1. **State the problem:** Find the centroid $(\bar{x}, \bar{y})$ of the composite area consisting of a rectangle, a right triangle, and a semicircle. 2. **Identify individual shapes and their dimensions:** - Rectangle: width $b=4$ m, height $h=6$ m - Right triangle: base $b=4$ m, height $h=3$ m - Semicircle: radius $R=2$ m 3. **Calculate area of each shape:** - Rectangle area $A_\text{rect} = b \times h = 4 \times 6 = 24$ m$^2$ - Triangle area $A_\text{tri} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 4 \times 3 = 6$ m$^2$ - Semicircle area $A_\text{semi} = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi \times 2^2 = 2\pi \approx 6.283$ m$^2$ 4. **Locate centroids of each shape relative to origin O:** - Rectangle centroid $g_2$ at center: $\left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3)$ m - Triangle centroid $g_1$ (right triangle with right angle at origin, base along x-axis, height along y-axis): centroid at $\left(\frac{b}{3}, \frac{h}{3}\right) = \left(\frac{4}{3}, 1\right)$ m - Semicircle centroid $g_3$ (centroid of semicircle from flat edge is $\frac{4R}{3\pi}$ above the base): - x-coordinate: center of semicircle diameter $= 2$ m (same as rectangle top center) - y-coordinate: rectangle top edge $+ \frac{4R}{3\pi} = 6 + \frac{4 \times 2}{3\pi} = 6 + \frac{8}{3\pi} \approx 6 + 0.849 = 6.849$ m 5. **Calculate composite centroid coordinates:** - Total area $A = 24 + 6 + 6.283 = 36.283$ m$^2$ - $\bar{x} = \frac{A_\text{rect} x_\text{rect} + A_\text{tri} x_\text{tri} + A_\text{semi} x_\text{semi}}{A} = \frac{24 \times 2 + 6 \times \frac{4}{3} + 6.283 \times 2}{36.283}$ Calculate numerator: $24 \times 2 = 48$ $6 \times \frac{4}{3} = 8$ $6.283 \times 2 = 12.566$ Sum: $48 + 8 + 12.566 = 68.566$ So, $$\bar{x} = \frac{68.566}{36.283} \approx 1.89 \text{ m}$$ - $\bar{y} = \frac{A_\text{rect} y_\text{rect} + A_\text{tri} y_\text{tri} + A_\text{semi} y_\text{semi}}{A} = \frac{24 \times 3 + 6 \times 1 + 6.283 \times 6.849}{36.283}$ Calculate numerator: $24 \times 3 = 72$ $6 \times 1 = 6$ $6.283 \times 6.849 \approx 43.03$ Sum: $72 + 6 + 43.03 = 121.03$ So, $$\bar{y} = \frac{121.03}{36.283} \approx 3.34 \text{ m}$$ 6. **Final answer:** The centroid of the composite area is approximately $$\boxed{(\bar{x}, \bar{y}) = (1.89, 3.34) \text{ meters}}$$