1. **Problem statement:** A semi-circular sheet of metal with diameter 28 cm is bent into an open conical cup. Find the depth and capacity of the cup. 2. **Understanding the problem:** The semi-circular sheet forms the curved surface of the cone. The diameter of the semi-circle is 28 cm, so the radius $r = \frac{28}{2} = 14$ cm. 3. **Formula for the cone:** - The slant height $l$ of the cone equals the radius of the semi-circle, so $l = 14$ cm. - The circumference of the base of the cone equals the length of the arc of the semi-circle. 4. **Calculate the arc length of the semi-circle:** $$\text{Arc length} = \pi \times r = \pi \times 14 = 14\pi \text{ cm}$$ 5. **Base circumference of the cone:** $$2\pi R = 14\pi \implies R = \frac{14\pi}{2\pi} = 7 \text{ cm}$$ where $R$ is the radius of the base of the cone. 6. **Find the height $h$ of the cone using Pythagoras theorem:** $$h = \sqrt{l^2 - R^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3} \text{ cm}$$ 7. **Calculate the capacity (volume) of the cone:** $$V = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi \times 7^2 \times 7\sqrt{3} = \frac{1}{3} \pi \times 49 \times 7\sqrt{3} = \frac{343}{3} \pi \sqrt{3} \text{ cm}^3$$ **Final answers:** - Depth of the cup $h = 7\sqrt{3}$ cm (approximately 12.12 cm) - Capacity of the cup $V = \frac{343}{3} \pi \sqrt{3}$ cm³ (approximately 619.92 cm³)