Let's solve the problem **step by step** to find the centroid \((\bar{x}, \bar{y})\) of the composite area. --- ## **Step 1: Divide the Composite Area** Let’s break the shape into simple areas (subtracting holes): 1. **Triangle (Main body)** 2. **Rectangle (Base hole)** 3. **Rectangle (Top hole)** --- ## **Step 2: Find Area and Centroid of Each Part** ### 1. **Triangle (Main Body)** - **Base = 14 m**, **Height = 13 m** - **Area, \(A_1\):** \[ A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 13 = 91\, \text{m}^2 \] - **Centroid:** - For a triangle with base on \(x\)-axis: - \( \bar{x}_1 = \frac{14}{2} = 7 \) m (center horizontally) - \( \bar{y}_1 = \frac{13}{3} \approx 4.33 \) m (from base up) ### 2. **Rectangle (Base Hole)** - **Width = 8 m**, **Height = 2 m** - **Area, \(A_2\):** \[ A_2 = 8 \times 2 = 16\, \text{m}^2 \] - **Centroid:** - Horizontally: starts at 3 m from left (since triangle base is 14 m, triangle isosceles, so left gap is (14-8)/2 = 3 m) - \( \bar{x}_2 = 3 + \frac{8}{2} = 7 \) m - Vertically: \( \bar{y}_2 = 1 \) m (since height is 2 m, centroid is at 1 m above x-axis) ### 3. **Rectangle (Top Hole)** - **Width = 2 m**, **Height = 5 m** - **Area, \(A_3\):** \[ A_3 = 2 \times 5 = 10\, \text{m}^2 \] - **Centroid:** - Horizontally: starts at \( \frac{14-2}{2} = 6 \) m from left - \( \bar{x}_3 = 6 + 1 = 7 \) m - Vertically: from bottom, up \( 13 - 5 - 3 = 5 \) m (since 5 m hole, with 3 m margin from base) + \(2.5\) m (half height) - \( \bar{y}_3 = 11.5 \) m (from base: 3 m + 5 m + 2.5 m) --- ## **Step 3: Subtract Holes from Main Area** - The holes are **subtracted** (their area is negative). So, assign: - \(A_1 = +91\) m², \( (x_1, y_1) = (7, 4.33)\) - \(A_2 = -16\) m², \( (x_2, y_2) = (7, 1)\) - \(A_3 = -10\) m², \( (x_3, y_3) = (7, 11.5)\) --- ## **Step 4: Apply Centroid Formula** \[ \bar{x} = \frac{\sum (A_i x_i)}{\sum A_i} \] \[ \bar{y} = \frac{\sum (A_i y_i)}{\sum A_i} \] Let's compute: ### **Sum of Areas:** \[ A_{\text{total}} = 91 - 16 - 10 = 65\, \text{m}^2 \] ### **Sum \(A_i x_i\):** \[ (91 \times 7) + (-16 \times 7) + (-10 \times 7) = (637) + (-112) + (-70) = 455 \] \[ \sum (A_i x_i) = 455 \] ### **Sum \(A_i y_i\):** \[ (91 \times 4.33) + (-16 \times 1) + (-10 \times 11.5) \\ = (394.03) + (-16) + (-115) = 394.03 - 131 = 263.03 \] --- ## **Step 5: Calculate Centroid Coordinates** ### **\(\bar{x}\):** \[ \bar{x} = \frac{455}{65} \approx 7\, \text{m} \] ### **\(\bar{y}\):** \[ \bar{y} = \frac{263.03}{65} \approx 4.05\, \text{m} \] --- ## **Final Answer** ### **Centroid Coordinates:** \[ \boxed{ (\bar{x}, \bar{y}) = (7\,\text{m},\, 4.05\,\text{m}) } \] --- ### **Explanation:** - The centroid is near the center horizontally (\(x = 7\)), and a bit above the base (\(y \approx 4.05\)), slightly lower than the centroid of the triangle alone, because the holes remove area above and below. - We used the subtraction method, treating holes as negative areas. --- **If you have any questions about these steps, let me know!**