Let's solve the problem step by step: ## **Step 1: Understanding the Shape** The shaded area consists of: - A large quarter circle of radius \( 30 \) units. - A small quarter circle (hole) of radius \( 15 \) units. Both are centered at the origin \((, )\), and the quarter circles are in the first quadrant. --- ## **Step 2: Area Calculation** ### **Area of Large Quarter Circle** \[ A_1 = \frac{1}{4} \pi R^2 = \frac{1}{4} \pi (30)^2 = \frac{1}{4} \pi (900) = 225\pi \] ### **Area of Small Quarter Circle (hole)** \[ A_2 = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (15)^2 = \frac{1}{4} \pi (225) = 56.25\pi \] ### **Shaded Area** \[ A_{\text{shaded}} = A_1 - A_2 = 225\pi - 56.25\pi = 168.75\pi \] --- ## **Step 3: Centroid Calculation** For a quarter circle of radius \( R \) in the first quadrant centered at the origin, the centroid is given by: \[ (\bar{x}, \bar{y}) = \left(\frac{4R}{3\pi}, \frac{4R}{3\pi}\right) \] ### **Centroid of Each Shape** - **Large Quarter Circle:** \[ (\bar{x}_1, \bar{y}_1) = \left(\frac{4 \times 30}{3\pi}, \frac{4 \times 30}{3\pi}\right) = \left(\frac{120}{3\pi}, \frac{120}{3\pi}\right) \] - **Small Quarter Circle (hole):** \[ (\bar{x}_2, \bar{y}_2) = \left(\frac{4 \times 15}{3\pi}, \frac{4 \times 15}{3\pi}\right) = \left(\frac{60}{3\pi}, \frac{60}{3\pi}\right) \] ### **Shaded Area Centroid (Using Composite Areas)** \[ \bar{x} = \frac{A_1\bar{x}_1 - A_2\bar{x}_2}{A_1 - A_2} \] \[ \bar{x} = \frac{225\pi \cdot \frac{120}{3\pi} - 56.25\pi \cdot \frac{60}{3\pi}}{225\pi - 56.25\pi} \] \[ = \frac{225 \cdot 40 - 56.25 \cdot 20}{168.75} \] \[ = \frac{900 - 1125}{168.75} = \frac{7875}{168.75} = 46.67 \] So, \[ \boxed{\bar{x} = \bar{y} = 46.7 \text{ units (rounded to 1 decimal place)}} \] --- ## **Step 4: Moment of Inertia About x and y Axes** For a quarter circle about the origin: \[ I_{x,1} = I_{y,1} = \frac{1}{4} \left(\frac{\pi}{4}\right) R^4 = \frac{\pi}{16} R^4 \] Similarly for the hole: \[ I_{x,2} = I_{y,2} = \frac{\pi}{16} r^4 \] ### **Subtract the Hole** \[ I_x = I_{x,1} - I_{x,2} \] \[ = \frac{\pi}{16} (30^4 - 15^4) \] \[ 30^4 = 810,000,\quad 15^4 = 50,625 \] \[ I_x = \frac{\pi}{16}(810,000 - 50,625) = \frac{\pi}{16}(759,375) \] \[ = 47,460.94\pi \] \[ = 149,074.2 \text{ units}^4 \] --- ## **Step 5: Radius of Gyration** \[ k_x = \sqrt{\frac{I_x}{A}} \] \[ A = 168.75\pi \approx 530.14 \] \[ k_x = \sqrt{\frac{149,074.2}{530.14}} \approx \sqrt{281.23} \approx 16.78 \] --- # **Summary of Answers** | Quantity | Value | |----------------------|------------------------------------------------| | **Shaded Area** | \( 168.75\pi \approx 530.1 \) units\(^2\) | | **Centroid** | \( (46.7, 46.7) \) units | | **Moment of Inertia**| \( I_x = I_y = 149,074 \) units\(^4\) | | **Radius of Gyration** | \( 16.8 \) units | --- **Let me know if you want further breakdowns or calculations!**