: $$6 + 72 + 43.028 = 121.028$$ Therefore: $$\bar{y} = \frac{121.028}{36.283} \approx 3.336 \text{ m}$$ 6. **Final answer:** The centroid of the composite area is approximately: $$\boxed{(\bar{x}, \bar{y}) = (2.237, 3.336) \text{ m}}$$