1. **Problem Statement:** Calculate the position of the centre of gravity (centroid) of a composite shape made of two areas, A4 (a right triangle) and A2 (a rectangle), with respect to the X-X and Y-Y axes, using point P as the origin. 2. **Given Data:** - A4: Right triangle with base $b=10$ in and height $h=10$ in. - A2: Rectangle with width $w=10$ in and height $h=10$ in. - Point P is at the intersection of X-X and Y-Y axes (origin). 3. **Formulas and Rules:** - Area of triangle: $A_\triangle = \frac{1}{2}bh$ - Area of rectangle: $A_{rect} = w \times h$ - Centroid of triangle from right angle vertex: $\left(\frac{b}{3}, \frac{h}{3}\right)$ - Centroid of rectangle from bottom-left corner: $\left(\frac{w}{2}, \frac{h}{2}\right)$ - Composite centroid coordinates: $$ \bar{x} = \frac{\sum A_i x_i}{\sum A_i}, \quad \bar{y} = \frac{\sum A_i y_i}{\sum A_i} $$ 4. **Calculate Areas:** - $A_4 = \frac{1}{2} \times 10 \times 10 = 50$ in$^2$ - $A_2 = 10 \times 10 = 100$ in$^2$ 5. **Locate Centroids of Each Area:** - For A4 (triangle), centroid from P: - $x_4 = \frac{10}{3} \approx 3.33$ in - $y_4 = \frac{10}{3} \approx 3.33$ in - For A2 (rectangle), centroid from P: - The rectangle is to the right of the triangle, so its base starts at $x=10$ in. - $x_2 = 10 + \frac{10}{2} = 15$ in - $y_2 = \frac{10}{2} = 5$ in 6. **Calculate Composite Centroid Coordinates:** - Total area: $A = 50 + 100 = 150$ in$^2$ - $\bar{x} = \frac{50 \times 3.33 + 100 \times 15}{150} = \frac{166.5 + 1500}{150} = \frac{1666.5}{150} = 11.11$ in - $\bar{y} = \frac{50 \times 3.33 + 100 \times 5}{150} = \frac{166.5 + 500}{150} = \frac{666.5}{150} = 4.44$ in 7. **Final Answer:** The centre of gravity is located at $$ \boxed{\bar{x} = 11.11 \text{ in}, \quad \bar{y} = 4.44 \text{ in}} $$ from point P along the X-X and Y-Y axes respectively.