1. **Problem (a):** The area of the square is 100 cm². Find the area of the circle in terms of $\pi$. 2. The square is inscribed in the circle, so the diagonal of the square equals the diameter of the circle. 3. The side length $s$ of the square is found by $s^2 = 100 \Rightarrow s = 10$ cm. 4. The diagonal $d$ of the square is $d = s\sqrt{2} = 10\sqrt{2}$ cm. 5. The diameter of the circle is $d = 10\sqrt{2}$, so the radius $r$ is $r = \frac{d}{2} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ cm. 6. The area $A$ of the circle is $A = \pi r^2 = \pi (5\sqrt{2})^2 = \pi \times 25 \times 2 = 50\pi$ cm². --- 1. **Problem (b):** The volume of a sphere is $36\pi$ cm³. Find the radius of the sphere. 2. The volume $V$ of a sphere is given by $V = \frac{4}{3} \pi r^3$. 3. Substitute $V = 36\pi$: $$36\pi = \frac{4}{3} \pi r^3$$ 4. Divide both sides by $\pi$: $$36 = \frac{4}{3} r^3$$ 5. Multiply both sides by $\frac{3}{4}$: $$36 \times \frac{3}{4} = r^3 \Rightarrow \cancel{36} \times \frac{3}{\cancel{4}} = r^3 \Rightarrow 27 = r^3$$ 6. Take the cube root: $$r = \sqrt[3]{27} = 3 \text{ cm}$$ --- 1. **Problem (c):** When the sphere is fully immersed in water in a cylinder, the water level rises by 2.25 cm. Work out the radius of the cylinder. 2. The volume of water displaced equals the volume of the sphere, which is $36\pi$ cm³. 3. The volume of water displaced is also the volume of the cylinder segment raised, given by $V = \pi R^2 h$, where $R$ is the radius of the cylinder and $h = 2.25$ cm is the height of water rise. 4. Set volumes equal: $$36\pi = \pi R^2 \times 2.25$$ 5. Divide both sides by $\pi$: $$36 = R^2 \times 2.25$$ 6. Divide both sides by 2.25: $$\frac{36}{2.25} = R^2 \Rightarrow \cancel{36} \div \cancel{2.25} = R^2 \Rightarrow 16 = R^2$$ 7. Take the square root: $$R = \sqrt{16} = 4 \text{ cm}$$ **Final answers:** (a) Area of circle = $50\pi$ cm² (b) Radius of sphere = 3 cm (c) Radius of cylinder = 4 cm